Kvpy Sx 8 Year Tp 3r4639

R 0 Zer

(B) E(r) =

4 0R 3

r

=0r>R Ze  1 r  (C) E(r) = 4  2 – 3  r < R r R   0 =0 r>R Ze (D) E(r) = 4 r 2 0

r    1 –  , for all r R 

CLASS-XII_STREAM-SB+2_PAGE # 12

100.

A particular ideal gas is supplied 10.61 J of heat at a constant pressure of 1.01 × 105 Pa. If the volume of the gas increases by 3 × 10–5 m3 , the gas consists of (A) monatomic molecules. (B) diatomic molecules. (C) polyatomic molecules. (D) a mixture of mono and diatomic molecules.

CHEMISTRY 101.

Optically pure 3-bromopent-1-ene upon addition of 1 mole of Br2 produces a tribromo compound. The number of stereoisomers in the product is (A) 2

102.

(B) 4

(C) 6

(D) 3

The plot that is not valid for an ideal gas where P is the pressure and V is the volume of the gas, is -

(A) PV

(B) PV

1/V

P

(C)

V

(D)

P

1/P

103.

V

The volume of 0.1M acetic acid (pKa = 4.76) that should be added to 10ml of 0.2M sodium acetate solution to prepare a buffer solution of pH 4.91 is (A) 14.2 ml

104.

(B) 4.0 ml

(C) 20.0 ml

(D) 70.0 ml

A hydrated salt of a bivalent metal when heated strongly produces a mixture of gases which when bubbled through Ba(OH)2 solution produces a white solid. This solid on treatment with dilute HCl dissolves a part of the solid leaving another part insoluble which is filtered. The filtrate on treatment with Br2 water slowly precipitates another white solid, which is not soluble in HCl. The gas mixtures are (A) CO2 and CO

105.

(B) SO3 and SO2

(C) SO3 and CO2

(D) SO2 and CO

An electron is accelerated from rest through a potential (V) and then diffracted on a Ni crystal to measure its wavelength (). The wavelength is related to V as (A)   V

(B)   1/V

(C)  V1/2

(D)   1/V1/2


106.

A dilute aqueous solution of a polymer of known molecular weight shows an elevation of boiling point of water by 0.052 K. The molar boiling point elevation constant of water is 0.52 K kg mol–1. Using these data(A) it is possible to calculate the relative lowering of vapour pressure of water and the magnitude of freezing point depression, but not the osmotic pressure at 300 K. (B) it is possible to calculate the relative lowering of vapour pressure of water, but neither the magnitude of freezing point depression nor the osmotic pressure at 300K. (C) it is possible to calculate the relative lowering of vapour pressure but neither the magnitude of freezing point depression nor the osomotic pressure at 300 K. (D) it is possible to calculate the relative lowering of vapour pressure and the osmotic pressure at 300K but not the magnitude of freezing point depression.

107.

An organic compound fused with metallic sodium dissolves in water and the solution is divided into two parts. One part is treated with FeSO4, boiled and filtered. In the filtrate addition of FeCl3 does not produce any precipitate. To the other part, addition of sodium nitro prusside produces violet color. The organic compound contains.

108.

(A) only nitrogen

(B) both bromine and nitrogen

(C) only sulphur

(D) both sulphur and nitrogen

The main product III obtained in the following reaction sequence is -

NH2

(CH3CO)2O

Me

+

I

Br2/CH3COOH

II

H2O/H

NH2

NHCOCH3

(A)

(B) Me

Br

Me

Br

NH2

NHCOCH3

(C)

(D) Me Br

109.

III

Br

A water droplet spreads on a clean gold surface. The droplet, however, does not spread onto a gold surface which is pre-treated with a dilute solution of hexanethiol [CH3 – (CH2)5-SH] and washed with excess water. This is because (A) chemisorption of hexanethiol on gold surface renders it hydrophilic (B) chemisorption of hexanethiol onto gold surface renders it hydrophobic (C) physisorption of hexanethiol onto gold surface renders it hydrophobic (D) physisorption of hexanethiol onto gold surface renders it hydrophilic.

110.

Ir (CO)Cl (PPh3) when reacts with O2, the oxidation state and coordination number of Iridium, respectively, become. (A) +1 and 5

(B) + 2 and 5’

(C) + 4 and 6

(D) + 3and 6


BIOLOGY 111.

If our body fluid were alcohol instead of water, which of the following would happen to our body temperature if everthing else in the body remains the same (A) Temperature control would be maintained as well as it happens now (B) We will maintain a constant temperature of 37ºC (C) Fluctuations of body temperature will be more since alcohol has a lower specific heat capacity (D) We will feel feverish all the time

112.

A gene has two alleles ‘A’ and ‘a’ with a frequency of p and q respectively the genotypic frequencies would have the proportion : (A) (p + q)2 = 1 (B) p2 – q2 – 1 (C) pq = 1/2 (D) (p – q)2 = 1

113.

At low concentrations sodium ions are good for the heart, but excess of the same results in hypertension or high blood pressure. This is because. (A) The balance between Na and K ions is destroyed (B) Blood becomes thicker with excess of Na+ ions (C) Na+ ions narrow down the artery (D) Water retention in kidneys increases with excess Na ions and heart requires to pump more

114.

When Edward Jenner inoculated Joseph with the cow pox pustules from a milkmaid, the small boy did not get small pox because (A) Pustules contained lgG against small pox which gave ive immunity to Joseph (B) Pustules contained IFN which killed the small pox virus (C) Pustules contained cow pox virus which gave protection against small pox (D) Pustules contained immune cell which gave protection against small pox

115.

‘Camoulflage and Mimicry are adaptations in animals. Which of the following statement is NOT correct ? (A) Camouflaging represents the ability of the animal to blend with surrounding. (B) Mimicry means one specie imitates or resembles another species to gain some benefit. (C) Mimicry helps to escape from predators (D) Camouflaging is only meant for desert animals.

116.

Heparin secreted by the mast cells present in the inner walls of the blood vessels is known to inhibit blood coagulation. Which of the following statement is NOT correct ? (A) It help neutralizing the action of thrombin (B) It prevents conversion of prothrombin to thrombin (C) It prevents conversion of fibronogen to fibrin (D) It was originally isolated from liver cells

117.

During winter, frost often kills plants because. (A) It leads to crystallization of cellular water (B) Low temperature decreases the rates of catabolic pathways to an extent that the energy produced is inadequate to sustain life (C) Low temperatures decreases the rate of anabolism to an extent that the rate of ATP production is far lower than the rate of ATP consumed (D) The water in soil is frozen and the roots are unable to transport it to various parts of the plant


118.

If the two genes ‘A’ and ‘B’ on a chromosome are located next to each other (A) Chances of crossing over between them are very low (B) Chances of crossing over between them are very high (C) The two genes serve as single unit when it comes to accumulation of mutations within them (D) Then of the two genes one has to be nonfunctional gene for the crossover to occur in between the two

119.

Contractile vacuoles in amoeba (A) Accumulate water and burst upon fusion with the cell membrane releasing water from the cell (B) Are organelles that accumulate nutrients (C) Are responsible for locomotion of amoeba (D) Provide a surface for attachment of ribosomes

120.

During the course of intensive exercise, muscle fatigue occurs because. (A) During exercise entire deposit of glucose is converted to CO2 and H2O (B) O2 deficiency results in accumulation of lactic acid in the muscles (C) O2 deficiency results in accumulation of citric acid in the muscles (D) Excessive utilization of glucose results excessive production of acetyl Co A which is not fully utilized by the TCA cycle (Krebs cycle).

*****


GENERAL INSTRUCTIONS FOR KVPY TEST PAPER, STREAM- SB+2 YEAR : 2007 Time : 3Hr.

1.

Max. Marks : 160

In Year 2007, Question Number 1-80 each carries 1 (one) mark, Maths(1-20), Physics (21-40), Chemistry(41-60), Biology(61-80) & Question Number 81 to 125 each carries 2 marks, Maths(81-90),Physics(91-100), Chemistry(101-110), Biology(111-120)

2.

Blank papers, clip boards, log tables, slide rule, calculators, mobile or any other electronic gadgets in any form is not allowed.

3.

Write your Name and Roll No. in the space provided in the bottom of this booklet.

4.

Before answering the paper, fill up the required details in the blank space provided in the answer sheet.

5.

Do not forget to mention your roll number neatly and clearly in the blank space provided in the answer sheet.

6.

In case of any dispute, the answer filled in the OMR sheet available with the institute shall be final.


1.

Max. Marks : 160

In Year 2008, Question Number 1-80 each carries 1 (one) mark, Maths(1-20), Physics (21-40), Chemistry(41-60), Biology(61-80) & Question Number 81 to 125 each carries 2 marks, Maths(81-90), Physics(91-100), Chemistry(101-110), Biology(111-120)

2.


3.


4.


5.


6.



1.

Max. Marks : 160


2.


3.


4.


5.


6.



Max. Marks : 160

1.

The Question paper consists of two parts (both contain only multiple choice question) for 100 marks. There will be four sections in Part-I (each containing 20 questions) four sections in Part-II (each containing 10 questions). Answer any THREE of the four sections in Part-I and any TWO of the four sections in Part-II.

2.

The composition of the question paper is given in the table below : S No. 1 2 3 4

Subject

Part-I

Part-II

Mathematics

20 questions - 1 mark each


Physics



Chemistry



Biology



3.

The answer paper is machine readable. Do not forget to mention your paper code and Roll Number neatly and clearly in the blank space provided in the Objective Response Sheet (ORS)/Answer Sheet.

4.

For each question, indicate your answer by filling the corresponding oval with a HB pencil only.

5.

For each question there will be four choices given and only one of them is the correct answer. Fill only one oval per question. If you mark more than one oval, it will be considered as an incorrect answer.

6.

There is negative marking for wrong answers. Unanswered questions will not be evaluated and will not be penalized as a wrong answer.

7.

In Part-I each correct answer gets 1 mark and for each incorrect answer 0.25 mark will be deducted. In Part-II each correct answer gets 2 marks and for each incorrect answer 0.5 mark will be deducted.

8.

You are permitted to use a non programable calculator.

9.

Candidates are permitted to carry their question paper after the examination.


Max. Marks : 160

1.


2.


Subject

Part-I

Part-II

Mathematics



Physics



Chemistry



Biology



3.


4.


5.


6.


7.


8.


9.



Max. Marks : 160

1.


2.


Subject

Part-I

Part-II

Mathematics



Physics



Chemistry



Biology



3.


4.


5.


6.


7.


8.


9.


YEAR-2008 (KVPY-STREAM-SB+2) PART-I (1 Mark) MATHEMATICS 1.

The equation z2 = z , where z is a complex number, has (A) 4 solution (B) 2 solution (C) no solution (D) infinitely many solutions

10

2.

x 1 x 1     is : The coefficient of the term independent of x in the expansion of  2 / 3 1/ 3  x  1 x  x 1/ 2  x (A) 35

3.

(B) 70

(C) 105

If no two of the real numbers a, b, c are equal and (A) abc = 1

(B) a + b + c = 1

(D) 210

a a2

a3  1

b b2

b3  1

2

c3 1

c c

= 0, then :

(C) a + b + c = 0

(D) ab + bc + ca = 0

4.

If the slopes of one of the lines represented by ax2 – 6xy + y2 = 0 is the square of the other for some positive value of a, then a is (A) 1 (B) 2 (C) 4 (D) 8

5.

The number of distinct points common to the curves x2 + 4y2 = 1 and 4x2 + y2 = 4 is (A) 0 (B) 1 (C) 2 (D) 4

6.

An ellipse has its major axis equal to the diameter of a circle and the area of the ellipse is one-third the area of the circle. The eccentricity of the ellipse is : (A)

7.

9.

(B)

2 3

(C)

2 2 3

(D)

1 3

Two perpendicular chords AB and CD of a circle meet at P. If PA = 2, PB = 18 and PC = 4, then the diameter of the circle is : (A)

8.

2 3

5 17 2

(B) 5 17

(C) 20

The value of cos 5º + cos 10º + cos 15º + .......+ cos 355º is (A) 0 (B) 1 (C) – 1

(D) 10 5

(D) 71 cos 5arc

In a square ABCD, points P on BC and Q on CD are such that AP = 4, PQ = 3 and QA = 5. The area of ABCD is (A)

256 15

(B) 16

(C)

256 17

(D)

256 18


10.

In a triangle ABC, AD bisects A. Suppose AC = 2, BD = 2, DC = 1. The value of cos B is (A)

11.

12.

6 7

(B)

7 8

(C)

8 9

(D)

9 10

The sides of a triangle are 9x + 1, 6x + 2, 3x + 3, where x is a positive integer. If the area is also an integer, the number of issible values of x in the set {1, 2, 3........., 20} is : (A) 3 (B) 4 (C) 6 (D) 12

f (1  h) Suppose f is a real function defined on R and hlim exists. Then : 0 h (A) f is not continuous at 1 (B) f is not continuous at 1 but not differentiable at 1 (C) f is differentiable at 1 (D) f is differentiable at 0

13.

14.

Consider the following statements : (I) The derivative of an odd differentiable function is always even. (II) If f(x) is differentiable at a point x0 and g(x) is not differentiable at x0, then f(x) g(x) is not differentiable at x0. Which of the following is true ? (A) I and II are both true (B) I is true and II is false (C) I is false and II is true (D) I and II are both false

 sin[ x ] , if [ x ]  0  If f(x) =  [ x ] . (Here [x] denotes the integer part of x.) Then xlim f(x) 0  0, if [ x ]  0 (A) is –1

15.

(B) is 0

Let f : [0, )  R be defined by f(x) = x – (A) [ – 1/2, ) x

16.

Let f(x) =

(B) [0, )

x . Then the range of f is

(C) { – 1/4, )

(D) R

t

(B) (– 2, – 1) 1

18.

4

0

The value of

(A)

x +

(D) does not exist

 e (t  1)(t  2) dt .Then f(x) decreases in the interval

(A) (– , – 2) 17.

(C) is sin 1



1

(C) (1, 2)

(D) (2, )

x | x |3 / 2 dx is :

4 7

(B)

4 5

(C)

4 3

(D) 0

Let fn(x) = log log ....log(x), where log is repeated n times. Then

 x f ( x)f (x)......f 1

2

(A) f11(x) + c

1dx

10 ( x )

(B)

is equal to :

f11( x ) c 11

(C) 10f10(x) + c

(D) 11f11(x) + c


19.

The least value of x12 + x22 + x32 where x1, x2, x3 are real numbers satisfying x1 + 2x2 + 3x3 = 4 is 8 (A) (B) 1 (C) 4 (D) 2 7

20.

A coin is tossed until one observes a sequence of exactly three tails. The probability that the experiment comes to an end at 7-th toss is . (A)

7 128

(B)

1 128

(C)

1 32

(D)

5 128

PHYSICS 21.

Super cooled steam suddenly freezes to water. In this process, (A) entropy of the system (steam) decreases, but entropy of the universe increases (B) entropy of the system as well as entropy of environment decreases (C) entropy of the system as well as entropy of the environment increases (D) entropy of the system increases but entropy of the environment decreases

22.

Isothermal compressibility  of a substance is defined as : =–

1  V    V  P  T

where the derivative is taken keeping T constant. V : Volume P : Pressure T : Temperature For an ideal gas,  equals (A) R (Gas constant) (B) P (C) P–1 (D) P ( is the ratio of specific heats at constant pressure and constant volume  = (/ Cv) 23.

Consider two cases : (i) A point charge q at the origin (ii) A uniformly charged solid sphere of radius R (wth its centre at the origin) and total charge q. Given below are graphs of a property X versus distance r from the origin for the two cases. The graphs coincide for r  R.

(A) X is electric potential due to point charge/charged sphere. (B) X is magnitude of electric field due to point charge/charged sphere (C) X is electrostatic potential energy of point charge/charged sphere (D) X is charge density in space.


24.

One end of a cylindrical solid rod of length L and radius r is clamped in a fixed positon. The other end is turned by an external torque  resulting in a twist . The sheer modulus is given by . The twist angle  is proportional to : (A) r1/L2 (B) r4/L (C)  L/r4 (D)  L /r4

25.

 A particle of charge q and mass m moves in a cirular orbit with angular momentum given by l . The ratio  of its magnetic moment  to orbital angular moment l is given by (A)

26.

q m

(B) q

(C)

q 2m

(D) m q

Thermal motion of atoms in a gas causes the spectral line emitted by the atoms to be shifted at random towards both the red and blue. This leads to Doppler broadening of the spectral lines. If f is the frequency of the spectral line and  f is a measure of broadening of the line, the ratio

f is proportional to f

(Proportionality constant is given to be dimensionless)

m (A)

kT

(B)

kT m

(C)

mkT

(D)

1 kT c m

27.

Consider a sinusoidal travelling wave along a string, with amplitude A and wave velocity v. The power carried by the wave is proportonal to (A) Av2 (B) AV (C) A2v2 (D) A2 v (The proportionality constant is not dimensionless)

28.

Monochromatic light falls on a pair of slits mounted 24 cm in front of a photographic film. After exposure, the film shows a series of bright bands spaced 0.20 cm apart. The separation between the slits is 6.0 × 10-3 cm. The wavelength of light is : (A) 600 nm (B) 400 nm (C) 500 nm (D) 560 nm

29.

Two clean mercury droplets when pushed into spontaneously coalesce to form a single droplet. The single droplet so formed will : (A) Be slightly warmer than the separate pair of droplets (B) Be slightly cooler than the separate pair droplets (C) Have the same temperature as the separate pair of droplets (D) Be warmer or cooler depending on the size of each initial droplet.

30.

The speed of a comet at perihelion (closest to the sun) is 5.6 × 104 ms–1, while its distance from the sun is 9.0 × 1010m. At aphelion (farthest from the sun) its distance is 5.6 × 1012m. The comet’s aphelion speed is : (A) 1.0 km s–1 (B) 900 ms–1 (C) 504 ms–1 (D) 5.6 × 104 ms –1

31.

Consider two wave forms y1 = A cos (k x – t) and y2 = A cos (k x + t) whose superposition forms a standing wave. Here x is in metres and t is in seconds. A node is found at x = 10 m. The longest wavelength possible for this situation is (A) 1 m (B) 40 m (C) 20 m (D) 10 cm


32.

The displacement x of a damped oscillator in one dimension is given by x(t) = A e – t sin (t + ) where x is in metres. The SI units of the constants A, , ,  are respectively (A) m, s–1, s–1, dimensionless (B) m, s, s–1, m (C) m, s–1, s, dimensionless (D) dimensionless, s–1, s–1, dimensionless

33.

In the following reaction, a proton bombards a lithium nucleus at rest producing two -particles : Li7 + 1H1  2 2He4 3 The total rest mass of the products is less than the total rest mass of the reactants by 0.01864 atomic mass unit (amu). By mass-energy conversion formula, 1 amu = 931 MeV approximately. This means that (A) The kinetic energy of the proton 17.4 MeV (B) The total kinetic energy of two - particles is more than the kinetic energy of the proton by 17.4 Mev (C) Each -particle has a kinetic energy of 8.7 MeV. (D) The proton’s kinetic energy is 8.7 MeV more than that of each -article

34.

The radius of the first (n = 1) orbit in Bohr’s model of hydrogen atom is 0.53 × 1010 m. (Planck’s constant h = 6.63 × 10–34 Js, mass of electron = 9.11 × 10–31 kg). The speed of the electron in this orbit is approximately (A) 3.0 × 1010 m s–1 (B) 1.4 × 107 m s –1 (C) 3.5 × 107 m s –1 (D) 2.2 ×107 m s –1

35.

In an electric motor, two parallel wires 60 cm long are separated by a distance of 1.5 cm. The wires have non-magnetic insulation between them. A short circuit in the motor results suddenly in a large current of 4000 A. The currents flow in opposite directions in the two wires. Then (A) There will be a repulsive force of 128 N on each wire (B) There will be an attractive force of 128 N on each wire (C) There will be no force experienced by any wire, since they are insulated. (D) There will be a force along the length of each wire, of magnitude128 N.

36.

Let  be the typical de Broglie wavelength associated with an He atom in helium gas at room temperature T (20ºC) and pressure P(1 atmosphere).Let d be the mean separation between helium atoms under these conditions.Then (A)  is greater than d under the given T and P. (B)  is less than d under the given T and P. (C)  is equal to d under the given T and P (D) The ratio of  to d has no relation to T or P.

37.

In a plane electromagnetic wave of frequency f traveling in free space with speed c, the electric field magnitude E and magnetic field amplitude B are related ( In S.I. units) by (A) B = fE

38.

(C) B =

E c

(D) E = fB

A resistor R and a capacitor C are connected in series to a 220 V, 50 Hz a.c. source. Let VR and Vc be the r.m.s. voltages across R and C respectively. Then (A) VR = Vc = 220 V

39.

(B) B = cE

(B) VR – Vc = 220 V

(C) Vc – VR = 220 V

(D)

VR2  VC2 = 220 V

Two identical point sources P and Q vibrating in phase with the same amplitude generate sinusoidal waves on a water surface.The sources are 6.5 cm apart. Two nearest points (from P) of destructive interference along PQ are found to be at 0.7 cm and 2.4 cm from P. The total number of points of destructive interference on the line segment PQ is (A) 4 (B) 3 (C) 2 (D) 5


40.

An ideal gas is subjected to an isothermal -isochoric cycle 1 – 2 – 3 – 4 – 1 as shown. V

3

4

2

1

T

On the pressure - density (P-) graph, this cycle is represented by

P

P

2

4 3

3

1 4

1

2

(A)

(B)

P

P 2

2 3

3 1

1 4 4

(C)

(A) Graph (A)

(D)

(B) Graph (C)

(C) Graph (B)

(D) Graph (D)

CHEMISTRY 41.

Li metal is a better reducing agent than Na metal because (A) The ionization enthalpy of Li is lower than that of Na (B) The hydration enthalpy of Li is lower than that of Na (C) The ionization enthalpy of Li is higher than that of Na (D) The hydration enthalpy of Li is higher than that of Na

42.

One mole each of the two gases X and Y are stored separately in two cylinders at 25ºC at pressures 1 atm. and 2 atm, respectively. The difference in the compressibilities of the two gases. (kx – ky) is (A) 0.1 atm–1 (B) 0.5 atm–1 (C) 1.0 atm–1 (D) 2.0 atm–1

43.

(S)-lactic acid is OH

OH (A)

(B)

CO2H

OH

OH (C)

CO2H

(D)

CO2H

CO2H


44.

45.

The reaction between p-methylbenzaldehyde and NaOH is an example of (A) Aldol condensation reaction

(B) Cannizzarro reaction

(C) Disproportionation reaction

(D) Hydrolysis reaction

A saturated solution of BaSO4 is heated from 25ºC to 35ºC and the conductance of the solution and the solubility of BaSO4 are measured. It is found that : (A) both conductance and solubility increase (B) both conductance and solubility decrease (C) conductance increases but solubility decreases (D) conductance decreases but solubility increases.

46.

An ideal gas is subjected to a cyclic change as shown in the P-V diagram below :

A P

C

B

V The step in which the gas will cool down is along (A) AB 47.

(B) BC

(A) (3Z)-hept-3-en-1-ol

(C) (3Z)-hept-4-en-7-ol

(D) (3E)-hept-4-en-7-ol

is

(B) (3E)-hept-3-en-1-ol

The Lewis acid strength of BF3, BCl3 and BBr3 follows the order. (A) BBr3 > BCl3 > BF3

49.

(D) both BC and CA

The IUPAC name of

OH

48.

(C) both AB and CA

(B) BF3 > BCl3 > BBr3

(C) BCl3 >BBr3 > BF3

(D) BCl3 > BF3 > BBr3

The total number of possible geometrical and optical isomers for [CoCl2(en)2]+ (en = 1, 2-diaminoethane) is (A) 1

50.

(B) 2

(C) 3

(D) 4

In balancing the reaction. xH2S + 2NaNO3 + 2HCl = y S + zNO + kNaCl + 4H2O one would get x,y, z and k, respectively, as (A) 3,3,2 and 2

51.

(B) 2,2,3 and 3

(C) 3,3,4 and 4

(D) 4,4,3 and 3

The overall order of a reaction involving two reactants, X and Y, which follows the rate expression,Rate = k [X]1/3 [Y]2/3 . Where k is the specific rate and [ ] represents concentration, is (A) 2/3

(B) 0

(C) 1/3

(D) 1


52.

In a Daniel cell in operation : (A) Electrons flow externally from copper to zinc while anions flow from zinc to copper in solution. (B) Electrons flow externally from zinc to copper while anions flow from zinc to copper in solution. (C) Electrons flow externally from zinc to copper while anions flow from copper to zinc in solution. (D) Electrons flow externally from coppper to zinc while anions flow from copper to zinc in solution.

53.

In the pressure vs. molefraction of benzene curves/lines shown below, the total vapour pressure of an ideal mixture of benzene and toluene will follow the curve/line.

0.5 2 P(bar)

1

3 4

0.0 0.0

(A) 1 54.

(D) [Fe(H2O)6]2+

(B) SO2

(C) SO3

(D) O2

(B) O22– > O2– > O2+

(C) O2– > O22– > O2+

(D) O2+ > O22– > O2–

(B) Polyvinylchloride

(C) Teflon

(D) Bakelite

(B) 200 s

(C) 0.05 s

(D) 100 s

(C) 6

(D) 8

The number of facial atoms in a fcc unit shell is (A) 2

60.

(C) [FeCl4]2–

The half-life of ammonia adsorbed on a Ni surface if the rate of desorption is 6.93 × 10–3 s–1 , is (A) 0.01 s

59.

(B) [Fe(CN)6]3–

A naturally occuring polymer is (A) Amylose

58

(D) 4

The interatomic distance in O2+, O2– , and O22– follow the order (A) O2+ > O2– > O22–

57.

(C) 3

The gas formed when conc. H2SO4 is added to a mixture of NaCl and MnO2, is (A) Cl2

56.

(B) 2

sp3d2 hybridization explains the bonding in (A) [FeCl4]–

55.

1.0

molefraction (benzene)

(B) 3

The maximum amount of work produced by a heat engine operating between 200 K and 800 K, if 100 J of heat is absorbed from the hot reservoir, is (A) 100 J

(B) 75 J

(C) 50 J

(D) 25 J


BIOLOGY 61.

62.

Embryonic stem cells are derived from, (A) Inner cell mass of the blastocyst (C) Cells from morula

(B) Outer cell mass of the blastocyst (D) Cells from the placenta

The phase in which meiotic recombination occurs (A) Diplotene (B) Zygotene (C) Pachytene

(D) Diakinesis

63.

SCID (Severe Combined Immuno Defficiency Syndrome) patients do not have (A) T-lymphocytes (B) Platelets (C) Monocytes (D) Erythrocytes

64.

What is the maximum number of hydrogen bonds possible involving a water molecule ? (A) 0 (B) 1 (C) 2 (D) 3

65.

A strand of a nucleic acid having 60% of base A and 30% of base G is (A) B-DNA (B) A-DNA (C) Should be an RNA

(D) Single stranded

66.

Tetracycline, cyclohexamide, and streptomycin inhibit : (A) Transcription (B) Translation (C) Splicing (D) mRNA editing

67.

In the genetic code, a nonsense codon codes for (A) An antisense amino acid (B) A sensible amino acid (C) Does not code for any amino acid (D) Methionine

68.

The dinosaurs became extinct at the end of the following period : (A) Cambrian (B) Cretaceous (C) Ordovician

(D) Silurian

The anticoagulant most commonly used in blood banks is : (A) EDTA (B) Heparin (C) Sodium Citrate

(D) Hirudin

69.

70.

Totipotent cell is (A) A cell which can be differentiated into most of the cell types. (B) A cell which can be differentiated to all cell types to form a complete organism (C) A cell which can be differentiated into only a specific cell type (D) A cell which does not differentiated at all

71.

You have an infection of pneumococcus (a bacteria). Which part of the bacteria may be toxic to your body and causes immunological reaction ? (A) The outer cell wall lipoplysaccharide (B) The plasma membrane of the bacteria (C) bacterial nucleus (D) Polysomes

72.

Three nucleotides form a codon, which can code for a single amino acid. However, the same amino acid can be coded by three different codons. These three codons are collectively called (A) Puncture codons (B) Degenerat codons (C) Nonsense codons (D) Termination codons

73.

The reason for doctor’s advice for vaccination of newborns is : (A) To increase the innate immunity of the baby (B) To increase the acquired immunity of the baby (C) To increase both, the innate and acquired immunity of the baby (D) To protect the baby only against bacterial diseases


74.

If you start a bacterial culture with 100 E. coli cells and allow the culture to grow for 2 hours, approximately what would be the total number of E. coli in the culture considering E.coli has a doubling time of 20 minutes. (A) 102 (B) 103 (C) 105 (D) 107

75.

Bacterial Flagellar movement requires (A) Energy through ATP hydrolysis (B) Nutrient gradient (C) Proton gradient coupled to transport (D) Energy through GTP hydeolysis

76.

Which of the following bio-molecules is responsible for the “prion” infectious diseases (A) DNA (B) Protein (C) RNA (D) Lipid

77.

Charles Darwin’s Theory of natural selection was heavily influenced by : (A) Thomas Malthus “An Essay on the Principle of Population” (B) Jean-Bapiste Lamark “Philosophie Zoologique” (C) Issac Newton “Principia” (D) Gregor Mendel “Experiments on Plant Hybridization”

78.

Which one of the following chemical groups is not present in the nascent polypeptide chain of a protein ? (A) Methine (B) Methylene (C) Amide (D) Phosphate

79.

Identify the evolutionarily related proteins : (A) Pepsin & Papain (C) Myoglobin & Hemoglobin

80.

(B) Collagen & Collagenase (D) Lysozyme & Ribozyme

How many peptide linkages are present in a protein with 176 residues ? (A) 174 (B) 175 (C) 176

(D) 177

PART-II (2 Mark) MATHEMATICS 81.

Suppose a, b, c are in arithmetic progression and a2, b2, c2 are in geometric progression. If a < b < c and a + b + c = 3/2, the value of a is :

1 (A)

82.

2

(B)

1

1 1 – 2 2

(C)

3

(D)

1 1 – 3 3

Let y = g(x) be a function whose derivative g'(x) has the following graph. Which of the following values of g is the largest ?

2 1 –1 –2 –3 –4

(A) g(2)

(B) g(3)

1 2 3 4 5

(C) g(4)

(D) g(5)


83.

Let ABCDEFGHI be a regular nanogon (9-sided polygon). If AB = x, AC = y, AD = z, then x, y, z are related by (A) z2 = x2 + xy + y2 (B) z = x + y (C) y2 = xz (D) 2y = x + z

84.

If A is a 10 × 10 matrix with entries from the set {0, 1, 2, 3} and if AAT is of the form :

0 * *  *    * 0 *  * * * 0  *           * * *  0 the number of such matrices A is : (A) (43)10 (B) (42)10 85.

(A)

 1 5 2

(C)

5 2 2

(B)

88.



0

[ x ]e  x dx equals :

e e 1

(B)

 What is the value of  

/2



0

 2006

(A)

(C) 1 –

2

e 1

  cos1003 xdx    

(B)

x x



0

 2007

/2



0

1 e

(D)

1 e 1

(D)

 2009

 cos1004 xdx  ? 

(C)

 2008

e f ( t ) f ( x ) dt is

(A) 0

90.

e2

Let f(x) and g(x) be real polynomials of degree 4 and 3 respectively with leading coefficients 4 and 3 respectively. Then

lim g( x )

89.

2 3 2

(D) not uniquely determinable

The integral

(A)

87.

(D) 1

In an ellipse, O is the centre, AB is the major axis and CD is the minor axis. Suppose the focus between A and O is the ortho-centre of the triangle ACD. The eccentricity of the ellipse is



86.

(C) 410

(B) 

(C)

3 16

(D)

The equation of the curve through the origin satisfying the differential equation (A) log 1  tan

(x  y) =y 2

(B) log |1 + tan(x + y)| = x

(C) log 1  tan

(x  y) =x 2

(D) log |1 + tan(x + y)| = y

4 13

dy = sin(x + y) + cos(x + y) dx

Let A = {1, 2, 3, 4, 5, 6} and f : A  A be a bijection. What is the probability that f o f = id, i.e., f is its own inverse ? (A)

19 180

(B)

13 48

(C)

1 720

(D)

5 16


PHYSICS 91.

Two stars of masses M1 and M2 form a binary system. The distance between the centres of the stars is d. The orbital period is given by (A)

(B)

(C)

(D)

2 d 3 G(M1  M2 ) 2d3 / 2 G(M1  M2 ) 2d3 / 2 G

M1 M2 , where  is the ‘reduced’ mass  = M  M 1 2

2 d 3 G

92.

A particle is stuck on the rim of a wheel of radius 50 cm. The wheel is rotating with an angular acceleration of 20 rad s–2. If at an instant the angular speed of the wheel is 10 rad s–1, the total linear acceleration of the particle is : (A) 10 ms–2 (B) 50 ms–2 (C) 51 ms–2 (D) 60 ms–2

93.

A beam of 30 keV electrons strikes different targets in different experiments. The lowest wave length cut -off of the continuous spectrum of X-rays generated by beam for any target is

94.

(A) 1.0 × 10– 10 m

(B) 3.0 × 10– 10 m

(C) 4.14 × 10– 11 m

(D) dependent on the nature of the target

A bat flying towards a wall with a speed 9.0 m s–1 emits ultrasound of frequency 90 kHz. (Speed of ultrasonic waves is 340 ms–1) The frequency of ultrasound received by the bat after reflection from the wall is : (A) 90 kHz

95.

(B) 94.9 kHz

(C) 85.4 kHz

(D) 99 kHz

A cylindrical glass vessel of height 18 cm and diameter 8 cm is 4 mm thick. It is covered with a1 mm thick copper lid (Thermal conductivity of copper is 400 W K–1m–1) In a cold environment the water in the vessel has got frozen into 0ºC ice. It is then immersed into a tank of 15ºC water. (Density of ice is 0.92 ×103 kgm–3 and latent heat of fusion of water is 333 x 103 J kg–1) The time taken for the ice to melt completely is (A) 9.2 s

96.

(B) 46 s

(C) 18.4 s

(D) 136 s

An upright cylinder of large base area is filled with water up to height H. Water is flowing out through holes (1, 2 and 3) of equal diameter on the side of the cylinder at heights H/4, H/2 and 3H/4 respectively. Let x1, x2 and x3 be the respective horizontal distance covered by the water flowing out of the holes before hitting the ground. Then (A) x3 > x2 > x1

97.

(B) x3 < x2 < x1

(C) x2 > x1 = x3

(D) x2 > x1 > x3

Ocean tides on the Earth are caused by the gravitational effects of the Moon (for lunar tides) as well as the Sun (for solar tides). If the diameter of the earth were to increase by 20%, then (A) Lunar tides would be strengthened but solar tides would be weakened (B) Lunar tides would be weakened but solar tides would be strengthened (C) Both lunar and solar tides would be strengthened with lunar tides strengthening more than solar tides (D) Both lunar and solar tides would be weakened with lunar tides weakening more than solar tides.


98.

The wavelength of radiation emitted by a hydrogen atom when it de-excites from its first excited state to ground state (n = 2 to n = 1) is 121.7 nm. For the analogous transition in a positronium atom (bound state of electron and its antiparticle called positron) the wavelength of radiation will be approximately (A) 243.4 (B) 121.7 (C) 60.85 (D) 0.53

99.

The objective lens of a telescope has a diameter of 12.2 cm. The angular resolution of the telescope at the wavelength of 500 nm is (A) 2 × 10 –9 rad (B) 1.22 × 10 –7 rad (C) 4 × 10 –5 rad (D) 5 × 10 –6 rad

100.

A conducting rod of length 80 cm rotates with its one end fixed at the centre of a circular metallic ring and the other end in with the ring. The angular frequency of the rod is 300 s–1. There is a uniform and constant magnetic field of 1.0 T parallel to the axis of rotation. The emf developed between the centre and the ring is : (A) 48 V (B) 192 V (C) 36 V (D) 96 V

CHEMISTRY 101.

The numbers of lone pairs of electrons in XeF2 and XeF4 respectively, are (A) 3 and 3 (B) 2 and 3 (C) 2 and 2 (D) 3 and 2

102.

When 4 moles of N2 gas reacts with 16 moles of H2 gas in 10 lit vessel, 4 moles of ammonia gas is produced in the equilibrium mixture. The equilibrium constant Kc for this reaction in mol–2 lit2 is (A) 0.4 (B) 0.2 (C) 0.8 (D) 1.6

103.

The freezing point of pure benzene is 5.5º C. When 2.9 g of butane is dissolved in 200g of benzene, the freezing point of benzene decreases to 4ºC. To lower the freezing point of benzene by another 1.5ºC, the amount of butane that has to be added to mixture is (A) 5.8 g (B) 2.9 g (C) 1.5 g (D) 8.7 g

104.

X and Y in the following reactions, respectively are SO3H

O

MgBr (i) +

cat

X

Y

(ii) H3O

OH

(A) X =

Q

Q

Y= Q

OH

(B) X =

Q

Y=

Q

OH

(C) X =

Q

Y=

OH Q

OH

(D) X =

Q

Y=

Q = Phenyl group

O


105.

The configuration at C-2 and C-5 in the following compound, respectively, are OH 2

5 OH

(A) 2R and 5R 106.

(B) 2S and 5S

(C) 2R and 5S

The cell potential of an electrochemical cell with the cell reaction Zn + 2Ag+ (0.0001M)  Zn2+ (0.1 M) + 2Ag, is (A) 1.25 V (B) 1.35 V (C) 1.45 V Given that the standard cell potential is 1.56 V

(D) 2S and 5R

(D) 1.55 V

107.

Methyl chloride reacts with X at high temperature in the presence of a catalyst Y to give Me2SiCl2 as one of the products. The compound can also be formed by reaction of silicon tetrachloride with an organometallic reagent, Z. X, Y, and Z respectively,are (A) Si, Cu, MeMgBr (B) SiO2, Cu, Me2Zn (C) Si, Ni, MeMgBr (D) SiO2, Fe, Me2Zn

108.

A chemical reaction takes place at 625 K with an activation energy barrier, E/R = 500 K. A catalyst when used in the reaction reduces the activation energy barrier to 400 K. The temperature at which the rate constant with and without the catalyst will be same is (A) 300 K (B) 400 K (C) 500 K (D) 600 K

109.

A divalent transition metal ion with valence electron configuration 3d8 forms a complex with cyanide ion. The correct identity of the metal ion, the complex and its geometry, respectively, are (A) Fe2+, [Fe(CN)6]4– , and octahedral (B) Ni2+, [Ni (CN)4]2– and tetrahedral (C) Fe2+, [Fe(CN)6]3– and octahedral (D) Ni2+, [Ni(CN)4]2– and square planar

110.

The number of  and  particles to be emitted by 238U92 to give 206Pb82 , respectively, are (A) 8 and 6 (B) 4 and 3 (C) 6 and 8 (D) 3 and 4

BIOLOGY 111.

In a double-stranded DNA coding for a protein, in principle, how many codon reading frames are possible? (A) 1 (B) 2 (C) 3 (D) 6

112.

Which of the following is advantage of meiotic recombination in diploids ? (A) Helps in maintaining chromosome length (B) Ensures chromosome segregation (C) Helps chromosomes to attach to the spindle microtubules (D) Ensures new combinations of genetic traits

113.

Plants obtain their nitrogen supply through bacteria of the soil because : (A) Nitrogen is absent in the air and present in the soil (B) Plants do not have a mechanism to absorb gaseous Nitrogen (C) Bacteria are Nitrogen rich (D) Getting nitrogen is a side-effect of a bacterial infection


114.

The number of genes in man (Homo sapiens) is X-fold more than in a fly (Drosophila melanogaster). The value of X is (A) 2 (B) 10 (C) 250 (D) 10000

115.

You are supposed to mix a 3000 bp DNA fragment and a 600 bp DNA fragment in 1 : 5 molar ratio. If you take 100 ng of the 3000 bp fragment how much of the 600 bp fragment do you require to get desired ratio (A) 10 ng (B) 100 ng (C) 500 ng (D) 2500 ng

116.

An enzyme that cleaves DNA recognizes an 8 base-pair unique DNA sequence. The probable number of times this enzyme will cleave a 70 kilobase pair random DNA sequence is (A) 1 (B) 2 (C) 3 (D) 4

117.

Mendel’s law of independent assortment when interpreted in a modern context indicates : (A) Alleles are present on the same chromosome, but they assort independently due to recombination (B) Alleles are present on independent chromosomes and these separate and assort independently (C) Alleles are present on independent chromosomes and they can be sorted during meiosis. (D) Alleles are present on the same chromosome and their combination is dependen

118.

Fat absorption in the microvilli is by : (A) Endocytosis (B) Simple diffusion through the plasma membrane (C) Facilitated diffusion (D) Active transport

119.

During the process of photosynthesis (A) Glucose is synthesized during the dark reaction and ATP during the light reaction (B) Glucose and ATP are produced during light and dark reaction (C) Glucose and ATP are produced during the dark reaction (D) Glucose is synthesized during the light reaction and ATP during the dark reaction

120.

Identify the protein with more than one polypeptide chain. (A) Myoglobin (B) Trypsin (C) Immunoglobulin

(D) Lysozyme



1.

Max. Marks : 160


2.


3.


4.


5.


6.



1.

Max. Marks : 160


2.


3.


4.


5.


6.



1.

Max. Marks : 160


2.


3.


4.


5.


6.



Max. Marks : 160

1.


2.


Subject

Part-I

Part-II

Mathematics



Physics



Chemistry



Biology



3.


4.


5.


6.


7.


8.


9.



Max. Marks : 160

1.


2.


Subject

Part-I

Part-II

Mathematics



Physics



Chemistry



Biology



3.


4.


5.


6.


7.


8.


9.



Max. Marks : 160

1.


2.


Subject

Part-I

Part-II

Mathematics



Physics



Chemistry



Biology



3.


4.


5.


6.


7.


8.


9.



Suppose the sequence a1, a2, a3 ...... is an arithmetic progression of distinct numbers such that the sequence a1, a2, a4, a8...... is a geometric progression. The common ratio of the geometric progression is : (A) 2 (B) 4 (C) a1 (D) not determinable

2.

The positve integer k for which (101)k/2/k! is a maximum is : (A) 9 (B) 10 (C) 11

3.

(D) 101

Let p(x) = a0 + a1x + .......+ anxn be a zero polynomial with integer coefficients. If p( 2  3  6 ) = 0, the smallest possible value of n is : (A) 8 (B) 6

(C) 4

(D) 2

4.

Three players play a total of 9 games. In each game, one person wins and the other two lose ; the winner gets 2 points and the losers get 1 each. The number of ways in which they can play all the 9 games and finish each with a zero score is : (A) 84 (B) 1680 (C) 7056 (D) 0

5.

In a triangle, two vertices are (2, 3) and (4, 0), and its circumcentre is (2, z) for some real number z. The circumradius is :

6 (A)

6.

2  13

(B)

5

(C) 2

(D)

13 6

Consider an ellipse with foci at (5, 15) and (21, 15). If the x-axis is a tangent to the ellipse, then the length of its major axis equals : (A) 17

(B) 34

(C) 13

(D)

416

7.

Let the line 2x + 3y = 18 intersect the y-axis at B. Suppose C ( B), with coordinates (a, b), is a point on the line such that PB = PC, where P = (10, 10). Then 8a + 2b equals : (A) 60 (B) 62 (C) 66 (D) 79

8.

If cosec2( + ) – sin2( – ) + sin2(2 – ) = cos2( – ), where ,   (0, /2) then sin( – ) is equal to : (A)

9.

If sinx + siny =

(A)

10.

1 2

7 25

(B)

1 2

 3 2

(D)

3 2

(D)

24 25

7 1 and cos x + cosy = , the sin(x + y) equals : 5 5

(B)

24 25

The number of solutions to sin x = (A) 1

(C)

(B) 6

(C)

7 25

6 with 0  x  12 is : x (C) 10

(D) 12


11.

Define a function f : R  R given by :

 sin x 2  , for x  0 f(x) =  2 x x  ax  b, for x  0  Suppose f(x) differentiable on R. Then, (A) a =0, b = 0 (B) a = 1, b = 0 12.

(C) a = 0, b = 1

(D) a = 1, b = 1

The shortest distance from (0, 3) to the parabola y2 = 4x is : (A) 2

(B)

(C) 5

2

(D)

5

13.

Ten trucks, numbered 1 to 10, are carrying packets of sugar. Each packet weighs either 999 gms or 1000 gms and each truck carries only the packets equal weights. The combined weight of 1 packet selected from the first truck, 2 packets from the second, 4 packet from the third, and so on and 29 packets from the tenth truck is1022870 gms. The trucks that have the lighter bags are : (A) 1, 3, 5 (B) 2, 4, 5 (C) 1, 9 (D) 2, 8

14.

What is the value of

1

(A) 1

 cos( x ) cos([2x]) dx ? (Here | t | denotes the integral part of the real number t) 0

(B) – 1 1

15.

The value of the limit lim

n 0

(A) 0 16.

(B)

2 

(D)

 2

(D) 1

2 

x10 sin(nx )dx equals : 1 10!

(C)

The area bounded by the parabolas y = x2 and y = 1 – x2 equals : (A)

17.



(C)

2 3

(B)

2 2 3

(C)

1 3

(D)

2 3

      A vector which bisects the angle between a  3 i  4k and b = 5 j  12k is :

   (A) 39 i – 25 j  8k

  5 (B) 3 i  5 j  k 8

   (C) 39 i  25 j  8k

  5 (D) 3 i  5 j  k 8

18.

An envelope has space for at most 3 stamps. If you are given three stamps of denomination 1, and three stamps of denomination a (a > 1), the least positive integer for which there is no stamp value is : (A) 7 (B) 8 (C) 9 (D) 10

19.

If m, n are positive integers such that m < n and

 d =  d (here d | k moeans d is a positive of k), then : d/n

d/m

1

(A)

1

d < d

d/m

d/n

1

(C) 20.

1

d = d

d/m

d/n

1

d > d

d/m

1

(B)

(D) no relationship can be determined

d/n

The number of relation R from an m-element set A to an n-element set B satisfying the condition : (a, b1) R, (a, b2)  R  b1 = b2 for a  A, b1, b2  is : (A) nm (B) 2m + n – 2m – 2n (C) mn (D) (n + 1)m


PHYSICS 21.

The relation – Cv = R(, Cv : Molar specific heats at constant pressure, volume) is exactly true for : (A) an ideal mono-atomic gas (B) any ideal gas, whether mono-dia polyatomic (C) any real gas above its critical temperature (D) all real gases

22.

The molecules of air in the room that you are sitting are all experiencing the force of gravity tending to bring them down. The molecules are also frequently and randomly undergoing collisions, which tend to oppose the effect of fall under gravity. The density of air is nearly uniform throughout the room because (A) the mass of the molecules is very small (B) the gravitational potential energy mgh is such lesser than the average thermal energy kT. (C) the gravitational potential enrgy mgh is mush greater the the average thermal energy kT. (D) mgh is nearly of the same magnitude as kT, which results in the cancellation of the two opposing factors. A parallel plate capacitor is charged fully by using a battery. Then without disconnecting the battery, the plates are moved further apart. Then, (A) the charge on the capacitor increases (B) the voltage difference between the plates decreases (C) the capacitance increases (D) the electrostatic energy stored in the capacitor decreases

23.

24.

     The five sides of a regular pentagon are represented by vectors A 1 , A 2 , A 3 , A 4 and A 5 , in cyclic order      as shown. Corresponding vertices are represented by B1 , B 2 , B 3 , B 4 and B 5 drawn from the centre of the pentagon. A2

A1 B1 B5

B2 A5

B4

B3

A3

A4

    Then B 2 + B 3 + B 4 + B 5 =   (A) A 1 (B) – A 1 25.

 (C) B1

 (D) – B1

Four metallic plates each of surface area (of one side) A, are placed at a distance d apart from each other. The two outer plates are connected to a point P and the two inner plates to another point Q as shown in figure. Q

P

Then the capacitance of the system is (A)  0

A 2d

(B)  0

A d

(C) 2 0

A d

(D) 3 0

A d

26.

A progressive wave travelling in positive x-direction given by y = a cos (kx – t) meets a denser surface at x = 0, t = 0. The reflected wave is then given by (A) y = – a sin(kx – ax) (B) y = – a cos(kx + ax) (C) y = a sin (t – kx) (D) y = a cos (kx – t)

27.

A charge Q is spread non uniformly on the surface of a hollow sphere of radius R, such that the charge density is given by  = 0 (1 – sin), where  is the usual polar angle. The potential at the centre of the sphere is

Q (A) 2 R 0

Q (B)  R 0

Q (C) 8 R 0

Q (D) 4 R 0


28.

An ideal diatomic gas is heated at constant pressure. The ratio of the work done to the heat supplied is (A)

29.

2 5

(C)

2 7

(D)

4 7

5 27

(B)

5 48

(C)

27 5

(D)

1 3

Two identical conducting spheres carry identical charges. If the spheres are set at a certain distance apart, they repel each other with a force F. A third conducting sphere, identical to the other two, but initially uncharged, is then touched to one sphere, and then to the other before being removed. The forcebetween the original two spheres is now (A)

31.

(B)

In the hydrogen spectrum, the ratio of the wavelengths for Lyman  – radiation to Balmer –  radiation is (A)

30.

3 5

F 2

(B)

F 4

(C)

3F 4

(D)

3F 8

A small rectangular loop of wire in the plane of the paper is moved with uniform speed across a limited region of uniform magnetic field perpendicular to the plane of the paper, as shown. Uniform magnetic field Initial position

x x x x x x x x

x x x x x x x x

x x x x x x x x

x x x x x x x x

x x x x x x x x

x x x x x x x x

x x x x x x x x

x x x x x x x x

x x x x x x x x

x x x x x x x x

Final position

Which graph would best represent the variation of the electric current, I, in the wire with time t ?

I

(A)

(B)

t

I t

I

(C)

32.

I

(D)

t

t

I=0

The moment of inertia of a solid disc made of thin metal of radius R and mass M about one of its diameter is given by

MR 2 . What will be the moment of inertia about this axis if the disc is folded in half about this 4

diameter ? (A)

33.

MR 2 8

(B)

MR 2 2

(C)

MR 2 4

(D) MR2

A plane electromagnetic weve propagating in the direction of the unit vector nˆ with a speed c is described   by electric and magnetic field vectors E and B , respectively. Which of the following relations ( in SI units) s)   between E and B can be ruled out on dimensional grounds alone ?    nˆ  E nˆ  B      (A) E = (B) E  cnˆ  B (C) B  (D) nˆ  E  B  0 c c


34.

A point electric dipole placed at the origin has a potential given by V(r, ) =

p cos  4 0r 2

where  is the angle

made by the position vector with the direction of the dipole. Then (A) since the potential vanishes at  =

  , the electric field is zero everywhere on the  = plane 2 2

(B) the electric field everywhere on the  =

 plane is normal to the plane. 2

(C) the electric field everywhere on the  =

 plane is along the plane 2

(D) the electric field vanishes on the  = 0 35.

A uniform non-deformable cylinder of mass m and radius R is rolling without slipping on a horizontal rough surface. The force of friction is (A)  mg, where  is the coefficient of sliding friction (B) zero (C) increases with time (D) decreases with time

36.

Consider a one-dimensional potential V(x) as shown in the figure below. A classical particle of mass m moves under its influence and has total energy E as shown.

V(x) E

x

The motion is (A) non-periodic (C) periodic but not simple harmonic

(B) stationary (D) simple harmonic

37.

A source of frequency f is emitting sound waves. If temperature of the medium increases, then (A) wavelength of the sound wave increases (B) speed of the sound wave decreases (C) wavelength of the sound wave decreases (D) amplitude of the sound wave increases

38.

A block of mass m is stationary on a rough plane of mass M inclined at an angle  to the horizontal while the whole set up is accelerating upwards at an acceleration a. If the coefficient of friction between the block and the plane is , then the force that the plane exerts on the block is (A) m (g + a) upwards (B) mg cos  normal to the plane (C) resultant of mag cos  normal to the plane and  mg cos  along the plane (D) resultant of m (g + a) cos  normal to the plane and mg cos  along the plane.


39.

A stream of charged particles enter into a region with crossed electric and magnetic fields as shown in the figure. On the other side is a screen with a hole that is right on the original path of the particles then, E

B

Charged Particles

(A) no particle can get through the hole (B) all particles can get through the hole (C) only positively charged particles with speed (D) all particles with speed 40.

E can get through the hole B

E can get through the hole. B

A small body is released from a height H of an inclined plane. At the bottom of the plane is a loop of radius R as shown.

H

O R x

Ignoring friction, the minimum H required for the body to just complete the loop ( that is, reach the point O) is (A) 2R

(B)

5R 2

(C) 3R

(D)

7R 2

CHEMISTRY 41.

The gas that has the slowest rate of diffusion among O2, H2, CO2 and CH4 is (A) O2 (B) H2 (C) CO2 (D) CH4

42.

Assuming ideal behaviour the ratio of kinetic energies of 3 g of H2 and 4g of O2 at any temperature is (A) 3 : 4 (B) 1 : 16 (C) 4 : 3 (D) 12 : 1

43.

The IUPAC name for the compound O

Cl

is

Me Me

44.

(A) 1-chloro - 3-methyl- 4-pentanone (C) 5-chloro - 3 - methyl- 2-pentanone

(B) 1-chloro - 2-methyl- 4-pentanone (D) 5-chloro - 2-methyl- 3-pentanone

The shape of the molecule CIF3 is (A) triangular (B) pyramidal

(C) T-shape

(D) linear


45.

Among CO32–, OH–, NH3 and HCO3–, the species that acts as a Bronsted acid as well as a Bronsted base is (A) Na2CO3 (B) NH3 (C) OH– (D) HCO3–

46.

The ratio of the heat capacities /Cv for one mole of a gas is 1.67. The gas is : (A) He (B) H2 (C) CO2 (D) CH4

47.

The ion that is isoelectronic with CO is (A) O2+ (B) O2–

48.

49.

(D) N2+

(C) CN–

Among CH4, CO2, H2O and SO2, the bond angle is the highest in (A) CH4 (B) CO2 (C) H2O

(D) SO2

The solvent of choice for carrying out a Grignard reaction is (A) diethyl ether (B) chloroform (C) ethyl acetate

(D) ethanol

50.

The reaction of butanal with n-propylmagnesium bromide gives a (A) chiral secondary alcohol (B) achiral secondary alcohol (C) chiral tertiary alcohol (D) achiral tertiary alcohol

51.

The hybridization of Ni centre in Ni[(PPh3)2Cl2] and [NiCl4]2– respectively are (A) dsp2 and sp3 (B) dsp2 and sp2d (C) sp3 and sp3 (D) sp3 and dsp2

52.

Oxalic acid when treated with potassium permanganate in the presence of an acid, produces (A) O2 (B) C (C) CO (D) CO2

53.

The equilibrium constant for the reaction N2 + 3H2 2NH3 at 400 K is 41. The equilibrium constant for the reaction 1/2N2 + 3/2H2 NH3 at the same temperature will be closest to (A) 41 (B) 20.5 (C) 6.4 (D) 1681

54.

In a one component second order reaction, if the concentration of the reactant is reduced to half, the rate (A) increases two times (B) increases four times (C) decreases to one half (D) decreases to one fourth

55.

The conjugate bases for HCO3– and NH3, respectively, are (A) H2CO3 and NH4+ (B) CO32– and NH2– (C) H2CO3 and NH2–

56.

(D) CO32– and NH4+

Among the following

S CH2 I II the aromatic compounds are (A) I & II (B) I & III 57.

IV

III (C) II & III

(D) II & IV

Among the compounds O

N H I the order of basicity is (A) I > III > II > IV

N

N H

N H

II

III

IV

(B) II > IV > I > III

(C) III > I > IV > II

(D) II > I > III > IV


58.

The Newman projection of Me H

Me

H

H H

is known as the (A) eclipsed conformer

(B) staggered conformer (C) skewed conformer

(D) gauche conformer

59.

The half-life of a first order reaction is 30 min. The time required for 75% completion of the same reactionwill be (A) 45 min (B) 60 min (C) 75 min (D) 90 min

60.

The hydrogen ion concentration in a mixture of 10 ml of 0.1 M H2SO4 and 10 ml of 0.1 M KOH solution in water, is (A) 0.1 M (B) 0.05 M (C) 0.2 M (D) 0.02 M

BIOLOGY 61.

62.

During photosynthesis the chemical conversion of water is termed : (A) photolysis (B) hydrolysis (C) hydration

(D) condensation

In the organisms muscle, oxygen is carried by : (A) albumin (B) myosin (C) myoglobin

(D) hemoglobin

63.

Enzymes do the following : (A) make products and reactants of equal energy (B) help the chemical processes by lowering the energy of products (C) reduce the activation barrier and speed up chemical processes (D) hydrolyze all the biopolymers indiscriminately

64.

Glycolysis is : (A) biosynthesis of glucose (C) degradation of glucose

65.

(B) biosynthesis of glycine (D) reaction of glucose with proteins

Plants are attracted to light through the hormonal action of : (A) Gibberelic acid (B) Auxin (C) Chlorophyll

(D) Thiamine

66.

During development, unspecified cells become cells having unique functions. This process is called : (A) evolution (B) differentiation (C) translation (D) replication

67.

The chromosomal attachment site of the spindle microtubule is : (A) centrosome (B) liposome (C) centromere

(D) telomere

Which fo the following diseases is NOT sexually transmitted ? (A) Syphilis (B) Gonorrhoea (C) AIDS

(D) Tuberculosis

68.

69.

This cell organelle consists of two granule-like centrioles and is found in animal cells only. It helps in cell division. What is it called ? (A) centrosome (B) chromosome (C) centromere (D) chromatids


70.

Nucleotides are monomers of DNA. What does each nucleotide consist of ? (A) A nitrogenous base and a pentose sugar (B) A nitrogenous base and a phosphate group (C) A pentose sugar and a phosphate group (D) A nitrogenous base, a pentose sugar and a phosphate group

71.

Fetilization in humans usually takes place in : (A) Uterus (B) Graafian follicle

(C) Ovary

(D) Fallopian tube

72.

ELISA, the standard screening test for HIV, detects which of the following ? (A) HIV DNA (B) HIV RNA (C) HIV proteins (D) Antibodies to HIV proteins

73.

Sickle cell anemia is caused by : (A) complete absence of the haemoglobin gene (C) increased affinity of haemoglobin for oxygen

(B) point mutation of the haemoglobin gene (D) truncation of the haemoglobin protein

The natural source of Ti plasmid is : (A) bacteria (B) virus

(C) plants

74.

(D) animals

75.

Earthworms are bisexual but still cross-fertilization is common. This is because : (A) spermatozoa of different earthworms are different (B) spermatozoa and ova mature at different times in the same earthworm (C) ova from other earthworms may be larger (D) sperm and ova from the same earth worm cannot fertilize

76.

One difference between blood and lymph is that : (A) blood contains WBC and lymph contains RBC (B) blood contains RBC and WBC and lymph contains only WBC (C) blood contains RBC and lymph contains WBC (D) blood is liquid while lymph is solid

77.

The abnormal development of which of the following lymphoid organs would result in the most severe immunodeficiency ? (A) Spleen (B) Thymus (C) Tonsil (D) Lymph node

78.

Mitochondria are associated with all of the following functions, EXCEPT : (A) ATP synthesis (B) DNA syntheis (C) Protein synthesis

(D) Protein glycosylation

79.

The probability of having a girl child with blood group O when the parents have blood group A and B is : (A) 0% (B) at least 50% (C) at most 25% (D) exactly 75 %

80.

Wooden doors and windows swell up in the rainy season by : (A) a special type of diffusion called imbibition (B) evaporation of stored water wood (C) conduction of water from walls (D) transpiration



Let p(x) = a0 + a1x + ........+ anxn. If p(– 2) = – 15, p(– 1) = 1, p(0) = 7, p(1) = 9, p(2) = 13 and p(3) = 25, then the smallest possible value of n is : (A) 5 (B) 4 (C) 3 (D) 2

82.

Let a, b, c be the sides of triangle. If t denotes the expression (a2 + b2 + c2)/(ab + bc + ca), the set of all possible values of t is : (A) {x  R| x > 1} (B) {x  R | 1  x < 2} (C) {x R | 1  x < 2} (D) {x  R | 1  x  2}

83.

The area of the region bounded by y = | x  3 |  4 – 5 and the x-axis is (A) 24.5

84.

(C) 49

1 ( 6  2) 2

(B)

1 ( 5  1) 2

Define a sequence {an}n  0 by an =

87.

(C)

(D)

3

2

1  a n1 for n  1. a0 = cos1. 2

Then lim n  4n(1 – an) equals (A) 2 (B) 2/2 86.

(D) 35 2

The lengths of the sides and the diagonal of an isosceles trapezium from a two-element set {a, b}. If a > b, then a/b equals (A)

85.

(B) 37

(C) /2

The range of the function f(x) = (sin x)sinx defined on (0, ) is (A) (0.1) (B) (e– 1/e , 1) (C) [e– 1/e, 1)

(D) 

(D) [e–1/e , 1]

Let A denote the area bounded by the curve y = 1/x and the lines y = 0, x = 1, x = 10, Let B = 1 + 1 1 1 1 , and let C = + + ......+ . Then 9 3 10 2 (A) C < B < A (C) C < A < B and A – C < B – A

1 + 2

....+

88.

(B) A < C < B (D) C < A < B and B – A < A – C

Two points are randomly choses on the circumference of a circle of radius r. The probability that the distance between the two points is at least r is equal to (A)

2 

(B) sin r

(C)

2 

(D)

2 3

89.

Consider al natural numbers whose decimal expansion has only then even digits 0, 2, 4, 6, 8. Suppose these are arranged in increasing order. If an denotes the n-th number in this sequence, then lim n  log an /log n equals : (A) 0 (B) log510 (C) log210 (D) 2

90.

The sum of all absolute values of the differences of the numbers 1, 2, 3......., n, taken two at a time, i.e.

 i j

equals :

jin

(A)

n 1 3

 

(B)

n 3



(C)

n 1 3

 

(D)

n2 3

 


PHYSICS 91.

A spherical cavity of radius r is carved out of a uniform solid sphere of radius R as shown in the figure.

R r

The distance of the center of mass of the resulting body from that of the solid sphere is given by (A)

92.

(B)

Rr 2

(C) 0

(D)

r3 R 2  Rr  r 2

A plano-convex lens made of material of refractive index  with radius of curvature R is silvered on the curved side. How far away from the lens-mirror must you place a point object so that the image coincides with the object ? (A)

93.

Rr 2

R 

(B) R

(C)

R  1

(D) R

 n2a   P  n moles of a van der Waal gas obeying the equation of state  (V – nb) = nRT, where a and b are V 2   gas dependent constants, is made to undergo a cyclic process that is depicted by a rectangle in the PV diagram as shown in the figure. What is the heat absorbed by the gas in one cycle? P P1

P2 V1

94.

V2

V

(A) n(P1 – P2) (V2 – V1)

(B) (P1 – P2) (V2 – V1)

 n 2a n 2 a   (C)  P1  2  P2  2  (V1 – V2) V1 V2  

 n 2a n 2 a   (D)  P1  2  P2  2  (V2 – V1) V1 V2  

For what value of the resistor X will the equivalent resistance of the two circuits shown be the same ?

R

R 6x

(A) R

6x

R

R

6x

(B) 6R

R x

R 6x

(C) 2R

6x

R

R

6x

(D)

6x

R x

5 1 R 2


95.

2 MR2 rolls down a plane 5 inclined at an angle  to the horizontal starting from rest. The coefficient of static friction between the sphere and plane is . Then, (A) the sphere will always roll without slipping (B) the sphere will always slide

A solid uniform sphere having a mass M, radius R, and moment of inertia of

(C) the sphere will roll without slipping only if   sin

7 2

(D) the sphere will roll without slipping only if   tan–1

96.

A cubical box of side a sitting on a rough table-top is pushed horizontally with a gradually increasing force until the box moves. If the force is applied at a height from the table top which is greater than a critical height H, the box topples first. If it is appliced at a height less than H, the box starts sliding first. Then the coefficient of friction between the box and the table top is : (A)

97.

a 2H

(B)

2H a

(C)

a H

(D)

H a

A vehicle is moving with speed v on a curved road of radius r. The coefficient of friction between the vehicle and the road is . The angle  of banking needed is given by (A) tan =

98.

7 2

v2  r g v2  r g

v2  r g

(B) tan =

v2  r g

(C) tan =

v2  r g r g   v2

(D) tan =

r g  v2 r g   v2

Two small identical speakers are connected in phase to the same source. The speakers are 3m apart and at ear level. An observer stands at P, 4m in front of one speaker as shown. The sound she hears is least intense when the wavelength is 1 and most intense when the wavelength is 2.

4m

3m

Then possible values of 1 and 2 are : (A) 1 = 1m and 2 = 2m (C) 1 = 2m and 2 = 1m 99.

(B) 1 = 4m and 2 = 3m (D) 1 = 0.5m and 2 = 0.25m

Two small blocks slide without losing with the surface along two frictionless tracks 1 and 2, starting at the same time with same initial speed v. Track 1 is perfectly horizontal, while track 2 has a dip in the middle, as shown. V

1

V

2 Start

Finish

Which block reaches the finish line first ? [Hint : Use velocity-time graph to solve] (A) Block on track 1 reaches the finish line first (B) Block on track 2 reaches the finish line first (C) Both blocks reach the finish line at the same time (D) It depends on the length of the dip in the second track, relative to the total length of the tracks.


100.

Consider 1 kg of liquid water undergoing change in phase to water vapour at 100ºC. At 100ºC, the vapour pressure 1.01 × 105 Nm2 and the latent heat of vaporization is 22.6 × 105 J kg–1. The density of liquid water is 103 kg m–3 and that of vapour is change is nearly. (A) 1.8 × 105 J kg–1

(B) 20.8 × 105 J kg–1

1 kg m–3. The change in internal energy in this phase 1 .8

(C) 22.6 × 105 J kg–1

(D) 11.3 × 105 J kg–1

CHEMISTRY 101.

If the pH of a mixture of 10 ml of 0.1 M NH4OH and 10 ml of 1 M NH4Cl solution is 8, the pKb value of NH4OH is then closest to (A) 3 (B) 5 (C) 7 (D) 9

102.

A cylinder of cooking gas in a household contains 11.6 kg of butane.The thermochemical reaction for the combustion of butane is 2C4H10(g) + 13O2(g)  8CO2(g) + 10H2O(); H = – 2658 KJ/mol. If the household needs15000 KJ of energy per day, the cooking gas cylinder will last for about (A) 64 days (B) 45 days (C) 20 days (D) 35 days

103.

The addition of 0.643 g of a compound to 50 ml of benzene (density = 0.879 g ml–1) lowers the freezing point from 5.51ºC to 5.03ºC. If the freezing point constant, Kf for benzene is 5.12 K Kg mol –1, the molar mass of the compound is approximately (A) 156 g mol–1 (B) 88 g mol–1 (C) 60 g mol–1 (D) 312 g mol–1

104.

Consider the following electrochemical cell : Zn(s) + 2Ag+(0.04 M)  Zn2+(0.28M)+ 2Ag(s). If Ecellº = 2.57 V, then e.m.f. of the cell at 298 K, is (A) 2.5 V (B) 1.5 V (C) 0.5 V (D) – 0.5 V

105.

When Co(II) chloride is dissolved in concentrated HCl a blue solution is obtained. Upon dilution with water, the color changes to pink because (A) CoCl64 – is converted to CoCl63 – (B) CoCl42 – is converted to Co(OH2)62 + 2+ 3+ (C) Co(OH2)6 is converted to Co(OH2)6 (D) CoCl42 – is converted to Co(OH2)63 +

106.

The rate constant for the reaction COCl2(g)  CO(g) + Cl2(g) is given by ln [k/(min–1)] = – 11067/T K + 31.33. The temperature at which the rate of this reaction will be doubled from that at 25ºC is (A) 75ºC (B) 100ºC (C) 31ºC (D) 50ºC

107.

Some reactions and their equilibrium constants given below CuCl42– + Br – CuCl3Br2– + Cl – K1 CuCl3Br2– + Br – CuCl2Br22– + Cl – K2 2– – 2– – CuCl2Br2 + Br CuClBr3 + Cl K3 CuClBr32 – + Br – CuBr42– + Cl – K4 The equilibrium constant, K for the reaction CuCl42– + 3Br – CuClBr32– + 3 Cl–, is (B) K1K2K3K4 (C) K1 + K2 + K3 (A) K1K2K3

(D) 1/(K1K2K3)


OH Br2 in CS2

108.

NaOH

X

Me-I

Y

In the above sequence of reactions, the major products X and Y are

OH

OMe Br

(A) X =

OH Br

(C) X =

(B) X =

OMe

Y=

OH

Y= Br

(D) X = Br

109.

234

110.

In the following transformation

OH

Br

Y=

OH

OH

OH

Y= Br

OH

Th90 gets converted to 206Pb82 through a series of radioactive decay processes. The number of alpha and beta particles lost in this transformation respectively, are (A) 6 and 6 (B) 6 and 7 (C) 4 and 2 (D) 7 and 6

O O

OH

Me

CO2H

Reagent 1 Reagent 2 OMe

OMe

reagents 1 and 2 are : (A) H2SO4 ; alkaline KMnO4 (C) H3PO4 ; CHCl3/KOH

(B) AlCl3 ; l2 / NaOH (D) KOH;CHCl3 / KOH

BIOLOGY 111.

112.

The mode of action of penicillin is as follows : (A) It inhibits viral replication (C) It inhibits bacterial cell wall synthesis

(B) It enhances immunity (D) It inhibits transcription

Which of the following statements is true for meiosis ? (A) One round of chromosome duplication and one round of cell division (B) One round of chromosome duplication and two rounds of cell division (C) Two rounds of chromosome duplication and one round of cell division (D) Two rounds of chromosome duplication and two rounds of cell division


113.

Instead of 3, if 2 bases code for an amino acid, the degeneracy of codons coding for the same amino acid would have : (A) increased (B) decreased (C) remained the same (D) been uncertain

114.

Gregor Mendel showed that unit factors exist in pairs and exhibit as dominant-recessive relationship. These unit factors, in modern terminology, are called : (A) genes (B) alleles (C) Ioci (D) determinants

115.

E.coli has optimal growh temperature of 37ºC. Which of the following in an INCORRECT explanation for this ? (A) The membrane is most permeable at this temperature (B) DNA synthesis makes the least mistakes at this temperature (C) Most enzymes in the cell have the highest activity at this temperature (D) Protein synthesis is most efficient at this temperature

116.

Male offsprings of which of the following couples have the highest chance of haemophilia ? (A) Haemophiliac father and normal, non-carrier mother (B) Haemophiliac father and normal, carrier mother (C) Normal father and normal, carrier mother (D) Normal father and haemophiliac mother

117.

The effect of consumption of excess protein by normal individuals would result in : (A) excretion of excess protein in urine (B) increase in the amount of adipose tissue (C) increase in the synthesis of muscle protein (D) increase in the circulatory plasma proteins

118.

The condition varicose veins in swelling of veins, that occurs due to : (A) loss of elasticity of the muscular layer (B) condition of high blood pressure (C) condition of low blood pressure (D) condition of anoxia

119.

Greatest proportion of photosyntheis in the world is carried out by : (A) trees in the rain forests of the world (B) trees in the temperate forests of the world (C) algae in oceans (D) irrigated crop fields

120.

Energetically unfavourable reactions occur in human cells through : (A) heat energy supplied by the body (B) heat energy released through exercise (C) coupling of energetically favourable reactions with unfavourable ones (D) photosynthesis



1.

Max. Marks : 160


2.


3.


4.


5.


6.



1.

Max. Marks : 160


2.


3.


4.


5.


6.



1.

Max. Marks : 160


2.


3.


4.


5.


6.



Max. Marks : 160

1.


2.


Subject

Part-I

Part-II

Mathematics



Physics



Chemistry



Biology



3.


4.


5.


6.


7.


8.


9.



Max. Marks : 160

1.


2.


Subject

Part-I

Part-II

Mathematics



Physics



Chemistry



Biology



3.


4.


5.


6.


7.


8.


9.



Max. Marks : 160

1.


2.


Subject

Part-I

Part-II

Mathematics



Physics



Chemistry



Biology



3.


4.


5.


6.


7.


8.


9.


YEAR-2010 (KVPY-STREAM-SB+2) PART-I (1 Mark) MATHEMATICS

1.

0 i  1 0  , where i2 = –1, and let I denote the identity matrix   . Then Let denote the matrix   i 0  0 1

I + A + A2 + .............+ A2010 is 0 0  (A)  0 0

2.

0 i   (B)   i 0

1 i   (C)   i 1

– 1 0   (D)   0 – 1

Suppose the sides of a triangle from a geometric progression with common ratio r. Then r lies in the interval

 – 1 5    (A)  0, 2  

 1 5 2  5   (B)  2 , 2   

 – 1 5 1 5   ,  (C)  2 2  

2 5   (D)  2 ,   

3.

The number of rectangles that can be obtained by ing four of the twelve vertices of a 12-sided regular polygon is (A) 66 (B) 30 (C) 24 (D) 15

4.

Let 1,  and 2 be the cube roots of unity. The least possible degree of a polynomial, with real coefficients, having 22, 3 + 4, 3 + 42 and 5 – –2 as roots is (A) 4 (B) 5 (C) 6 (D) 8

5.

A circle touches the parabola y2 = 4x at (1, 2) and also touches its directrix. the y-coordinate of the point of of the circle and the directrix is (A) 2

6.

(B) 2

(C) 2 2

(D) 4

Let ABC be an equilateral triangle; let KLMN be a rectangle with K, L on BC, M on AC and N on AB. Suppose AN / NB = 2 and the area of triangle BKN is 6. The area of the triangle ABC is

7.

(A) 54

(B) 108

(C) 48

(D) not determinable with the above data

Let P be an arbitrary point on the ellipse

x2 a2



y2 b2

 1 , a > b > = 0. Suppose F and F are the foci of the 1 2

ellipse. The locus of the centroid of the triangle PF1F2 as P moves on the ellipse is (A) a circle 8.

(B) an ellipse

(C) a parabola

(D) a hyperbola

The number of roots of the equation cos7  – sin6 = 1 that lie in the interval [0, 2] is (A) 2

(B) 3

(C) 4

(D) 8


9.

The product (1 + tan 1º) (1 + tan 2º) (1 + tan 3º) ... (1 + tan 45º) equals (A) 221

10.

(B) 222

(C) 223

(D) 224

Let f : R  R be a differentiable function such that f (a) = 0 = f (b) and f’(a) f’(b) > 0 for some a < b. Then the minimum number of roots of f’(x) = 0 in the interval (a, b) is (A) 3

11.

(B) 2

(C) 1

(D) 0

The roots of (x – 41)49 + (x – 49)41 + (x – 2009)2009 = 0 are (A) all necessarily real (B) non-real except one positive real root (C) non-real except three positive real roots (D) non-real except for three real roots of which exactly one is positive

12.

The figure shown below is the graph of the derivative of some function y = ƒ (x)

Then (A) f has local minima at x = a, b and a local maximum at x = c (B) f has local minima at x = b, c and a local maximum at x = a (C) f has local minima at x = c, a and a local maximum at x = b (D) the given figure is insufficient to conclude any thing about the local minima and local maxima of f


13.

The following figure shows the graph of continuous function y = ƒ (x) on the interval [1, 3]. The points A, B, C have coordinates (1, 1), (3, 2), (2, 3) respectively, and the lines 1 and 2 are parallel, with 1 being tangent to the curve at C. If the area under the graph of y = ƒ (x) from x = 1 to x = 3 is 4 square units, then the area of the shaded region is :

(A) 2

(B) 3

(C) 4

(D) 5

e

14.

n Let  n  (log x ) dx, where n is a non-negative integer..

 1

Then I2011 + 2011 I2010 is equal to

15.

(A) 1000 + 999 998

(B) 889 + 890 891

(C) 100 + 100 99

(D) 53 + 54 52

Consider the regions A = {(x, y) | x2 + y2  100} and B = {(x, y) | sin (x + y) > 0} in the plane. Then the area of the region A  B is (A) 10 

16.

(B) 100

(C) 100 

(D) 50 

Three vertices are chosen randomly from the seven vertices of a regular 7–sided polygon. The probability that they form the vertices of an isosceles triangle is (A)

17.

(B)

1 3

(C)

3 7

(D)

3 5

  Let u = 2î – ˆj + kˆ , v = – 3 ˆj + 2 kˆ be vectors in R3 and w be a unit vector in the xy-plane. Then the    maximum possible value of |( u × v ) . w | is (A)

18.

1 7

5

(B)

12

(C) 13

(D) 17

How many six-digit numbers are there in which no digit is repeated, even digits appear at even places, odd digits appear at odd places and the number is divisible by 4? (A) 3600

(B) 2700

(C) 2160

(D) 1440


19.

The number of natural number n in the interval [1005, 2010] for which the polynomial 1 + x + x2 + x3 +.......+ xn–1 divides the polynomial 1 + x2 + x4 + x6 + .........+ x2010 is (A) 0

20.

(B) 100

(C) 503

(D) 1006

Let a0 = 0 and an = 3 an–1 + 1 for n  1. Then the remainder obtained on dividing a2010 by 11 is (A) 0

(B) 7

(C) 3

(D) 4

PHYSICS 21.

A pen of mass m is lying on a piece of paper of mass M placed on a rough table. If the coefficients of friction between the pen and paper, and, the paper and the table are 1 and 2 respectively, then the minimum horizontal force with which the paper has to be pulled for the pen to start slipping is given by :

22.

(A) (m + M) (1 + 2)g

(B) (m 1 + M 2)g

(C) (m 1 + (m + M)2)g

(D) m(1 + 2)g

Two masses m1 and m2 connected by a spring of spring constant k at rest on a frictionless surface. If the masses are pulled apart and let go, the time period of oscillation is :

1  m1m 2 (A) T = 2 k  m  m 2  1

23.

 m1  m 2 (B) T = 2 k  m m 1 2 

  

  (C) T = 2 m1 (D) T = 2 m 2 k k 

A bead of mass m is attached to the mid–point of a taut, weightless string of length  and placed on a frictionless horizontal table.

Under a small transverse displacement x, as shown, if the tension in the string is T, then the frequency of oscillation is :

(A)

24.

1 2T 2 m 

(B)

1 4T 2 m 

(C)

1 4T 2 m

(D)

1 2T 2 m

A comet (assumed to be in an elliptical orbit around the sun) is at a distance of 0.4 AU from the sun at the perihelion. If the time period of the comet is 125 years, what is the aphelion distance ? AU : Astronomical Unit. (A) 50 AU

(B) 25 AU

(C) 49.6 AU

(D) 24.6 AU


25.

The circuit shown consists of a switch (S), a battery (B) of emf E, a resistance R, and an inductor L.

The current in the circuit at the instant the switch is closed is : (A) E/R 26.

(B) E/(R(1 – e))

(C) 

(D) 0

Consider a uniform spherical volume charge distribution of radius R. Which of the following graphs correctly represents the magnitude of the electric field E at a distance r from the center of the sphere ?

(A)

27.

(B)

(C)

(D)

A charge +q is placed somewhere inside the cavity of a thick conducting spherical shell of inner radius R1 and outer radius R2. A charge +Q is placed at a distance r > R2 from the center of the shell. Then the electric field in the hollow cavity.

28.

(A) depends on both +q and +Q

(B) is zero

(C) is only that due to +Q

(D) is only that due to +q

The following travelling electromagnetic wave Ex = 0, Ey = E0sin(kx + t), Ez = –2E0sin(kx + t) is :

29.

(A) elliptically polarized

(B) linearly polarized

(C) circularly polarized

(D) unpolarized

A point source of light is placed at the bottom of a vessel which is filled with water of refractive index  to a height h. If a floating opaque disc has to be placed exactly above it so that the source is invisible from above, the radius of the disc should be : h

h (A) 30.

 1

(B)

2

 1

h

h (C)

(D)

2

 1

2  1

Three transparent media of refractive indices 1,2, 3, respectively, are stacked as shown. A ray of light follows the path shown. No light enters the third medium.

Then : (A) 1 < 2 < 3

(B) 2 < 1 < 3

(C) 1 < 3 < 2

(D) 3 < 1 < 2


31.

A nucleus has a half–life of 30 minutes. At 3PM its decay rate was measured as 120,000 counts/sec. What will be the decay rate at 5 PM ?

32.

(A) 120,000 counts/sec.

(B) 60,000 counts/sec.

(C) 30,000 counts/sec.

(D) 7,500 counts/sec.

A block is resting on a shelf that is undergoing vertical simple harmonic oscillations with an amplitude of 2.5 cm. What is the minimum frequency of oscillation of the shelf for which the book will lose with the shelf ? (Assume that g = 10 m/s2) (A) 20 Hz

33.

(C) 125.6 Hz

(D) 10 Hz

 n2a   P  A vander Waal's gas obeys the equation of state  (V – nb) = nRT. Its internal energy is given by V 2  

U = CT –

34.

(B) 3.18 Hz

n2a . The equation of a quasistatic adiabat for this gas is given by : V

(A) TC/nRV = constant

(B) T(C+nR)/nRV = constant

(C) TC/nR (V – nb) = constant

(D) P(C+nR)/nR (V – nb) = constant

An ideal gas is made to undergo a cycle depicted by the PV diagram alongside. The curved line from A to B is an adiabat.

Then : (A) The efficiency of this cycle is given by unity as no heat is released during the cycle (B) Heat is absorbed in the upper part of the straight line path and released in the lower part (C) If T1 and T2 are the maximum and minimum temperatures reached during the cycle, then the efficiency

T2 is given by 1 – T 1 (D) The cycle can only be carried out in the reverse of the direction shown in the figure

35.

A bus driving along at 39.6 kmph is approaching a person who is standing at the bus stop, while honking repeatedly at an interval of 30 seconds. If the speed of sound is 330 ms–1, at what interval will the person hear the horn. (A) 31 sec.

(B) 29 sec.

(C) 30 sec.

(D) The interval will depend on the distance of the bus from the enger


36.

Velocity of sound measured at a given temperature in oxygen and hydrogen is in the ratio : (A) 1 : 4

37.

(B) 4 : 1

(C) 1 : 1

(D) 32 : 1

In Young's double slit experiment, the distance between the two slits is 0.1 mm, the distance between the slits and the screen is 1m and the wavelength of the light used is 600 nm. The intensity at a point on the screen is 75% of the maximum intensity. What is the smallest distance of this point from the central fringe? (A) 1.0 mm

38.

(B) 2.0 mm

(C) 0.5 mm

(D) 1.5 mm

Two masses m1 and m2 are connected by a massless spring of spring constant k and unstretched length . The masses are placed on a frictionless straight channel – which we consider our x-axis. They are initially at rest at x = 0 and x = , respectively. At t = 0, a velocity of v0 is suddenly imparted to the first particle. At a later time t, the center of mass of the two masses is at :

39.

m2 (A) x = m  m 1 2

m2 v 0 t m1 (B) x = m  m  m  m 1 2 1 2

m2 v 0 t m2  (C) x = m  m  m  m 1 2 1 2

m1v 0 t m2  (D) x = m  m  m  m 1 2 1 2

A charged particle of charge q and mass m, gets deflected through an angle  upon ing through a square region of side a which contains a uniform magnetic field B normal to its plane. Assuming that the particle entered the square at right angles to one side, what is the speed of the particle ? (A)

40.

qB a cot  m

(B)

qB a tan  m

(C)

qB a cot 2  m

(D)

qB a tan 2  m

A piece of hot copper at 100ºC is plunged into a pond at 30ºC. The copper cools down to 30ºC, while the pond, being huge, stays at its initial temperature. Then : (A) copper loses some entropy, the pond stays at the same entropy (B) copper loses some entropy, and the pond gains exactly the same amount of entropy (C) copper loses entropy, and the pond gains more than this amount of entropy (D) both copper and the pond gain in entropy

CHEMISTRY 41.

The number of isomers of Co(diethylene triamine) Cl3 is. (A) 2

42.

(C) 4

(D) 5

(B) NH3

(C) CN–

(D) I–

(B) 3

(C) 1.5

(D) 1

Among the following, the -acid ligand is : (A) F–

43.

(B) 3

The bond order in O22– is (A) 2


44.

The energy of a photon of wavelength  = 1 meter is (Planck's constant = 6.626  10–34 J.s, speed of light = 3  108 ms–1) (A) 1.988  10–25 J

45.

(B) 1.988  10–30 J

(C) 1.988  10–28 J

(D) 1.988  10–31 J

The concentration of a substance undergoing a chemical reaction becomes one half of its original value after time t, regardless of the initial concentration. The reaction is an example of a :

46.

(A) zero order reaction

(B) second order reaction

(C) first order reaction

(D) third order reaction

The shape of the molecule CIF3 is : (A) trigonal planar

47.

48.

49.

(C) pyramidal

(D) Y-shaped

Friedal–Crafts acylation is : (A) -acylation of a carbonyl compound

(B) acylation of phenols to generate esters

(C) acylation of aliphatic olefins

(D) acylation of aromatic nucleus

The order of acidity of compounds I–IV, is

I

II

III

IV

(A) I < III < II < IV

(B) III < I < II < IV

(C) IV < I < II < III

(D) II < IV < III < I

(C)

(D)

(C) 11H

(D) 12 H

The most stable conformation for n-butane is

(A)

50.

(B) T-shaped

(B)

In the nuclear reaction 234 90 Th

234  91 Pa + X

X is : (A)

51.

0 1e

(B) 10 e

A concentrated solution of copper sulphate, which is dark blue in colour, is mixed at room temperature with a dilute solution of copper sulphate, which is light blue. For this process (A) Entropy change is positive, but enthalpy change is negative. (B) Entropy and enthalpy changes are both positive. (C) Entropy change is positive and enthalpy does not change. (D) Entropy change is negative and enthalpy change is positive.


52.

53.

Increasing the temperature increacses the rate of reaction but does not increase the : (A) number of collisions

(B) activation energy

(C) average energy of collisions

(D) average velocity of the reactant molecules

In metallic solids, the number of atoms for the face centred and the body-centered cubic unit cells, are, respectively. (A) 2,4

54.

(B) 2,2

(C) 4,2

From equation 1 and 2 CO2

CO +

H2 O

H2 +

1 O 2 2 1 O 2 2

[ K c = 9.1  10–12 at 1000°C]

(eq 1)

[ Kc

(eq 2)

1

2

find out Kc for the following equation CO2 + H2 (A) 0.78 55.

(B) 2.0

CO + H2O (C) 16.2

(D) 1.28

(B) [R0] (e–kt)

(C) [R0e]–kt

(D) [R0] (–ekt)

(C)

(D)

The correct structure of PCl3F2 is

(A)

57.

= 7.1  10–12 at 1000°C]

For a first order reaction R  P, the rate constant is k. If the initial concentration of R is [R0], the concentration of R at any time 't' is given by the expression. (A) [R0] ekt

56.

(D) 4,4

(B)

The enantiomeric pair among the following four structures is

I

II

III

IV

(A) I & II

(B) I & IV

(C) II & III

(D) II & IV


58.

59.

Consider the reaction : 2NO2 (g)  2NO (g) + O2(g). In the figure below, identify the curves X,Y and Z associated with the three species in the reaction

(A) X = NO, Y = O2, Z = NO2

(B) X = O2, Y = NO, Z = NO2

(C) X = NO2, Y = NO , Z = O2

(D) X = O2, Y= NO2, Z = NO

The aromatic carbocation among the following is

(A)

60.

(B)

(C)

(D)

Cyclohexene is reacted with bromine in CCl4 in the dark. The product of the reaction is

(A)

(B)

(C)

(D)


BIOLOGY 61.

Ribouncleic Acid (RNA) that catalyze enzymatic reactions are called ribozymes. Which one of the following acts as a ribozyme? (A) Ribosome

62.

(B) Amylase

(C) tRNA

(D) Riboflavin

In 1670, Robert Boyle conducted an experiment wherein he placed a viper (a poisonous snake) in a chamber and rapidly reduced the pressure in that chamber. Which of the following would be true? (A) Gas bubble developed in the tissues of the snake (B) The basal metabolic rate of the snake increased tremendously (C) The venom of the snake was found to decrease in potency (D) The venom of the snake was found to increase in potency

63.

Bacteria can survive by absorbing soluble nutrients via their outer body surface, but animals cannot, because (A) Bacteria cannot ingest particles but animals can (B) Bacteria have cell walls and animals do not (C) Animals have too small a surface area per unit volume as compared to bacteria (D) Animals cannot metabolize soluble nutrients

64.

A horse has 64 chromosomes and a donkey has 62. Mules result from crossing a horse and donkey. State which of the following is INCORRECT? (A) Mules can have either 64, 63 or 62 chromosomes (B) Mules are infertile (C) Mules have well defined gender (male/female) (D) Mules have 63 chromosomes

65.

If the total number of photons falling per unit area of a leaf per minute is kept constant, then which of the following will result in maximum photosynthesis?

66.

67.

(A) Shining green light

(B) Shining sunlight

(C) Shining blue light

(D) Shining ultraviolet light

Path-finding by ants is by means of (A) Visually observing landmarks

(B) Visually observing other ants

(C) Chemical signals between ants

(D) Using the earth's magnetic field

Sometimes urea is fed to ruminates to improve their health. It works by (A) Helping growth of gut microbes that break down cellulose (B) Killing harmful microorganisms in their gut (C) Increasing salt content in the gut (D) Directly stimulating blood cell proliferation


68.

If you compare adults of two herbivore species of different sizes, but from the same geographical area, the amount of faeces produced per kg body weight would be (A) More in the smaller one than the larger one (B) More in the larger one than the smaller one (C) Roughly the same amount in both (D) Not possible to predict which would be more

69.

Fruit wrapped in paper ripens faster than when kept in open air because (A) Heat of respiration is retained better (B) A chemical in the paper helps fruit ripening (C) A volatile substance produced by the fruit is retained better and helps in ripening (D) The fruit is cut off from the ambient oxygen which is an inhibitor to fruit ripening.

70.

When a person is suffering form high fever, it is sometimes observed that the skin has a reddish tinge. Why does this happen? (A) Red colour of the skin radiates more heat (B) fever causes the release of a red pigment in the skin (C) There is more blood circulation to the skin to keep the body warm (D) There is more blood circulation to the skin to release heat from the body

71.

Bacteriochlorophylls are photosynthetic pigments found in phototrophic bacteria. Their function is distinct from the plant chlorophylls in that they (A) do not produce oxygen (B) do not conduct photosynthesis (C) absorb only blue light (D) function without a light source

72.

Athletes often experience muscle cramps. Which of the following statements is true about muscle cramps? (A) Muscle cramp is caused due to conversion of pyruvic acid into lactic acid in the cytoplasm (B) Muscle cramp is caused due to conversion of pyruvic acid into lactic acid in the mitochondria (C) Muscle cramp is caused due to nonconversion of glucose to pyruvate in the cytoplasm (D) Muscle cramp is caused due to conversion of pyruvic acid into ethanol in the cytoplasm

73.

A couple went to a doctor and reported that both of them are "carriers" for a particular disorder, their first child is suffering from that disorder and that they are expecting their second child. What is the probability that the new child would be affected by the same disorder? (A) 100% (B) 50% (C) 25% (D) 75%

74.

Of the following combinations of cell biological processes which one is associated with embryogenesis? (A) Mitosis and Meiosis (B) Mitosis and Differentiation (C) Meiosis and Differentiation (D) Differentiation and Reprogramming

75.

Conversion of the Bt toxin produced by Bacillus thuringienesis to its active form in the gut of the insects is mediated by (A) acidic pH of the gut (B) alkaline pH of the gut (C) lipid modification of the protein (D) cleavage by chymotrypsin


76.

If you dip a sack full of paddy seeds in water overnight and then keep it out for a couple of days, it feels warm. What generates this heat? (A) Imbibation (B) Exothermic reaction between water and seed coats (C) Friction among seeds due to swelling (D) Respiration

77.

Restriction endonucleases are enzymes that cleave DNA molecules into smaller fragments. Which type of bond do they act on? (A) N-glycosidic Bond (B) Phosphodiester bond (C) Hydrogen bond (D) Disulfide bond

78.

The fluid part of blood flows in and out of capillaries in tissues to exchange nutrients and waste materials. Under which of the following conditions will fluid flow out from the capillaries into the surrounding tissue? (A) When arterial blood pressure exceeds blood osmotic pressure (B) When arterial blood pressure is less than blood osmotic pressure (C) When arterial blood pressure is equal to blood osmotic pressure (D) Arterial blood pressure and blood osmotic pressure have nothing to do with the outflow of fluid from capillaries

79.

The distance between two consecutive DNA base pairs is 0.34 nm. If the length of a chromosome is 1 mm the number of base pairs in the chromosome is approximately (A) 3 million (B) 1.5 million (C) 30 million (D) 6 million

80.

Estimate the order of the speed of propagation of an action potential or nerve impulse (A) nm/s (B) micron/s (C) cm/s (D) m/s

PART-II (2 Mark) MATHEMATICS n

81.

1   1/ 2 Arrange the expansion of  x  1/ 4  in decreasing powers of x. Suppose the coefficients of the first 2x   three form an arithmetic progression. Then the number of in the expansion having integer powers of x is (A) 1

82.

(B) 2

(C) 3

(D) more than 3

Let r be a real number and n  N be such that the polynomial 2x2 + 2x +1 divides the polynomial (x + 1)n– r. Then (n, r) can be (A) (4000, 41000)

1   (B)  4000, 1000  4  

 1000 1  , 1000  (C)  4 4  

1    (D)  4000, 4000  


83.

Suppose a, b are real numbers such that ab  0. Which of the following four figures represents the curve (y – ax – b) (bx2 + ay2 – ab) = 0 ?

(A) Fig.1 84.

(B) Fig.2

(C) Fig.3

(D) Fig.4

Among all cyclic quadrilaterals inscribed in a circle of radius R with one of its angles equal to 120º, consider the one with maximum possible area. Its area is (A)

85.

2 R2

(B) 2 R2

(C)

3 R2

(D) 2 3 R 2

The following figure shows the graph of a differentiable function y = ƒ (x) on the interval [a, b] (not containing 0).

Let g(x) = ƒ (x) / x. Which of the following is a possible graph of y = g (x)?

(A) Fig.1

(B) Fig.2

(C) Fig.3

(D) Fig.4


86.

Let V1 be the volume of a given right circular cone with O as the centre of the base and A as its apex. Let V2 be the maximum volume of the right circular cone inscribed in the given cone whose apex is O and whose base is parallel to the base of the given cone. Then the ratio V2 / V1 is (A)

3 25

(B)

4 9

(C)

4 27

(D)

8 27

x

87.

Let ƒ : R  R be a continuous function satisfying ƒ( x )  x  ƒ( t ) dt , for all x  R. Then the number of

 0

elements in the set S = {x  R ƒ (x) = 0 } is (A) 1

(B) 2

(C) 3

(D) 4

2

88.

The value of

 min| x –  |, cos

–1



(cos x ) dx is

0

(A)

89.

2 4

(B)

2 2

(C)

2 8

(D) 2

Let ABC be a triangle and P be a point inside ABC such that PA  2PB  3PC  0 . The ratio of the area of triangle ABC to that of APC is (A) 2

90.

(B)

3 2

(C)

5 3

(D) 3

Suppose m, n are positive integers such that 6m + 2m+n 3m + 2n = 332. The value of the expression m2 + mn + n2 is (A) 7

(B) 13

(C) 19

(D) 21

PHYSICS 91.

A ball is dropped vertically from a height of h onto a hard surface. If the ball rebounds from the surface with a fraction r of the speed with which it strikes the latter on each impact, what is the net distance travelled by the ball up to the 10th impact ? (A) 2h

92.

1  r 10 1 r

(B) h

1  r 20 1 r 2

(C) 2h

1  r 22 1 r 2

h

(D) 2h

1  r 20 1 r 2

h

A certain planet completes one rotation about its axis in time T. The weight of an object placed at the equator on the planet's surface is a fraction f (f is close to unity) of its weight recorded at a latitude of 60º. The density of the planet (assumed to be a uniform perfect sphere) is given by : (A)

4  f 3 1  f 4GT 2

(B)

4  f 3 1  f 4GT 2

(C)

4  3f 3 1  f 4GT 2

(D)

4  2f 3  1  f 4GT 2


93.

Three equal charges +q are placed at the three vertices of an equilateral triangle centred at the origin. They are held in equilibrium by a restoring force of magnitude F(r) = kr directed towards the origin, where k is a constant. What is the distance of the three charges from the origin ?  1 q2   (A)   6 0 k 

94.

1/ 3

 3 q2   (B)  12 0 k 

1/ 3

 1 q2   (C)   6 0 k 

2/3

 3 q2   (D)   4  0 k 

2/3

Consider the infinite ladder circuit shown below :

For which angular frequency  will the circuit behave like a pure inductance ?

LC (A) 95.

2

R(2  ) 2(  1)

LC

C

(B)

R(2  ) 2(  1)

(C)

R(2  ) 2(  1)

(D)

R(2  ) 2(  1)

x 1 2 kx – V0cos   , a 2 where V0,k, a are constants. If the amplitude of oscillation is much smaller than a, the time period is given by

ma 2 ka 2  V0

(B) 2

m k

(C) 2

ma 2 V0

(D) 2

ma 2 ka 2  V0

An ideal gas with heat capacity at constant volume CV undergoes a quasistatic process described by PV in a P–V diagram, where  is a constant. The heat capacity of the gas during this process is given by : (A) CV

98.

LC

2L (D)

A particle of mass m undergoes oscillations about x = 0 in a potential given by V(x) =

(A) 2 97.

1 (C)

A narrow parallel beam of light falls on a glass sphere of radius R and refractive index  at normal incidence. The distance of the image from the outer edge is given by : (A)

96.

2 (B)

(B) CV +

nR 1 

(C) CV + nR

(D) CV +

nR 1 2

3 nR is made to carry out 2 a cycle that is depicted by a triangle in the figure given below : The following statement is true about the cycle : An ideal gas with constant heat capacity CV =

P1V1 (A) The efficiency is given by 1 P V 2

2

1 P1V1 (B) The efficiency is given by 1 2 P V 2

2

(C) Net heat absorbed in the cycle is (P2 – P1) (V2 – V1) (D) Heat absorbed in part AC is given by 2(P2V2 – P1V1) +

1 (P V – P2V1) 2 1 2


99.

Two identical particles of mass m and charge q are shot at each other from a very great distance with an initial speed v. The distance of closest approach of these charges is :

q2 (A)

100.

q2 (B)

8 0mv 2

q2

4 0mv 2

(C)

2 0mv 2

(D) 0

At time t = 0, a container has N0 radioactive atoms with a decay constant . In addition, c numbers of atoms of the same type are being added to the container per unit time. How many atoms of this type are there at t = T ? (A)

c exp(–T) – N0exp(–T) 

(B)

c exp(–T) + N0exp(–T) 

(C)

c (1 – exp(–T)) + N0exp(–T) 

(D)

c (1 + exp(–T) + N0exp(–T) 

CHEMISTRY 101.

2.52 g of oxalic acid dihydrate was dissolved in 100 mL of water. 10 mL of this solution was diluted is 500 mL. The normality of the final solution and the amount of oxalic acid (mg/mL) in the solution are respectively. (A) 0.16 N, 5.04

102.

(B) 0.08 N, 3.60

(C) 0.04 N, 3.60

(D) 0.02 N, 10.08

Two isomeric compounds I and II are heated with HBr.

I

II

The products obtained are

(A)

(B)

(C)

(D)


103.

The number of possible enantiomeric pair(s) produced from the bromination of I and II, respectively, are

(A) 0, 1

104.

105.

(B) 1 , 0

(C) 0, 2

(D) 1, 1

For the reaction A  B, H° = 7.5 kJ mol–1 and S° = 25 J mol–1, the value of G° and the temperature, at which the reaction reaches equilibrium are, respectively, (A) 0 kJ mol–1 and 400 K

(B) – 2.5 kJ mol–1 and 400 K

(C) 2.5 kJ mol–1 and 200 K

(D) 0 kJ mol–1 and 300 K

The solubility product of Mg(OH)2 is 1.0  10–12. Concentrated aqueous NaOH solution is added to a 0.01 M aqueous solution of MgCl2. The pH at which precipitation occurs is (A) 7.2

106.

(B) 7.8

(C) 8.0

(D) 9.0

A metal with an atomic radius of 141.4 pm crystallizes in the face centred cubic structure. The volume of the unit cell in pm3 is (A) 2.74  107

107.

(C) 2.19  107

(D) 9.20  107

Identify the cyclic silicate ion given in the figure below

(A) [Si6O24]24– 108.

(B) 6.40  107

(B) [Si6O18]18–

(C) [Si6O18]12–

Diborane is formed the elements as shown in equation (1) 2B (s) + 3H2(g)  B2H6(g) Given that H2O ()  H2O (g) 2B + 3/2 O2 (g)  B2O3(s) B2H6(g) + 3 O2 (g)  B2O3 (s) + 3 H2O (g) H2(g) + 1/2 O2 (g)  H2O () the H0 for the reaction (1) is : (A) 36 kJ (B) 520 kJ (C) 509 kJ

(D) [Si6O24]12–

... (1) H10 = 44 kJ H20 = – 1273 kJ H30 = – 2035 kJ H40 = 286 kJ (D) – 3550 kJ


109.

The Crystal Field stabilization Energy (CFSE) and the spin only magnetic moment in Bohr Magneton (BM) for the complex K3[Fe(CN)6] are, respectively. (A) 0.0  and

35 BM

(C) – 0.4  and

110.

(B) – 2.0  and

24 BM

3 BM

(D) – 2.4  and 0 BM

A solution containing 8.0 g of nicotine in 92 g of water freezes 0.925 degrees below the normal freezing point of water. If the molal freezing point depression constant, Kf = 1.85ºC mol–1 then the molar mass of nicotine is : (A) 16

(B) 80

(C) 320

(D) 160

BIOLOGY 111.

A host cell has intracellular bacterial symbionts. If the growth rate of the bacterial symbiont is always 10% higher than that of the host cell, after 10 generations of the host cell the density of bacteria in host cells will increase (A) by 10%

112.

(B) two-fold

(C) ten-fold

(D) hundred-fold

In a diploid organism, there are three different alleles for a particular gene. Of these three alleles one is recessive and the other two alleles exhibit co-dominance. How many phenotypes are possible with this set of alleles? (A) 3

113.

(B) 6

(C) 4

(D) 2

Two students are given two different double stranded DNA molecules of equal length. They are asked to denature the DNA molecules by heating. The DNA given to student A has a following composition of base (A:G:T:C::35:15:35:15). While that given to student B is (A : G : T : C : : 12 : 38 : 12 : 38) Which of the following statements is true? (A) Both the DNA molecules would denature at the same rate (B) The information given is insufficient to draw any conclusion (C) DNA molecule given to student B would denature faster than that of student A (D) DNA molecule given to student A would denature faster than that given to student B


114.

The amino acid sequences of a bacterial protein and a human protein carrying out similar function are found to be 60% identical. However, the DNA sequences of the genes coding for these proteins are only 45% identical. This is possible because (A) Protein sequence does not depend on DNA sequence (B) DNA codons having different nucleotides in the third position can code for the same amino acids (C) DNA codons having different nucleotides in the second position can code for the same amino acids (D) Same DNA codons can code for multiple amino acids.

115.

The following DNA sequence (5´  3´) specifies part of a protein coding sequence, starting from position 1. Which of the following mutations will give rise to a protein that is shorter than the full-length protein?

1

2

3

4

5

6

7

8

9

10

11

12

13

14

15

A

T

G

C

A

A

G

A

T

A

T

A

G

C

T

(A) Deletion of nucleotide 13 (B) Deletion of nucleotide 8 (C) Insertion of a single nucleotide between 3 and 4 (D) Insertion of a single nucleotide between 10 and 11

116.

Which of the following correctly represents the results of an enzymatic reaction? Enzyme is E, Substrate is S and Products are P1 and P2

117.

(A) P1 + S  P2 + E

(B) E + S  P1 + P2

(C) P1 + P2 + E  S

(D) E + S  P1 + P2 + E

Four species of birds have different egg colors: [1] while with no markings, [2] pale brown with no markings, [3] grey-brown with dark streaks and spots, [4] pale blue with dark blue-green spots. Based on egg color, which species is most likely to nest in a deep tree hole? (A) 1

118.

(B) 2

(C) 3

(D) 4

Consider a locus with two alleles, A and a, If the frequency of AA is 0.25, what is the frequency of A under Hardy-Weinberg equilibrium? (A) 1

(B) 0.25

(C) 0.5

(D) 0


119.

Which of the following graphs accurately represents the insulin levels (Y-axis) in the body as a function of time (X-axis) after eating sugar and bread/roti?

120.

(A)

(B)

(C)

(D)

You marked two ink-spots along the height at the base of a coconut tree and also at the top of the tree. When you examine the spots next year when the tree has grown taller, you will see (A) the two spots at the top have grown more apart than the two spots at the bottom (B) the top two spots have grown less apart than the bottom two spots (C) both sets of spots have grown apart to the same extent (D) both sets of spots remain un-altered.

*****



1.

Max. Marks : 160


2.


3.


4.


5.


6.



1.

Max. Marks : 160


2.


3.


4.


5.


6.



1.

Max. Marks : 160


2.


3.


4.


5.


6.



Max. Marks : 160

1.


2.


Subject

Part-I

Part-II

Mathematics



Physics



Chemistry



Biology



3.


4.


5.


6.


7.


8.


9.



Max. Marks : 160

1.


2.


Subject

Part-I

Part-II

Mathematics



Physics



Chemistry



Biology



3.


4.


5.


6.


7.


8.


9.



Max. Marks : 160

1.


2.


Subject

Part-I

Part-II

Mathematics



Physics



Chemistry



Biology



3.


4.


5.


6.


7.


8.


9.



Suppose logab + logb a= c. The smallest possible integer value of c for all a,b > 1 is (A) 4 (B) 3 (C) 2 (D) 1

2.

Suppose n is a natural number such that | i + 2i2 + 3i3 + ....+ nin | = 18 1–1 . Then n is. (A) 9 (B) 18 (C) 36

3.

2 . Where i is the square root of

Let P be an m × m matrix such that P2 = P . Then (I+P)n equals (A) I + P (B) I + nP (C) I + 2nP

(D) 72

(D) I + (2n–1)P

4.

Consider the cubic equation x3 + ax2 + bx + c = 0 , where a,b,c are real numbers . Which of the following statements is correct ? (A) If a2 – 2b < 0 , then the equation has one real and two imaginary roots (B) If a2– 2b  0, then the equation has all real roots (C) If a2– 2b > 0, then the equation has all real and distinct roots (D) If 4a3– 27b2 > 0 , then the equation has real and distinct roots

5.

All the points (x,y) in the plane satisfying the equation x2 + 2x sin(xy) + 1 = 0 lie on (A) a pair of straight lines (B) a family of hyperbolas (C) a parabola (D) an ellipse

6.

Let A = (4,0) , B = (0,12) be two points in the plane . The locus of a point C such that the area of triangle ABC is 18 sq.units is (A) (y + 3x + 12)3 = 81 (B) (y + 3x + 81)2 = 12 2 (C) (y + 3x – 12) = 81 (D) (y + 3x – 81)2 = 12

7.

In a rectangle ABCD , the coordinates of A and B are (1,2) and (3,6) respectively and some diameter of the circumscribing circle of ABCD has equation 2x – y + 4 = 0. Then the area of the rectangle is (A) 16

(B) 2 10

(C) 2 5

(D) 20

8.

In the xy–plane , three distinct lines 1,2,3 concur at a point (,0). Further the lines 1,2,3 are normals to the parabola y2 = 6x at the points A = (x1 , y1) , B (x2 , y2) , C = (x3 , y3) respectively . Then we have (A)  < – 5 (B)  > 3 (C) – 5 <  < – 3 (D) 0 <  < 3

9.

Let f(x) = cos5x + A cos 4x + Bcos3x + Ccos2x + Dcosx + E. and   2   3   8   9   – f  + ..............+ f   – f  Then T T = f(0) – f   + f  5  5   5   5   5  (A) depends on A,B,C,D,E (B) depends on A,C,E, but independent of B and D (C) depends on B,D , but independent of A,C,E (D) is independent of A,B,C,D,E

10.

In triangle ABC , we are given that 3sinA + 4 cosB = 6 and 4sinB + 3cosA = 1. Then the measured of the angle C is (A) 30º (B) 150º (C) 60º (D) 75º

11.

Which of the following intervals is possible domain of the function f(x) = log{x} [x] + log[x] {x} , where [x] is the greatest integer not exceeding x and {x} = x – [x] ? (A) (0,1) (B) (1,2) (C) (2,3) (D) (3,5) CLASS-XII_STREAM-SB+2_PAGE # 68

12.

If f(x) = (2011 + x)n , where x is a real variable and n is a positive integer , then the value of f(0) + f(0) +

f " (0 ) f (n –1)(0) + ......+ is 2! (n – 1)! (A) (2011)n 13.

(B) (2012)n

(C) (2012)n – 1

(D) n(2011)n

The minimum distance between a point on the curve y = ex and a point on the curve y = loge x is

1 (A)

(B)

2

(C)

2

(D) 2 2

3

8

14.

Let f : (2,)  N be defined by f(x) = the largest prime factor of [x] , Then

 f ( x) dx is equal to 2

(A) 17 15.

(B) 22

(C) 23

(D) 25

Let [x] denote the largest integer not exceeding x and {x} = x – [x]. Then

2012

e cos(  { x })



e cos( { x })  e – cos(  { x })

0

is equal to (A) 0

16.

 1  The value of lim  2 n 4 n – 1  (A)

17.

1 4

(B)

(C) 2012

1 4n 2 – 4

 .......... .. 

 12

(C)

(D) 2012 

  is 2 2  4n – n  1

 4

(D)

 6

Two players play the following game : A writes 3,5,6 on three different cards ; B writes 8 ,9,10 on three different cards . Both draw randomly two cards from their collections . Then A computes the product of two numbers he/she has drawn , and B computes the sum of two numbers he/she has drawn . The player getting the larger number wins. What is the probability that A wins ? (A)

18.

(B) 1006

dx

1 3

(B)

5 9

(C)

4 9

(D)

1 9

        Let a, b, c be three vectors in the xyz space such that a  b = b  c = c  a  0 . If A,B,C are points with  position vectors a, b, c respectively , then the number of possible positions of the centriod of triangle ABC

is (A) 1

(B) 2

(C) 3

(D) 6

19.

The sum of (12 – 1 + 1) (1!) + (22 – 2 + 1)(2!) + .... + (n2 – n + 1)(n!) is (A) (n + 2) ! (B) ( n – 1) ((n + 1) !) + 1 (C) (n + 2)! – 1 (D) n((n + 1)!) – 1

20.

Let X be a nonempty set and let P(X) denote the collection of all subsets of X. Define f : X × P(X)  R by 1, x  A f(x,A) =  .Then f(x , A B) equals 0, x  A

(A) f(x, A) + f(x ,B) (B) f(x ,A) + f (x ,B) – 1 (C) f (x , A) + f(x , B) – f (x ,A) f( x ,B) (D) f( x ,A) + | f( x ,A) – f( x ,B) |


PHYSICS 21.

A narrow but tall cabin is falling freely near the earth’s surface. Inside the cabin, two small stones A and B are released from rest (relative to the cabin). Initially A is much above the centre of mass and B much below the centre of mass of the cabin. A close observation of the motion of A and B will reveal that : (A) both A and B continue to be exactly at rest relative to the cabin (B) A moves slowly upward and B moves slowly downward relative to the cabin (C) both A and B fall to the bottom of the cabin with constant acceleration due to gravity (D) A and B move slightly towards each other vertically.

22.

Two plates each of mass m are connected by a massless spring as shown :

A weight W is put on the upper plate which compresses the spring further. When W is removed, the entire assembly jumps up. The minimum weight W needed for the assembly to jump up when the weight is removed is just more than : (A) mg (B) 2mg (C) 3mg (D) 4mg 23.

If the speed (v) of the bob in a simple pendulum is plotted against the tangential acceleration (a), the correct graph will be represented by :

(A)

(B)

(C)

(D)

(A) I

(B) II

(C) III

(D) IV

24.

A container with rigid walls is covered with perfectly insulating material. The container is divided into two parts by a partition. One part contains a gas while the other is fully evacuated (vacuum). The partition is suddenly removed. The gas rushes to fill the entire volume and comes to equilibrium after a little time. If the gas is not ideal , (A) the initial internal energy of the gas equals its final internal energy (B) the initial temperature of the gas equals its final temperature (C) the initial pressure of the gas equals to its final pressure (D) the initial entropy of the gas equals to its final entropy

25.

Two bulbs of identical volumes connected by a small capillary are initially filled with an ideal gas at temperature T. Bulb 2 is heated to maintain a temperature 2T while bulb 1 remains at temperature T. Assume throughout that the heat conduction by the capillary is negligible. Then the ratio of the final mass of the gas in bulb 2 to the initial mass of the gas in the same bulb is close to : (A) 1/3 (B) 2/3 (C) 1/3 (D) 1 CLASS-XII_STREAM-SB+2_PAGE # 70

26.

Two rods, one made of copper and the other steel of the same length and cross sectional area are ed together. (The thermal conductivity of copper is 385 J.s–1. m–1. K–1 and steel is 50 J.s–1.m–1.K–1.) If the copper end is held at 100°C and the steel end is held at 0°C, what is the junction temperature (assuming no other heat losses) ? (A) 12°C (B) 50°C (C) 73°C (D) 88°C

27.

Jet aircrafts fly at altitudes above 30,000 ft where the air is very cold at –40°C and the pressure is 0.28 atm. The cabin is maintained at 1 atm pressure by means of a compressor which exchanges air from outside adiabatically. In order to have a comfortable cabin temperature of 25°C, we will require in addition : (A) a heater to warm the air injected into the cabin (B) an air conditionaer to cool the air injected into the cabin (C) neither a heater nor an air -conditioner : the compressor is sufficient (D) alternatively heating and cooling in the two halves of the compressor cycle.

28.

A speaker emits a sound wave of frequency f0. When it moves towards a stationary observer with speed u, the observer measures a frequency f1. If the speaker is stationary and the observer moves towards it with speed u, the measured frequency is f2. Then : (A) f1 = f2 < f0 (B) f1 > f2 (C) f1 < f2 (D) f1 = f2 > f0

29.

A plane polarized light ed through successive polarizers which are rotated by 30° with respect to each other in the clockwise direction. Neglecting absorption by the polarizers and given that the first polarizer’s axis is parallel to the plane of polarization of the incident light, the intensity of light at the exit of the fifth polarizer is closest to : (A) same as that of the incident light (B) 17.5% of the incident light (C) 30% of the incident light (D) zero

30.

At 23ºC, a pipe open at both ends resonates at a frequency of 450 hertz. At what frequency does the same pipe resonate on a hot day when the speed of sound is 4 percent higher than it would be at 23º C ? (A) 446 Hz (B) 454 Hz (C) 468 Hz (D) 459 Hz

31.

In a Young’s double slit set-up, light from a laser source falls on a pair of very narrow slits separated by 1.0 micrometer and bright fringes separated by 1.0 millimeter are observed on a distant screen. If the frequency of the laser light is doubled, what will be the separation of the bright fringes ? (A) 0.25 mm (B) 0.5 mm (C) 1.0 mm (D) 2.0 mm

32.

For a domestic AC supply of 220 V at 50 cycles per second, the potential difference between the terminals of a two -pin electric outlet in a room is given by :

33.

(A) V(t) = 220 2 Cos(100t)

(B) V(t) = 220Cos(50t)

(C) V(t) = 220Cos(100t)

(D) V(t) = 220 2 Cos(50t)

In the circuit shown below the resistances are given in ohms and the battery is assumed ideal with emf equal to 3.0 volts. The resistor that dissipates the most power is

(A) R1 34.

(B) R2

(C) R3

(D) R4

An electron collides with a free molecules initially in its ground state. The collision leaves the molecule in a excited state that is metastable and does not decay to the ground state by radiation. Let K be the sum 

of the initial kinetic energies of the electron and the molecule, and P the sum of their initial momenta. Let 

K’ and P' represent the same physical quantities after the collision. Then       (A) K = K’, P  P' (B) K’ < K, P  P' (C) K = K’, P  P'





(D) K’ < K, P  P'


35.

In the circuit shown, the switch is closed at time t = 0

Which of the graphs shown below best reprsents the voltage across the inductor, as seen on an oscilloscope?

(I)

(II)

(III)

(IV)

(A) I 36.

(B) II

(C) III

(D) IV

Given below are three schematic graphs of potential energy V(r) versus distance r for three atomic particles electron (e), proton (p+) and neutron (n), in the presence of a nucleus at the origin O. The radius of the nucleus is r0. The scale on the V–axis may not be the same for all figures. The correct pairing of each with the corresponding atomic particle is :

(A) (1, n), (2, p+), (3, e–) (C) (1, e–), (2, p+), (3, n)

(B) (1, p+), (2, e–), (3, n) (D) (1, p+), (2, n), (3, e–)

37.

Due to transitions among its first three energy levels, hydrogenic atom emits radiation at three discrete wavelength 1,2 and 3(1 < 2 < 3). Then : (A) 1 = 2 + 3 (B) 1 + 2 = 3 (C) 1/1 + 1/2 = 1/3 (D) 1/1 = 1/2 + 1/3

38.

The total radiative power emitted by spherical blackbody with radius R and temperature T is P. If the radius is doubled and the temperature is halved then the radiative power will be : (A) P/4 (B) P/2 (C) 2P (D) 4P


39.

The Quantum Hall Resistance RH is a fundamental constant with dimensions of resistance. If h is Planck’s constant and e the electron charge, then the dimension of RH is the same as : (A) e2/h (B) h/e2 (C) h2/e (D) e/h2

40.

Four students measure the height of a tower. Each student uses a different method and each measures the height many different times. The data for each are plotted below. The measurement with highest precision is :

(A) I

(B) II

(C) III

(D) IV

CHEMISTRY 41.

Ther hybridizations of Ni (CO)4 and [Cr(H2O)6]2+, respectively (A) sp3 and d3sp2 (B) dsp2 and d2sp3 (C) sp3 and d2sp3 (D) dsp2 and sp3 d2

42.

Extraction of silver is achieved by initial complexation of the ore (Argentite) with X followed by reaction with Y, X and Y, respectively, are (A) CN– and Zn (B) CN– and Cu – (C) Cl and Zn (D) Br – and Zn

43.

Assuming ideal behaviour, the enthalpy and volume of mixing of two liquids, respectively, are (A) zero and zero (B) +ve and zero (C) –ve and zero (D) –ve and –ve

44.

At 298 K, the ratio of osmotic pressures of two solutions of a substance with concentrations of 0.01 M and 0.001 M, respectively, is (A) 1 (B) 100 (C) 10 (D) 1000

45.

The rate of gas phase chemical reactions generally increases rapidly with rise in temperature. This is mainly because : (A) the collision frequency increases with temperature. (B) the fraction of molecules having energy in excess of the activation energy increases with temperature. (C) the activation energy decreases with temperature. (D) the average kinetic energy of molecules increases with temperature.


46.

Among i-iv :

the compound that does not undergo polymerization under radical initiation, is : (A) i (B) ii (C) iii (D) iv 47.

Two possible stereoisomers for

are : (A) enantiomers

(B) diastereomers

(C) conformers

(D) rotamers

48.

For a process to occur spontaneously. (A) only the entropy of the system must increase. (B) only the entropy of the surrounding must increase. (C) either the entropy of the system or that of the surroundings must increase. (D) the total entropy of the system and the surroundings must increase.

49.

When the size of a spherical nanoparticle decreases from 30 nm to 10 nm, the ratio surface area/volume becomes (A) 1/3 of the original (B) 3 times the original (C) 1/9 of the original (D) 9 times the original

50.

The major product of the following reaction is :

H

 

(A)

51.

(B)

(C)

(D)

(C) H3O+

(D) H2/Pt

For the transformation

 the reagent used is : (A) LiAlH4

(B) H3PO2

52.

The values of the limiting molar conductivity (º) for NaCl, HCl and NaOAc are 126.4. 425.9 and 91.0 S cm2 mol–1 respectively. For HOAC. º in S cm2 mol–1 is : (A) 390.5 (B) 299.5 (C) 208.5 (D) 217.5

53.

To obtain a diffraction peak, for a crystalline solid with interplane distance equal to wavelength of incident X-ray radiation, the angle of incidence should be : (A) 90º (B) 0º (C) 30º (D) 60º

54.

The standard Gibbs free energy change (Gº in kJ mol–1), in a Daniel cell (Eºcell = 1.1 V), when 2 moles of Zn(s) is oxidized at 298 K, is closed to : (A) – 212.3 (B) – 106.2 (C) – 424.6 (D) – 53.1


55.

All the products formed in the oxidation of NaBH4 by I2, are : (A) B2H6 and NaI (B) B2H6, H2 and NaI (C) BI3 and NaH

(D) NaBI4 and HI

56.

The spin-only magnetic moments of [Mn(CN)6]4– and [MnBr4]2– in Bohr Magneton, respectively, are : (A) 5.92 and 5.92 (B) 4.89 and 1.73 (C) 1.73 and 5.92 (D) 1.73 and 1.73

57.

In a zero-order reaction, if the initial concentration of the reactant is doubled, the time required for half the reactant to be consumed : (A) increases two-fold (B) increases four-fold (C) decreases by half (D) does not change

58.

The adsorption isothermal for a gas is given by the relation x = ap/(1 + bp) where x is moles of gas adsorbed per gram of the adsorbent, p is the pressure of the gas, and a and b are constants. Then x : (A) increases with p (B) remains unchanged with p (C) decreases with p (D) increases with p at low pressure and the same at high pressure.

59.

The reaction

NaOH / Heat

+ CHCl3      H

is known as : (A) Perkin reaction (C) Reimer-Tiemann reaction 60.

(B) Sandmeyer reaction (D) Cannizzaro reaction

Among i-iii

the boiling point follows the order (A) ii < i < iii (B) iii < ii < i

(C) i < ii < iii

(D) ii < iii < i


BIOLOGY 61.

The major constituents of neurofilaments are (A) microtubules (C) actin filaments

(B) intermediate filaments (D) protofilaments

62.

In which phase of the cell cycle are sister chromatids available as template for repair ? (A) G1 phase (B) G2 phase (C) S phase (D) M phase

63.

A person has difficulty in breathing at higher altitudes because (A) oxygen is likely to diffuse from lungs to blood. (B) oxygen is likely to diffuse from blood to lungs (C) partial pressure of O2 is lower than partial pressure of CO2 (D) overall intake of O2 by the blood becomes low.

64.

In humans , the composition of a zygote that will develop into a female is (A) 44A +XX (B) 44A + XY (C) 22 + X (D) 23 A

65.

If you fractionate all the organelles from the cytoplasm of a plant cell. In which one of the following sets of fractions will you find nucleic acids ? (A) nucleus, mitochondria, chloroplast, cytoplasm (B) nucleus, mitochondria, chloroplast, glyoxysome (C) nucleus, chloroplast , cytoplasm and peroxisome (D) nucleus, mitochondria, chloroplast, Golgi bodies

66.

A protein with 100 amino acid residues has been translated based on triplet genetic code. Had the genetic code been quadruplet. the gene that codes for the protein would have been. (A) same in size (B) longer in size by 25% (C) longer in size by 100% (D) shorter in size

67.

If the sequence of base in DNA is 5'- ATGTATCTCAAT- 3', than the sequence of bases in its transcript will be : (A) 5' - TACATAGAGTTA - 3' (B) 5' - UACAUAGAGUUA - 3' (C) 5' - AUGUAUCUCAAU - 3' (D) 5' - AUUGAGAUACAU - 3'

68.

The Na+/K+ pump is present in the plasma membrane of mammalian cells where it (A) expels potassium from the cell (B) expels sodium and potassium from the cell. (C) pumps sodium into the cell. (D) expels sodium from the cell.

69.

The CO2 in the blood is mostly carried (A) by haemoglobin in RBCs (B) in the cytoplasm of WBCs (C) in the plasma as bicrbonate ions (D) by plasma proteins


70.

Patients who have undergone organ transplants are given anti-rejection medications to (A) minimize infection (B) stimulate B- macrophage cell interaction (C) prevent T- lymphocyte proliferation (D) adopt the HLA of donor

71.

Saline drip is given to a Cholera patient because (A) NACl kills Vibrio cholera (B) NACl generates ATP (C) Na+ ions stops nerve impulse and hence sensation of pain (D) Na+ ions help in retention of water in body tissue

72.

A water molecule can form a maximum of hydrogen bonds. (A) 1 (B) 2 (C) 3

(D) 4

73.

Circadian Rhythm is an endogenously driven cycle for biochemical, physiological and behavioral processes. In humans, the approximate duration of this 'biological clock' is : (A) 1 Hours (B) 6 Hours (B) 12 Hours (D) 24 Hours

74.

Modern evolutionary theory consists of the concepts of Darwin modified by knowledge concerning : (A) population statistics (B) Mendel's laws (C) the idea of the survival of the fittest (D) competition

75.

Soon after the three germ layers are formed in a developing embryo, the process of organogenesis starts. The human brain is formed from the (A) ectoderm (B) endoderm (C) mesoderm (D) partly endoderm and partly mesoderm

76.

Puffs in the polytene chromosomes of Drosophila melanogaster salivary glands represent (A) transcriptionally active genes (B) transcriptionally inactive genes (C) heterochromatin (D) housekeeping genes

77.

The process of cell death involving DNA cleavage in cells is known as (A) necrosis (B) apoptosis (C) cytokinesis (D) endocytosis

78.

According to the original model of DNA. as proposed by Watson & Crick 1953, DNA is a (A) left handed helix (B) helix that makes a full turn every 70 nm. (C) helix where one turn of DNA contains 20 basepairs (D) two stranded helix where each strand has opposite polarity.

79.

At which stage of meiosis I does crossing over occur ? (A) lepoptene (B) zygotene (C) pachytene (D) diplotene

80.

An electrode is placed in the axioplasm of a mammalian axon and another electrode is placed just outside the axon. The potential difference measured will be (A) 0 (B) – 70mV (B) –70V (D) +70 V



Let A and B be any two n × n matrices such that the following conditions hold : AB = BA and there exist positive integers k and l such that Ak = I (the identity matrix) and B = 0 (the zero matrix), Then (A) A + B = l (B) det (AB) = 0 (C) det (A + B)  0 (D) (A + B)m = 0 for some integer m

82.

The minimum value of n for which (A) is 101

83.

2 2  4 2  6 2  .......  ( 2n)2 1 2  3 2  5 2  .......  ( 2n – 1)2

(B) is 121

(C) is 151

< 1.01 (D) does not exist

The locus of the point P = (a , b) where a , b are real numbers such that the roots of x3 + ax2 + bx + a = 0 are in arithmetic progression is

84.

(A) an ellipse

(B) a circle

(C) a parabola whose vertex is on the y – axis

(D) a parabola whose vertex is on the x – axis

The smallest possible positive slope of a line whose y–intercept is 5 and which has a common point with he ellipse 9x2 + 16y2 = 144 is (A)

85.

3 4

(B) 1

(C)

4 3

(D)

9 16

Let A = {  R | cos2(sin) + sin2(cos) = 1} and B = {   R | cos(sin) sin(cos) = 0 }. Then A  B (A) is the empty set (B) has exactly one element (C) has more than one but finitely many elements (D) has infinitely many elements

86.

Let f(x) = x3 + ax2 + bx + c , where a,b,c are real numbers. If f(x) has local minimum at x = 1 and a local

maximum at x = –

1 1 and f(2) = 0 , then f ( x ) dx equals 3



–1

(A)

87.

14 3

(B)

– 14 3

(C)

7 3

(D)

–7 3

Let f(x) = x12 – x9 + x4 – x + 1. Which of the following is true ? (A) f is one –one (B) f has a real root (C) f never vanishes (D) f takes only positive values


1

88.

 x n (1 – x )n    For each positive integers n, define fn(x) = minimum  n! , n!  , for 0  x  1. Let In = fn ( x ) dx, n  1 .   0





Then  I n is equal to n1

(A) 2 e – 3

(B) 2 e – 2

(C) 2 e – 1

(D) 2 e

89.

The maximum possible value of x2 + y2 – 4x – 6y , x,y real subject to the condition | x + y | + | x – y | = 4 (A) is 12 (B) is 28 (C) is 72 (D) does not exist

90.

The arithemetic mean and the geometric mean of two distinct 2– digit numbers x and y are two integers one of which can be obtained by reserving the digits of the other (in base 10 representation). Then x + y equals (A) 82 (B) 116 (C) 130 (D) 148

PHYSICS 91.

An isolated sphere of radius R contains uniform volume distribution of positive charge. Which of the curves shown below correctly illustrates the dependence of the magnitude of the electric field of the sphere as a function of the distance r from its centre ?

(A) I 92.

(B) II

(C) III

(D) IV

The surface of a planet is found to be uniformly charged. When a particle of mass m and no charge is thrown at an angle from the surface of the planet, it has a parabolic trajectory as in projectile motion with horizontal range L. A particle of mass m and charge q, with the same initial conditions has a range L/2. The range of particle of mass m and charge 2q with the same initial conditions is : (A) L (B) L/2 (C) L/3 (D) L/4


93.

Figure below shows a small mass connected to a string, which is attached to a vertical post.If the ball is released when the string is horizontal as shown , the magnitude of the total acceleration (including radial and tangential) of the mass as a function of the angle  is

(A) g Sin

(B) g

3 Cos 2   1

(C) g Cos 

(D) g

3Sin 2   1

94.

One mole of an ideal gas at intial temprature T, undergoes a quasi–static process during which the volume V is doubled .During the process the internal energy U obeys the equation U = aV3, where a is a constant .The work done during this process is (A) 3RT / 2 (B) 5RT / 2 (C) 5RT / 3 (D) 7RT / 2

95.

A constant amount of an ideal gas undergoes the cyclic process ABCA in the PV diagram shown below.

The path BC is an isothermal .The work done by the gas during one complete cycle , beginning and ending at A is nearly (A) 600 KJ (B) 300 KJ (C) – 300 KJ (D) – 600 KJ 96.

A material is embedded between two glass plates . Refractive index n of the material varies with thickness as shown below .The maximum incident angle (in degrees) on the material for which beam will through the material is

(A) 60.0

(B) 53.1

(C) 43.5

(D) 32.3


97.

At a distance  form a uniformly charged long wire , a charged particle is thrown radially outward with a velocity u in the direction perpendicular to the wire . When the particle reaches a distance 2 from the wire its speed is found to be

2u . The magnitude of the velocity , when it is a distance 4l away from the wire,

is (ignore gravity) (A) 3u 98.

(B) 2 u

(C) 2 2 u

(D) 4 u

A rectangular loop of wire shown below is coplanar with a long wire carrying current I .

The loop is pulled to the right as indicated .What are the directions of the induced current in the loop and the magnetic forces on the left and the right sides of the loop ? Inducded Current Force on left side Force on right side (A) Counterclockwise To the left To the right (B) Clockwise To the left To the right (C) Counter clockwise To the right To the left (D) Clockwise To the right To the left 99.

Two batteries V1 and V2 are connected to three resistors as shown below

If V1 = 2V and V2 = 0V , the current I = 3mA. If V1 = 0V and V2 = 4V , the current I = 4 mA. Now if V1 = 10 V and V2 = 10V , the current I will be (A) 7 mA (B) 15 mA (C) 20 mA (D) 25 mA 100.

y2 + = 1. At point (0, b), the x-component a2 b2 of velocity is u. the y-component of acceleration at this point is : (A) – bu2 / a2 (B) – u2 / a2 (C) – au2 / b2 (D) – u2 / a A particle moves in a plane along an eliptic path given by

x2

CHEMISTRY 101.

XeF6 hydrolyses to give an oxide. The structure of XeF6 and the oxide, respectively are (A) octahedral and tetrahedral (B) distorted octahedral pyramidal (C) octahedral and pyramidal (D) distorted octahedral and tetrahedral

102.

MnO4– oxidizes (i) oxalate ion in acidic medium at 333 K and (ii) HCl. For balanced chemical equations, the rations [MnO4– : C2O42–] in (i) and [MnO4– : HCl] in (ii), respectively, are : (A) 1 : 5 and 2 : 5 (B) 2 : 5 and 1 : 8 (C) 2 : 5 and 1 : 5 (D) 5 : 2 and 1 : 8

103.

If E º Fe 2 / Fe = – 0.440 V and Eº Fe3  / Fe 2  = 0.770 V, then E º Fe3  / Fe is : (A) 0.330 V (B) – 0.037 V (C) – 0.330 V (D) – 1.210 V

104.

The electron in hydrogen atom is in the first Bohr orbit (n = 1). The ratio of transition energies, E(n = 1  n = 3) to E (n = 1  n = 2), is (A) 32/27 (B) 16/27 (C) 32/9 (D) 8/9


105.

In the following conversion, (i) MeMgBr

NaOH / I

(ii) H3O 

H3O

2      X    Y

the major products X and Y, respectively, are :

106.

(A) i (B) ii The reaction sequence,

HNO2    X

(C) iii

(D) iv

(C) iii

(D) iv

Y

the major products X and Y, respectively, are :

(A) i 107.

(B) ii

Optically active (S)--methoxyacetaldehyde on reaction with MeMgX gave a mixture of alcohols. The major diastereomer ‘P’ on treatment with Me1/K2CO3 gave an optically inactive compound. P is :

(A) i

(B) ii

(C) iii

(D) iv


108.

At 300 K the vapour pressure of two pure liquids, A and B are 100 and 500 mm Hg, respectively. If in a mixture of A and B, the vapour pressure is 300 mm Hg, the mole fractions of A in the liquid and in the vapour phase, respectively, are : (A) 1/2 and 1/10 (B) 1/4 and 1/6 (C) 1/4 and 1/10 (D) 1/2 and 1/6

109.

The crystal field stabilization energies (CFSE) of high spin and low spin d6 metal complexes in of 0, respectively, are : (A) – 0.4 and – 2.4 (B) – 2.4 and – 0.4 (C) – 0.4 and 0.0 (D) – 2.4 and 0.0

110.

Emulsification of 10 ml of oil in water produces 2.4 × 1018 droplets. If the surfaces tension at the oil-water interface is 0.03 Jm–2 and the area of each droplet is 12.5 × 10–16 m2, the energy spent in the formation of oil droplets is : (A) 90 J (B) 30 J (C) 900 J (D) 10 J

BIOLOGY 111.

Which sequence of events gives rise to flaccid guard cells and stomatal closure at night ? (A) low [Glucose ]  low osmotic pressure  low pH  high pCO2 (B) low pH  high pCO2  low [Glucose ]  low osmotic pressure (C) low osmotic pressure  high pCO2  low pH  low [Glucose] (D) high pCO2  low pH  low [Glucose]  low osmotic pressure

112.

Rice has a diploid genome with 2n = 24. If crossing over is stopped in a rice plant and then selfed seeds are collected, will all the offsprings be genetically identical to the parent plant ? (A) yes, because crossing-over is the only source of genetic variation (B) no, because stopping of crossing over automatically increases rate of point mutation (C) yes, only if the parent plant was a completely inbred line (D) yes, only if the parent plant was a hybrid between two prue-bred lines

113.

Rodents can distinguish between many different types of odours. The basis for odour discrimination is that (A) they have a small number of odorant receptors that bind to many different odorant molecules (B) the mechanoreceptors in the nasal cavity are activated by different odorant molecules found in the air ing through the nostrils (C) the part of the brain that processes the sense of smell has many different receptors for odorant molecules (D) a large number of different chemoreceptors are present in the nasal cavity that binds a variety of odorant molecules

114.

Although blood flows through large arteries at high pressure, when the blood reaches small capillaries the pressure decreases because (A) the valves in the arteries regulate at he rate of blood flow into the capillaries (B) the volume of blood in the capillaries is much lesser than that in the arteries (C) the total cross-sectional area of capillaries arising form an artery is much greater than that of the artery (D) elastin fibers in the capillaries help to reduce the arterial pressure

115.

E.coli about to replicate was pulsed with tritiated thymidine for 5 min and then transferred to normal medium . After one cell division which one of the following observations would be correct ? (A) both the strands of DNA will be radioactive (B) one strand of DNA will be radioactive (C) none of the strands will be radioactive (D) half of one strand of DNA will be radioactive


116.

Selection of lysine auxotroph (bacteria which requires lysine for growth) from a mixed population of bacteria can be done growing bacterial population in the presence of (A) lysine (B) penicillin (B) lysine and penicillin (D) glucose

117.

Increasing the number of measurements of an experimental variable will (A) increase the standard error of the sample (B) increase the mean of the sample (C) decrease the standard error of the sample (D) result in all of the above

118.

119.

For a human male what is the probability that all the maternal chromosomes will end up in the same gamete ? (A) 1/23 (B) 223 46 (C) 2 (D) (½)23 Nocturnal animal have retinas that contain (A) a high percentage of rods to increase sensitivity to low light conditions (B) a high percentage of cones so that nocturnal color vision can be improved in low light conditions (C) an equal number of rods and cones so that vision can be optimized (D) retinas with the photoreceptor layer present in the front of the eye to increase light sensitivity

120.

The length of one complete turn of a DNA double helix is

 (A) 34 A

(B) 34 nm

 (C) 3.4 A

(D) 3.4 m

*****



1.

Max. Marks : 160


2.


3.


4.


5.


6.



1.

Max. Marks : 160


2.


3.


4.


5.


6.



1.

Max. Marks : 160


2.


3.


4.


5.


6.



Max. Marks : 160

1.


2.


Subject

Part-I

Part-II

Mathematics



Physics



Chemistry



Biology



3.


4.


5.


6.


7.


8.


9.



Max. Marks : 160

1.


2.


Subject

Part-I

Part-II

Mathematics



Physics



Chemistry



Biology



3.


4.


5.


6.


7.


8.


9.



Max. Marks : 160

1.


2.


Subject

Part-I

Part-II

Mathematics



Physics



Chemistry



Biology



3.


4.


5.


6.


7.


8.


9.



Three children, each accompanied by a guardian, seek ission in a school. The principal wants to interview all the 6 persons one after the other subject to the condition that no child is interviewed before its guardian. In how many ways can this be done? (A) 60 (B) 90 (C) 120 (D) 180

2.

In the real number system, the equation (A) no solution (C) exactly four distinct solutions

x  3 – 4 x – 1 + x  8  6 x  1 = 1 has (B) exactly two distinct solutions (D) infinitely many solutions

3.

The maximum value M of 3x + 5x – 9x + 15x – 25x, as x varies over reals, satisfies (A) 3 < M < 5 (B) 0 < M < 2 (C) 9 < M < 25 (D) 5 < M < 9

4.

Suppose two perpendicular tangents can be drawn from the origin to the circle x2 + y2 – 6x – 2py + 17 = 0, for some real p. Then |p| is equal to (A) 0 (B) 3 (C) 5 (D) 17

5.

Let a, b, c be numbers in the set {1, 2, 3, 4, 5, 6} such that the curves y = 2x3 + ax + b and y = 2x3 + cx + d have no point in common. The maximum possible value of (a – c)2 + b – d is (A) 0 (B) 5 (C) 30 (D) 36

6.

Consider the conic ex2 + y2 – 2e2x – 22y + e3 + 3 = e. Suppose P is any point on the conic and S1, S2 are the foci of the conic, then the maximum value of (PS1 + PS2) is (A) e

(B)

e

(C) 2 

sin( x  a)  sin( x  a) , then cos( x  a) – cos( x  a) (A) f(x + 2) = f(x) but f(x + )  f(x) for any 0 < < 2 (B) f is a strictly increasing function (C) f is strictly decreasing function (D) f is a constant function

7.

Let f(x) =

8.

The value of tan 81 – tan 63 – tan27 + tan9 is (A) 0 (B) 2 (C) 3

9.

(D) 2 e

The mid–point of the domain of the function f(x) = (A)

1 4

(B)

3 2

(D) 4

4  2x  5 for real x is (C)

2 3

(D) 

2 5

10.

Let n be a natural number and let a be a real number. The number of zeros of x2n+1 – (2n + 1) x + a = 0 in the interval [–1, 1] is (A) 2 if a > 0 (B) 2 if a < 0 (C) at most one for every vale of a (D) at least three for every value of a

11.

Let f : R  R be the function f(x) = (x – a1) (x – a2) + (x – a2) (x – a3)+ (x – x3) (x –x1) with a1, a2, a3 R. Then f(x)  0 if and only if (A) at least two of a1, a2, a3 are equal (B) a1 = a2 = a3 (C) a1, a2, a3 are all distinct (D) a1, a2, a3 are all positive and distinct CLASS-XII_STREAM-SB+2_PAGE # 85

/2

12.

The value

 (sin x )

2 1

 (sin x )

2 1

dx

0 /2

is

dx

0

2 1

(A)

2 1

(B)

2 –1

2 1

2 1

(C)

(D) 2 – 2

2

2012

13.

 (sin( x

The value

3

)  x 5  1) dx is

– 2012

(A) 2012

14.

(B) 2013

(C) 0

(D) 4024

Let [x] and {x} be the integer part and fractional part of a real number x respectively. The value of the 5

integral

 [ x] { x} dx is 0

(A) 5/2

15.

Let Sn =

(B) 5 n



k 1

(C) 34.5

(D) 35.5

k denote the sum of the first n positive integers. The numbers S1, S2, S3, ....S99 are

written on 99 cards. The probability of drawing a cards with an even number written on it is (A)

16.

(B)

49 100

(C)

49 99

(D)

48 99

A purse contains 4 copper coins and 3 silver coins. A second purse contains 6 coins and 4 silver coins. A purse is chosen randomly and a coin is taken out of it. What is the probability that it is a copper coin? (A)

17.

1 2

41 70

(B)

31 70

(C)

27 70

(D)

1 3

Let H be the orthocentre of an acute–angled triangle ABC and O be its circumcenter. Then HA  HB  HC (A) is equal to HO

(B) is equal to 3HO

(C) is equal to 2HO

(D) is not a scalar multiple of HO in general

18.

The number of ordered pairs (m,n) where m, n  {1, 2, 3, ......, 50}, such that 6m + 9n is a multiple of 5 is (A) 1250 (B) 2500 (C) 625 (D) 500

19.

Suppose a1, a2, a3, ....., a2012 are integers arranged on a circle. Each number is equal to the average of its two adjacent numbers. If the sum of all even indeced numbers is 3018, what is the sum of all numbers? (A) 0 (B) 1509 (C) 3018 (D) 6036

20.

Let S = 1, 2, 3, ...., n} and A = {(a, b) | 1  a, b  n} = S  S . A subset B of A is said to be a good subset if (x, x)  B for every x  S. Then the number of good subsets of A is (A) 1

(B) 2n

(C) 2n(n–1)

(D) 2 n2 CLASS-XII_STREAM-SB+2_PAGE # 86

PHYSICS 21.

An ideal monatomic gas expands to twice its volume. If the process is isothermal, the magnitude of work done by the gas is W i . If the process is adiabatic the magnitude of work done by the gas is W a . Which of the following is true (A) W i = W a > 0 (B) W i > W a > 0 (C) W i > W a = 0 (D) W a > W i = 0

22.

The capacitor of capacitance C in the circuit shown is fully charged initially, Resistance is R.

After the switch S is closed, the time taken to reduce the stored energy in the capacitor to half its initial value is : (A)

23.

RC 2

(B) RC ln 2

(C) 2RC ln 2

(D)

RC ln 2 2

A liquid drop placed on a horizontal plane has a near spherical shape (slightly flattened due to gravity). Let R be the radius of its largest horizontal section. A small disturbance causes the drop to vibrate with frequency v about its equilibrium shape. By dimensional analysis the ratio

c

can be (Here  is

 R 3 surface tension,  is density, g is acceleration due to gravity, and k is an arbitrary dimensionless constant) (A)

24.

kgR 2 

(B)

k R 3 g

(C)

k R 2 g

k (D) g

Seven identical coins are rigidly arranged on a flat table in the pattern shown below so that each coin touches its neighbours. Each coin is a thin disc of mass m and radius r. Note that the moment of inertia of an individual coin about an axis ing through center and perpendicular to the plane of the coin is mr 2 . 2

The moment of inertia of the system of seven coins about an axis that es through the point P (the centre of the coin positioned directly to the right of the central coin) and perpendicular to the plane of the coins is (A)

55 mr2 2

(B)

127 mr2 2

(C)

111 mr2 2

(D) 55 mr2


25.

A planet orbits in an elliptical path of eccentricity e around a massive star considered fixed at one of the foci. The point in space where it is closest to the star is denoted by P and the point where it is farthest is denoted by A. Let vP and vA be the respective speeds at P and A. Then

vP 1 e (A) v  1  e A 26.

vP (B) v = 1 A

(C)

vP 1 e2  vA 1 e

(D)

vP 1 e2  v A 1  e2

In a Young's double slit experiment the intensity of light at each slit is 0. Interference pattern is observed along a direction parallel to the line S1 S2 on screen S.

The minimum, maximum, and the intensity averaged over the entire screen are respectively. (A) 0, 40 , 20 (B) 0 , 20 , 30/2 (C) 0, 40 , 0 (D) 0, 20 , 0 27.

A loop carrying current  has the shape of a regular polygon of n sides. If R is the distance from the centre  to any vertex, then the magnitude of the magnetic induction vector B at the centre of the loop is (A) n

28.

0  tan 2 R n

(B) n

0 2 tan 2 R n

(C)

0 2R

(D)

0  tan R n

A conducting rod of mass m and length  is free to move without friction on two parallel long conducting rails, as shown below. There is a resistance R acorss the the rails. In the entire space around, there is a uniform magnetic filed B normal to the plane of the rod and rails. The rod is given an impulsive velocity v0.

1 mv02 2 (A) will be converted fully into heat energy in the resistor (B) will enable rod to continue to move eith velocity v0 since the rails are frictionless (C) will be converted fully into magnetic energy due to induced current (D) will be converted into the work done against the magnetic field Finally, the initial energy


29.

A steady current  flows through a wire of radius r, length L and resistivity  . The current produces heat in the wire. The rate of heat loss in a wire is proportional its surface area. The steady temperature of the wire is independent of (A) L (B) r (C)  (D) 

30.

The ratio of the speed of sound to the average speed of an air molecule at 300K and 1 atmospheric pressure is close to (A) 1

31.

(B)

300

1 300

(C)

(D) 300

In one model of the election, tho electron of mass me is thought to be a uniformly charged shell of radius R and total charge e, whose electrostatic energy E is equivalent to its mass me via Einstein's mass energy relation E = mec2 . In this model, R is approximately (me = 9.1 × 10–31 kg, c = 3 × 108 m.s–1,

1  = 9 × 109 Farads m–1, magnitude of the electron charge = 1.6 × 10–19 C) 4 0 (A) 1.4 × 10–15 m (B) 2 × 10–13 m (C) 5.3 × 10–11 m (D) 2.8 × 10–35 m 32.

A body is executing simple harmonic motion of amplitude a and period T about the equilibrium position x = 0. Large numbers of snapshots are taken at random of this body in motion. The probability of the body being found in a very small interval x to x + |dx| is highest at (A) x = ± a

(B) x = 0

(C) x = ± a/2

(D) x = ±

a 2

33.

Two identical bodies are made of a material for which the heat capacity increases with temperature One of these is held at a temperature of 100°C while the other one is kept at 0ºC. If the two are brought into , then, assuming no heat loss to the environment, the final temperature that they will reach is (A) 50°C (B) more than 50°C (C) less than 50°C (D) 0°C

34.

A particle is acted upon by a force given by F = –x3 – x4 where  and  are positive constants. At the point x = 0, the particle is (A) in stable equilibrium (B) in unstable equiibrium (C) in neutral equilibrium (D) not in equilibrium

35.

The potential energy of a point particle is given by the expression V(x) = x +  sin (x / ). A dimensionless combination of the constants ,  and  is (A)

36.

 

(B)

2 

(C)

 

(D)

 

A ball of mass m suspended from a rigid by an inextensible massless string is released from a height h above its lowest point. At its lowest point it colIides elastically with a block of mass 2m at rest on a frictionless surface. Neglect the dimensions of the ball and the block. After the collision the ball rises to a maximum height of

(A)

h 3

(B)

h 2

(C)

h 8

(D)

h 9


37.

A particle released from rest is falling through a thick fluid under gravity. The fluid exerts a resistive force on the particle proportional to the square of its speed. Which one of the following graphs best depicts the variation of is speed v with time t?

(A)

(B)

(C)

(D)

38.

A cylindrical steel rod of length 0.10m and thermal conductivity 50 W.m–1.K–1 is welded end to end to copper rod of thermal conductivity 400 W.m–1.K–1 and of the same area of cross section but 0.20 m long. The free end of the steel rod is maintained at 100°C and that of the copper rod at 0°C. Assuming that the rods are perfectly insulated from the surrounding the temperature at the junction of the two rods is (A) 20°C (B) 30°C (C) 40°C (D) 50°C

39.

A parent nucleus X is decaying into daughter nucleus Y which in turn decays to Z. The half lives of X and Y are 4000 years and 20 years respectively. In a certain sample, it is found that the number of Y nuclei hardly changes with time. If the number of X nuclei in the sample is 4 × 1020 , the number of Y nuclei present in it is (A) 2 × 1017 (B) 2 × 1020 (C) 4 × 1023 (D) 4 × 1020

40.

An unpolarized beam of light of intensity 0 es through two linear polarizers making an angle of 30° with respect to each other. The emergent beam will have an intensity. (A)

3 0 4

(B)

3 0 4

(C)

3 0 8

(D)

0 8

CHEMISTRY 41.

42.

43.

44.

Among the following, the species with the highest bond order is : (A) O2 (B) F2 (C) O2+

(D) F2–

The molecule with non-zero dipole moment is : (A) BCl3 (B) BeCl2

(D) NCl3

(C) CCl4

For a one-electron atom, the set of allowed quantum numbers is : (A) n = 1,  = 0, m = 0, ms = +½

(B) n = 1,  = 1, m = 0, ms = +½

(C) n = 1,  = 0, m = –1, ms = –½

(D) n = 1,  = 1, m = 1, ms = –½

In the reaction of benzene with an electrophile E+, the structure of the intermediate -complex can be represented as :

(A)

(B)

(C)

(D)


45.

The most stable conformation of 2, 3-dibromobutane is :

(A)

(B)

(C)

(D)

46.

Typical electronic energy gaps in molecules are about 1.0 eV. In of temperature, the gap is closest to: (A) 102 K (B) 104 K (C) 103 K (D) 105 K

47.

The major final product in the following reaction is : 1) CH MgBr

3 CH3CH2CN     

2) H3O

(A)

(B)

(C)

(D)

48.

A zero-order reaction, A  Product, with an initial concentration [A]0 has a half-life of 0.2 s. If one starts with the concentration 2[A]0, then the half-life is : (A) 0.1 s (B) 0.4 s (C) 0.2 s (D) 0.8 s

49.

The isoelectronic pair of ions is : (A) Sc2+ and V3+ (B) Mn3+ and Fe2+

50.

(C) Mn2+ and Fe3+

(D) Ni3+ and Fe2+

(C)

(D)

(C)

(D)

The major product in the following reaction is : NaNH

2  

(A)

51.

(B)

The major product of the following reaction is : Conc.HBr

 

(A)

(B)


52.

The oxidation state of cobalt in the following molecule is :

(A) 3

(B) 1

(C) 2

(D) 0

53.

The pKa of a weak acid is 5.85. The concentrations of the acid and its conjugate base are equal at a pH of: (A) 6.85 (B) 5.85 (C) 4.85 (D) 7.85

54.

For a tetrahedral complex [MCl4]2–, the spin-only magnetic moment is 3.83 BM. The element M is : (A) Co (B) Cu (C) Mn (D) Fe

55.

Among the following graphs showing variation of rate (k) with temperature (T) for a reaction, the one that exhibits Arrhenius behavior over the entire temperature range is :

(A)

56.

57.

58.

(B)

(C)

(D)

The reaction that gives the following molecule as the major product is

(A)

(B)

(C)

(D)

The C–O bond length in CO, CO2 and CO32– follows the order : (A) CO < CO2 < CO32– (B) CO2 < CO32– < CO (C) CO > CO2 > CO32–

(D) CO32– < CO2 < CO

The equilibrium constant for the following reactions are K1 and K2, respectively. 2P(g) + 3Cl2(g) 2PCl3(g) PCl3(g) + Cl2(g) PCl5(g) The the equilibrium constant for the reaction 2P(g) + 5Cl2(g) 2PCl5(g) is : (A) K1K2 (B) K1K22

(C) K12K22

(D) K12K2 CLASS-XII_STREAM-SB+2_PAGE # 92

59.

The major product of the following reaction is : AlCl

3  + (CH3C)2CHCH2Cl 

60.

(A)

(B)

(C)

(D)

Doping silicon with boron produces a : (A) n-type semiconductor (C) p-type semiconductor

(B) metallic conductor (D) insulator

BIOLOGY 61.

The disorders that arise when the immune system destroys 'self' cells are called autoimmmune disorders. Which of the following would be classified under this (A) rheumatoid arthritis (B) asthma (C) rhintis (D) eczema

62.

Which of the following class of immunoglobulins can trigger the complement cascade ? (A) IgA (B) IgM (C) IgD (D) IgE

63.

Diabetes insipidus is due to (A) hypersecretion of vasoperssin (C) hypersecretion of insulin

(B) Hyposecretion of insulin (D) hyposecretion of vasopressin

64.

Fossils are most often found in which kind of rocks ? (A) meteorites (B) sedimentary rocks (C) igneous rocks (D) metamorphic rocks

65.

Peptic ulcers are caused by (A) a fungus, Candida albicans (B) a virus, cytomegalo virus (C) a parasite, Trypanosoma brucei (D) a bacterium, Helicobacter pylori

66.

Transfer RNA (tRNA) (A) is present in the ribosomes and provides structural integrity (B) usually has clover leaf-like structure (C) carries genetic information from DNA to ribosomes (D) codes for proteins

67.

Some animals excrete uric acid in urine (uricotelic) as it requires very little water. This is an adaptation to conserve water loss. Which animals among the following are most likely to be uricotelic? (A) fishes (B) amphibians (C) birds (D) mammals

68.

A ripe mango, kept with unripe mangoes causes their ripening. This is due to the release of a gaseous plant hormone (A) auxin (B) gibberlin (C) cytokinine (D) ethylene


69.

Human chromosomes undergo structural changes during the cell cycle. Chromosomal structure can be best visualized if a chromosome is isolated from a cell at (A) G1 phase (B) S phase (C) G2 phase (D) M phase

70.

By which of the following mechanisms is glucose reabsorbed from the glomerular filtrate by the kidney tubule (A) osmosis (B) diffusion (C) active transport (D) iver transport

71.

In mammals, the hormones secreted by the pituitary, the master gland, is itself regulated by (A) Hypothalamus (B) median cortex (C) pineal gland (D) cerebrum

72.

Which of the following is true for TCA cycle in eukaryotes (A) takes place in mitochondrion (B) produces no ATP (C) takes place in Golgi complex (D) independent of electron transport chain

73.

A hormone molecule binds to a specific protein on the plasma membrane inducing a signal. The protein it binds to it called (A) ligand (B) antibody (C) receptor (D) histone

74.

DNA mutations that do not cause my functional change in the protein product are known as (A) nonsense mutations (B) missense mutations (C) deletion mutations (D) silent mutations

75.

Plant roots are usually devoid of chlorophyll and cannot perform photosynthesis. However, three are exceptions. Which of the following plant root can perform photosynthesis (A) Arabidopsis (B) Tinospora (C) Rice (D) Hibiscus

76.

Vitamin A deficiency leads to night-blindness. Which of the following is the reason for the disease ? (A) rod cells are not converted to cone cells (B) rhodopsin pigment of rod cells is defective (C) melanin pigment is not synthesized in cone cells (D) cornea of eye gets dried

77.

In Dengue virus infection, patients often develop haemorrhagic fever due to internal bleeding. This happens due to the reduction of (A) platelets (B) RBCs (C) WBCs (D) lymphocytes

78.

If the sequence of bases in sense strand of DNA is 5'-GTTCATCG-3, then the sequence ofd bases in its RNA transcript would be (A) 5'-GTTCATCG-3' (B) 5'GUUCAUCG-3 (C) 5'CAAGTAGC-3' (D) 5'CAAGUAGC -3

79.

A refilex aciton is a quick involuntary response to stimulus. Which of the following is an example of BOTH, unconditioned and conditioned reflex (A) knee jerk reflex (B) secretion of saliva in response to the aroma of food (C) sneezing reflex (D) contration of the pupil in response to bright light

80.

In a food chain such as grass deer  lion, the energy cost of respiration as a proportion of total assimilated energy at each level would be (A) 60%- 30%-20% (B) 20%- 30%-60% (C) 20%- 60%-30% (D) 30%- 30%-30% CLASS-XII_STREAM-SB+2_PAGE # 94

PART-II (2 Marks)

MATHEMATICS 81.

Suppose a, b, c are real numbers, and each of the equations x2 + 2ax + b2 = 0 and x2 + 2bx + c2 = 0 has two distinct real roots. Then the equation x2 + 2cx + a2 = 0 has (A) two distinct positive real roots (B) two equal roots (C) one positive and one negative root (D) no real roots

82.

The coefficient of x2012 in

(1  x 2 )(1  x )

(A) 2010

(B) 2011

83.

325 36

(B)

 6

(B)

36 325

(C)

13 25

5 6

(D)

25 13

1  cosx = sin2(x + ) is 2 4

(C) 

(D) 2

(B) P(1)  0

(C) P(1)  0

(D)

Define a sequence (an) by a1 = 5, an = a1a2 ... an–1 + 4 for n > 1. Then lim

n

(A) equals

1 2

(B) equal 1



87.

(D) 2013

A polynomila P(x) with real coefficients has the property that P(x)  0 for all x. Suppose P(0) = 1 and P(0) = – 1. What can you say about P(1)? (A) P(1)  0

86.

(C) 2012

The sum of all x  [0, ] which satisfy the equation sinx + (A)

85.

is

Let (x, y) be a variable point on the curve 4x2 + 9y2 – 8x – 36y + 15 = 0. Then min (x2 – 2x + y2 – 4y + 5) + max(x2 – 2x + y2 – 4y + 5) is (A)

84.

1 x

The value of the integral

cos 2 x

 1 a

x

(C) equals

2 5

1 –1 < P(1) < 2 2

an an 1 (D) does not exists

dx , where a > 0, is

–

(A)  88.

(B) a

(C)

 2

(D) 2

Consider L=

3

2012 +

3

2013 + ... +

3

3011

R=

3

2013 +

3

2014 + ... +

3

3012

3012

and I =



3 x dx .

2012

(A) L + R < 2I

(B) L + R = 2I

(C) L + R > 2I

(D)

LR = I


89.

A man tosses a coin 10 times, scoring 1 point for each head and 2 points for each tail. Let P(K) be the

1 is 2 (D) 17

probability of scroing at least K points. The largest value of K such that P(K) > (A) 14

90.

(B) 15

(C) 16

x 1 for all x  1. Let f1(x) = f(x), f2(x) = f (f(x)) and generally fn(x) = f(fn–1(x)) for n > 1. Let P = x 1 f1(2) f2(3) f3(4) f4(5) which of the following is a multiple of P (A) 125 (B) 375 (C) 250 (D) 147 Let f(x) =

PHYSICS 91.

The total energy of a black body radiation source is collected for five minutes and used to heat water. The temperature of the water increases from 10.0°C to 11.0°C. The absolute temperature of the black body is doubled and its surface area halved and the experiment repeated for the same time. Which of the following statements would be most nearly correct? (A) The temperature of the water would increase from 10.0°C to a final temperature of 12°C (B) The temperature of the water would increase from 10.0°C to a final temperature of 18°C (C) The temperature of the water would increase from 10.0°C to a final temperature of 14°C (D) The temperature of the water would increase from 10.0°C to a final temperature of 11°C

92.

A small asteroid is orbiting around the sun in a circular orbit of radius r0 with speed V0. A rocket is launched from the asteroid with speed V = V0 where V is the speed relative to the sun. The highest value of  for which the rocket will remain bound to the solar system is (ignoring gravity due to the asteroid and effect of other planets) (A)

93.

(B) 2

2

(C)

(D) 1

3

A radioactive nucleus A has a single decay mode with half life A. Another radioactive nucleus B has two decay modes 1 and 2. If decay mode 2 were absent, the half life of B would have been A/2. If decay mode

B 1 were absent, the half life of B would have been 3 A. If the actual half life of B is B , then the ratio  A is (A)

3 7

(B)

7 2

(C)

7 3

(D) 1

94.

A stream of photons having energy 3 eV each impinges on a potassium surface. The work function of potassium is 2.3 eV. The emerging photo-electrons are slowed down by a copper plate placed 5 mm away. If the potential difference between the two metal plates is 1V, the maximum distance the electrons can move away from the potassium surface before being turned back is (A) 3.5 mm (B) 1.5 mm (C) 2.5 mm (D) 5.0 mm

95.

Consider three concentric metallic spheres A, B and C of radii a, b, c respectively where a
96.

bc c b

(B) + 40V

bc c b

(C) – 40V

ac ca

(D) zero

On a bright sunny day a diver of height h stands at the bottom of a lake of depth H. Looking upward, he can see objects outside the lake in a circular region of radius R. Beyond this circle he sees the images of objects lying on the floor of the lake. If refractive index of waler is 4/3, then the value of R is (A)

3(H  h) 7

(B) 3h 7

(C)

(H  h) 7 3

(D)

(H  h) 5 3


97.

As shown in the figure below, a cube is formed with ten indentical resistance R (thick lines) and two shorting wires (dotted lines) along the arms AC and BD.

Resistance between point A and B is 5R R (A) (B) 6 2 98.

(C)

3R 4

(D) R

A standing wave in a pipe with a length L = 1.2 m is described y

 2    2   y (x, t) = y0 sin  L  x  sin  L  x  4        Based on above information, which one of the following statement is incorrect. (Speed of sound in air is 300 ms–1) (A) A pipe is closed at both ends (B) The wavelength of the wave could be 1.2 m (C) There could be a nods at x = 0 and antinode at x = L/2 (D) The frequency of the fundamental mode of vibrations is 137.5 Hz 99.

Two block (1 and 2) of equal mass m are connected by an ideal string (see figure below) over a frictionless pulley. The blocks are attached to the ground by springs having spring constants k1 and k2 such that k1 > k2 .

Initially, both springs are unstretched. The block 1 is slowly pulled down a distance x and released. Just after the release the possible value of the magnitudes of the acceleration of the blocks a1 and a2 can be (k 1  k 2 ) x  k x k x     or  a1  1  g and a 2  2  g  (A) either  a1  a 2  2 m m m     (k 1  k 2 ) x    only (B)  a1  a 2  2m  

(k 1  k 2 ) x    only (C)  a1  a 2  2m  

  (k 1k 2 ) x (k 1  k 2 ) x    g   or  a1  a 2  (D) either  a1  a 2  (k 1  k 2 )m 2m     CLASS-XII_STREAM-SB+2_PAGE # 97

100.

A simple pendulum is released from rest at the horizontally stretched position. When the string makes an angle  with the vertical, the angle  which the acceleration vector of the bob makes with the string is given by

(A)  = 0

 tan    (B)  = tan–1   2 

(C)  = tan–1 (2 tan )

(D)  =

 2

CHEMISTRY 101.

The final major product obtained in the following sequence of reactions is :

(A)

(B)

(C)

(D)

102.

In the DNA of E. Coli the mole ratio of adenine to cytosine is 0.7. If the number of moles of adenine in the DNA is 350000, the number of moles of guanine is equal to : (A) 350000 (B) 500000 (C) 225000 (D) 700000

103.

(R)-2-bromobutane upon treatment with aq. NaOH gives

104.

(A)

(B)

(C)

(D)

Phenol on treatment with dil. HNO3 gives two products P and Q. P is steam volatile but Q is not. P and Q are respectively

(A)

(B)

(C)

(D)


105.

A metal is irradiated with light of wavelength 660 nm. Given that the work function of the metal is 1.0 eV, the de-Broglie wavelength of the ejected electron is close to : (A) 6.6 × 10–7 m (B) 8.9 × 10–11 m (C) 1.3 × 10–9 m (D) 6.6 × 10–13 m

106.

The inter-planar spacing between the (2 2 1) planes of a cubic lattice of length 450 pm is : (A) 50 pm (B) 150 pm (C) 300 pm (D) 450 pm

107.

The H for vaporization of a liquid is 20 kJ/mol. Assuming ideal behaviour, the change in internal energy for the vaporization of 1 mole of the liquid at 60ºC and 1 bar is close to : (A) 13.2 kJ/mol (B) 17.2 kJ/mol (C) 19.5 kJ/mol (D) 20.0 kJ/mol

108.

Among the following, the species that is both tetrahedral and diamagnetic is : (A) [NiCl4]2– (B) [Ni(CN)4]2– (C) Ni(CO)4 (D) [Ni(H2O)6]2+

109.

Three moles of an ideal gas expands reversibly under isothermal condition from 2 L to 20 L at 300 K. The amount of heat-change (in kJ/mol) in the process is : (A) 0 (B) 7.2 (C) 10.2 (D) 17.2

110.

The following data are obtained for a reaction, X + Y  Products. Expt. [X0]/mol [Y0]/mol rate/mol L–1 s–1 1 0.25 0.25 1.0 × 10–6 2 0.50 0.25 4.0 × 10–6 3 0.25 0.50 8.0 × 10–6 The overall order of the reaction is : (A) 2 (B) 4 (C) 3

(D) 5

BIOLOGY 111.

When hydrogen peroxide is applied on the wound as a disinfectant, there is frothing at the site of injury, which is due to the presence of an enzyme in th skin that uses hydrogen peroxide as a substrate to produce (A) hydrogen (B) carbon dioxide (C) water (D) oxygen

112.

Persons suffering from hypertension (high blood pressure) are advised a low-salt diet because (A) more salt is absorbed in the body of a patient with hypertension (B) high salt leads to water retention in the blood that further increases the blood pressure (C) high salt increases nerve conduction and increases blood pressure (D) high salt causes adrenaline release that increases blood pressure

113.

Insectivorous plants that mostly grow on swampy soil use insects as a source of (A) carbon (B) nitrogen (C) phosphorous (D) magnesium

114.

In cattle, the coat colour red and white are two dominant traits, which express equally in F1 to produce roan (red and white clour in equal proportion). If F1 progeny are self-bred, the resulting progeny in F2 will have phenotypic ratio (red:roan:white) is (A) 1 : 1 : 1 (B) 3 : 9 : 3 (C) 1 : 2 : 1 (D) 3 : 9 : 4

115.

The restriction endonuclease EcoR-I recognises and cleaves DNA sequence as shown below 5’ – G A A T T C – 3’ 3’ – C T T A A G – 5’ What is the probable number of cleavage sites that can occur in a 10 kb long random DNA sequence? (A) 10 (B) 2 (C) 100 (D) 50

116.

Which one of the following is true about enzyme catalysis ? (A) the enzyme changes at the end of the reaction (B) the activation barrier of the process is lower in the presence of an enzyme (C) the rate of the reaction is retarded in the presence of an enzyme (D) the rate of the reaction is independent of substrate concentration


117.

Vibrio cholerae causes cholera in humans. Ganga water was once used successfully to combat the infection. The possible reason could be (A) high salt content of Ganga water (B) low salt content of Ganga water (C) presence of bacteriophages in Ganga water (D) presence of antibiotics in Ganga water

118.

When a person begins to fast, after some time glycogen stored in the liver is mobilized as a source of glucose. Which of the following graphs best represents the change of glucose level (y-axis) in his blood, starting from the time (x-axis) when he begins to fast ?

(A)

(B)

(C)

(D)

119.

The following sequence contains the open reading frame of a polypeptide. How many amino acids will the polypeptide consist of ? 5’ – AGCATATGATCGTTTCTCTGCTTTGAACT–3 (A) 4 (B) 2 (C) 10 (D) 7

120.

Insects constitute the largest animal group on earth. About 25-30% of the insect species are known to be herbivores. In spite of such huge herbiore perssure, globally, green plants have persisted. One possible reason for this persistence is : (A) food preference of insects has tended to change with time (B) herbivore insects have become inefficient feeders of green plants (C) herbivore population has been kept in control by predators (D) decline in reproduction of herbivores with time


KISHORE VAIGYANIK PROTSAHAN YOJANA - 2013 Date : 27-10-2013

Duration : 3 Hours

Max. Marks : 160

STREAM - SB/SX GENERAL INSTRUCTIONS •

The Test Booklet consists of 120 questions.

•

There are Two parts in the question paper. The distribution of marks subjectwise in each part is as under for each correct response.

MARKING SCHEME : PART-I : MATHEMATICS Question No. 1 to 20 consist of ONE (1) mark for each correct response. PHYSICS Question No. 21 to 40 consist of ONE (1) mark for each correct response. CHEMISTRY Question No. 41 to 60 consist of ONE (1) mark for each correct response. BIOLOGY Question No. 61 to 80 consist of ONE (1) mark for each correct response.

PART-II : MATHEMATICS Question No. 81 to 90 consist of TWO (2) marks for each correct response. PHYSICS Question No. 91 to 100 consist of TWO (2) marks for each correct response. CHEMISTRY Question No. 101 to 110 consist of TWO (2) marks for each correct response. BIOLOGY Question No. 111 to 120 consist of TWO (2) marks for each correct response.

YEAR-2013 (KVPY-STREAM-SB+2)

PART-I One Mark Questions MATHEMATICS 1.

The sum of non-real roots of the polynomial equation x3 + 3x2 + 3x + 3 = 0. (A) equals 0 (B) lies between 0 and 1 (C) lies between –1 and 0 (D) has absolute value bigger than 1

2.

Let n be a positive integer such that log2log2log2log2log2(n) < 0 < log2log2log2log2(n). Let  be the number of digits in the binary expansion of n. Then the minimum and the maximum possible values of  are (A) 5 and 16 (B) 5 and 17 (C) 4 and 16 (D) 4 and 17

3.

Let  be a cube root of unity not equal to 1. Then the maximum possible value of | a + bw + cw2 | where a, b, c  {+1, –1} is (A) 0

(B) 2

(C)

(D) 1+ 3

3

4.

If a, b are positive real numbers such that the lines ax + 9y = 5 and 4x + by = 3 are parallel, then the least possible value of a + b is (A) 13 (B) 12 (C) 8 (D) 6

5.

Two line segments AB and CD are constrained to move along the x and y axes, respectively, in such a way that the points A, B, C, D are concyclic. If AB = a and CD = b, then the locus of the centre of the circle ing through A, B, C, D in polar coordinates is (A) r 2 

a2  b 2 4

(B) r 2 cos 2 

a2  b2 4

(C) r2 = 4(a2 + b2)

(D) r2 cos2 = 4(a2 – b2)

6.

Consider a triangle ABC in the xy -plane with vertices A = (0,0), B = (1,1) and C = (9, 1). If the line x = a divides the triangle into two parts of equal area, then a equals (A) 3 (B) 3.5 (C) 4 (D) 4.5

7.

Let ABC be an acute-angled triangle and let D be the midpoint of BC. If AB = AD, then tan(B)/tan(C) equals (A)

2

(B)

(C) 2

3

(D) 3

8.

The angles , ,  of a triangle satisfy the equations 2sin + 3cos = 3 2 and 3sin + 2cos = 1. Then angle  equals (A) 150° (B) 120° (C) 60° (D) 30°

9.

Let f : R  R be a function such that lim f(x) = M > 0. Then which of the following is false?

10.

x 

(A) xlim xsin(1/x) f(x) = M 

(B) xlim sin(f(x)) = sin M 

(C) xlim xsin(e–x) f(x) = M 

(D) lim

x 

sin x .f ( x )  0 x

For x, t  R let pt(x) = (sint)x2 – (2cost)x + sint 1

be a family of quadratic polynomials in x with variable coefficients. Let A(t) =

 p ( x)dx . Which of the t

0

following statements are true? (I) A(t) < 0 for all t. (III) A(t) = 0 for infinitely many t. (A) (I) and (II) only (B) (II) and (III) only

(II) A(t) has infinitely many critical points. (IV) A’(t) < 0 for all t. (C) (III) and (IV) only (D) (IV) and (I) only CLASS-XII_STREAM-SB+2_PAGE # 101

11.

Let f(x) =

2  x  x 2 and g(x) = cos x. Which of the following statements are true? (I) Domain of f((g(x))2) = Domain of f(g(x)) (II) Domain of f(g(x)) + g(f(x)) = Domain of g(f(x)) (III) Domain of f(g(x)) = Domain of f(g(x)) (IV) Domain of g((f(x))3) = Domain of f(g(x)) (A) Only (I) (B) Only (I) and (II) (C) Only (III) and (IV) (D) Only (I) and (IV) x

12.

For real x with – 10  x  10 define f(x) =

2

[t]

dt , where for a real number r we denote by [r] the largest

10

integer less than or equal to r. The number of points of discontinutiy of f in the interval (–10,10) is (A) 0 (B) 10 (C) 18 (D) 19 13.

For a real number x let [x] denote the largest integer less than or equal to x and {x} = x – [x]. The n

smallest possible integer value of n for which

 xxdx exceeds 2013 is 1

(A) 63 14.

(B) 64

4 2    (B)  8   2  

4 2    (C)  4   2  

4 2    (D)  4   2  

A box contains coupons labeled 1, 2, 3....n. A coupon is picked at random and the number x is noted. The coupon is put back into the box and a new coupon is picked at random. The new number is y. Then the probability that one of the numbers x, y divides the other is (in the options below [r] denotes the largest integer less than or equal to r) (A)

16.

(D) 91

The area bounded by the curve y = cos x, the line ing (–/4, cos(–/4)) and (0, 2) and the line ing (/4, cos(/4)) and (0, 2) is

4 2    (A)  8   2   15.

(C) 90

1 2

(B)

1

n

n   2 n k 1  k 



(C) 

1 1  n n2

n

n k  k 1  



(D) 

1 2  n n2

n

n

  k  k 1

Let n  3. A list of numbers 0 < x1 < x2 < ... < xn has mean  and standard deviatiion . A new list of numbers is made as follows : y1 = 0, y2 = x2, ....,yn–1 = xn–1, yn = x1 + xn. The mean and the standard deviation of the new list are ˆ and ˆ . Which of the following is necessarily true?

17.

(A)  = ˆ ,   ˆ

(B)  = ˆ ,   ˆ

(C)  = ˆ

(D)  may or may not be equal to ˆ









Let 1,  2 ,  3 ,  4 be unit vectors in the xy -plane, one each in the interior of the four quadrants. Which of the following statements is necessarily true? 







(A)         0 1 2 3 4 



(B) There exist i, j with 1  i < j  4 such  i   j is in the first quadrant 







(C) There exist i, j with 1  i < j  4 such that  i .  j < 0 (D) There exist i, j with 1  i < j  4 such that  i .  j > 0 18.

The number of integers n with 100  n  999 and containing at most two distinct digits is (A) 252 (B) 280 (C) 324 (D) 360 CLASS-XII_STREAM-SB+2_PAGE # 102

19.

For an integer n let Sn = {n + 1, n + 2, ....., n + 18}. Which of the following is true for all n  10? (A) Sn has a multiple of 19 (B) Sn has a prime (C) Sn has at least four multiples of 5 (D) Sn has at most six primes

20.

Let P be a closed polygon with 10 sides and 10 vertices (assume that the sides do not intersect except at the vertices). Let k be the number of interior angles of P that are greater than 180°. The maximum possible value of k is (A) 3 (B) 5 (C) 7 (D) 9

PHYSICS 21.

Consider an initially neutral hollow conducting spherical shell with inner radius r and outer radius 2r. A point charge +Q is now placed inside the shell at a distance r/2 from the centre. The shell is then grounded by connecting the outer surface to the earth. P is an external point at a distance 2r from the point charge +Q on the line ing through the centre and the point charge +Q as shown in the figure.

The magnitude of the force on a test charge +q placed at P will be

1 qQ (A) 4 2 0 4r 22.

1 9qQ (B) 4 2 0 100r

1 4qQ (C) 4 2 0 25r

(D) 0

Consider the circuit shown in the figure below :

All the resistors are identical. The charge stored in the capacitor, once it is fully charged, is (A) 0 23.

(B)

5 CV 13

(C)

2 CV 3

(D)

5 CV 8

A nuclear decay is possible if the mass of the parent nucleus exceeds the total mass of the decay particles. If M(A, Z) denotes the mass of a single neutral atom of an element with mass number A and atomic number Z, then the minimal condition that the  decay

X AZ  YZA1   –  v e will occur is (me denotes the mass of the  particle and the neutrino mass mv can be neglected) : (A) M(A, Z) > M(A, Z + 1 ) + me (B) M(A, Z) > M(A, Z + 1) (C) M(A, Z) > M(A, Z + 1) + Zme (D) M(A,Z) > M(A, Z + 1) – me


24.

The equation of state of n moles of a non-ideal gas can be approximated by the equation 2    P  n a V  nb  nRT  V 2   where a and b are constants characteristic of the gas. Which of the following can represent the equation of a quasistatic adiabat for this gas (Assume that CV, the molar heat capacity at constant volume, is independent of temperature)?

25.

(A) T( V  nb)R / C V = constant

(B) T( V  nb)C V / R = constant

ab   R / CV  constant (C)  T  2 V  nb  V R 

 n 2ab   V  nbCV / R  constant T  (D)  2  V R  

A blackbox (BB) which may contain a combination of electrical circuit elements (resistor, capacitor or inductor) is connected with other external circuit elements as shown below in the figure (a). After the switch (S) is closed at time t = 0, the current (I) as a function of time (t) is shown in the figure (b).

From this we can infer that the blackbox contains (A) A resistor and a capacitor in series (C) A resistor and an inductor in series 26.

(B) A resistor and a capacitor in parallel (D) A resistor and an inductor in parallel

In a photocell circuit the stopping potential, V0, is a measure of the maximum kinetic energy of the photoelectrons. The following graph shows experimentally measured values of stopping potential versus frequency v of incident light.

The values of Plank’s constant and the work function as determined from the graph are (taking the magnitude of electronic charge to be e = 1.6 × 10–19 C) (A) 6.4 × 10–34 Js, 2.0 eV (B) 6.0 × 10–34 Js, 2.0 eV –34 (C) 6.4 × 10 Js, 3.2 eV (D) 6.0 × 10–34 Js, 3.2 eV 27.

An engine moving away from a vertical cliff blows a born at a frequency f. Its speed is 0.5% of the speed of sound in air. The frequency of the reflected sound received at the engine is (A) 0.990 f (B) 0.995 f (C) 1.005 f (D) 1.010 f


28.

An arangement with a pair of quarter circular coils of radii r and R with a common centre C and carrying a current I is shown.

The permeability of free space is 0. The magnetic field at C is (A) 0 I(1/r – 1/R)/8 into the page (B) 0 I (1/r – 1/R)/8 out of the page (C) 0 I(1/r + 1/R)/8 out of the page (D) 0 I(1/r + 1/R)/8 into the page 29.

The circuit shown has been connected for a long time. The voltage across the capacitor is

(A) 1.2 V 30.

(B) 2.0 V

(C) 2.4 V

(D) 4.0 V

A wheel of radius R with an axle of radius R/2 is shown in the figure and is free to rotate about a frictionless axis through its centre and perpendicular to the page. Three forces (F, F, 2F) are exerted tangentially to the repective rims as shown in the figure.

The magnitude of the net torque acting on the system is nearly (A) 3.5FR (B) 3.2 FR (C) 2.5 FR

(D) 1.5 FR

31.

Two species of radioactive atoms are mixed in equal number. The disintegration of the first species is  and of the second is /3. After a long time the mixture will behave as a species with mean life of approximately (A) 0.70/ (B) 2.10/ (C) 1.00/ (D) 0.52/

32.

The bulk modulus of a gas is defined as B = –VdP/dV. For an adiabatic process the variation of B is proportional to Pn. For an idea gas, n is (A) 0

(B) 1

(C)

5 2

(D) 2 CLASS-XII_STREAM-SB+2_PAGE # 105

33.

Photons of energy 7 eV are incident on two metals A and B with work functions 6 eV and 3 eV respectively. The minimum de Broglie wavelengths of the emitted photoelectrons with maximum energies are A and B, respectively where A/B is nearly (A) 0.5 (B) 1.4 (C) 4.0 (D) 2.0

34.

An electron enters a chamber in which a uniform magnetic field is present as shown. Ignore gravity. During its motion inside the chamber. (A) the force on the electron remains constant (B) the kinetic energy of the electron remains constant (C) the momentum of the electron remains constant (D) the speed of the electron increases at a uniform rate

35.

A ray of light incident on a glass sphere (refractive index

36.

Young-Laplace law states that the excess pressure inside a soap bubble of radius R is given by P = 4/R where  is the coefficient of surface tension of the soap. The number E0 is a dimensionless number that is used to describe the shape of bubbles rising through a surrounding fluid. It is a combination of g, the acceleration due to gravity, , the density of the surrounding fluid,  and a characterstic length scale L which could be the radius of the bubble. A possible expression for E0 is

3 ) suffers total internal reflection before emerging out exactly parallel to the incident ray. The angle of incidence was (A) 75° (B) 30° (C) 45° (D) 60°

(A) 37.

g L3

L2 (B) g

gL2 (C) 

gL2 (D) 

A plank is resting on a horizontal ground in the northern hemisphere of the Earth at a 45° latitude. Let the angular speed of the Earth be  and its radius re. The magnitude of the frictional force on the plank will be (A) mre2

(B)

mre 2 2

(C)

mre 2 2

(D) Zero

38.

The average distance between molecules of an ideal gas at STP is approximately of the order of (A) 1 nm (B) 100 nm (C) 100 cm (D) 1 m

39.

A point particle of mass 0.5 kg is moving along the x-axis under a force described by the potential energy V shown below. It is projected towards the right from the origin with a speed v. What is the minimum value of v for which the particle will escape infinitely fasr away from the origin ?

(A) 2 2 ms–1 (C) 4 ms–1

(B) 2 ms–1 (D) The particle will never escape


40.

The figure below shows pressure variation in two different sound waves in air with time at a given position. Both the figures are drawn to the same scale.

Which of the following statements is true? (A) Wave 1 has lower frequency and smaller amplitude compared to wave 2 (B) Wave 1 has higher frequency and greater amplitude compared to wave 2 (C) Wave 1 has shroter wavelength and greater amplitude compared to wave 2 (D) Wave 1 has shorter wavelength and smaller amplitude compared to wave 2

CHEMISTRY 41.

Among the following, the set of isoelectronic ions is : (A) Na+, Mg2+, F—, Cl— (B) Na+, Ca2+, F—, O2— (C) Na+, Mg2+, F—, O2—

(D) Na+, K+, S2—, Cl—

42.

For a zero-order reaction with rate constant k, the slope of the plot of reactant concentration against time is: (A) k/2.303 (B) k (C) –k/2.303 (D) – k

43.

The compound which reacts with excess bromine to produce 2, 4, 6-tribromophenol, is : (A) 1, 3-cyclohexadiene (B) 1, 3-cyclohexanedione (C) salicylic acid (D) cyclohexanone

44.

Ethyl acetate reacts with NH2NHCONH2 to form : (A) CH3CONHCONHNH2 (C) CH3CONHNHCONH2

45.

(B) CH3CON(NH2)CONH2 (D) CH3CH2NHNHCONH2

The variation of solubility of four different gases (G1, G2, etc.) in a given solvent with pressure at a constant temperature is shown in the plot.

The gas with the highest value of Henry’s law constant is (A) G4 (B) G2 (C) G3

(D) G1 CLASS-XII_STREAM-SB+2_PAGE # 107

46.

For the reaction, the concentration of A decreases from 0.06 to 0.03 mol L–1 and that of B –1 rises from 0 to 0.06 mol L at equilibrium. The values of n and the equilibrium constant for the reaction, respectively, are : (A) 2 and 0.12 (B) 2 and 1.2 (C) 3 and 0.12 (D) 3 and 1.2

47.

The reaction of ethyl methyl ketone with Cl2/excess OH– gives the following major product : (A) ClCH2CH2COCH3 (B) CH3CH2COCCl3 (C) ClCH2CH2COCH2Cl (D) CH3CCl2COCH2Cl

48.

The compound that readily tautomerizes is : (A) CH3COCH2CO2C2H5 (B) CH3COCH2CH2CH3

(C) CH3COCH2CH2CH3

(D) (CH3)3CCOC(CH3)3

49.

Hydrolysis of BCl3 gives X which on treatment with sodium carbonate produces Y, X and Y, respectively, are (A) H3BO3 and NaBO2 (B) H3BO3 and Na2B4O7 (C) B2O3 and NaBO2 (D) B2O3 and Na2B4O7

50.

The numbers of lone pair(s) on Xe in XeF2 and XeF4 are, respectively : (A) 2 and 3 (B) 4 and 1 (C) 3 and 2

(D) 4 and 2

51.

The entropy change in the isothermal reversible expansion of 2 moles of an ideal gas from 10 to 100 L at 300 K is : (A) 42.3 JK–1 (B) 35.8 J K–1 (C) 38.3 J K–1 (D) 32.3 J K–1

52.

D-Glucose upon treatment with bromine-water gives :

(A)

(B)

(C)

(D)

53.

In the structure of borax, the numbers of boron atoms and B-O-B units, respectively, are : (A) 4 and 5 (B) 4 and 3 (C) 5 and 4 (D) 5 and 3

54.

The number of peptide bonds in the compound is :

(A) 1 55.

(B) 2

(C) 3

(D) 4

For the isothermal reversible expansion of an ideal gas : (A) H > 0 and U = 0 (B) H > 0 and U < 0 (C) H = 0 and U = 0

(D) H = 0 and U > 0

56.

If the angle of incidence of X-ray of wavelength 3Å which produces a second order diffracted beam from the (100) planes in a simple cubic lattice with interlayer spacing a = 6 Å is 30°, the angle of incidence that produ ces a first-order diffracted beam from the (200) planes is : (A) 15° (B) 45° (C) 30° (D) 60°

57.

The number of ions produced in water by dissolution of the complex having the empirical formular, COCl34NH3, is : (A) 1 (B) 2 (C) 4 (D) 3

58.

The spin-only magnetic moments of [Fe(NH3)6]3+ and [FeF6]3– in BM are, respectively, (A) 1.73 and 1.73 (B) 5.92 and 1.73 (C) 1.73 and 5.92 (D) 5.92 and 5.92


59.

The order of SN1 reactivity in aqueous acetic acid solution for the compounds is :

(A) 1 > 2 > 3 60.

(B) 1 > 3 > 2

(C) 3 > 2 > 1

(D) 3 > 1 > 2

An ionic compound is formed between a metal M and a non-metal Y. If M occupies half the octahedral voids in the cubic close-packed arrangement formed by Y, the chemical formula of the ionic compound is: (A) MY (B) MY2 (C) M2Y (D) MY3

BIOLOGY 61.

Human fetal haemoglobin differs from the adult haemoglobin in that it has (A) higher affinity for oxygen (B) lower affinity for oxygen (C) two subunits only (D) is glycosylated

62.

Nucleolus is an organelle responsible for the production of (A) carbohydrates (B) messenger RNA (C) lipids

(D) ribosomal RNA

63.

The sequences of four DNA molecules are given below: i. TATATATATATATA ii. TTTCCCGGGAAA ATATATATATATAT AAAGGGCCCTTT iii. TTGCGTTGCCC iv. GCCGGATCCGGC AACGCAACGGG CGGCCTAGGCCG Which one of these DNA moelcules will have the highest melting temperature (Tm)? (A) i (B) ii (C) iii (D) iv

64.

If DNA codons are ATG GAA, insertion of thymine after the first codon results in, (A) non-sense mutation (B) mis-sense mutation (C) frameshift mutation (D) silent mutation

65.

Genetic content of a cell reduces to half during (A) meiotic prophase I (B) mitotic prophase

66.

67.

(C) meiotic prophase II

(D) meiotic telophase

Which one of the following techniques is used for the detection of proteins ? (A) Northern blotting (B) Western blotting (C) Southern blotting

(D) In-situ hybridization

Fission yeasts are (A) Archaebacteria

(D) Eukaryotes

(B) Eubacteria

(C) Prokaryotes

68.

In green leaves, the light and dark reactions occur in (A) stroma and grana respectively (B) grana and stroma respectively (C) cristae and matrix respectively (D) both occur in cytoplasm

69.

According to Mendel, ..................................... segregate and ........................... assort independently. (A) alleles of a gene; alleles of different genes (B) alleles of different genes; alleles of a gene (C) dominanat traits; recessive traits (D) recessive traits; recessive traits

70.

The two enzymatic activities associated with RUBISCO are (A) oxidase and oxygenase (B) oxygenase and carboxylase (C) oxidase and carboxylase (D) oxygenase and carbamylation

71.

Chlorofluorocarbons (CFCs) are belived to be associated with cancers because, (A) CFCs react with DNA and cause mutations (B) CFCs react with proteins involved in DNA repair (C) CFCs destroy the ozone layer and permit harmful UV rays to reach the earth (D) CFCs react with DNA polymerase and reduce fidelity of DNA replication

72.

Morphogenetic movements take place predominantly during the following embryonic stage (A) blastula (B) Morula (C) Gastrula (D) Fertilized eggs CLASS-XII_STREAM-SB+2_PAGE # 109

73. 74.

75.

The only organ which is capable of producing Fructose in humans is (A) liver (B) pancreas (C) seminal vesicles

(D) muscle

Stroke could be prevented/treated with (A) balanced diet (B) clotting factors

(C) insulin

(D) blood thinners

In orange and lemon, the edible part of the fruit is (A) placenta (C) hairs of the ovary wall

(B) thalamus (D) succulent Mesocap

76.

Which one of the following statements about nitrogenase is correct? (A) It is sensitive to CO2 and therefore present in isolated nodules. (B) It requires O2 and therefore functional during the day (C) It is sensitive to O2 and therefore is functional in anaerobic environments (D) It is sensitive to light and therefore functions only in dark.

77.

Part of epidermis that keeps out unwanted particles is called (A) columnar epithelium (B) squamous epithelium (C) ciliated epithelium (D) cuboidal epithelium

78.

Species that are most effective at colonising new habitats show (A) low reproductive ability (B) high dispersal ability (C) slow growth and maturation (D) high competitive ability

79.

In a large isolated population, alleles p and q at a locus are at Hardy Weinberg equilibrium. The frequencies are p = 0.6 and q = 0.4. The proportion of the heterozygous genotype in the population is (A) 0.24 (B) 1 (C) 0.48 (D) 0.12

80.

In vertebrates ‘glycogen’ is stored chiefly in (A) heart and blood (B) spleen and stomach

(C) bones and lymph

(D) liver and muscles

PART-II Two Mark Questions MATHEMATICS 81.

82.

 1 Let f(x) be a non-constant polynomial with real coefficients such that f   = 100 and f(x)  100 for all 2 real x. Which of the following statements is NOT necessarily true ? (A) The coefficient of the highest degree term in f(x) is negative (B) f(x) has at least two real roots (C) If x  1/2 then f(x) < 100 (D) At least one of the coefficients of f(x) is bigger than 50.

Let a, b, c, d be real numbers such that n

 (ak

3

 bk 2  ck  d)  n 4

k 1

for every natural number n. Then | a | + | b | + | C | + | d | is equal to (A) 15 (B) 16 (C) 31 83.

(D) 32

The vertices of the base of an isosceles triangle lie on a parabola y2 = 4x and the base is a part of the line y = 2x – 4. If the third vertex of the triangle lies on the x-axis, its coordinates are 5  (A)  ,0  2 

7  (B)  ,0  2 

9  (C)  ,0  2 

 11  (D)  ,0  2 


84.

In a triangle ABC, let G denote its centroid and let M, N be points in the interiors of the segments AB, AC, respectively, such that M, G, N are collinear. If r denotes the ratio of the area of triangle AMN to the area of ABC then (A) r = 1/2 (B) r > 1/2 (C) 4/9  r < 1/2 (D) 4/9 < r

85.

Let XY be the diameter of a semicircle with centre O. Let A be a variable point on the semicircle and B another point on the semicircle such that AB is parallel to XY. The value of BOY for which the inradius of triangle AOB is maximum, is

  1  5  1  (A) cos  2   

86.

Let f(x) = 1+ (A) 0

87.

  1  5  1  (B) sin  2   

 3

(D)

x x2 x3 x 4    . The number of real roots of f(x) = 0 is 1! 2! 3! 4! (B) 1 (C) 2

 5

(D) 4

Suppose that the earth is a sphere of radius 6400 kilometers. The height from the earth’s surface from where exactly a fourth of the earth’s surface is visible, is (A) 3200 km

88.

(C)

(B) 3200 2 km

(C) 3200 3 km

(D) 6400 km

Let n be a positive integer. For a real number x, let [x] denote the largest integer not exceeding x and n 1

{x} = x – [x]. Then

 1

(A) loge(n)

x[ x ] dx is equal to x  (B)

1 n 1

(C)

n n 1

(D) 1 

1 1  ....  2 n

89.

A box contains coupons labelled 1,2,...,100. Five coupons are picked at random one after another without replacement. Let the numbers on the coupons be x1, x2,....,x5. What is the probability that x1 > x2 > x3 and x3 < x4 < x5? (A) 1/120 (B) 1/60 (C) 1/20 (D) 1/10

90.

In a tournament with five teams, each team plays against every other team exctly once. Each game is won by one of the playing teams and the winning team scores one point, while the losing team scores zero. Which of the following is NOT necessarily true? (A) There are at least two teams which have at most two points each. (B) There are at least two teams which have at least two points each. (C) There are at most three teams which have at least threee points each (D) There are at most four teams which have at most two points each

PHYSICS 91.

A bullet of mass m is fired horizontally into a large sphere of mass M and radius R resting on a smooth horizontal table.

The bullet hits the sphere at a height h from the table and sticks to its surface. If the sphere starts rolling without slippng immediately on impact, then (A)

h 4m  3M  R 2m  M

(B)

h m  3M  R m  2M

(C)

h 10m  7M  R 5m  M

(D)

h 4m  3M  R mM


92.

A small boy is throwing a ball towards a wall 6 in front of him. He releases the ball at a height of 1.4 m from the ground. The ball bounces from the wall at a height of 3 m, rebounds from the ground and reaches the boy’s hand exactly at the point of release. Assuming the two bounces (one from the wall and the other from the ground) to be perfectly elastic, how far ahead of the boy did the ball bounce from the ground? (A) 1.5m (B) 2.5 m (C) 3.5 m (D) 4.5 m

93.

In the P-V diagram below the dashed curved line is an adiabat.

For a process that is described by a straight line ing two points X and Y on the adiabat (solid line in the diagram) heat is : (hint : Consider the variations in temperature from X to Y along the straitght line) (A) absorbed throughtout from X to Y (B) released throughout from X to Y (C) absorbed from X up to an intermediate point Z (not shown in the figure) and then released from Z to Y (D) released from X up to an intermediate point Z (not shown in the figure) and then absorbed from Z to Y 94.

A singly ionized helium atom in an excited state (n = 4) emits a photon of energy 2.6 eV. Given that the ground state energy of hydrogen atom is –13.6 eV, the energy (Et) and quantum number (n) of the resulting state are respectively, (A) Et = – 13.6eV, n = 1 (B) Et = – 6.0 eV, n = 3 (C) Et = – 6.0 eV, n = 2 (D) Et = – 13.6 eV, n = 2

95.

The figure below shows a circuit and its input voltage vi as function of time t.

Assuming the diodes to be ideal, which of the following graphs depicts the output voltage v0 as function of time t ?

(A)

96.

(B)

(C)

(D)

A ball is rolling without slipping in a spherical shallow bowl (radius R) as shown in the figure and is executing simple harmonic motion. If the radius of the ball is doubled, the period of oscillation

(A) increases slightly (C) is increased by a factor of 2

(B) is reduced by a factor of 1/2 (D) decreases slightly


97.

A solid sphere rolls without slipping, first horizontal and then up to a point X at height h on an inclined plane before rolling down, as shown.

The initial horizontal speed of the sphere is (A) 98.

10gh / 7

7gh / 5

(B) 10 J

5gh / 7

(D)

2gh

(C) 20 J

(D) 30 J

A block of mass m slides from rest at a height H on a frictionless inclined plane as shown in the figure. It travels a distance d across a rough horizontal surface with coefficient of kinetic friction , and compresses a spring of spring k by a distance x before coming to rest momentarily. Then the spring extends and the block travels back attaining a final height of h. Then

(A) h = H –2(d + x) (C) h = H –2d + kx2/mg 100.

(C)

The three processes in a thermodynamic cycle shown in the figure are : Process 1  2 is isothermal; Process 2  3 is isochoric (volume remains constant); Process 3  1 is adiabatic. The total work done by the ideal gas in this cycle is 10 J. The internal energy decreases by 20 J in the isochoric process. The work done by the gas in the adiabatic process is –20 J. The heat added to the system in the isothermal process is

(A) 0 J 99.

(B)

(B) h = H + 2(d + x) (D) = H – 2(d + x) + kx2/ 2mg

A metallic prong consists of 4 rods made of the same material, cross-section and same lengths as shown. The three forked ends are kept at 100° C and the handle end is at 0°C. The temperature of the junction is

(A) 25°C

(B) 50°C

(C) 60°

(D) 75°C


CHEMISTRY 101.

The major product obtained in the reaction of aniline with acetic anhydride is NHCOCH3

NH2

NHCOCH3

(B) H3C

(A)

(D) H3C

(C)

O

O

102.

The maximum number of isomers that can result from monobromination of 2-methyl-2-pentene with Nbromosuccinimide in boiling CCl4 is (A) 1 (B) 2 (C) 3 (D) 4

103.

The compound X (C7H9N) reacts with benzensulfonyl chloride to give Y (C13H13NO2S) which is insoluble in alkali. The compound X is

NH2 (A)

NH2

NH2

NH CH3

(B)

CH3

(C)

(D)

CH3

H3C

104.

In 108 g of water, 18 g of a non-volatile compound is dissolved. At 100°C the vapor pressure of the solution is 750 mm Hg. Assuming that the compound does not undergo association or dissociation, the molar mass of the compound in g mol–1 is (A) 128 (B) 182 (C) 152 (D) 228

105.

The standard electrode potential of Zn2+/Zn is –0.76 V and that of Ca2+/Cu is 0.34 V. The emf (V) and the free energy change (kJ mol–1), respectively, for a Daniel cell will be (A) – 0.42 and 81 (B) 1.1 and –213 (C) –1.1 and 213 (D) 0.42 and –81

106.

Consider the equilibria (1) and (2) with equilibrium constants K1 and K2, respectively SO2(g) +

1 O (g) 2 2

2SO3(g)

SO3(g)

.... (1)

2SO2(g) + O2(g)

.... (2)

K1 and K2 are related as (A) 2K1 = K22

(B) K12 =

1 K2

(C) K22 =

1 K1

2 (D) K2 =

K 12

107.

Aqueous solution of metallic nitrate X reacts with NH4OH to form Y which dissolves in excess NH4OH. The resulting complex is reduced by acetaldehyde to deposit the metal. X and Y, respectively, are (A) Cs(NO3) and CsOH (B) Zn(NO3)2 and ZnO (C) AgNO3 and Ag2O (D) Mg(NO3)2 and Mg(OH)2

108.

The density of eq. wt of a metal are 10.5 g cm–3 and 100, respectively. The time required for a current of 3 amp to deposit a 0.005 mm thick layer of the same metal on an area of 80 cm2 is closest to (A) 120 s (B) 135 s (C) 67.5 s (D) 270 s

109.

The amount of Na2S2O3.5H2O required to completely reduce 100 mL of 0.25 N iodine solution, is (A) 6.20 g (B) 9.30 g (C) 3.10 g (D) 7.75 g

110.

In aqueous solution, [Co(H2O)6]2+ (X) reacts with molecular oxygen in the presence of excess liquor NH3 to give a new complex Y. The number of unpaired electrons in X and Y are, respectively (A) 3, 1 (B) 3, 0 (C) 3, 3 (D) 7, 0


BIOLOGY 111.

109 bacteria were spread on an agar plate containing penicillin. After incubation overnight at 37°C, 10 bacterial colonies were observed on the plate. That the colonies are likely to be resistant to penicillin can be tested by (A) measuring their growth rate (B) observing the colour of the colonies (C) checking their ability to grow on another plate containing penicillin. (D) checking their ability to cause disease

112.

Watson and Crick model of DNA is (A) B-form DNA with a spiral length of 34 Å and a diameter of 20 Å (B) A-form DNA with a spiral length of 15 Å and a diameter of 20 Å (C) Z-form DNA with a spiral length of 34 Å and a diameter of 20 Å (D) B-form DNA with a spiral length of 28 Å and a diameter of 14 Å

113.

Eco RI and Rsa I restriction endonucleases require 6 bp and 4 bp sequences respectively for cleavage. In a 10 kb DNA fragment how many probable cleavage sites are present for these enzymes (A) 0 Eco RI and 10 Rsa I (B) 1 Eco RI and 29 Rsa I (C) 4 Eco RI and 69 Rsa I (D) 2 Eco RI and 39 Rsa I

114.

From an early amphibian embryo the cells that would give rise to skin in adults were transplanted into the developing brain region of another embryo. The transplanted cells developed into brain tissue in the recipient embryo. What do you infer from this experiment? (A) Cell fate is permanently determined during early embryonic development. (B) Developmental fate of donor cells is influenced by the surrounding cells. (C) Developmental fate of donor cells is not influenced by recipient cells. (D) Any cell which is transplanted into another embryo always develops into a brain.

115.

Presence of plastids in Plasmodium suggests (A) it is a plant species (B) it is a parasite with a cynobacterium as an endosymbiont (C) it is a parasite with a archebacterium as an endosymbiont (D) it is a plant species with a archebacterium as an endosymbiont

116.

The figure below demonstrates the growth curves of two organisms A and B growing in the same area. What kind of relation exists between A and B?

Increase in biomass

organism A only organism B only organism B in presence of A organism A in presence of B

Time

(A) Competition

(B) Symbiosis

(C) Commensalisms

(D) Mutualism


117.

A scientist has cloned an 8 Kb fragment of a mouse gene into the Eco RI site of a vector of 6 Kb size. The cloned DNA has no other Eco RI site within. Digestions of the cloned DNA is shown below.

Which one of the following sets of DNA fragments generated by digestion with both Eco RI and Bam HI as shown in (iii) is from the gene? (A) 1 Kb and 4 Kb (B) 1 Kb and 2.5 Kb (C) 1 Kb and 3 Kb (D) 1 Kb and 3.5 Kb 118.

Brown fat is a specialised adipose tissue with abundant mitochondria and rich blood supply. Brown fat (A) insulates animals that are acclimatised to cold. (B) is the major source of heat production of birds. (C) provides energy to muscles. (D) produces heat without producing ATP.

119.

In some species, individuals forego reproduction and help bring up another individual’s offspring. Such altruistic behaviour CANNOT be explained by which of the following? (A) An individual helps relatives only and gets indirect genetic benefits. (B) The individual benefits because it can later inherit the breeding position. (C) The individual benefits because it gets access to resources, such as food and security from predators, in return. (D) The species benefits from a reduction in competition among offspring.

120.

Lions in India are currently restricted to Gir, Gujarat. Efforts are being made to move them to other parts of the country. This is because they are MOST susceptible to extinction due to infectious diseases under the following conditions when present as (A) several small, isolated populations (B) one large population (C) several large, connected populations (D) several large, isolated populations


KISHORE VAIGYANIK PROTSAHAN YOJANA - 2014 Date : 02-11-2014

Duration : 3 Hours

Max. Marks : 160

STREAM - SB/SX GENERAL INSTRUCTIONS •

The Test Booklet consists of 120 questions.

•

There are Two parts in the question paper. The distribution of marks subjectwise in each part is as under for each correct response.

MARKING SCHEME : PART-I : MATHEMATICS Question No. 1 to 20 consist of ONE (1) mark for each correct response. PHYSICS Question No. 21 to 40 consist of ONE (1) mark for each correct response. CHEMISTRY Question No. 41 to 60 consist of ONE (1) mark for each correct response. BIOLOGY Question No. 61 to 80 consist of ONE (1) mark for each correct response.

PART-II : MATHEMATICS Question No. 81 to 90 consist of TWO (2) marks for each correct response. PHYSICS Question No. 91 to 100 consist of TWO (2) marks for each correct response. CHEMISTRY Question No. 101 to 110 consist of TWO (2) marks for each correct response. BIOLOGY Question No. 111 to 120 consist of TWO (2) marks for each correct response.

YEAR-2014 (KVPY-STREAM-SB/SX)

PART-I One Mark Questions MATHEMATICS 1.

Let C0 be a circle of radius 1.for n  1 , let Cn be a circle whose area equals the area of a square inscribed in Cn-1.Then





i0

Area  Ci  equals :

(A) 2 2.

(B)

(C)

1 2

(D)

2 2

For a real number r we denote by [r] the largest integer less than or equal to r. If x,y are real numbers with x,y  1 then which of the following statements is always true ? (A) [x+y]  [x]+[y]

3.

2 2

(C) [2x]  2[x]

(B) [xy]  [x][y]

x x (D)  y   y  

 n   For each positive integer n, let A n  max  r  | 0  r  n  . Then the number of elements n is   

1,2,3,........20 An for which 1.9  A  2 is : n 1 (A) 9 4.

(B) 10

(D) 12

Let b, d>0. The locus of all points P(r,) for which the line OP (where O is the origin) cuts the line rsin=b in Q such that PQ=d is : (A) (r  d)sin   b

5.

(C) 11

(B) (r  d)sin   b

(C) (r  d)cos   b

(D) (r  d)cos   b

Let C be the circle x2 + y2 = 1 in the xy-plane. For each t  0 , let Lt be the line ing through (0,1) and (t,0). Note that Lt intersects C in two points, one of which is (0,1). Let Qt be the other point. As t varies between 1 and 1  2 , the collection of points Qt sweeps out an arc on C. The angle subtended by this are at (0,0) is : (A)

6.

 8

(B)

 4

(C)

 3

(D)

3 8

In a ellipse, its foci and the ends of its major axis are equally spaced, if the length its semi-minor axis is

2 2 , then the length of its semi-major axis is : (A) 4 7.

(C) 10

(D) 3

Let ABC be a triangle such that AB=BC. Let F be the midpoint of AB and X be a point on BC such that FX is perpendicular to AB. If BX=3XC then the ratio BC/AC equals : (A)

8.

(B) 2 3

3

(B)

2

(C)

3 2

4 The number of solutions to the equation cos x 

(A) 6

(B) 4

(C) 2

(D) 1 1 1  sin4 x  in the interval  0,2 is : 2 cos x sin2 x (D) 0 CLASS-XII_STREAM-SB+2_PAGE # 117

9.

Consider the function

x  5  f x   x  2 1

 if x  2   if x  2 

Then f(f(x)) is discontinuous (A) at all real numbers (C) at exactly one value of x 10.

(B) at exactly two values of x (D) at exactly three values of x

For a real number x let [x] denote the largest number less than or equal to x. for x  R let f(x)  [x]sin x . Then : (A) f is differentiable on R (B) f is symmetric about the line x=0

(C)

3

 f  x  dx  0 3

(D) For each real , the equation f(x)-=0 has infinitely many roots. 11.

Let f : 0,  R be defined as sin x, if x is irrational and x  0,   f x   2  tan x, if x is rational and x   0,  

The number of points in[0,  ] at which the function f is continuous is : (A) 6 (B) 4 (C) 2

(D) 0

1

12.

Let f : [0,1]  [0,  ) be a continuous function such that

 f  x  dx  10 . Which of the following state0

ments is NOT necessarily true ? 1

1 x (A)  e f  x  dx  10

(B)

f x

 1  x 

2

dx  10

0

0

1

1 2

(C) 10   sin 100x  f  x  dx  10

(D)  f  x  dx  100

0

0

x

13.

A continuous function f : R  R satisfies the equation f  x   x   f  t  dt . 0

Which of the following options is true ? (A) f(x+y) = f(x) + f (y) (C) f(x+y) = f(x) + f(y) + f(x)f(y) 14.

(B) f (x+y) = f(x)f(y) (D) f(x+y) = f(xy)

For a real number x let [x] denote the largest integer less than or equal to x and {x} = x - [x]. Let n be a n

positive integer. Then  cos  2[x] x dx is equal to : 0

(A) 0

(B) 1

(C) n

(D) 2n-1

15.

Two persons A and B throw a (fair) die (six-faced cube with faces numbered from 1 to 6) alternately, starting with A. The first person to get an outcome different from the previous one by the opponent wins. The probability that B wins is : (A) 5/6 (B) 6/7 (C) 7/8 (D) 8/9

16.

Let n  3 .A list of numbers x1,x 2 ,.........xn has mean  and standard deviation  . A new list of numbers

y1,y 2 ,..........yn is made as follows : y1 

x1  x 2 x1  x 2 , y2  and y j  x for j  3,4,.......n . The mean 2 2

and the standard deviation of the list are ˆ and ˆ . Then which of the following is necessarily true ? (A)   ˆ and   ˆ

(B)   ˆ and   ˆ

(C)   ˆ

(D)   ˆ CLASS-XII_STREAM-SB+2_PAGE # 118

17.

What is the angle subtended by an edge of a regular tetrahedron at its center ?  1  1  (B) cos    2

1  1  (A) cos    2

1  1  (C) cos    3 

 1  1   (D) cos   3

18.

Let S = {(a,b):a,b  Z,0  a,b  18}. The number of elements (x,y) in S such that 3x + 4y + 5 is divisible by 19 is : (A) 38 (B) 19 (C) 18 (D) 1

19.

For a real number r let [r] denote the largest integer less than or equal to r. Let a > 1 be a real number which is not an integer, and let k be the smallest positive integer such that [ak]>[a]k. then which of the following statements is always true ? (A) k  2  a   1

20.

2

(B) k  a   1

4

(C) k  2[a ] 1

(D) k 

1 1 a  [a]

Let X be a set of 5 elements. The number d of ordered pairs (A,B) of subsets of X such that

A  ,B  ,A  B   satisfies : (A) 50  d  100

(B) 101  d  150

(C) 151  d  200

(D) 201  d

PHYSICS 21.

A uniform thin rod of length 2L and mass m lies on a horizontal table. A horizontal inpulse J is given to the rod at one end. There is no friction. The total K.E. of the rod just after the impulse will be : (A) J2/2m (B) J2/m (C) 2J2/m (D) 6J2/m

22.

A solid cylinder P rolls without slipping from rest down an inclined plane attaining a speed vp at the bottom. Another smooth solid cyclinder Q of same mass and dimensions slides without friction from rest

 vp  down the inclined plane attaining a speed vq at the bottom. The ratio of the speeds  v  is :  q  3 (A)  4   

 3 (B)  2   

 2 (C)  3   

 4 (D)  3   

23.

A body moves in a circular orbit of radius R under the action of a central force. Potential due to the central force is given by V(r) = kr (k is a positive constant). period of revolution of the body is proportional to : (A) R1/2 (B) R-1/2 (C) R-3/2 (D) R-5/2

24.

A simple pendulum is attached to a block which slides without friction down an inclined plane (ABC) having an angle of inclination  as shown

while the block is sliding down the pendulum oscillates in such a way that at its mean position the direction of the string is : (A) at angle  to the perpendicular to the inclined plane AC. (B) parallel to the inclined plane AC. (C) vertically downwards. (D) perpendicular to the inclined plane AC. CLASS-XII_STREAM-SB+2_PAGE # 119

25.

Water containing air bubbles flows without turbulence through a horizontal pipe which has a region of narrow cross- section. In this region the bubbles : (A) move with greater speed and are smaller than in the rest of the pipe (B) move with greater speed and are larger in size than in the rest of the pipe (C) move with lesser speed and are smaller than in the rest of the pipe. (D) move with lesser speed and are of the same size as in the rest of the pipe

26.

A solid expands upon heating because : (A) the potential energy of interaction between atoms in the solid is asymmetric about the equilbrium positions of atoms (B) the frequency of vibration of the atoms increases (C) the heating generates a thermal gradient between opposite sides (D) a fluid called the caloric flows into the interatomic spacing of the solid during heating thereby expanding it.

27.

Consider two thermometers T1 and T2 of equal length which can be used to measure temperature over the range 1 to 2. T1 contains mercury as thermometric liquid while T2 contains bromine. The volumes of the two liquids are the same at the temperature 1.The volumetric coefficients of expansion of mercury and bromine are 18 × 10-5 K-1 and 108 × 10-5 K-1, respectively. The increase in length of each liquid is the same for the same increase in temperature. If the diameters of the capillary tubes of the two thermometers are d1 and d2 respectively, then the ratio d1:d2 would be closest to : (A) 6.0 (B) 2.5 (C) 0.5 (D) 0.4

28.

An ideal gas follows aprocess described by PV2=C from (P1,V1,T1) to (P2,V2,T2)(C is a constant). Then (A) if P1 > P2 then T2 > T1 (B) if V2 > V1 then T2 < T1 (C) if V2 > V1 then T2 > T1 (D) if P1 > P2 then V2 > V2

29.

A whistle emitting a loud sound of frequency 540 Hz is whirled in a horizontal circle of radius 2m and at a constant angular speed of 15rad/s. The speed of sound is 330 m/s. The ratio of the highest to the lowest frequency heard by a listener standing at rest at a large distance from the center of the circle is : (A) 1.0 (B) 1.1 (C) 1.2 (D) 1.4

30.

Monochromatic light es through a prism. Compares to that in air, inside the prism the light’s (A) speed and wavelength are different but frequency remains same. (B) speed and frequency are different but wavelength remains same. (C) frequency and wavelength are different but speed remains same. (D) speed,wavelength and frequency are all different.

31.

The flat face of a plano-convex lens of focel length 10 cm is silvered. A point source placed 30 cm in front of the curved surface will produce a : (A) real image 15 cm away from the lens (B) real image 6 cm away from the lens (C) virtual image 15 cm away from the lens (D) Virtual image 6 cm away from the lens

32.

Two identical metallic square loops L1 and L2 are placed next to each other with their sides parallel on a smooth horizontal table. Loop L1 is fixed and a current which increases as a function of time is ed through it.Then loop L2 : (A) rotates about its center of mass (B) moves towards L1 (C) remains stationary (D) moves away from L1

33.

An electron enters a parallel plate capacitor with horizontal speed  and is found to deflect by angle  on leaving the capacitor as shown. It is found that tan = 0.4 and gravity is negligible

If the initial horizontal speed is doubled,then tan will be : (A) 0.1 (B) 0.2 (C) 0.8

(D) 1.6 CLASS-XII_STREAM-SB+2_PAGE # 120

34.

Consider a spherical shell of radius R with a total charge +q uniformly spread on its surface (center of the chell lies at the origin x = 0). Two point charges, +q and -q are brought, one after the other, from far away and placed at x = -a/2 and x = +a/2 (a
(A)  Q  q  / 4 0 a 35.

(C) q2 / 40 a

(B) zero

(D) Qq / 40 a

two identical parallel plate capacitors of capacitance C each are connected in series with abattery of emf, E as shown, If one of the capacitors is now filled with a dielectric of dielectric constant k, the amount of charge which will flow through the battery is (neglect internal resistance of the battery) k 1 (A) 2  k  1 CE

(C)

k 1 (B) 2  k  1 CE

k2 CE k2

(D)

k2 CE k2

36.

A certain p-n junction, having a depletion region of width 20  m,was found to have a breakdown voltage of 100 V.If the width of the depletion region is reduced to 1  m duringits production, then it can be used as azener diode for voltage regulation of : (A) 5 V (B) 10 V (C) 7.5 V (D) 2000 V

37.

The Half life of a particle of mass 1.6×10-26 kg is 6.9 s and a stream of such particles is travelling with the K.E. of a particle being 0.05eV. The fraction of particles which will decay when they travel a distance of 1 m is : (A) 0.1 (B) 0.01 (C) 0.001 (D) 0.0001

38.

A 160 watt light source is radiating light of wavelength 6200 A uniformly in all directions. The photon flux at a distance of 1.8 m is of the order of (Plank’s constant 6.63×10-34 J-s) : (A) 102 m-2s-1 (B) 1012 m-2s-1 (C) 1019 m-2s-1 (D) 1025 m-2s-1

39.

The wavelength of the first Balmer line caused by a transition from the n = 3 level to the n = 2 level in

0

hydrogen is 1 . the wavelength of the line caused by an electronic transition from n=5 to n=3 is : (A)

40.

375 1 128

(B)

125 1 64

(C)

64 1 125

(D)

128 1 375

The binding energy per nucleon of 5 B10 is 8.0 MeV and that of 5 B11 is 7.5 MeV. The Energy required to remove a neutron from 5 B11 is (mass of electron and proton are 9.11×10-31 kg and 1.67×10-27 kg) : (A) 2.5 MeV

(B) 8.0 MeV

(C) 0.5 MeV

(D) 7.5 MeV

CHEMISTRY 41.

When 1.88 g of AgBr(s) is added to a 10-3 M aqueous solution of KBr, the concentration of Ag+ is 5×10-10 M. If the same amount of AgBr(s) is added to a 10-2 M aqueous solution of AgNO3, the concentration of Br- is: (A) 9.4 × 10-9 M (B) 5 × 10-10 M (C) 1 × 10-11 M (D) 5 × 10-11 M

42.

Aniline reacts with excess Br2/H2O to give the major product : NH2

NH2 NH 2

Br Br

(A)

Br Br

Br

NH2

Br

Br

(B)

(C) Br

(D)

Br

Br

Br

Br


43.

The Metal with the highest oxidation state present in K2CrO4, NbCl5 and MnO2 is : (A) Nb (B) Mn (C) K (D) Cr

44.

The number of geometrical isomers of [CrCl2(en)(NH3)2], where en = ethylenediamine, is : (A) 2 (B) 3 (C) 4 (D) 1

45.

The element that combines with oxygen to give an amphoteric oxide is : (A) N (B) P (C) Al

46.

47.

48.

The Arrhenius plots of two reactions, I and II are shown graphically : The graph suggests that (A) EI > EII and AI > AII

(B) EII > EI and AII > AI

(C) EI > EII and AII > AI

(D) EII > EI and AI > AII

(D) Na I ln k II 1/T

Ni(CO)4 is : (A) tetrahedral and paramagnetic (C) tetrahedral and diamagnetic

(B) square planar and diamagnetic (D) square planar and paramagnetic

In the following reactions :

1. ozonolysis

X

2. OH

the major product X is : O

O O

O

(A)

(B)

(C)

(D) O

49.

Given the structure od D-(+)-glucose as : CHO H

OH H

HO H

OH

H

OH CH2OH

The structure of L-(-)- glucose is : CHO

CHO

(A)

HO

H

HO

HO

H

H

H

OH

H

OH CH2OH

(B)

H OH

HO

H

HO

H CH2 OH

CHO

CHO

(C)

H

OH

HO

HO

H

H

H

OH H

HO CH2OH

(D)

H OH H

HO H

OH CH2OH


50.

In a cubic close packed structure, fractional contributions of an atom at the corner and at the face in the unit cell are, respectively : (A) 1/8 and 1/2 (B) 1/2 and 1/4 (C) 1/4 and 1/2 (D) 1/4 and 1/8

51.

The equilibrium constant Kc of the reaction 2A  B  C is 0.5 at 25°C and i atm. The reaction will proceed in the backward direction when concentrations [A], [B] and [C] are respectively : (A) 10-3, 10-2 and 10-2 M (B) 10-1, 10-2 and 10-2 M (C) 10-2, 10-2 and 10-3 M (D) 10-2, 10-3 and 10-3 M

52.

Major products formed in the reaction of t-butyl methyl ether with HI are :

(A)

(B)

(C)

(D)

53.

If the molar conductivities (in S cm2 mol-1) of NaCl, KCl and NaOH at infinite dilution are 126, 150 and 250 respectively, the molar conductivity of KOH (in S cm2 mol-1) : (A) 526 (B) 226 (C) 26 (D) 274

54.

4-Formylbenzoic acid on treatment with one equivalent of hydraine followed by heating with alcoholic KOH gives the major product :

(A)

(B)

(C)

(D)

55.

Two elements, X and Y, have atomic numbers 33 and 17 respectively. The milecular formula of a stable compound formed between then is (A) XY (B) XY2 (C) XY3 (D) XY4

56.

The number of moles of KMnO4 required to oxidize one equivalent of Kl in the presence of sulfuric acid is (A) 5 (B) 2 (C) 1/2 (D) 1/5

57.

Three successive measurements in an experiment gave value 10.9, 11.4042 and 11.42 the correct way of reporting the average value is (A) 11.2080 (B) 11.21 (C) 11.2 (D) 11

58.

The latent heat of melting of ice at 0 °C is 6 kJ mol-1. The entropy change during the melting in J K-1 is closest to (A) 22 (B) 11 (C) -11 (D) -22


59.

The major product of the following reaction

is

60.

(A)

(B)

(C)

(D)

The energies of dxy and d2z orbitals in octahedral and tetrahedral transition metal complexes are such that (A) E (dxy) > E (dz2) in both tetrahedral and octahedral complexes (B) E (dxy) < E (dz2) in both tetrahedral and octahedral complexes (C) E (dxy) > E (dz2) in tetrahedral abut E (dxy) < E (dz2) in octahedral complexes (D) E (dxy) < E (dz2) in tetrahedral abut E (dxy) < E (dz2) in octahedral complexes

BIOLOGY 61.

In which of the following types of glands is the decretion collected inside the cell and discharged by disintegration of intire gland ? (A) Apocrine (B) Merocrine (C) Holocrine (D) Epicrine

62.

Which one of the following interactions does NOT promote coevolution ? (A) Commensalism (B) Mutualism (C) Parasitism

(D) Interspecific competition

Stratification is more common in which of the following ? (A) Deciduous forest (B) Tropical rain forest (C) Temperate forest

(D) Tropical savannah

Where is the 3rd ventricle of the brain located? (A) Cerebrum (B) Cerebellum

(D) Diencephalon

63.

64.

(C) Pons varoli

65.

Which of the following is the final product of a gene ? (A) a polypeptide (B) an RNA only (C) either polypeptide or RNA (D) a nucleotide only

66.

Forelimbs of whales,bats,humans and cheetah are examples of which of the following processes? (A) Divergent evolution (B) Convergent evolution (C) Adaptation (D) Saltation

67.

Which of the following results from conjugation in paramecium ? (A) Cell death (B) Cell division (C) Budding

68.

(D) Recombination

In an experiment investigating photoperiodic response, the leaves of a plant are removed.what is the most likely outcome ? (A) Photoperiodism is not affected (B) Photoperiodic response does not occur (C) The plant starts flowering (D) The plant starts to grow taller


69.

70.

71.

72.

73.

74.

Testosterone is secreted by which endocrine part of testis? (A) Leydig cells (B) Seminiferous tubules (C) Tunica albugenia

(D) Sertoli cells

The mutation of a purine to a pyrimidine is known as: (A) transition (B) frame shift (C) nonsense

(D) transversion

Which of the following is secreted at the ends of an axon ? (A) Ascorbic acid (B) Acetic acid (C) Acetyl choline

(D) Acetyl CoA

A bacterial colony is produced from : (A) a single bacterium by its repetitive division (C) clumping of two to three bacteria

(B) multiple bacterium without replication (D) a single bacterium without cell division

Rhinoviruses are the causative agents of : (A) Diarrhoea (B) AIDS

(C) Dengue

What is the genetic material of Ebola virus ? (A) Single-Standard DNA (C) Single-Standard RNA

(D) Common Cold

(B) Double-Standard RNA (D) Double-Standard DNA

75.

Name the terminal acceptor of electrons in the mitochondrial electron transport chain : (A) Nitrate (B) Fumarate (C) Succinate (D) Oxygen

76.

Two tubes labeled ‘P’and ’Q’ contain food stuff. Tube ‘P’gave positive test with Benedict’s solution while tube ‘Q’ gave positive test with Nitric acid. Which of the foloowing is correct ? (A) Tube ‘P’ contains sugar;tube ‘Q’ contains protein (B) Tube ‘P’ contains protein;tube ‘Q’ contains sugar (C) Both tube ‘P’ and tube ‘Q’ contain sugar (D) Both tube ‘P’ and tube ‘Q’ contain protein

77.

How many linear DNA fragments will be produced when a circular plasmid is digested with a restriction enzyme having 3 sites ? (A) 4 (B) 5 (C) 3 (D) 2

78.

If the humidity of the atmosphere suddenly increases substantially, the water flow in the xylem will : (A) increase (B) decrease (C) remain unaltered (D) increase sharply and then reduce slowiny to the pre-existing level

79.

Which one of the following is the complementary sequence for the DNA with 5’-CGTACTA-3’ (A) 5’-TAGTACG-3’ (B) 5’-ATCATGC-3’ (C) 5’-UTCUTGC-3’ (D) 5’-GCUAGCA-3’

80.

A diploid plant has 14 chromosomes,but its egg cell has 6 chromosomes. Which one of the following is the most likely explanation of this ? (A) Non-disjunction in meiosis I and II (B) Non-disjunction in meiosis I (C) Non-disjunction in meiosis (D) Normal meiosis



Let n  3 be an integer. For a permutation  =(a1,a2,....an) of (1,2,...,n) we let f  (x) = anxn-1+an1+...+a x+a .Let S  be the sum of the roots of f  (x) = 0 and let S denote the sum over all permutation 2 1  of (1,2,....,n) of the number S  . Then (A) S < -n! (B) -n! < S < 0 (C) 0 < S < n! (D) n! < S

82.

If n is a positive integer and   1 is a cube root of unity, the number of possible values of

n

(A) 2

(B) 3

(C) 4

n k   is k  ek  0

 

(D) 6

83.

Suppose a parabola y = ax2 + bx + c has two x intercepts one positive and one negative, and its vertex is (2,2). Then which of the following is true (A) ab > 0 (B) bc > 0 (C) ca > 0 (D) a + b + c > 0

84.

Let n  3 and let C1, C2,....Cn, be circles with radii r1,r2,...,rn, respectively. Assume that Ci and Ci+1 touch externally for 1  i  n-1. It is also given that the x-axis and the line y = 2 2 x+10 are tangential to each of the circles. Then r1,r2,.....,rn are in (A) An arithmetic progression with common difference 3+ 2 (B) A geometric progression with common ratio 3+ 2 (C) An arithetic progression with common diffrence 2+ 3 (D) A geometric progression with common ratio 2+ 3

85.

86.

The under of integers n for which 3x3 - 25x+n = 0 has three real roots is (A) 1 (B) 25 (C) 55

An ellipse inscribed in a semi-circle touches the circular are at two distinct points and also touches the bounding diameter. its major axis is parallel to the bounding diameter. When the ellipse has the maximum possible area, its eccentricity is

1

(A)

(B)

2 / 2

87.

Let In =

x

n

0

 /2 (A) e  1 

88.

(D) Infinite

1 2

(C)

1

(D)

3

2 3

I    I cos x dx, where n is a non negative int eger . Then  n  2  n  n  2  equal  n! (n  2)! 

 2

(B) e / 2  1

 /2 (C) e 

 2

(D) e /2

For a real number x let [x] denote the largest integer less than or equal to x. The smallest positive n

integer n for which the integral

 [x][

x ]dx exceeds 60 is

1

(A) 8 89.

(C) 10

(D) [602/3]

Choose a number n uniformly at random from the set {1,2,...100}. Choose one of the first seven days of the year 2014 at random and consider n consencutive days among the chosen n days, the number of Sunday is different from the number of Mondays? (A)

90.

(B) 9

1 2

(B)

2 7

(C)

12 49

(D)

43 175

Lets S = {(a,b)}Ia,b  Z, 0  a,b,  18}. The number of lines in R2 ing through (0,0) and exactly one other point in S is (A) 16 (B) 22 (C) 28 (D) 32 CLASS-XII_STREAM-SB+2_PAGE # 126

PHYSICS 91.

A solid sphere spinning about a horizontal axis with an angular velocity  is placed on a horizontal surface. Subsequently it rolls without slipping with an angular velocity of (A) 2  /5 (B) 7  /5 (C) 2  /7 (D) 

92.

Consider the system shown below F

X

Y

A horizontal force F is applied to a block X of mass 8 kg such that the block Y of mass 2kg adjacent to it does not slip downwards under gravity. There is no friction between the horizontal plane and the base of the block X. The coefficient of friction between the surfaces of blocks X and Y is 0.5. Take acceleration due to gravity to be 10ms-2. The minimum value of F is (A) 200N (B) 160N (C) 40N (D) 240N 93.

The maximum value attained by the tension in the string of a swinging pendulum is four times the minimum value it attains. There is no slack in the string. The angular amplitude of the pendulum is (A) 90° (B) 60° (C) 45° (D) 30°

94.

One mole of a monoatomic ideal gas is expanded by a process described by PV3 = C where C is a constant. The heat capacity of the gas during the process is given by (R is the gas constant)

5 3 R (C) R (D) R 2 2 A concave mirror of radius of curvature R has a circular outline of radius r. A circular disk is to be placed normal to the axis at the focus so that it collects all the light that is reflected from the mirror from a beam parallel to the axis. for r<
95.

(A) 96.

(B)

r 4 4R 2

(C)

r 5 4R 3

(D)

r 4 R2

The angles of incidence and refraction of a monochromatic ray of light of wavelength  at an air glass inteface are i and r, respectively. A parallel beam of light with a small spread   in wavelenth about a mean wavelenth  is refracted at the same air-glass interface. The refractive index  of glass depends on the wavelength  as  (  ) = a+b/  2where a and b are constant. Then the angular spread in the angle of refraction of the beam is (A)

97.

r 6 4R 4

(B)

sin i   cosr 3

(B)

2b  3

(C)

2b tanr  a 3  b

(D)

2b(a  b /  2 )sin i  3

What are the charges stored in the 1  F and 2  F capacitors in the circuit below, once the currents s become steady?

(A) 8  C and 4  C respectively (C) 3  C and 6  C respectively

(B) 4  C and 8  C respectively (D) 6  C and 3  C respectively


98.

A 1.5 KW (kilo-watt) laser beam of wavelength 6400 A is used to levitate a thin aluminium disk of same area as the cross section of the beam. The laser light is reflected by the aluminium disk without any absorption. The mass of the foil is close to (A) 10-9 kg (B) 10-3 kg (C) 10-4 kg (D) 10-6 kg

99.

When ultraviolet radiation of a certain frequency falls on a potassium target, the photoelectrons released can be stopped competely by a retarding potential of 0.6 V. If the frequency of the radiation is increased by 10%, this stopping potential rises to 0.9V. The work functin of potassium is (A) 2.0eV (B) 2.4eV (C) 3.0eV (D) 2.8 eV

100.

The dimensions of stefan-Boltzmann constant  can be written in of Planck’s constanth’ Boltzmann constant KB and the speed of light c as  = h KBC. Here (A)  = 3,  = 4 and  = -3 (B)  = 3,  = -4 and  = 2 (C)  = -3,  = 4 and  = -2 (D)  = 2,  = -3 and  = -1

101.

101.

CHEMISTRY In the raction sequence

X and Y are,respectively

(A)

(B)

(C)

(D)

102.

The density of acetic acid vapor at 300 K and 1 atm is 5 mg cm-3. The number of acetic acid molecules in the cluster that is formed in the gas phase is closest to (A) 5 (B) 2 (C) 3 (D) 4

103.

The milar enthalpy change for H2O() H2O(g) at 373 K and 1 atm is 41 kJ/mol. Assuming ideal behavior, the internal energy change for vaporization of 1 mol of water at 373 K and 1 atm in kJ mol-1 is (A) 30.2 (B) 41.0 (C) 48.1 (D) 37.9


104.

The equlibrium constants (Ke) of two reactions H2 + I2 2HI and N2 + 3H2 2NH3 are 50 and 1000, respectively. The equilibrium constant of the reaction N2 + 6HI 2NH3+3I2 is closest to (A) 50000 (B) 20 (C) 0.008 (D) 0.005

105.

Given that the bond energies of : N  N is946 kJ mol-1 H-H is 435 kJ mol-1,N-N is 159kJ mol-1, and N-H is 389 kJ mol-1, the heat of formation of hydrazine in the gas phase in kJ mol-1 is (A) 833 (B) 101 (C) 334 (D) 1264

106.

The radius of K+ is 133 pm and that of Cl- is 181 pm. The volume of the unit cell of KCl expressed in 10-22 cm3 is (A) 0.31 (B) 1.21 (C) 2.48 (D) 6.28

107.

The reaction, K2Cr2O7 + m FeSO4 + n H2SO4  Cr2(SO4)3 + p Fe2(SO4)3+K2SO4 + q H2O when balanced, m, n, p,and q are, respectively (A) 6,14,3,14 (B) 6,7,3,7 (C) 3,7,2,7 (D) 4,14,2,14

108.

The standard free energy change (in J) for the reaction 3Fe2+(aq) + 2Cr (s)  2Cr3+(aq) + 3Fe(s), given 0 0 EFe 2 / Fe = – 0.44 V and ECr 3  / Cr = – 0.74 V is (F = 96500C)

(A) 57,900 109.

110.

(B) – 57,900

(C) – 173,700

(D) 173,700

Calcium butanoate on heating followed by tratment with 1,2-ethanediol in the presence of catalytic amount of acid, produces a major product which is

(A)

(B)

(C)

(D)

XeF6 on complete hydrolysis yields ‘X’. The molecular formula of X and its geometry, respectively, are (A) XeO2 and linear (B) XeO3 and trigonal planar (C) XeO3 and pyramidal (D) XeO4 and tetrahedral

BIOLOGY 111.

Following the cell cycle scheme given below, what is the probability that a cell would be in M-Phase at any given time ?

(A) 1/24

(B) 1/12

(C) 1/6

(D) 1/2 CLASS-XII_STREAM-SB+2_PAGE # 129

112.

A flower with Tt genotype is cross-pollinated by TT pollens. What will the genotypes of the resulting endosperm and embro,respectively, be? (A) TTT,(TT+Tt) (B) (TTT+TTt),TT (C) TTt,Tt (D) TTt,(TT+Tt)

113.

A new life from discovered on a distant has a genetic code consisting of five unique nucleotides and only one stop codon.If each codon has four bases,what is the maximum number of unique amino acids this life from can use ? (A) 624 (B) 20 (C) 124 (D) 3124

114.

A spontaneous mulation results in a couple having only female progeny.when the daughter marries and has children,none of them are males.However,in the third generation there are few male offspring. What is the most likely explanation of this observation ? (A) The mutation reverses spontaneously in the third generation (B) The mutation occurs on the X chromosome and is both recessive and ethal (C) The mutation occurs on the X chromosome and is both recessive and dominant (D) The mutation occurs on an autosome and is dominant

115.

A circular plasmid of 10,000 base pairs (bp) is digested with two restriction enzymes,A and B, to produce a 3000 bp and a 2000 bp bands when visualised on an agarose gel. when digested with one enzyme at a time,only one band is visible at 5000 bp. If the first site for enzyme A(A1) is present at the 100th base, the order in which the remaining sites(A2,B1 and B2) are present is : (A) 3100,5100,8100 (B) 8100,3100,5100 (C) 5100,3100,8100 (D) 8100,5100,3100

116.

After meiosis-II,daughter cells differ from the parent cells and each other in their genotypes. This can occur because of which one of the following mechanism(s)? (A) Only synaptic (B) Only crossing over and independent assortment of chromosomes (C) only crossing over and chromosomal segregation (D) crossing over, independent assortment and segregation and chromosomes

117.

A desert lizard (an ectotherm) and a mouse (an endotherm) are placed inside a chamber at 15°C and their body temperatures [T(L) for the lizard and T(M) for the mouse] and metabolic rates [M(L) for the lizard and M(M) for the mouse] are monitored. Which one of the following is correct ? (A) T(L) and M(L) will fall while T(M) M(M) will increase (B) T(L) and M(L) will increase while T(M) and M(M) will fall (C) T(L) and M(L) will fall, T(M) will remain same and M(M) will increase (D) T(L) and M(L) will remain same and T(M) and M(M) will decrease

118.

In Griffith’s experiments mice died when injected with : (A) heat killed S-strain (B) heat killed S-strain combined wuth R-strain (C) heat killed R-strain (D) live R-strain

119.

Human height is a multigenic charecter. If the Heights of all the individuals living in a metropoils are measured and he percentages of the population belonging to a specific height are plotted as shown below,which of the plots would represent the most realistic distribution ?

(A)

120.

(B)

(C)

(D)

If mitochondria isolated from a cell are first placed without carbon source in a buffer at pH 8.0 and then transferred to a buffer at pH 4, it will lead to (A) an increase in intra-mitochondrial acidity (B) a decrease in intra-mitochondrial acidity (C) blockage of ATP synthesis (D) synthesis os ATP


HINTS & SOLUTIONS (YEAR-2007) ANSWER KEY Que s.

1

2

3

4

5

6

7

8

9

10

11

Ans.

D 16

B 17

C 18

C 19

D 20

D 21

A 22

B 23

B 24

A 25

C

D

D

D

B

Que s.

26

27

28

29

30

Ans.

A

B

B

D

B

B

D

A

B

B

Que s.

31

32

33

34

35

36

37

38

39

40

A 41

D 42

B 43

A 44

A 45

Ans.

B 46

B 47

B 48

D 49

D 50

C 51

D 52

B 53

B 54

A 55

D 56

C 57

A 58

59

D 60

B 61

C 62

B 63

C 64

B 65

A 66

B 67

B 68

A 69

B 70

C 71

A 72

A 73

C 74

B 75

C 76

B 77

A 78

B 79

C 80

D 81

D 82

D 83

C 84

C 85

C 86

A 87

B 88

B 89

C 90

Ans. B Que s. 91 A Ans. Que s. 106 C Ans.

A 92 D

B 93 D

C 94 A

A 95 C

B 96 D

B 97 A

C 98 A

A 99 C

A 100 D

B 101 B

A 102 D

C 103 A

B 104 B

B 105 D

107 C

108 D

109 B

110 B

111 C

112 A

113 A

114 D

115 D

116 B

117 A

118 A

119 A

120 B

Que s. Ans. Que s. Ans. Que s.

12

13

14

15

PART-I (1 Mark) MATHEMATICS

1.

2  1 x2  x3     

7

 1  2 r 7 General term = (– 1) Cr  x    Tr + 1

r 7

= (– 1) Cr r 7

= (– 1) Cr r 7

= (– 1) Cr r 7

= (– 1) Cr

7 r

 2  3 × x   

r

7 r 2r x 2 x3 7 r 2r  x 2 3 213r  4r x 6 21r x 6

21 r =4 6 21 + r = 24 r = 24 – 21 r=3 = (– 1)3 7C3

For the coefficient of x4,



T3 + 1

=–1×

765 32

= – 35.


z 2.

z



z z = 1 and | z | = 1

Let

z = x + iy

Then,

z = x – iy x2  y2 = 1

|z|=

x2 + y2 = 1

or

x  iy x  iy  x  iy x  iy = 1

x 2  y 2  2ixy  x 2  y 2  2ixy x2  y2 2( x 2  y 2 ) x2  y2

=1

=1

2x2 – 2y2 = x2 + y2 x2 = 3y2 x=± Now,

3y x +y =1 3y2 + y2 = 1 4y2 = 1 2

y2 =

2

1 4

y=± Then, x = ±

1 2 3 2

 Different complex number are : z = 3.

1 1 1 1 3 3 3 3 +i ,z= –i ,z=– + i and z = – –i . 2 2 2 2 2 2 2 2

Let A– 1 = B Therefore AB =  35 37   1 0 A   =    41 43  0 1 R1 

R1 35

37    1   1 35   35 0 A   =  0 1 41 43    R 2  R 2  41R1 37    1  1 35   35 0   12  =  41 A    0  1   35  35  R2  

35 R2 12


 1  0  37    35 1  35  =  41 35  A      0 1  12   12

R1  R1 

37 R2 35

 43  12 1 0   A   =  41  0 1  12

 43  12 A =  41   12

37  12  35    12 

37  12  35    12 

  43 37  12 A =    41  35 

det (12A) = | 43 × 35 – 41×37| = 1505 – 1517 = – 12. 

4.

1

n

Given :

2

2 6

=

n 1



1  3  5  ...  ( 2n – 1)

1

3

n 1

 23  3 3  .....  n3

n2



 n (n  1) 2

2

n 1

4 

4

 (n  1)

2

n 1



4

1



 1  2

2

n 1



4

 1

  n n 1

2



1

   .........   1 3   2

  1 

 2    4  6  1 .  


5.

4ax2 + xy + 4y2 = 0 m1 + m2 =

2h b

m1 + m2 =

1 4

m 1m 2 =

a b

4a =a 4 Given : m2 = m12 m 1m 2 =

1 4 4m12 + 4m1 = – 1 4m12 + 4m1 + 1 = 0 (2m1 + 1)2 =0 m 1 + m 12 =

1 2 m1 m2 = a m1 × m12 = a m 13 = a m1 =

3

  1   =a  2  a=

 6.

1 . 8

y = 2x2 + x – 2 A

B

Let point A is (t1, 2t12 + t1 – 2 ) and point B is (t2, 2t22 + t2 – 2) Given : Mid point of AB is (0, 0) So,

t1  t 2 = 0  t1 + t2 = 0 2

2

2

2t 1  t 1  2  2t 2  t 2  2 =0 2 2(t12 + t22) + (t1 + t2) – 4 = 0 2[(t1 + t2)2 – 2t1t2] + 0 – 4 = 0 2[0 – 2t1t2] = 4 – 4t1t2 = 4 t1 t2 = – 1 (t1 – t2)2 = (t1 + t2)2 – 4t1t2 = 0 – 4(– 1) =4 t1 – t2 = 2. Distance AB : 2

2

AB =

( t 1  t 2 ) 2  ( 2t 1  t 1  2  2t 2  t 2  2)2

AB =

( t1  t 2 )2  2( t1  t 2 )  ( t1  t 2 )



2

2

2




2

AB =

( 2)2  2( t 1  t 2 )( t1  t 2 )  ( t 1  t 2 )

AB =

4  2(0)  ( 2)

AB =

44 =

2

8=2 2.

7.

0, – c 4 3x + 4y + c =

1 (1, 1)

0

1

1 1

–c , 0 3 Perperndicular from (1, 1) on line 3x + 4y + c = 0

34c

=1 9  16 |7+c|=5 7+c= ±5 c=–5–7 c = – 12 | c | = 12.

8.

R P p cos2   + r cos2   2 2

p

s(s  r ) s(s  p) +r pq qr

=

s [s – r + s – p] q

=

s [2s – r – p] q

=

s [p + q + r – r – p] q

=s = 9.

pqr . 2

3cos2x sin2x – sin4x – cos2x = 0  3(1 –sin2x) sin2x – sin4x – (1 – sin2x) = 0  3 sin2x – 3sin4x – sin4x –1 + sin2x = 0  – 4sin4x + 4sin2x – 1 = 0  4sin4 – 4sin2x + 1 = 0  (2 sin2x – 1)2 = 0  2sin2x = 1




sin2x =



sinx = ±

1 2 1

2 sin x has 4 solution in the interval [0, 2].

 10.

=

 1 1 sin–1   4 2

sin2 =

1 4



2sincos =

1 4

1 8 Value of : 64sin + 64cos – 8sec – 8cosec + tan + cot



sincos =

 sin  cos     = 64(sin + cos) – 8 (sec + cosec)+   cos  sin   2 2 1   sin   cos    1    = 64(sin + cos) – 8 +  cos  sin    sin  cos  

 sin   cos   1  + = 64(sin + cos) – 8   sin  cos   sin  cos  (sin   cos ) +8 1/ 8 = 64(sin + cos) – 64 (sin + cos) + 8 = 8.

= 64(sin + cos) – 8

11.

 lim   1  1   1  2 n  x 0  x sin   x sin 2   ....  x sin n  L = nlim   x x   x     1  1   1  lim x sin  lim x 2 sin 2   .........  lim x n sin n   L = nlim   x 0 x 0 x x 0 x   x   L = nlim 0  0  0  .......... .  0  L=0

12.

 x 3 for x  1  f(x) =  2 ax  bx  c for x  1 f’(x) = 3x2 = 2ax + b [f”(x)] x = 1 = 6x = 2a  6 = 2a a = 3.

13.

f(x) = x3 – 3ax2 + b f’(x) = 3x2 – 6ax For the f(x) to be increasing f’(x) > 0. 3x2 – 6ax > 0 3x (x – 2a) > 0 (x – 2a) > 0 [ x > 0] x > 2a So, a  0.


3

14.





 xcos 2 x  xdx 1

2

3

  = cos 2 x  1dx +   1





 2 cos 2 x  2dx



2

2

=

15.

    sin 2 x 1          2

=

2 4 [1  0]  [1  0]  

=

6 . 

1

3

    sin 2 x  2      + 2    2

2

Let A =

1 {(2n + 1) (2n + 2) ------- (2n + n)}1/n n

log A =

( 2n  n)   ( 2n  1) ( 2n  2) ( 2n  3 ) 1 ..... lim log  n n n   n n n 

log A =

 1  2  2  n  1 lim log  2  n  2  n  2  n ..... 2  n  n        n  1

log A =

 log( 2  x )dx 0

1

log A = x log( 2  x) x  2 log(2  x) 0 log A = [log 3 – 1 + 2 log 3 – 2 log2] log A = [log 3 – loge + log 9 – log4] log A = [log (27/4e)] A = (27/4e). 16.

V1 + V2 + V3 = 0

V1  V2 = ? V1  V2 = – V3

V  V  . V  V  =  V  .  V  1

2

2

V1

1

2

3

2

3

2

+ V2 + 2V2 . V2 = V3

1 + 1 + 2V2 . V2 = 1 2V2 . V2 = – 1 V2 . V2 =

1 2


2

V1  V2

2

= V1

2

+ V2 – 2 V1 . V2

1 2 =1+1+1=3 =1+1–2×

2



V1  V2

=3

V1  V2 =

17.

Total number of possible cases = 4 + 3 + 2 + 1 = 10 Favourable cases = {(H T T H T),(T H T H T ), (T T H H T ), (T T T H H)} Total number of favourable cases = 4 

18.

3.

p(4th toss is a head) =

4 10

=

2 . 5

a + b + c = even Case 1 : All taken are even :  5C3 = 10 Case 2 : One even + 2 odd : 5C1 × 5C2 = 50 Total = 10 + 50 = 60.

19.

2log(x – 2y) = logx + logy log(x – 2y)2 = log(xy) (x – 2y)2 = xy x2 + 4y2 – 4xy = xy x2 + 4y2 – 5xy = 0 x2 – 4xy – xy + 4y2 = 0 x(x – 4y) – y (x – 4y) = 0 (x – y)(x – 4y) = 0 x = y (N.P.), x = 4y

x y = 4. 20.

Let x, y and z are 2n, n2 – 1 & n2 +1. Let x = 2n, y = n2 – 1 and z = n2 +1. Option (A) : 2 does not divide x . Clearly 2 divides x. So option (A) is not true. Ex : 8, 15 and 17 are pythogorean triplet. 2 divides 8. Option (B) : 2 does not divide z(x + y) (n2 + 1)(2n + n2 – 1) which is an odd number so, 2 does not divides z(x + y).  So, option (B) is true. Option (C) : 4 divides x + y + z Ex : 5, 12, 13 are the pythogorean triplet. Clearly 5 + 12 + 13 = 30, and 4 does not divides 30. So, option (C) is not true. Option (D) : 8 divides x + y + z Ex : 8, 15, 17 are the pythogorean triplet. Clearly 8 + 15 + 17 = 40, and 8 divides 40.  So, option (B) is true. CLASS-XII_STREAM-SB+2_PAGE # 138

PHYSICS 21.

By work energy theorem, W = K mgh – wf = 0 wf = mgh

22.

E = –13.6 ×

z2 n

2

= – 13.6 ×

4 = 54.4eV Ans (D) 1

2E 23.

E VFB deflected in –x direction (Ans.B) P P1 I

25.

P2

45º

I

90º

By Malus's law, I=cos245º initensity after P1

I 2 initensity after P2 I = Icos245 =

I2 =

I1 I cos245 = Ans : B 2 4

26.

Due to inertia ballon displace in the direction of motion of the bus.

27.

(D) As for hydrogen like atom

 1 1 1  R.z 2  2  2    n1 n2  So,

28.

(Here z = 1, same for deuterium and hydrogen)

1 will be same for deuterium as that for hydrogen. 

Bulk modulus, B = 

P  V     V 

As in isothermal process, PV = K On differentiating, PdV + VdP = 0

V P  V P  P  Hence B =   P  = P  P 


29.

2 From brewster’s law tan =  1

30.

Cheaking dimensionally all the formula g L   [L] 2   [L2 T  2 ] 2 T 

and

g  [MLT  2 ] [L1T  2 ]      [ML3 ]   [L]

1/ 2

= [L2T–2]

(dimension of v2)

2 [M1L1T 2 ][L1]  = [L2T–2]  [L][ML3 ] Only (A) option is correct dimensionally 33.

As v & d depends on frame of referance but change in KE and hence heat does not depend. V

34.

V>> 2gr

B

v

36.

x

2v

A

B

v

C 2v

 2v  vB = v0 1  c    vC = vo 1  

v c 

 2v cos   VA = V0 1   c   VA > VC > VB (Answer : C)

37.

By dalton's law of partion pressure.

38.

Ein = 0 Eout =

(Ans : D)

KQ r2

KQ KQ = const. Vout = R r X is for electric field versus distance graph Y is for potential versus distance graph Ans. is (B) Vin =

kq 39.

Enet = 2 Enet =

2

( x  y 2 )3 / 2

[ x]

2kq x2


CHEMISTRY 41.

For weak acid [H+] = C [H+] = 0.1 × 0.1 [H+] = 10–2 pH = –log[H+] pH = –log[10–2] pH = 2

42.

N

O O

 it has two donor side [O & N] –

43.

H3O+ sp3 hybridisation, due to presence of 1 p structure is trigonal pyramidal.

45.

(D) Ba > Ca > Mg > Be Solubility increases down the group due to increase of size of cation.

46.

C > Be > B > Li Zeff  I E 

47.

Dipole moment µ = e × d F

Be µ=0

F

48.

[NiCl4]2– is tetrahedral.

50.

mvr =

51.

(A) both valencies of one double bonded C-atom are occupied by the same group (H).

52.

p-nitrophenol –I & –M

nh 2

where n = 1, 2, 3 -------

-o-nitrophenol –I & –M ( intramolecular H-bond)

CH2 – O – COR

54.

(A) CH – O – COR

m-nitrophenol –I

CH2 – OH + 3NaOH

CH2 – O – COR

3RCOONa + CH – OH Salt of fatty acid CH2 – OH glycerol

55.

(B) Branching  surface area  vanderwaals’ forces b.p. 


56.

(C) Zn/NaOH 2

58.

NO2

Zn/NaOH

N=N

CH3 – CH2 – CH(OH) – COOH

OH COOH

59.

(C)

[contains phenolic group]

60.

(B) stability of carbanions 

–  & –M   & M


ab = 2 (a + b) ab 1  ab 2 1 1 1   a b 2

...... (i)

bc = 3( b + c) bc 1  bc 3 1 1 1   b c 3

...... (ii)

ca = 4( c + a) ca 1  ca 4 1 1 1   c a 4

...... (iii)

Add (i), (ii) and (iii)  1 1 1 1 1 1 2       a b c  2 3 4 1 1 1 643    a b c 24 1 1 1 13    a b c 24

...... (iv)

From (i) and (iv) c = 24


From (ii) and (iv) a=

24 5

From (iii) and (iv) b=

24 7

So, value of 5a + 7b + c = 24 + 24 + 24 = 72. 82.

x2 + ax + b = 0  +  = – a,  = b x4 + ax3 + cx2 + dx + e = 0 Roots are  + ,  – , –  +  and –  –  Sum of roots = – a  + +  – –  + –  – = – a a = 0   + = 0 Sum of roots taken two at a time = + c ( + ) ( – ) + ( – ) (– + ) + (– + )(– – ) + (– – ) ( + ) + ( + ) (– + ) + ( – ) (– + ) = c 0 – ( + )2 + 0 + 0 + 0 – ( + )2 = c – 2( – )2 = c – 2[2 +  2 – 2] = c – 2[( + )2 – 2 – 2] = c – 2[– 4] = c c = 8 c = 8b  0 Product of roots taken three at a time = – d ( + ) ( – ) (– + ) + ( + ) (– + )(– – ) + (– – ) ( + ) + ( + ) (– + ) (– + ) (– + ) = – d 0+0+0+0=–d d=0 Product of roots = e ( + )( – )(– + )(– – ) = e e=0 So, c = 0 is the false statement

83. y = 3x Q y = 2x P O

PQ = 5 Let point Q is (x, 3x)


Given QP = 5 and QP  line 2x – y = 0

2x  3 x

=5

4 1 x= 5 5

Therefore point Q is ( 5 5 , 15 5 ) 2

OQ =



2

5 5   15 5 

=

125  1125

=

1250

= 25 2 In OPQ OQ2 = OP2 + PQ2



2

25 2 

= OP2 + (5)2

1250 – 25 = OP2 OP2 = 1225 OP = 35 cm. 84.

s(s  b) (s  a )(s  c )

cot

B = 2

cot

s(s  c ) C = ( s  a)(s  b) 2

cot



s(s  b) (s  a)(s  c )

B C cot = 2 2

s(s  c ) (s  a)(s  b)

s 2 (s  b)(s  c )

cot

B C cot = 2 2

cot

B C cot = 2 2

cot

s B C cot = sa 2 2

(s  a)2 (s  b)(s  c ) s2 ( s  a) 2

Given : b + c = 3a



s=

a  3a abc = = 2a. 2 2

cot

2a 2a B C cot = = = 2. 2 a  a a 2 2


85.

D

C F E

B

A

Given : AE = EF = FC =

AC 3

Let,

AB = BC = CD = DA = x

Then,

AC =

2x

Let BE = a In AEB

2x 2  x 2  a2 9 cos 45º = 2x 2 .x 3

11x 2  a2 9 1 = 2 2x 2 2 3

2x 2 11x 2 = – a2 3 9

a2 =

11x 2 2x 2 – 9 3

a2 =

11x 2  6 x 2 5x 2 = 9 9

a=

5 x 3

Similarily BF = a =

5x . 3

In BEF BE = BF =

and

EF =

5x 3

2x 3


cos =

BE 2  BF 2  EF 2 2BE  BF

5 x 2 5 x 2 2x 2   9 9 9 cos = 5x 5x 2  3 3

8x 2 9 cos = 10 x 2 9

tan =

 86. D

A

x

=

8 4 = . 10 5

3 . 4 x

C

B

2007

v = Volume of box = (2007 – 2x)2 × x dv = 2(2007 – 2x)(– 2x) + (2007 – 2x)2 dx

= (2007 – 2x)[– 4x + 2007 – 2x] = (2007 – 2x) (2007 – 6x) For the maximum volume dv =0 dx

(2007 – 2x)(2007 – 6x) = 0 x= d2 v dx 2

2007 2007 or x = 6 2

= (2007 – 6x) (– 2) + (2007 – 2x) (– 6) = – 4014 + 12x – 12042 + 12x = 24x – 16056

 d2 v    2007  dx 2  2007 = 24 × – 16056  x 6 6

= 8028 – 16056 < 0 

Volume is maximum when x =

2007 669 = cm. 6 2


87.

S = y2 – 4x S (0, 2) = 4, S (1/2, 5/2) = 25/4 – 2 = 17/4, S (5/2, 9/2) = 81/4 – 20/2 = 41/4 and S (1, 3) = 9 – 4 = 5 It is clear that (0, 2) is closest to parabola y2 = 4x.

88.

or 2

e

e

2

e

or 2

I<0 J<0

I<0 J>0

J<0

So, I and J can be both negative but not both positive. 89.

|z1 z2 + z2 z3 + z3z1| = |z1z2z3|

1 1 1   z1 z 2 z 3

= z1  z2  z3

[ |z1| = |z2| = |z3| ]

= |z1 + z2 + z3|. 90.

So, the angles will be 99, 108, ....., 153, 162 Sum of largest two angles = 153 + 162 = 315.

PHYSICS 91.

Rotational inertia does not depend on speed so rotational inertia of the cylinder about its axis will be

1 MR2. 2 92.

Air column is closed at one end, so fundamental frequency n=

v 330  = 125 Hz 4 4  66  10  2

n1 : n2 : n3 ....... = 1 : 3 : 5 ........... So, possible frequencies are 125, 375, 675, 875, 1125 (> 1kHz) So only possible frequenceies are 3. 93.

Time period of pendulum on the surface of the earth T0 = 2p 2

 g

At a depth d,

........(i)

d  g = g  1   R  

and  = 1   

So new time period, T = 2

1   d   (1  )   = 2 = (1+ )½ (1– d/R)–½ = 1    1  2   2R  g g(1  d / R) 


94.

For first darkness, d sin =

95.

6  10 7 2  10  6

 0.3  = 17º

In LC circuit time period, T = 2 Lc t=

96.

 sin =

T  LC = 4 2

For insulated material capacity,

0 A C = (d  a  a /  ) r But of metal, r =  0 A (d  a)

So, C = 97.

Amax = g A2 = g Amax =

98.

(block and piston remains together)

g 

2

=

g 4 2 f 2





z = 0, as P and E in same direction

dE  105 dZ 99.

F = P

dE = 10–7 × 105 = 10–2 N dZ

Inside the atom, Electric field due to +v charge , E(r) =

Due to –ve charge, E(r) = –K

So, E(r) =

kz e r2

ze r r3

ze  1 r   2  3 4  0  r R 

r

For out side atom Qnet = 0 E(r) = 0 r>R 100.

Q = T f  f  =   1 RT =   1 PV 2 2    

f  Q 10.61   1 =  2 PdV   1.01 10 5  3  10  5

f = 4.3 So, gas is a mixture of mono and diatomic molecules.


CHEMISTRY 101.

CH2 = CH – CH – CH2 – CH3

Br2

H – C – Br H – C – Br CH2 – CH3

CH2Br H – C – Br Br –C – H CH2CH3

/////////////////////////

CH2Br

Br

/////////////////////////

Br

* * CH2 – CH – CH – CH2 – CH3 Br

Br

CH2Br Br – C – H Br – C – H CH2 – CH3

CH2Br Br – C – H H – C – Br CH2CH3

P

102. V

according to above graph P  V This contradict the Boyle’s law 103.

 Molarity of acetic acid = 0.1 M [Suppose volume of acetic acid = x]  no. of millimoles present in x ml of acetic acid = 0.1x  Molarity of sodium acetate = 0.2  no. of millimoles present in 10 ml of sodium acetate = 0.2 × 10 = 2 [Salt] pH = pKa + log [ Acid]

2 4.91 = 4.76 + log 10  x 0.1x 10  x 2 0.15 = log 0.1x

[  total volume = (10 + x) ml]

0.15 = log20 – logx 0.15 = 1.3010 – logx logx = 1.151 x = antilog (1.151) x = 14.16 ml x  14.2 ml


104.

Ba(OH)2 + SO2  BaSO3 + H2O (White ppt.) Ba(OH)2 + SO3  BaSO4 + H2O (White ppt.) BaSO3 is soluble in dil HCl but BaSO4 is not soluble in dil HCl . After filteration , filtrate contains BaSO3 Br2 + H2O  2HBr + O (Bromine water) BaSO3 + O  BaSO4 White ppt

1 105.



106.

Kb = 0.52 Kg K mol–1 (given) Tb = 0.052 K (given) m = molecular mass of given polymer (solute)

V according to question given potential energy = kinetic energy 1 eV = mv2 ----- (1) 2 P2 eV = [ P = mv] 2m h2 h eV = 2 [ p = ]   2m 2 h 2 = 2meV h = 2meV  e, m, h are constants 1 so   V

Tb =

1000 K b w 0 mW

It means we can calculate

Tb  m w0 = 1000K W b

---------------- (i)

Relative lowering of vapour pressure of water 

Po – Ps w M = 0 × ------(ii) Po W m

M = molecular mass of water = 18 m = molecular mass of polymer (solute) is given

w0 = by equation (i) W so we can calculate relative lowering of V.P. of water by equation (ii) n ST V But V is unknown, so we can’t find out the value of osmotic pressure. osmotic pressure  =

freezing point depression Tf =

1000 K f w o mW


107.

But value of Kf is not given so we can’t calculate the magnitude of freezing point depression. Test of Nitrogen Organic compound no ppt. Nitrogen Fused with Na + FeSO4 observed absent + FeCl3 Test of sulphur Organic compound fused violet S present with Na + sodium nitroprusside colour observed Thus according to question organic compound contains sulphur only.

NH2 108.

CH3

NHCOCH3

(CH3CO)2O CH3

NH COCH3

Br2/CH3COOH (I)

CH3

Br (II) H2O/H+ NH2

CH3 Br (III)

*****



1

2

3

4

5

6

7

8

9

10

11

12

13

14

15

Ans.

D 17

A 18

D 19

C 20

C 21

B 22

C 23

C 24

B 25

B

C

A

D

B

Que s.

A 16

26

27

28

29

30

Ans.

C

A

A

D

A

D

C

B

D

C

D

Que s.

31

32

33

34

35

36

37

38

39

40

41

A 42

C 43

B 44

B 45

Ans.

C 46

A 47

B 48

D 49

B 50

D 51

C 52

D 53

A 54

D 55

D 56

B 57

B 58

B 59

A 60

B 61

A 62

A 63

C 64

A 65

D 66

C 67

B 68

D 69

A 70

B 71

A 72

D 73

B 74

B 75

A 76

C 77

A 78

A 79

D 80

B 81

C 82

B 83

C 84

B 85

A 86

C 87

B 88

B 89

C 90

B Que s. 91 B Ans. Que s. 106 B Ans.

A 92 A

B 93 C

C 94 B

B 95 A

B 96 C

A 97 D

A 98 B

D 99 D

D 100 D

D 101 D

C 102 C

C 103 B

C 104 A

A 105 A

107 A

108 C

109 D

110 A

111 D

112 D

113 B

114 A

115 C

116 A

117 B

118 D

119 A

120 C

Que s. Ans. Que s. Ans. Que s. Ans.

PART-I (1 Mark) MATHEMATICS 1.

Sol.

The equation z2 = z , where z is a complex number, has (A*) 4 solution (B) 2 solution (C) no solution (D) infinitely many solutions z = a + ib

z = a – ib It is given : z2 = z (a + ib)2 = a – ib a2 – b2 + i 2ab = a – ib  a2 – b2 = a and 2ab = – b 2ab + b = 0 b(2a + 1) = 0 b = 0 or a = When b = 0,

So,

1 2

a2 – b2 = a a2 – 0 = a a (a – 1) = 0 a = 0, or a = 1 2

1   1  1 When a = ,   – b2 = 2 2  2 


1 1 – b2 = 4 2 1 1 3 + = 4 2 4

b2 =



3 2 z = 0 + 0i or 1 + 0i



z=

b=±

1 1 3 3 +i. or –i 2 2 2 2 There will be four solutions possible.



2.

x 1 x 1     2/3  1/ 3  x  1 x  x 1/ 2  x

10

 ( x1/ 3  1)( x 2 / 3  x1 / 3  1) ( x1/ 2  1)( x1/ 2  1)     ( x 2 / 3  x1/ 3  1) x1/ 2 ( x1/ 2  1)  

(x x

1/ 3

 1)  (1  x 1 / 2 )

10

10



10

1/ 3

 x 1 / 2 General Term



Tr + 1

= (– 1)r 10 Cr (x1/3)10 – r (x-1/2)r = (– 1)r 10 Cr x

For the independent term

10 r r  3 2

10  r r  =0 3 2

10  r r  3 2 20 – 2r = 3r 5r = 20 r = 4.

10! So, independent term = (– 1)4 10C4 = 4!6! = 210.

3.

a a2

a3  1

b b2

b3  1

c c2

c3 1

a a2

a3

a a2

1

b b

2

b

3

b b

2

1

c c

2

c

3

c c

2

1

+

=0

1 a a2 abc

1 b b2 1 c c2

=0

a a2 1 –1

b b2 1 c c2 1

=0


1 a a2 abc

1 b b

a 1 a2

2

1 c c2

+1

1 a a2 abc

1 b b

c 1 c2

=0

1 a a2

2

1 c c2

b 1 b2

–1

1 b b2 1 c c2

=0

1 a a2 1 b b2 1 c c2  4.

(abc – 1) = 0

abc = 1.

ax2 – 6xy + y2 = 0 m1 + m2 =

2h b

m1 + m2 =

6 1

m 1m 2 =

a b

a =a 1 Given : m2 = m12 m 12 + m 1 – 6 = 0 (m1 + 3) (m1 – 2) = 0 m1 = – 3 and m1 = 2 and m 1m 2 = a m1 × m12 = a m 13 = a a is positive, so a = (2)3 = 8. m 1m 2 =

5.

x2 + 4y2 =1 ....(i) 2 2 4x + y = 4 y2 = 4 – 4x2 Substitute the value of y2 in equation (i) x2 + 4(4 – 4x2) = 1 x2 + 16 – 16x2 = 1 16 – 1 = 15x2 15 = 15x2 x2 = 1 x=±1 y2 = 4 – 4x2 = 4 – 4(1) y2 = 0 y=0  (1, 0) and (– 1, 0) are two common point.


6.

Diameter = major axis = 2a radius = a Area of circle = a2

(0, b)

Let the equation of the ellipse be

y=

x2 a2



y2 b2

(– a, 0)

 1.

(0, – b) a

1 2 a = 4 × 3

Area of ellipse =

2



0

b a 2  x 2 dx a a

1 x b x a 2  x 2  a 2 sin 1  =4×  2 a 0 a 2 2 1 2 b  a    a = 4 × 3 a  4 

b 1  a 3

7.

2

 1 1   3

2

e=

b 1   a

PD =

PA  PB 18  2 = =9 PC 4

=

=

8 2 2 = . 9 3 B

N O 18 4 P 9 C 2 M A

1 BN = AB = 10 2 1 13 CD = 2 2 PM = MC – PC

D

MC =

13 5 –4= 2 2 5 ON = PM = 2 In ONB =

OB =

5 10 2    2

2

=

100 

25 = 4

425 5 17 = 4 2

5 17 × 2 = 5 17 cm. 2 cos 175 = cos (180 – 5) = – cos 5 cos 185 = cos (180 + 5) = – cos 5 cos 355 = cos(360 – 5) = cos 5  cos 5 + cos 10 + cos 15 + ......+ cos 355 = cos5 + cos 10 + .....+ cos 85 + cos 90º + cos(180º – 85) + cos (180º – 80º) + .......+ cos (180º – 5) + cos 180º + cos (180º + 5) + cos(180º+ 10) + .......+ cos (180º + 85) + cos 270 + cos(360 – 85) + ......+ cos(360 – 5) = cos 5 + cos 10 + .......+ cos 85 + 0 – cos 85 – cos 80 – ..........– cos 5 – 1 – cos 5 – cos 10 – cos 85 + 0 + cos 85 + cos 80 + ...........+ cos 5 =–1

Diameter = 2 × OB = 8.

2

a –x

(a, 0)

(0, 0)

b a2  x 2 a

b y= a


9.

By AA similarity Q ~ PBA

y D

Q a– y

C a–x

3

ax ay 3 = = a x 4

5

P 4

ax 3 = a 4

A

x B

a

a 4

x=

ay 3 = x 4 ay 3 = a/4 4

y=

13a 16

In PAB 42 = x2 + a2 2

a 16 =   + a2 4

16 =

17a 2 16

a2 =

16  16 256 = 17 17

Area of square ABCD = a2 =

10.

256 . 17

By Angle bisector theorem

A

cos B =

11.

s= s=

7 3 2  4 2  22 9  16  4 21 = = = . 8 23 4 24 24

4

x 2 = x=4 2 1

x=

2

B

2

D 1

C

18 x  6 = 9x + 3 2 (9 x  3)( 2)(3 x  1)( 6 x )

= 6 (3 x  1)( 3 x  1)( x ) = 6(3x + 1) x x is a perfect square, so in 1 to 20, 4 perfect square i.e. possible values are {1, 4, 9, 16}.


12.

f(1) = 0 as lim

h0

f (1  h) exists. h

f (1  h)  f (1) f ‘(1) exists. h 0 h So, f is differentiable at 1.  lim

13.

14.

Statement I is true that the derivative of an odd differentiable function is always even and statement II is also true that If f(x) is differentiable at a point x0 and g(x) is not differentiable at x0, then f(x) g(x) is not differentiable at x0.

x   

0,

0  x 1

 1,  1  x  0

 sin( 1)   sin1,  1  x  0 f(x) =   1  0, 0  x 1

lim f ( x )  sin1

x 0 

f (x)  0 and xlim 0   lim f ( x ) does not exists. x 0

15.

x–

x > 0 for all x > 1.

and –

x + 4 x > 0 for all x < 0 So, range of f is [0, ). x

16.

f(x) =

t

 e (t  1)(t  2) dt 0

f’(x) = ex(x – 1)(x – 2) For decreasing f’(x) < 0 ex(x – 1)(x – 2) < 0 (x – 1)(x – 2) < 0 x  (1, 2). 1

17.



x | x |3 / 2 dx

1 0

=



1 0

=

[ ex > 0]



1

x(  x )3 / 2 dx +

1



x 5 / 2 ( 1)3 / 2 dx + 0

= (– 1)

3/2

x( x )3 / 2 dx

0 1



x 5 / 2 dx

0

1

 x7 / 2   x7 / 2    +    7 / 2  –1  7 / 2  0

= (– 1)3/2

2 2 [0 – (– 1)7/2] + [1 – 0] 7 7

= – (– 1)5

2 2 2 2 4 + = + = . 7 7 7 7 7


18.

 x f ( x)f (x)......f 1

10 ( x )

2

1 dx

Let f11(x) = t

dx x f1( x )f2 ( x )......f10 ( x ) = dt

 1.dt

=t+C = f11(x) + C.

19.

Applying AM and GM condition : x1  2x 2  3 x 3  6 x1x 2 x 3 1 / 3 3 4  6 x1x 2 x 3 1/ 3 3

Cubing both sides

64  6 x1x 2 x 3 27 x1x 2 x 3 

32 81

Now also

x1  x 2  x 3 2 2 2  x1 x 2 x 3 3

2

2

2

2



2

2

x1  x 2  x 3  32    3  81   210 x1 + x2 + x3  3  38  2

2

2

2

2

2

1/ 3



2/3

1/ 3

   

x1 + x2 + x3 

38  2   9 9

x 12 + x 22 + x 32 

24  0 .6 9

1/ 3

x 12 + x 22 + x 32  1 . 6 So, according options least value is 2. 20.

12 3 4 5 6 7 at each position two posibility i.e. either H or T So total posible outcomes = 27 As we want three tails is sequence and it happened at 7th turn. So at position 5, 6, 7 we have tails and position 4 must have head otherwise our requirement is fullfilled at position 6. So, the position 1, 2, 3 can have anything i.e. H/T = 23 To get tail at 1, 2,3 position we have only one probability. So, total favourable cases = 23 – 1 = 7 required probability =

7 27

=

7 . 128


PHYSICS 22.

PV = nRT PdV + VdP = 0

1 dV 1   V dP P  = P–1 –

23.

(Answer is C)

1 for point charge r2 Force solid sphere

E

Inside the sphere E  r and out side the sphere E 

24.

=

r2

L r 4

So,  = 25.

1

L r 4

Angular momentum,  = mvr Magnetic moment,  = IA = nqr2 =

 qr 2 2

v vqr 2  = 2r qr  2 vqr q     2mvr 2m

or

26.

kT c m

(Answer is C)

2  2 1/ 2

ML T  = LT M 1

1/ 2

= [M0L0T0]

which dimensionally satisfies

1 kT f = c m f 27.

(Answer is D)

Average power,
=  Av2

28.

Wave length,  =

1 2 2 v Aw w 2 (Answer is A)

d 0.2  6  103  = 0.5 × 10–4 m D 24

 = 5 × 10–5 cm  = 500 nm

(Answer is B)

29.

(B)

30.

By angular momentum conservation m1v1r1 = m2v2r2 ( m1 = m2 )

So, v2 =

5.6  10 4  9  1010 5.6  1012

= 900 m/s (Answer is B)


31.

y = y1 + y2 = 2A cos (2Kx) cos(2cost) x = 0, y = 0 (Ans. C)

  20m

32.

A  m (some as x)  

1 sec–1 t

1   sec–1 t  = angle(dimensionless) 33.

34.

(Answer is A)

KE = 0.01864 × 931 = 17.4 Mev KEf – KEi = 17.4 Mev As proton is at rest (B) option is correct

mv 2 kq1q2  2 r r kq1q2 mr

v=

v = 1.6 × 10–19

9  10 9 9.11 10 31  0.53  10 10

v = 2.24 × 106 m/s

(Answer is D)

I

35.

F

F=

F=

F

0 1 2  2d 4  10 7  4000  4000  60  10 2 2  1.5  10 2

F = 128N repulsion

(Answer is B)

37.

(C)

38.

Phase different between VR and VC is 90º so,

VR2  VC2 = 220

(Answer is D)

39.

Distance between to consecutive destructive interference is 1.7cm, so total number of points of destructive interference on the line PQ is 4. (Answer is A)

40.

(D)


CHEMISTRY 41.

The ionization energy or enthalpy of Na is less than that of Li but the hydration enthalpy of Li is higher than that of Na. Difference between ionization enthalpy of both is less than difference between their hydration enthalpy so overall oxidation potential of Li is greater than Na. Therefore Li metal is a better reducing agent than Na metal.

42.

Compressibility K =  Kx – Ky =

1 P

1 1 – 1 2

= 0.5 atm–1 OH

43. CO2H

44.

2CH3

CHO

conc. NaOH

CH3

CH2OH + CH3

COOH

Cannizzarro reaction 45.

On heating solution becomes unsaturated  solubility and conductance both are increase.

46.

According to graph, segment BC represents isobaric process. It means pressure is constant so according to Charle’s law V  T [on constant pressure] From B to C volume decreases so temperature will also decreases.

H 6

H 3

4

1

47. 7

5

OH

2

when same groups present in same side of the double bond then geometrical isomer is known as Z – isomer. Due to back donation electron defficiency of [B] almost neutral.

+

+ Cl

Cl

Cl 49.

en

Co Cl trans

en

en

Co en cis/d

/////////////////////////

48.

+

Cl Cl Co

en

en


50.

xH2S + 2NaNO3 +2HCl  yS + zNO + k NaCl + 4H2O Put the following values of variables in above equation x = 3, y = 3, z = 2, k = 2 3H2S + 2NaNO3 + 2HCl  3S + 2NO + 2NaCl + 4H2O

51.

 R = k [X]1/3 [Y]2/3 order of reaction =

1 2 + =1 3 3

52.

Because Zn has higher oxidation potential than Cu,So Zn loses its electrons at an anode and convert in Zn2+ ions and come into the solution. These electrons flow externally from zinc to Cu by wire.

53.

Because ideal mixture of benzene and toluene follows Raoult’s law according to Raoult’s law Pmix = PA + PB Pmix = XAPAº + XBPBº Pmix = XAPAº + (1 – XA) PBº Pmix = XA(PAº – PBº) + PBº This is a linear equation, by comparing with y = mx + c y = Pmix x = XA m = PAº – PBº c = PBº according to above data graph will be like following -

PBº Pmix PAº XA = 0 XA = 1 XA molefraction (benzene )

54.

sp3d2 hybridization explains the bonding of complex which has C.N. (co-ordination number) 6 so answer may be either [Fe(CN)6]3– or [Fe(H2O)6]2+. But H2O is a weak ligand and CN– is a strong ligand. Because weak ligand form an outer orbital complex with metal ion and this is possible when sp3d2 hybridization takes place.

55.

2NaCl + H2SO4  Na2SO4 + 2HCl MnO2 + 4HCl  MnCl2 + 2H2O + Cl2

56.

Bond order =

1 [N – Na] 2 b


 B.O. of O2+ =

1 [10 – 5] = 2.5 2

Nb = no. of electrons present in BMO

1 [10 – 7] = 1.5 Na = no. of electrons present in ABMO 2 1 B.O. of O22– = [10 – 8] = 1 2 1  Inter atomic distance  B.O  order of inter atomic distance ; O22– > O2– > O2+ B.O. of O2– =

57.

amylose [it is a part of starch]

58

t1/ 2 = t1/2 =

59.

60.

0.693 K 0.693

6.93  10 – 3

t1/2 = 100 s Because unit cell has six faces and every facial atom is a part of two unit cell. It means only half part of one facial atom belongs to one unit cell. 1 so, total no. of facial atoms in fcc unit cell = 6 × =3 2 =



W T1 – T2 = Q1 T1

W 800 – 200  800 100

W = 75 J


a=A– d b=A c=A+d a + b + c = 3A = A=



3 2

1 =b 2

a+b+c= a+c= b2 =

3 2

3 3 1 –b= – =1 2 2 2

....(i)

a 2c 2 2

 1   = ± ac 2 When

ac = c=

1 4

1 4a

...(ii)


From (i) & (ii) 1 =1 4a 4a2 – 4a + 1 = 0 (2a – 1)2 = 0 2a – 1 = 0

a+

1 2 By this we get a=b=c But a < b < c, so we take a=

When

– ac =

1 4

1 4a

c=

....(iii)

from (i) & (iii) a–

1 =1 4a

4a2 – 4a – 1 = 0 a=

=

4  16  16 8

44 2 1 2 = 8 2

as b > a 

a=

1 2 2

82.

g’(x) is changing its slope from positive to negative as it es through g(2). So, g(2) is largest.

83.

92  × 180º = 140º B =   9 

1 = 2 =

D

x

180 º 140 º = 20º 2

z 3

C

3 = 140 – 2 = 120º In ACD

y 2  x 2  z2 cos 120º = 2xy

2

y A

x

1 x

B

y 2  x 2  z2 1 = 2xy 2 – xy = y2 + x2 – z2 z2 = x 2 + y2 + xy


84.

0 * *  *    * 0 *  * * * 0  * T AA =           * * *  0 As the diagonal element of resulting matrix are zero.  multiplication R1 of A and C1 of AT (i.e. = R1 of A) = 0 It is possible only when all element in R1 of A is zero In the same way we can say that element of all 10 rown of A is zero so for the above condition we can formm only 1 matrix i.e. A = null matrix.

(0, b) C E

85.

(– a, 0) A

(0, 0) O

(–ae, 0)

F

B

D (0, – b)

(i)

As line DE and AC are  so product of there slope = – 1.

b  0 b × =–1 0  ae a b b × =–1 ae a 2

b e=   a (ii) and line CF and AD are also  b0 b  0 × =–1 0  ae 0a



b b × =–1 ae a 2

b e=   a (iii) and line AO is  to CD 00 2b × –1 0a 0



2

b From (i) & (ii) case e =   a But in (iii) case ‘e’ cannot be determined. So, we can not determined uniquely. 

86.



0

[ x ]e  x dx =

= 0 +  e x



1

 0e 0

2 1

x

2

dx +



1

(1) e  x dx +

3

 2e 2

x

dx + ......

x 3 2

 +  2e  + .........

= – (e–2 – e–1) – 2(e–3 – e–2) – 3(e–4 – e–3) ......... = e–1 + e– 2 + e–3 ............

=

1 e1

+

1 e2

+

1 e3

+ ..........



S =

1 e 1 1 e

=

1 . e 1


87.

  

/2



0

=

  cos1003 xdx    

lim g( x )

x





lim

lim x

= 89.

=

 2008

e f ( t ) dt

e

g' ( x )

=

 2  1004

0 f(x)

x

=

=

e f ( t ) f ( x ) dt x

=

 cos1004 xdx  

0

g( x )

=

0

(1002  1000  998 .....2) (1003  1001  .....1)   × 1003  1001  ....... 1 1004  1002  ....... 2 2

x

88.

/2



lim 0 

x 

lim 0 

x

x



0

e f ( t )  g( x )e f ( x )

f ' ( x )e f ( x )

g( x ) f ' (x) 3 x 3  ..... 16 x 3  ....

3 . 16

dy = sin(x + y) + cos(x + y) dx let x + y = u Differentiating wrt x

1+

dy du = dx dx

dy du = –1 dx dx



dy = sin(x + y) + cos(x + y) dx

du – 1 = sinu + cosu dx du = sinu + cosu + 1 dx du  dx sin u  cos u  1 Integrating both sides du

 sin u  cos u  1   dx


du



2 tan u / 2 2

1  tan u / 2



1  tan 2 u / 2 2

1  tan u / 2

 dx 1



sec 2 (u / 2)du

 2 tan u / 2  1  tan

2

u / 2  1  tan 2 u / 2



sec 2 (u / 2)du  dx 2 tan u / 2  2



sec 2 (u / 2)du  dx 2(tan u / 2  1)

 dx







Put 1+ tanu/2 = t (1/2)sec2u/2du = dt



dt  dx t



log |t| = x + c log |1+ tanu/2| = x + c log 1  tan

(x  y) =x+c 2

It es through origin.  log1 = 0 + c c = 0. So, log 1  tan 90.

(x  y) = x. 2

Total mappings of bijection is 6 !. for self inverse 4! 6 C 4 2! 2! 2!   

6! 2! 2! 2! 3! +  2 pairing

P=

2 going to their own value

+

6 C2  1     4 going to their own value

= 76

76 19 = . 720 180


PHYSICS d 91.

M2

M1 r1

r2

M1r1 = M2r2 and r1 + r2 = d

M2 d so r1 = M  M 1 2 2r1 T = v 1

........(i)

........(ii)

Here

M1v12 = r1

GM1M2 d2

GM2r1

or v1 =

d

From (ii) T=

T=

T=

2r1d GM2r1

2d r1 GM2 2d

T=

GM2

M2d M1  M2

2d3 / 2 G(M1  M2 )

92.

a = r = 20 × 0.5 = 10 m/s2

93.

For min

(Answer is B)

hc  min = K.E. min =

hc K.E.

6.6  10 34  3  10 8 min = 30  10 3  1.6  10 19 = 4.14 × 10–111 m


94.

95.

 V  V0   340  9  App. frequency, N =  V  V  N   340  9   90   S   N = 94.9 kHz

Mass of water, m =

kg 10 3 m =  × 18×10–2 × (3.96)2 × 10–4 × 0.92 ×103 mLf =

t = 96.

  18  (3.96 )2  0.92

mL f  t = KA (   ) 1 2

KA (1  2 ) t 

  18(3.96)2  0.92  333  103  10 3

t = 9.2 s

103  400    (3.96)2  15

(Answer is A)

Range, R = 2 (H  h) h (i) h = H/4 HH  X1 = 2  H   44 

3H/4 H/2 2

3 X1 = H 2 (ii) h = H/2 X2 = H (iii) h = 3H/4 X3 = H

3

H/4

1 X2

3 2

X1= x3

X2 > X1 = X3 (Answer is C)

CHEMISTRY 101.

Xe contains 8 electrons in its outermost shell or valence shell. In XeF2, Xe uses 2 electrons for  bonds so it contains 3 p. In XeF4., Xe uses 4 electrons for  bonds, so it contains 2 p.

N2 + 3H2

102. initial moles

4

2NH3

16

0

at equilibrium

[according to question ammonia gas is produced = 4 mol]  2 = 4 put the value of  = 2 4–2

16 – 3 × 2

2

10

4

concentration 2 at equilibrium V

10 V

4 V

moles at equilibrium

2×2


KC

=

[NH3 ]2 [N2 ][H2 ]3 2

4   V =  2  10 3     V  V  = =

42 2  10 3 8

× V2

× (10)2

10 3

(V = 10L]

= 0.8 mol–2 lit2 103.

Kw 0  Depression in freezing point TF = m W 0 In given two cases m0, W and K are constants so TF  w0 ( TF )1 ( w 0 )1 = ( TF )2 ( w 0 )2 5.5 – 4 2.9 = 5.5 – 2.5 2.9  x 1.5 2.9 = 3 2.9  x x = 2.9 g I

105.

OH II 2

2R and 5R III

'

III ' 5

II

OH I '

106.

Zn + 2Ag+ (0.0001M)  Zn2+ (0.1M) + 2Ag by nernst equation Ecell = Eº –

[ Zn2  ][ Ag]2 0.059 log10 [ Ag ]2 [ Zn] n

 [Ag] = [Zn] = 1  Ecell = Eº –

[ Zn2  ] 0.059 log10 [ Ag ]2 2

 Ecell = 1.56 –

(0.1) 0.059 log10 ( 0 . 0001)2 2

 Ecell = 1.56 – 0.2065  Ecell  1.35 V 107.

Cu(Y)

(CH3)2SiCl2 2CH3Cl + Si  (X)

SiCl4 + 2CH3MgBr  (CH3)2SiCl2 + 2Mg(Br)Cl (Z) CLASS-XII_STREAM-SB+2_PAGE # 170

108.

according to Arrhenius theory K = Ae

–

E RT

according to question K is same in both cases, so

E1 E2 RT1 = RT2 E1 and T1 are activation energy and temp. in absence of catalyst

400 500 = T2 625

E2 and T2 are activation energy and temp. in presence of catalyst

T2 = 500K 109.

28

Ni  [Ar]3d84s24p0

Ni2+  [Ar]3d8 4s0 4p0

[Ar] because CN– is a strong ligand ,so pairing of electrons takes place.

[Ar] dsp2 hybridization

[Ar] 2

dsp hybridized Unhybridized orbitals orbitals

In dsp2 hybridization geometry of complex is square planar.

110.

238 206  82 Pb 92 U 

 x 24He  y –01e

by balancing of mass no. 238 = 206 + 4x + 0y x=8 by balancing of nuclear charge 92 = 82 + 2x – 1y 92 = 82 + 2 × 8 – y y=6

*****


HINTS & SOLUTIONS (YEAR-2009) ANSWER KEY Que s. Ans. Que s.

1 A 16

2

3

B 17

4

C 18

5

B 19

6

D 20

B 21

7 D 22

8 A 23

9 A 24

10

11

12

13

14

15

C 25

B

B

D

C

A

26

27

28

29

30

Ans.

B

C

B

C

D

B

B

D

D

C

C

Que s.

31

32

33

34

35

36

37

38

39

40

41

D 42

C 43

A 44

D 45

Ans.

D 46

C 47

A 48

B 49

B 50

C 51

A 52

A 53

A 54

B 55

C 56

D 57

C 58

C 59

D 60

A 61

C 62

B 63

A 64

B 65

A 66

D 67

C 68

D 69

B 70

D 71

A 72

D 73

B 74

B 75

A 76

C 77

C 78

C 79

B 80

B 81

C 82

D 83

A 84

D 85

D 86

D 87

B 88

A 89

B 90

Ans. B Que s. 91 D Ans. Que s. 106 C Ans.

B 92 A

D 93 B

C 94 C

A 95 D

C 96 A

B 97 C

C 98 C

B 99 B

B 100 B

D 101 B

C 102 D

D 103 A

B 104 A

C 105 B

107 A

108 C

109 D

110 B

111 C

112 B

113 A

114 B

115 A

116 D

117 B

118 A

119 C

120 C

Que s. Ans. Que s. Ans. Que s.


Given : a1, a2, a3 .........AP and a1, a2, a4, a8 ......GP. Let common difference of A.P. = d a2 = a1 + d a4 = a1 + 3d a8 = a1 + 7d 

a8 a2 a4 = = a1 a2 a4 = r a1  d a1  3d a1  7d a1 = a1  d = a1  3d = r (a1 + d)2 = a1(a1 + 3d) a12 + d2 + 2 a1d = a12 + 3a1 d d2 = a1 d

(d  0)

d = a1

....(i)

a2 a1  d Hence, a = r ; a =r 1 1 a1  a1 =r a1

(using (i))

r = 2.


2.

Tk 101  till k = 10 Tk –1 k Tk > Tk–1 Let k = 11 T11 < T10  T10 is maximum at k = 10.

3.

x=

2+ 3+ 6 2

2

x  2  =  3  6 

x2 + 2 – 2 2 = 9 + 6 2 x2 – 7 = 8 2 (x2 – 7)2 = 64 × 2 So, smallest possible value of n is 4. 4.

Let the three players are A, B, C. Now, each player get 0 score after playing 9 games. It happened only when each player wins 3 games and loss 6. So, A win 3 games out of 9  9C3 B win 3 games out of remaining 6  6C3 C win 3 games out of remaining 3  3C3 So, required way = 9C3 × 6C3 × 3C3 =

9  8  7  6! 6  5  4  3! × ×1 3  2  1 6! 3  2  1 3!

= 1680. A (2, 3)

5.

O (2, z) B (4, 0)

C (x, y)

O is circumcenter  OA = OB = OC = circumradius (2 – 2)2 + (z– 3)2 = (4 – 2)2 + (0 – z)2 z2 + 9 – 6z = 4 + z2 9 – 6z = 4 5 = 6z 5 =z 6

 Circumcenter =

5 13 . ( z  3 ) 2  ( 2  2) 2 = | z – 3 | = 6  3 = 6


6. (5, 15) L(21, 15) P

P'

A

 5  21 15  15  ,  Mid point of PP’ =  2   2 L = (13, 15)  Point A will be (13, 0) By property PA + PA’ = 2a

PA =

(13  5)2  (0  15 )2 =

64  225

=

289 = 17 cm

PA’ = =

(13  21) 2  (0  15) 2 64  225

=

289 = 17 cm  2a = PA + PA’ 2a = 17 + 17 2a = 34 cm So, length of major axis = 2a = 34 cm. 7.

P (10, 10)

B (0, 6)

2x + 3y = 18

C (a, b)

PB = PC (10 – 0)2 + (10 – 6)2 = (a – 10)2 + (b – 10)2 100 + 16 = a2 + 100 – 20a + b2 + 100 – 20b a2 + b2 – 20a – 20b + 84 = 0 ....(i) Also (a, b) i.e. on 2x + 3y = 18 2a + 3b = 18

3b 2 Using equation (i) a=9–

2

3b   3b    – 20b + 84 = 0 9   + b2 – 20  9  2  2   

81 +

9b2 – 27b + b2 – 180 + 30b – 20b + 84 = 0 4

13b 2 – 17b – 15 = 0 4 13b2 – 68b – 60 = 0


13b2 – 78b + 10b – 60 = 0 13b(b – 6) + 10 (b – 6) = 0 10 13

b = 6 or b =



When b = 6, then a = 9 –

When b =

10 3  10 30 132 , then a = 9 + =9+ = 13 2  13 26 13

132 10 +2× 13 13

 8a + 2b = 8 ×

1056  20  79 . 13

=

8.

36 =0 2

cosec2( + ) – sin2( – ) + sin2(2 – ) = cos2( – ) cosec2( + ) + sin2(2 – ) =

cos 2 (  )  sin 2 (   ) 1

cosec2( + ) = 1 – sin2(2 – ) cosec2( + ) = cos2(2 – ) Minimum value of cosec2( + ) is 1 and maximum value of cos2(2 – ) is 1.  They will be equal for the value 1.

 2

.....(i)

2 –  = 0 By adding (i) & (ii)

.....(ii)

+=

 2

3 =



9.

=

 6

=

 3

sin( – ) = sin (

sinx + siny =

7 5

cos x + cosy =

   1 – ) = – sin ( ) = . 6 3 6 2

....(1) 1 5

....(2)

By (1)2 + (2)2 we get 2 + 2sinx siny + 2 cosx cosy = 2


sinx siny + cosx cosy = 0 cos(x – y) = 0  x – y = 90º By (1) × (2) we get sinx cosx + sinx cosy + siny cosx + siny cosy =

7 25

sin(90 + y)cosx + sin(x + y) + sin(x – 90) cos y =

7 25

cosy cosx + sin(x + y) – cosx cosy = sin(x + y) =

7 25

7 . 25

10.

6 x

y=

(1, 1) 1 0 y = sin x Clearly, curve meet each other twice in

 11.

2 – 3 4 – 5 6 – 7 8 – 9 10 – 11

Total 10 Times.

f(x) is differentiable on R. So, it will be contincous on R. Continuity at x = 0 LHL 2 lim– sin x x 0 x Put x = 0 – h, then h  0 2 lim sin(0  h) h0 h

lim sin h  h = 0 h h RHL h0

lim x2 + ax + b

x 0 

Put x = 0 + h, then h  0

lim h2 + ah + b = b

h0

Value of f(x) at x = 0 f(0) = b.


f(x) is contineous at x = 0 LHL = RHL = f(0) 0=b=b b=0 Differentiability at x = 0 LHD  

lim

f ( 0  h )  f ( 0) h

lim

sinh 2 b h h

h0

h0

2 lim sin h = 1 h0 h2 RHD

lim f (0  h)  f (0) h

h0

2 lim h  ah  b – b h0 h

lim h(h  a ) = a. h  f(x)is differentiable at x = 0, LHD = RHD a = 1. h0

12.

Let point p(x1, y1) is on the curve y2 = 4x. 2



y12 = 4x1  x1 =

y1 4

( x1  0) 2  ( y1  3 ) 2

PA =

AP2 = x12 + y12 – 6y1 + 9 AP2 = x12 + y12 – 6y1 + 9 Let AP = z z2 = x12 + y12 – 6y1 + 9 2

 y12    2 z =  4  + y12 – 6y1 + 9   4

z2 =

y1 + y12 – 6y1 + 9 16

Diff. w.r.t. y1 3 dz 4 y1 2z dy = + 2y1 – 6 16 1

3 dz y 2z dy = 1 + 2y1 – 6 1 4 3

=

y1  8 y1  24 4

dz 2z dy = (y1 – 2) (y12 + 2y1 + 12) 1


For the critical points

dz dy 1 = 0 (y1 – 2)(y12 + 2y1 + 12) = 0 y1 = 2 y12 = 4x1 (2)2 = 4x1 x1 = 1.

  

 dz   2   dy 1 

2

d2 z

+ 2z

2

dy 1

= (y12 + 2y1 + 12) + (y1 – 2) ( 2y1 + 2) = y12 + 2y1 + 12 + 2y12 – 4y1 + 2y1 – 4 = 3y12 + 8.

dz when y1 = 2 and dy = 0 1 d2 z 2

dy 1

>0

z is min at (1, 2)



Minimum distance = (1  0)2  (2  3)2 = 13.

1 1 =

2.

We can find the answer through option as the sum of weight of packet taken from trucks is 1022870 gm and its unit digit is 0. The truck that have heavier bags have unit digit 0. So, the truck have lighter bags in which the sum of weight of bags must have unit digit 0. So, according to option D. i.e. truck no. 2, 8 Track 2 have 21 bags and total weight = 21 × 999 gm = .......8 gm Truck have 27 bags and total weight = 27 × 999 = 128 × 999 gm = ......2 gm So, the unit digit of the weight contain by truck 2, 8 together is 0. 1

14.

 cos( x ) cos([2x]) dx 0

1/ 2

1

cos(  x ) cos 0 dx +

=



=



=

 sin x      

0 1/ 2

0



1/ 2

cos(  x ) cos  dx

1

cos(  x ) dx – 1/ 2



1/ 2

cos(  x ) dx 1

– 0

 sin x      

1/ 2

1  1  =   – 0     

=

2 . 


1

cos(nx )x 9 dx   cos(nx)x10  10   + n n   0 1

15.

IN =



0

=

1 1   10   sin(nx )x 9  9 sin(nx )x 8 dx     0+ n   n n   0 n 0  

= 

1  10  9   sin(nx )x 8 dx  2   n 0 





1  10 !   sin(nx )dx  = 10   n 0 



= 0 as Denom  

16.

y = x2 & y = 1 – x2 Point of intersections of graphs x2 = 1 – x2 2x2 = 1

1 x=±

2

 1 1  1 1 ,  and  ,  .  Point of intersections =   2 2  2 2 Area under graph : 1/ 2

x

=

2

 (1  x )2

1 / 2

1/ 2

=



1 /

1/ 2

 3  2x  1 =  2x  x   1/  3 2 2

2 =2×

=2 17.

6 2

26 6 2



1 2 4

=

2

3 2

=

2 2 . 3

      a  3 i  4k and b = 5 j  12k

  2 2 2 a  (3)2  4   5 and b  (5)  12   13

       Therefore, a vector which bisects the angle is 13 3 i  4k + 5 ( 5 j  12k ) = 39 i  25 j  8k .






19.

Let M  2 x1 .3 x 2 .5 x 3 ...... , N  2 y1 .3 y 2 .5 y 3 ...... xi & yi  w x1

 5 x3 – 1  2 y1 – 1  3 y 2 – 1        5 – 1  ....... =  2 – 1  3 – 1  .........     

 3 x2 – 1    3 –1    

 d =  d   22 ––11

d/m

d/N



x2   1  x1      – 1   1  – 1  2   3      ....  1 / 2 – 1  1 / 3 – 1        1/ d   1  y1   1  y 2  d/M    1/ d =   2  – 1   3  – 1         .... d/N  1 / 2 – 1  1 / 3 – 1       

 

(2 x1 – 1)(3 x 2 – 1).....



2 x1 3 x 2 .....

20.

m C0 

(2

y1

m C1 n 

– 1)(3

y2

– 1)......

=

N  1. M

2 y1 3 y 2

 m C2 n 2    mC m n m  (1 + n)m

only one element from A

PHYSICS 27.

Sphere is hollow so potential inside sphere will be same as that on surface.

28.

Heat supplied Q = dU + W PV = RT PdV = RdT

(at constant pressure)

PdV R Q = CVdT + PdV dT =

PdV + PdV R Work done at constant pressue, W = PdV Q = CV

Q  W

CV

PdV  PdV R PdV

Q CV  1 W R

(For diatomic gos, CV =

5 R) 2

Q 5R  1 W 2R Q 7  W 2



W 2  Q 7


29.

1 1  1  R 2  2  For lyman series  2  1 1 1   1  R 2  2  For balmer series  3  2 





(

1 1 94   ) 4 9 36

3/4 3 9 27 =   5 / 36 1 5 5

 5    27

A

B

q

q

30.

q q and 2 2

charge divides

Than, on touching

q sphere to q 2

Charge divides

q / 2  q 3q  2 4

force between

3q q R  4 2

f=

f=

unchanged

K 3q2 8R 2 3 kq2 3  F × 8 R2 8

31.

Intially block enters in the magnetic field rate of change of flux will be constant so constant current will induce, when it moves inside the magnetic field there is no change in magnetic flux, current I = 0, when it use the filed the rate of the change in flux will be again constant between in decreasing order so constant current will induced on opposite.

32.

No change in moment of inertia

34. E

–q

A

O

+q

Electric field at each point of OA obtained  to it and opposite to direction of dipole moment.


38.

a

m

Total force in upward direction m × (g + a) because mass m is stationary on inclined plane and whole system is accelerated with acceletration a in upward direction. 39.

Force of positve charge = Electric force + Magnetic force F = (qE + qVB) This force is in upward direction so no any particle will through the hole.

40.

Potential energy at H height = Kinetic energy at the lowest point of circular path. mgH =

1 mv2 2

To complete the circular motion minmum velocity at lowest point will be V = mgH =

H=

5gR

1 m (5gR) 2

5 R 2

CHEMISTRY 41.

According to Graham’s law

1 Rate of diffusion 

Molar mass

due to highest molar mass of CO2 rate of diffusion is slowest.

42.

Moles of H2 =

3 4 , Moles of O2 = 2 32

Kinetic energy of n moles of gas =

3 nRT 2

3 n1RT 2 Kinetic energy of hydrogen so, = 3 Kinetic energy of oxygen n 2RT 2 n1 = n 2 =

3/2 4 / 32

= 12 : 1 CLASS-XII_STREAM-SB+2_PAGE # 182

44.

ClF3  sp3d hybridisation, but due to presence of two lp on central atom Cl, according to VSEPR theory shape is ‘T’

45.

HCO3– + H+  H2CO3 Bronsted base HCO3–  H+ + CO32– Bronsted acid

47.

Isoelectronic means same no. of electrons CO has 6 + 8 = 14 electrons CN– has 6 + 7 + 1 = 14 electrons

48.

CO2, due to sp hybridisation bond angle = 180º

49.

Diethyl ether, because it is inert towards the Grignard reagent

50.

CH3 – CH2 – CH2 – CHO + CH3 – CH2 – CH2 – MgBr

H

H +

CH3 – CH2 – CH2 – C – CH2 – CH2 – CH3

H3O

CH3 – CH2 – CH2 – C – CH2 – CH2 – CH3

OMgBr 51.

[Ni (PPh3)2 Cl2]  [NiCl4]2–

52.



OH Achiral Secondary alcohol

dsp2 hybridisation, because PPh3 is strong ligand hence pairing of electrons takes place sp3 hybridisation, because Cl– is weak ligand hence pairing of electrons is not takes place

16H++ 2MnO4– + 5COO– COO–

53.

Suppose equilibrium constant for the following reaction is K1 N2 + 3H2

2NH3

; K1 =

[NH3 ]2

-------- (i)

[N2 ][H2 ]3

and equilibrium constant for the following reaction is K2 [NH3 ] 1 3 N2 + H2 NH3 ; K2 = [N ]1/ 2 [H ]3 / 2 2 2 2 2 square the both side of equation (ii)

K22 =

[NH3 ]2 [N2 ][H2 ]3

K22 = k1 K2 =

-------- (ii)

[by equation (i)

k1

K2 = 41 K2 = 6.4

[ K1 = 41]


54.

Suppose reaction is 2A  Product according to rate law Rate R = k [A]2

[ A ]12 R1 or R = [ A ]22 2 R1 [A]12 according to question R = 2 2  [A]1     2 

[ A ]1    [ A]2   2  

R1  R =4 2 R2 =

55.

R1 4

HCO3–  H+ + CO32– Conjugate base NH3  H+ + NH2– Conjugate base

56.

57.

(II) & (IV) Because both have close system of conjugated double bond and follow Huckel’s (4n+2)  e– rule.

••

N H p of N takes part in resonance with conjugated double bonds, so it is not easily available on N for the protonation.

N ••

p is not taken part in resonance so easily available for the protonation.

O ••

N H due to high E.N. of O availability of p on N decreases.

••

N H No extra effect, so availability of p on N increases.


58.

Gauche conformer. because angle between same groups is 60º

59.

Suppose initial quantity = No after 75% completion of the reaction remaining quantity N = No ×

T=

 No  2.303  log   N  K

T=

 No  2.303 log  N / 4  K  o 

T=

1.386 -------- (i) K

 T1/2 =

K=

25 N = o 100 4

0.693 K

0.693 -------- (ii) 30

so by equation (i) and (ii) T=

1.386 0.693 / 30

T = 60 min.

60.

Concentration of H+ ions in H2SO4 solution = 2 × 0.1 = 0.2 M So no. of moles of H+ ions in 10 ml H2SO4 solution =

0.2  10 = 0.002 1000

concentration of OH– ions in 0.1 KOH solution = 1 × 0.1 = 0.1 M So no. of moles of OH ions in 10 ml KOH –

solution =

0.1 10 1000

= 0.001 after mixing remaining moles of H+ ions

= 0.002 – 0.001 = 0.001

+

so concentration of H ions in mixture of solutions =

0.001 × 1000 = 0.05 M 10  10



p(x) = a0 + a1x + ........+ anxn p(0) = 7 a0 = 7 p(1) = a0 + a1 + a2 + ............+ an = 9 p(– 1) = a0 – a1 + a2 ........... = 1 p(2) = a0 + 2a1 + 4a2 + .......... = 13 p(– 2) = a0 – 2a1 + 4a2 ........... = – 15 p(1) + p(– 1) = 2[a0 + a2 + .........] = 10 a0 + a2 + a4 = 5 .....(1) 7 + a2 + a4 = 5 a2 + a4 = – 2 .....(2) p(2) + p(– 2) = 13 – 15 2(a0 + 4a2 + .........) = – 2 a0 + 4a2 + 16a4 = – 1 4a2 + 16a4 = – 8 .....(2) p(3) = 25 a0 + 3a1 + 9a2 + ........ = 25 a0 + 3a1 + 9a2 + 27a3 + 81a4 + 243a5 = 25 From (1) and (2)

.....(3)

4a2 + 4a4 = – 8 4a2 + 16a4 = – 8 – – + a4 = 0 and a2 = – 2  Smallest possible value of n is 3. 2

82.

 (a – b )

2

0 

abc

a  1  ab

|a – b| < c ...... (1) |b – c| < a ...... (2) |c – a| < b ...... (3) Squaring and adding a2 + b2 + c2 < 2ab + 2bc + 2ca

[Triangle inequalities]

 a2 2  ab So, b  [1,2). 83.

y = | x 3 | 4 – 5 When, x < – 1 y=|3–x–4|–5 y=–x–1–5 y=–x–6 When –1  x < 3 y=|3–x–4|–5 = | – x – 1| – 5 =x+1–5 =x–4 When, 3  x < 7 y = |x – 7| – 5 y=7–x–5 y=2–x

12 –6

–

3

–1

7

O

x

–

6 x

2– x

–4

x

–

12

–5


When, x  7 y = | x – 7| – 5 =x–7–5 = x – 12. Area bounded region 1

=



3

(  x  6 )dx



+

6

7

( x  4 )dx

+

1



12

( 2  x )dx +

3

1

3

 ( x  12)dx 7

7

12

  x2   x2    x2  x2  2 x  6 x  4 x   12 x         = + + + 2   2   6  2  1   2  7 3

49  9  1    36  9  1     144   49   6 –   36  +   12  –   4  + 14   144  –   84   – 6   +  =  2  2  2   2  2  2     2   2 

=

11 15 9 21 3 119 – 18 – – – – – 72 + 2 2 2 2 2 2

=

130 48 – – 90 2 2

= 65 – 24 – 90 = 49 sq. unit.

A 84.

b

a

b

C

a

D

cos =

B

b

b a2  b2  a2 = 2 a 2ab

cos(180º – ) =

– cos =

cos =

....(i)

b2  b2  a2 2b 2

2b 2  a 2 2b 2

a 2  2b 2

....(ii)

2b 2

From (i) & (ii) a 2  2b 2 2b

2

=

b 2a

3

a a   –2   –1=0 b b


a =x b

Let

x3 – 2x – 1 = 0 (x + 1)(x2 – x –1) = 0 x = – 1 or x =

1  1  4( 1)(1) 2



1 5 2 x cannot be negative



x=

=

85.

1 ( 5  1) 2

an =

1  a n1 2

a1 =

1  a0 2

a1 =

1  cos  2

2 cos 2 a1 =

2

a1 = cos

 2

1  a1 2

a2 =

=

 2

1  cos

 2

2

2 cos 2 =

 4

2 = cos

 22

an = cos

 2n

lim 4n(1 – a ) n

n 

   lim 4n 1  cos  2n  

n 




lim 4n × 2sin2

2 n 1

n 



2 2n  2 sin2

2 

lim

n 

 2n 1



 ×

2

n 1



 2

n 1

2n 1

sin2

2

n 1

 n 1

2 lim  ×   2  2n 1 2n 1

n 

2 . 2

= 86.

f(x) = (sin x)sinx f(x) = e sin x log sin x Minimum value of sinxlog(sinx) is 0. Maximum value of (sin x)sinx is e0 = 1. Maximum value of sinxlog(sinx) is – Minimum value of (sin x)sinx is 10

87.

1 . e

1 e

e .

1

 x dx 1

=

10

log x

1

= log10 = 2.303

1 1 + ....+ 9 2 = 1 + 0.5 + 0.33 + 0.25 + 0.20 + 0.16 + 0.14 + 0.12 + 0.11  2.81

B=1+

1 1 1 + ....+ +  2.81 – 1 + 0.1 =1.91 9 10 2 So, C < A < B and B – A  0.51, A – C = 2.303 – 1.91  0.40. So, B – A > A – C.

C=

r

A

88.

D

B

r

60º

r

r

60º

r

r

C

As we want the distance between two point is at least r. Now when the point A, B are at distance r.Then the angle made arc BA is 60º. Now as chord AB come closer to centre the length of chord AB is increased that is it is greater than r and the angle is also increases i.e. from 60º to 180º and now when chord AB move way from centre then the length of chord AB decreases , when chord AB reach CD the length of AB equal to r and the angle chang from 180º to 60º CLASS-XII_STREAM-SB+2_PAGE # 189

So, the angle required for desired conditions = 2(180 – 60) = 240 Total angle for all around the circle = 360º So, required propability = 89.

240 2 = 360 3

Let aN be pth digit no.

4(5P –1) 4(5P – 1)
2 × 10P–1 < aN 

So

8 (10P–1) 9

 log aN  10 P –1(ln10 )  (5P – 1) ln(2  10P1 )   = = 10 P –1 – 5P ln 5 ln(5P – 1)  logN min =

ln10 = log10 5 ln 5

8 P 10 ln 10  (5P 1 – 1) 8 ln (10P – 1) 9  log aN  9   = = 8 (10 P – 1) 5P 1 ln 5  log N max ln(5P 1 – 1) 9 = log10 5 By sandwitch therom limit is log10 5 .

90.

| i – j |

S=

1 i  j  n

(n – 1)n 2

= |1–1| + |1 – 2| + -------- |1–n| = |2 – 2| + -------- |2 – n| =

n – 2)(n – 1) ....... 2

N

S=

(n – r )(n – r  1) 2 r 1



= n  1C3

PHYSICS 91.

Mass of sphere of radius r

m=

Xcm =

Mr

   M      4 R3    3 

3

R3

M  0  m(R  r ) Mm

Mr 3 Xcm = 3 (R  r ) R

 4 3   r M   3   m  4 / 3R3     

3   M  R M   R3  


r 3 (R  r ) r 3 (R  r )  =  R3  r 3  (R  r )(R 2  rR  r 2 )  R3  3   R 

92.

 Xcm =

r3 2

R  rR  r 2

For plano convex lens

1 1  (  1) FL R

FL =

R  1

Refraction through lens

1  1 1   V R u This v must be centre of mirror –

1 1 1   R  R  u    r  1

 u = 

93.

In cyclic process u = 0 u =W = Area of loop = (P1 – P2) (V2 – V1)

94.

On comparing both the figures

R 

(R  x )(6R) (R  x )  6R x2 + xR – 6R2 = 0 x=

95.

x=

 x  R 2  ( 4  6R2 ) 2

x=

R  5R  2R 2

Torque about point A

f

2  2  MR 2  MR    mg sin R 5  

O A

5g sin  = 7R Applying Newton’s law, Ma = Mgsin – Mgcos gcos gsin– = tan–1

5g sin  7

gcos

2g sin  7

7 2

Velocity of sound =

RT M

T Velocity of sound 


96.

F H

mg

a/2

A

For height greater than H Balancing torque about point A

a 2 For height less than H F = gh From (1) and (2) F × H = mg

=

.........(1) .........(2)

a 2H

N

97.

2

mV /r mg   MV 2 mv 2 cos   mg sin    mg cos   sin  r r     V 2  V2  g   g      sin r  = cos  r   V 2  rg tan = V 2  rg 98.

S1

4

3

5

S2 Path difference = 5 – 4 = 1 m For constructive interference n1= 1m For destructive interference For n = 1, 1= 1m, 2= 2m

(2n  1) 2  1m 2 99.

(B*) Block on track to reaches the finish line first


100.

(1st law of thermodynamics )

Q = u + PV mv = u + 1.01 ×105 (

1 1  3) 1/ 1.8 10

u  20.8 × 105 J kg–1

CHEMISTRY 101.

Millimoles of NH4OH in 10 ml of 0.1 M NH4OH solution = 0.1 × 10 = 1 millimoles of NH4Cl in 10 ml of 1M NH4Cl solution = 1 × 10 = 10 [Salt] pOH = pKb + log [Base] 6 = pKb + log

10 / 10  10 1/ 10  10

[ pOH = 14 – pH = 14 – 8 = 6]

6 = pKb + log10 pKb = 6 – 1 pKb = 5

102.

2C4H10 + 13O2  8CO2 + 10H2O

H = – 2658 kJ/mol

Butane present in cylinder = 11.6 kg = 11600 g =

11600 mol 58

 Combustion of 1 mol of C4H10 gives = 2658 kJ energy  Combustion of 11600/58 mol of C4H10 gives =

2658  11600 kJ 58

= 531600 kJ energy energy consumes in 1 day = 15000 kJ so 531600 kJ energy will be consumed in =

103.

531600  35 days 15000

W = d × V = 0.879 × 50 = 43.95 , Kf = 5.12 Kg kg mol–1 = 5120K g mol–1 Tf =

Kf wo mW

5.51 – 5.03 =

5120  0 .643 m  43 .95

m =

3292.16 21.096

m = 156 g mol–1


104.

Zn + 2Ag+  Zn2+ + 2Ag (0.04M)

(0.28M)

[ Zn2  ][ Ag]2 0.059 log [ Ag ]2 [ Zn] n

Ecell

= Eº –

Ecell

= 2.57 –

2

0.28  1 0.059 log (0.04)2  1 2

{ [Ag] = [Zn] = 1}

by solving the equation we get  2.50 V Ecell

105.

2+

2–

Co + Con. HCl

CoCl4

HOH(excess) [Co(H2O)6]2+ Pink

106.

ln k =

–11067 + 31.33 T

2.303 log k =

–11067 + 31.33 T

suppose k1 and T1 are rate constant and temperature in case-I and k2 and T2 are rate constant and temperature in case-II So,

2.303 log k1 = –

11067 + 31.33 --------- (1) T1

2.303 log k2 = –

11067 + 31.33 --------- (2) T2

by subracting equation (2) from equation (1)

k1 2.303 log k = –11067 2

1 1  –  T T 2  1

k1 2.303 log 2k = –11067 1

2.303 × (–0.3010) =

 1 1   – 298 T 2  

[ k2 = 2k1]

–11067 11067  298 T2

by solving the above equation we get T2  303.7 K T2  31ºC


107.

K1 =

[CuCl3Br 2 – ][Cl – ] [CuCl4

2–

][Br – ]

------- (i)

2–

K2 =

[CuCl2Br2 ][Cl– ] [CuCl3Br 2 – ][Br – ]

------- (ii)

2–

K3 =

K4 =

[CuClBr3 ][Cl– ] 2–

[CuCl2Br2 ][Br – ] [CuBr4

2–

------- (iii)

][Cl– ]

2–

[CuClBr3 ][Br – ]

------- (iv)

equilibrium constant for given equation 2–

K=

[CuClBr3 ][Cl– ]3 [CuCl4

2–

][Br – ]3

------- (v)

by multiplying the right hand side of equation (i), (ii) and (iii) we get right hand side of equation (v) it means K = K1K2K3

OH 108.

OH

ONa NaOH

Br2 in CS2 –HBr

–H2O

Br (X)

Br –NaI Me–I OMe

Br (Y)

109.

234  206 90 Th  82 Pb

 x 24He  y –01e By comparing mass no. 234 = 206 + 4x + oy x=7 By comparing nuclear charge 90 = 82 + 2x 1y y = 82 + 2 × 7 – 90 y=6 O II O

110.

Me

OH

AlCl3

fries rearrangement

OH

O II Me

I2/NaOH

O II OH

Haloform reaction


HINTS & SOLUTIONS (YEAR-2010) ANSWER KEY Ques. 1 Ans. C Ques. 16 D Ans. Ques. 31 Ans. D Ques. 46 Ans. B Ques. 61 Ans. A Ques. 76 Ans. D Ques. 91 D Ans. Ques. 106 C Ans.

2

3

4

5

6

7

8

9

10 B 25 D 40

11 B 26 A 41

12 C 27

13 A 28

14 C 29

15 D 30

B 17 D 32

D 18 D 33

B 19 C 34

C 20 A 35

B 21 A 36

B 22 A 37

A 23 B 38

C 24 C 39

D 42

A 43

B 44

D 45

B 47

C 48

B 49

B 50

A 51

A 52

D 53

A 54

C 55

B 56

C 57

D 58

A 59

B 60

D 62

A 63

A 64

A 65

C 66

B 67

C 68

D 69

B 70

A 71

B 72

A 73

C 74

A 75

A 77

C 78

A 79

C 80

C 81

A 82

A 83

C 84

D 85

A 86

A 87

C 88

B 89

B 90

C 92 A 107 B

A 93 B 108 A

A 94 B 109 B

D 95 A 110 D

C 96 A 111 B

B 97 B 112 C

B 98 D 113 D

C 99 B 114 B

B 100 C 115 B

C 101

A 102 A 117 A

B 103 A 118 C

D 104 D 119 A

C 105 D 120 A

116 D


1.

 1 0 I=   0 1 0 i  A=    i 0 – 1 0  A2 =    0 – 1  0 – i A3 =   – i 0 

A4 = 



A4n = 

 + A + A2 + A3 = 0 = + A + A2 + A3 + ............+ A2010 = ( + A + A2 + A3) + A4 (+ A + A2 + A3) ..... + A2008 (+ A + A2) 0 i  0 i  =0+   =  .  i 0  i 0


2.

Let a, ar, ar2 are sides. Case 1 :

r>1 a + ar > ar2



r2 – r – 1 < 0



 1 5    r   2 ,   

Case 2 : Case 3 :



r = 1 equilateral triangle r<1 2 ar + ar > a

r2 + r – 1 > 0



 – 1 5 – 1 5    , r    2 2  

 1 5 2  5   r lies in the interval  2 , 2  .   3.

Number of diagonals ing through centre = 6  Number of rectangles = 6C2 = 15

4.

22, 3 + 4, 3 + 42, 5 –  – 2 –1–

3 i , 1 + 2 3 i , 1 – 2 3 i, 6

where  =

– 1 3 i 2 – 1 – 3 i , = 2 2

1 + 2 3 i , 1 – 2 3 i are conjugate of each other..  5.

Least possible degree = 5

Tangent at (2, 1) y.2 = 2 (x + 1)  y=x+1  x–y+1 =0 equation to circle (x – 1)2 + (y – 2)2 + (x – y + 1) = 0 putting x = – 1 4 + (y – 1)2 + (– y) = 0  y2 – 4y – y + 8 = 0 D=0  (+ 4)2 = 32 y=



(  4)  D 4

y= 

4 2 0 =±2 2 2


6.

BKN = 6 BN 1  BKN 1  AN 2       AB 3  APB 9  BN 1

9 ×6 1

APB =



=

ABC = 108.



7.

x2 a2



h=

a2

y2

=1, a>b>0

b2

x y , k= 3 3

(3h)2

8.

54 1



(3k )2 b2

= 1 which is ellipse

7 6 cos   1sin   1

1

LH.S is less than or equal 1 while R.H.S is greater or equal to 1 possible when

9.



cos  = 1



 = 0, 2

and sin  = 0

(1 + tan 1º) (1+ tan 2º) .........(1+ tan 45º) (1 + tan A) (1 + tan B) = 2

if A + B = 45º

Ans = 223

10.

f is differentiable function . f (a) f(b) > 0



f is increasing at x = a and b

or

f is decreasing at x = a and b

minimum number of roots of f (a) = 0 in (a, b) is 2,


11.

When x is less than 49 then f(x) has negative value. Which is not possible. when x > 49 then f(x) = 49 (x – 41)48 + 41 (x – 49)40 + 2009 (x – 2009)2008 so sing of f(x) does not change f(x) > 0 non real except one positive root.



12.

f (x) is given smooth curve hence differentiable i- given domain By first derivative test x = a, c is point of local maxima x = b is point of local minima 3

13.

 f ( x )dx  4 i

shaded area = Trapezium Area – Area under curve

5 7   (3 – 1) 2 2 =  –4=2 2

e

14.

n =

n

 (log x ) .dx 1

n = x (log x )

n

e

e 1

–

 i

n(log x )n –1 .x dx x

n = (e – 0) – n n–1


n + n n–1 = e 2011 + 2011 2010 = e 

15.

100 + 100 99 = e

x2 + y2  100



Area bounded = 100 

sin (x + y) > 0 Hence common area of region bounded = 50  as in half the region sin (x + y) > 0 and in other half sin (x + y) < 0

16.

n(S) = 7C3 = 35

n () = 7 + 7 + 7 = 21

î 17.

ˆj

P () =

21 3  35 5

kˆ

2 –1 1   = î (–2  3) – ˆj( 4 )  kˆ (–6) = u v = 0 –3 2

   (u  v ).w ,

53

greatest when x2 + y2 = 1 fmax. = 17

Ans. (D)

18.

Number of ways = 4P2 × 4P2 × 10 = 1440

19.

1 + x2 + x4 + .........+ x2010 =

1 – ( x 2 )1006 1– x2

=

1 – ( x1006 ) 2 (1  x )(1 – x )

=

(1  x1006 ) (1 – x1006 ) (1 – x ) (1  x )

= (1 + x1006)

(1 – x 503 ) (1  x 503 ) (1 – x ) (1  x )

= (1 + x1006) (1 + x + x2 + ...............+ x502) (1 – x + x2 – x3 + ............ + x502) 

n – 1 = 502

n = 503


20.

an = 3an–1 + 1 = 3(3an–2 + 1) + 1 = 32 an–2 + 3 + 1 = 32 (3an–3 + 1) + 3 + 1 = 33 an–3 + 32 + 3 + 1 aN = 3n a0 + 3N–1 + 3N–2 + .... 1 Method - I a2010 = (1 + 3 + 32 + 33 + 34) + ....... + 32009 = (1 + 3 + 32 + 33 + 34)+ 35 (1 + 3 + 32 + 33 + 34)+ 310 (1 + 3 + 32 + 34 + 35) +.....+32005 (1+ 3 + 32 + 33 + 34) each (1 + 3 + 32 + 33 + 34) is divisble by Method-II a2010 =

32010 – 1 35402 – 1 (243)402 – 1 (242  1)402 – 1    2 2 2 2

= 11 k (k  I) Hence remainder is 0

PHYSICS

21.

F – 1mg – 2(m + M)g = M.

1mg m

F = 1mg + 2(M + m)g + 1Mg = 1g(M + m) + 2g(m + M) = (1g + 2g) (M + m)

22.

 m1m2  1 T = 2  m  m  k 2   1 we can use concept of reduced mass.

23.

2T cos = – ma ;

If

x<<<<



=

4T m



2T m

x 2

=a

  x2 4

 2

2T x =a m    2



a= 

T = 2

m . 4T

4Tx m

f=

1 4T 2 m  CLASS-XII_STREAM-SB+2_PAGE # 201

24.

AU  = mean distance between Sun and Earth and

Time period = 125 year T 2  r3 (1 yr)2  (1AU)3  ra  0.4  (125)2     2 

3

 ra  0.4  25 =    2 



ra = 50 – 0.4 = 49.6. 25.

26.

Zero

KQr

E  r inside

R3 KQ r

E

2

1 r2

outside.

27.

There is no net effect of outside charge.

29.

tanC =

r h

from snell's law sinC =

1  h

r= 30.

2

 1

2 > 3 1sin i1 = 2 × sin r 1 < 2

from diagram

1 3  2 2

For condition of TIR 2 > 3 and 1 > 3 So, 2 > 1 > 3 . 31.

A =

A0 2

t/T

A=

120000 2

4

=

120  100  10 = 7500 2 2 2 2

(D).


32.

2 A = g

g A



f=

= 33.

1 2

1 2

g A 10 2.5  10  2

= 3.18

dQ = 0 –dU = dw

   nRT n 2a n 2a    CdT  2 dV  =   2  dV   V  V  nb V    nRT .dV V  nb

–CdT =



C nR

dT

dV



C nT = n(V – nb) + C nR

 T   V  nb

– n (V – nb) –

C nT = K nR

–[n(V – nb) + n TC/nR] = K ln(V nb) TC/nR = – K (V – nb) TC/nR = constant

Answer is (C) option.

35.

v0 = 39.6 kmph = 39.6 ×

5 = 2.2 × 5 = 11 m/s 18

t1 =

d 330

t2 =

d  11 30 d 11 30 = – 330 330 330

t = t1 – t2 = 1 time interval = 30 – 1 = 29 second


36.

RT M

v=

v0  vH

37.

MH  M0

2 1  32 4

d = 0.1 mm D = 1m  = 600 nm given IR = 75% of maximum = 75% of 4 = 30 R = 3 =  +  + 2    cos  =

 3

y=

1 600  10 9 D = 6d 6  0.1 10  3



x =

   dy = × = 2 3 6 D

y = 1 mm



38.

m1v 0  0 vcom = m  m 1 2



xi =

m1(0)  m 2 () m1  m 2

xcm = xi + vcom t

m1v 0 t m2 () xcm = m  m  m  m 1 2 1 2

39.

According to given information tan  =

a r

A

mv

a

B

C

r = a cot

qBr qBa cot  mv 2 = qvB v = v = m m r

r


CHEMISTRY 41.

[Co(dien)Cl3] dien = NH2– CH2 – CH2 – NH – CH2 – CH2 – NH2 [Co(dien)Cl3] have two geometrical isomers cis and trans. cis form is optically active but trans form is optically inactive.

42.

The ligands which allow back bonding to sufficient extent are called  acid acceptors. Example CO, CN–, NO etc.

43.

O22– : Total electron = 18 ; Bond order = 1. Peroxide (O22–) : (1s)2 (*1s)2 (2s)2 (*2s)2 (2pz)2 (2p2x = 2p2y ) (*2px2 = *2p2y)  NB – N A  10 – 8 Bond order =  = 1.  = 2   2

hc 6.62  10 34  3  108 = = 1.988 × 10–25 J.  1

44.

E=

45.

t1/2 (half-life) is independent of initial concentration so reaction is first order.

..

F F 87.5º

Cl

ClF3

..

46.

87.5º

F nearly 'T' shaped.

F F

47.

R – C – Cl || O

    AlCl3

Friedal craft acylation


48.

>

>

>

>

>

Order of stability of conjugate bases

>

49.

anti conformation is most stable

50.

234 234 0 Th  91 Pa  –1 e 90

51.

Entropy change is positive, but enthalpy does not change.

52.

On increasing temperature, change in activation energy is not significant.

53.

For FCC unit cell number of atoms per unit cell Z = 8[corner] ×

1 1 + 6 [face center] × = 4. 8 2

For body center unit cell number of atoms per unit cell Z = 8[corner] ×

54.

1 + 1 [body center] × 1 = 2. 8

CO2

CO +

H2 O

H2 +

1 O 2 2 1 O 2 2

;

;

K C1 = 9.1 × 10–12

.....(i)

K C2 = 7.1 × 10–12

.....(ii)

Target eq. CO2 + H2

CO + H2O

KC = ?

Target eq. can be deduce by

[Eq. (i) – Eq(ii)]



K C1 9.1 10 12 KC = K = = 1.28. C2 7.1 10 12


55.

R  P t=0

R0

0

t=t

(R0 – x)

x

Rate = –

d[R ] = K[R]1 dt x

t





1 dx = = K[R0 – x] = (R – x ) .dx = K.dt dt 0 0 0  R0  = ln  R – x  = Kt  0  [R] = (R0 – x) = R0e–Kt

56.

PCl3F2 :

57.

(I) & (IV) are enantiomeric pair

I 58.

IV

2NO2 (g)  2NO (g) + O2 (g) NO2 is reactant so its concentration is decrease with time while NO and O2 are product so their concentration increases with time. Formation of moles of NO is double than O2. So

59.

X = NO, Y = O2 and Z = NO2.

– Cyclic – Planar – Complete conjugation – (4n + 2) e– – Tropylium carbocation

60.

Anti addition takes place.


PART-II (2 Mark) MATHEMATICS n

81.

1  1   1/ 2 n/ 2  x  1/ 4  = x 1  3 / 4  2x  2x   

=x

n/2

n

2   1  n  1  1  n n  n   c 0  c 1 3 / 4   c 2  3 / 4   ....  c r  3 / 4 .........  2x   2x   2x   

n n

c0 ,

c1 n c 2 , are in AP 2 22 n n(n – 1) , are in AP 2 8



1,



n=1+



8(n – 1) – n (n – 1) = 0



(n – 1) (8 – n) = 0

but

n = 1 i.e. not posible



n=8

Now

1   1/ 2  x  1/ 4  2x  

n(n – 1) 8



8

r 0

n=8

 1  c r ( x1/ 2 )8 – r  1/ 4   2x  r

8

=

or

r

8

8

=

n=1



 r 0

8

 1  4– cr   x 2

3r 4

required number of = 3 (r = 0, 4, 8)

82.

Roots of 2x2 + 2x = 1 = 0 are ,

1 i 2

It will satisfies the (x + 1)n – r = 0 n



 – 1 i   1 = r   2 



 – 1 i   1 = r,   2 

n



n i  1    4   e  2  

rR

n

   =r 


  i e 4  

n

   will be real only if n is multiple of 4. 



For n = 4000



 r =    2

 1 

83.

y = ax + b

4000

=

4

x2 y2  1 a b

and

fig. 1 and 3 incorrect Now for a > 0 and 84.

1 1000

(a < 0 b the equation of line y = ax + b) b < 0 fig. 2 is correct

Area of cyclic quadratic and is maximum only if area of ABC and area of ADC is maximum  Area of ABC is maximum if ABC is equilateral . 

AC = 2R sin 60º =

3 R. 3 3 3 2 ( 3R)2 = R 4 4

maximum Area of ABC =

maximum area of ADC =

=

=

area of ABCD =

1 (AC) (DM) 2

1 ( 3R) 3R cot 60º   2  2  

3R 2 4

3 3 2 3 2 R  R  3R 2 4 4

85.

f(x) is monotonic increasing in [a, ) 

f(x) > 0  x  [a,)




f(x) is monotonic decreasing in (, b]

Now

g(x) =

g(x) =



f(x) < 0  x  (, b]

f (x) x xf( x ) – f ( x ) x2

h(x) = xf(x) – f(x) 

h(x) = f(x) + xf(x) – f(x)



h(x) = xf < 0



h(x) is M.D.

{ x > 0 and f(x) < 0 concavity}

So, at x = a it may be M.I. followed by M.D. till x = b

or

M.D. through out a But can’t be M.I. followed by M.D. So (B) is correct answer.

86.

Let radius of base of given cone is r and length as h



V1 =

1 2 r h 3

........(1)

Let radius of base 1 and R1 and height H. For the cane with apex O V2 =

Now

V=

1 R2H 3

r h  R h–H

..........(2)

H R  h r



i–



 R H = h 1 –  r 

R 1  R 2 h  1 –  r  3 


 dv 1 2R 2   h  2R –  dR 3 r  = 0 



R=

2r 3

For maximum volume R–

=

87.

2r 3

V2 =



R 1  R 2h  1 –  r  3 

1 4r 2  2 4 2 h 1 –  =  h 3 9  3 81

f(x) = 1 + f(x)

.......(2)\





dy =1+y dx



n (1 + y) = x + c



y = ex + c – 1



y = ex – 1



f(x) = ex – 1

V2 4  V1 27

..... (1)

x

Now

f(x) = x +

 f (t) dt 0

ex – 1 = x + et – (t)0x ex – 1 = x + ex – x –  

= 1



f(x) = ex – 1

for

f(x) = 0



x=0



ex – 1 = 0

88.

A=

89.

1  . 2 2

+

1  2 .  2 2 2

Let P is origin and p.v. of A, B, C are a, b, c respectively Given

    PA  2 PB  3 PC  0



   a  2b  3c  0

...... (i)


Area of ABC =

1       |a × b + b × c + c × a | 2      – 3c – a   – 3c  a  1     = | a ×   +   × c + c × a | 2 2 2    

Area of APC =



90.

=

1 3  1      | (c × a) – 0 – 0 + (c × a) + c × a| 2 2 2

=

3   |c × a | 2

...... (ii)

1   |c × a | 2 ABC 3 = APC 1

6m + 2m+n. 3m + 2n = 332 = 4 × 83 

2m–2. 3m + 2m+n–2 3m + 2n–2 = 83

.......(1)

83 is prime number so that this is possible only if m=2 32 + 2n 32 + 2n–2 = 83



by (1)



2n. 9 + 2n–2 = 74



2n–2 (36 + 1) = 74



2n–2 = 21



n=3

Hence. m2 + mn + n2 = 4 + 6 + 9 = 19

PHYSICS

91.

 V02 r 2 V02 r 4 V02     ........upto 10 th   h  g g   g =

V02 [1 + r2 + r4 + .............upto 10th] – h g

=

1  (r 2 )10  2h h 2   1  r 

=

 1  r 20  2h h . 2   1  r 


92.

GMm r

=

2

T=

mv 2  r

2r V

T = 2

2r T=

GM r

v=

r = 2

GM

r3 GM

r3 GM

geff = g – 2 Re cos2 f(g – 2 Re cos260º) = (g – 2Re cos20) f  fg – g = 2R   1 4 

g[f – 1] =

2R [f – 4] 4

4GM( f  1) 2

 (f  4)

= R3

T 2GM( f  1) = R3 2 ( f  4 )

93.

3 a = 2r × 2

Gm



R

2

2R (f – 4) 4

(f – 1) =

T 2 GM ( f  1) 

2 (f  4)

M = 4 R 3 3

= R3

3( f  4)

=

Kq2

2Kq2

q2

q2

4T 2G( f  1)

cos30º

= 3r kr = 2

kq2 q2

cos30º

2 3 k.r.3r2 = 4 .q2 × 0 2  q2 3   r =   12  k 0  

94.

1/ 3

Consider the infinite ladder circuit shown below :






Let the equivalent impedence of circuit be Z So, Z = wL + Z 1 Z C ZX C Z = L + Z  X = L + 1 Z C C On solving we get, Z=

LC  2L2C 2  4LC 2C

For Z to be purely inductive 2L2 C 2  4LC  0 = 0

2 =

Ans.

LC

95.

 1  1  V1  = R

  1  v1 R





v1 

R  1

1  1   Vf  (2R  V1 ) =  R 1 1     Vf R  R   2R     1    1    1 (  1)  = 1   R    2 R (2R  2R  R ) (  1)  (  2)       1   2      =  R  (  2 )   R    2 

96.

V(x) =

G= 



vf =

R(  2) 2(  1)

x 1 2 kx – V0cos   a 2

dV dx

x 1 = kx + V0 sin   . a a

Since, x <<<< a

EQ = kx +

=

T=

So,

sin

x ~– a

x a

V0   V0 x . = K  2  x a a a  

V    K  20  = a  

Ka 2  V0 a2

 K = m2

ma 2 2 = 2 . ka 2  v 0  CLASS-XII_STREAM-SB+2_PAGE # 214

97.

Q = U + W

nR(T2  T1 ) 1  Heat capacity in this process ndt = nCvdt +

= Cv + 98.

nR . 1 

W = Area of ABC

1 W = + (P2 –P1) (V2 –V1) 2 {It is clockwise} Heat released at C–A : It is at constant volume So, Q = nCVdT Q=n

P2 P1

C A

B

V2

V1

3 R (TA – T TC) 2

3 (P1 V2 – P2V2) 2 Heat released at A – B (Constant pressure) Q1 =

Q2 = ndT =

5 nR (TB – T TA) 2

5 (P V – P1V2) 2 1 1 Heat absorbed at B–C Q = du + W QBC + Q1 + Q2 = W QBC = W + Q1 + Q2 Q2 =

=

1 5 3 (P2 – P1)(V2–V1) – (P1V1 –P1V2) – (P1V2 –P2V2) 2 2 2

=

1 [(P2–P1)(V2–V1) – 5P1V1 + 5P1V2 – 3P1V2+ 3P2V2 ] 2

=

1 [P V – P2V1 – P1V2 + P1V1 – 5P1V1 + 5 P1V2 – 3P1V2 + 3P2V2] 2 2 2

=

1 [– 4P1V1 + 4P2V2 + P1V2 – P2V1] 2

Q = 2 [(P2V2 – P1V1) +

99.

du = 0

V

V

1 (P V – P2V1] 2 1 2



1   Kqq  2 mv 2   0 = 0 + 0   2    d 

 1 q2  . d =  4 2 0 mv 

 q2  =  4 0mv 2 


100.

N0  Initial nucleon N at t = 0 N0  

Add. at a constant rate (N – C) =  N

k



dt =



dN dt

dN 

  N  C 

N0

0

t= 

1 n( N  C)N N0 

1  N  C  t =   n N  C   0 

N  C e–t = N  C 0



N – C = e–t (N0 – C) N=

=

C C –(t) C e  t  (N0 – C) = + N0e–t – .e    

C [1 – e–(t)] + N0 e–dt . 

CHEMISTRY 101.

H2C2O4.2H2O (GMM = 126) Molarity (M) =

2.52  1000 = 0.2M 126  100

M1V1 = M2V2 0.2 × 10 = M2 × 500 M2 = 4 × 10–3. valence factor of H2C2O4.2H2O = 2. Normality = [M] × V.F. = 2 × [4 × 10–3] = 8 × 10–3 N Amount of oxalic acid (mg/mL) =

4  10 3  126  103  mg    = 0.504.  mL  103

Note : Options are not match with solution.

H

102.

Br 

 

HBr





+ CH3OH


103.

104.

A  B Hº = 7.5 kJ/mole

;

Gº = Hº – TSº

at equilibrium Gº = 0

Sº = 25 J/mole.

0 = 7.5 × 103 – T(25) T=

105.

7.5  10 3 = 300 K 25

[Mg2+] = 10–2 M K SP(Mg( OH )

2)

= [Mg2+] [OH–]2 = 1.0 × 10–12

= [OH–]2 =

1.0  10 12

10 2 [OH–] = 1 × 10–5 pOH = 5 ;

106.

= 10–10 pH = 9

For FCC unit cell 4r =

2a

 4  141 .4   = 400 pm a =  2   Volume of unit cell a3 = (400)3 = 6.4 × 107 pm3.

107.

Formula of cyclic silicate : [SinO3n]2n–

; Cyclic silicates : (SiO32–)n or (SiO3)n2n–


108.

Target eq., Given that

Target eq. :

109.

K3[Fe(CN)6] 3+x–6=0 Fe = 3d6 4s2 26 26

2B (s) + 3H2(g)  B2H6(g) H2O ()  H2O (g) H10 = 44 kJ 2B + 3/2 O2 (g)  B2O3(s) H20 = – 1273 kJ B2H6(g) + 3 O2 (g)  B2O3 (s) + 3 H2O (g) H30 = – 2035 kJ H2(g) + 1/2 O2 (g)  H2O () H40 = 286 kJ Eq.(ii) – Eq.(iii) + 3[Eq.(iv)] + 3 [Eq.(i)] = – 1273 – [– 2035] + 3[–286] + 3[44] = 36 KJ/mole.

.....(i) .....(ii) .....(iii) .....(iv)

x=+3

Fe3+  d5 with S.L.



t2g2,2,1, eg0,0

CFSE = –2.0 0 + 2P ~ –2.0 0

Number of unpaired electron = 1. So,

110.

=

n(n  2) =

3 B.M

Tf = Kf × molality  8  1000  = 0.925 = 1.85 ×    M  92  M = 173.9 ~ 174 Closest answer = 160.

*****



1

2

3

4

5

6

7

8

9

10

11

12

13

14

15

Ans.

C

C

D

A

A

C

A

B

C

B

C

C

B

B

B

Que s.

16

17

18

19

20

21

22

23

24

25

26

27

28

29

30

Ans.

D

C

A

B

C

A

B

A

A

B

D

B

C

C

C

Que s.

31

32

33

34

35

36

37

38

39

40

41

42

43

44

45

Ans.

D

A

A

B

D

A

D

A

B

A

C

A

A

C

B

Que s.

46

47

48

49

50

51

52

53

54

55

56

57

58

59

60

Ans.

D

A

D

B

C

B

A

D

C

B

C

A

D

C

C

Que s.

61

62

63

64

65

66

67

68

69

70

71

72

73

74

75

Ans.

B

C

D

A

A

B

B

D

C

C

D

D

D

A

A

Que s.

76

77

78

79

80

81

82

83

84

85

86

87

88

89

90

Ans.

A

B

D

C

B

B

C

C

B

A

B

D

A

B

C

92 C

93 D

94 D

95 C

96 B

97 A

98 B

99 D

100 A

101 B

102 C

103 B

104 A

105 C

107 B

108 D

109 B

110 A

111 D

112 C

113 D

114 C

115 B

116 A

117 C

118 D

119 A

120 A

Que s. 91 B Ans. Que s. 106 C Ans.


1.

(C) Let loga b = x C=x+

1 , x

then,

logba =

For a, b > 1

1 x logab > 1

By A.M.  G.M C= x

1 2 x

2.

(C) Hint n is multiple of = 4

3.

(D) Given P2 = P For n = 2 (I + P)2 = I + P2 + 2PI = I + P + 2P = I + 3P = I + (22 – 1) P

4.

(A) f ' (x) = 3x2 + 2ax + b D = 4a2 – 4(3b) = 4(a2 – 3b) If a2 – 2b < 0 then a2 – 3b < 0 So, D is negative, so f(x) has one real and two imaginary roots.


5.

1 1   1 x2 =    x 2 x   2x sin xy  1 or  –1  sin xy = 1 or –1 For x = –1 or 1 –sin y = 1 or sin y = –1

(A) sin (xy) =

  y =  2n   , n  I 2  (x, y) on pair of straight line.

6.

x y 1 1 (C) 4 0 1  18 2 0 12 1



7.

|–12x – y(4) + 48| = 36 (3x + y – 12)2 = 92 = 81

(A)

m1 m2 = –1  4t  6  6   6  2      = –1  2t  1  3   3  1   4t    (2) = –1  4t = –t + 2  2t  4 

 t=

2 5

 1 38  then C is   ,   5 5  2

 16  8     5   5

BC =

=

Area = xy = 8.

(B) h > 2a h>3

2

=

320 25

8 5 8  5 5 8 5

 2 5 = 16 here

4a = 6, 2a = 3


9.

(C) f(x) = cos5x + A cos 4x + Bcos3x + Ccos2x + Dcosx + E f( – x) = – cos5x + Acos4x – B cos3x + C cos2x – D cos x + E f (x) – f( – x) = 2 [A cos4x + C cos2x + E) for x = 0 f(0) – f() = 2(1 + B + D) ...... (i) at x =

at x =

  4   3   = 2 | –1 + B cos , f  – f + D cos | ...... (ii) 5 5 5 5 5    

 2   6  1  B cos     D cos 5   5  

 2   3  2  – f  =2 , f 5  5   5 

f( + x) = – cos5x + A cos4x – B cos5x + C cos2x – D cosx + E f ( – x) = f ( + x)

..... (iii)

...... (iv)

     9     2    8     3    7   T = f(0) – f() –  f  5   f  5   +  f  5   f  5   –  f  5   f  5   +               

  4   6     f    f     5   5 

 2   8   =f  from equation iind f  5    5    9   f  = f 3    3   3   7   4   6  = f  and f   = f  f  5   5   5   5 

     4     2    3   So T = f(0) – f() – 2  f  5  – f  5     f  5  – f  5             from equation 1, 2 & 3 T = 2(1 + B + D) – 2 [–1 + B cos

3  6 2 + Dcos | + 2 [ 1 + B cos + D cos ] 5 5 5 5

Will depend on B and D only 10.

(B) 3sinA + 4cosB = 6 ...... (i) 3 cosA + 4sinB = 1 .......(ii) Squaring and adding , we get 9 + 16 + 24 sin(A+B) = 37

1 2 1 sin( – C) = 2 sin(A+B) =

sin C =

11.

(C) [x] > 1  [x]  2 x2 But x cannot integers  Possible domain = (2,3)

12.

(C) f(0) + f(0) +

(2011) +

1 2

f " (0 ) f (n –1)(0) + ......+ 2! (n – 1)!

n(2011)n –1 n(n – 1)(2011)n – 2  n(n – 1)...2! (2011)'  ,  ......    n – 1! 1! 2!  

= nC02011 + nC12011n–1 + .... + nCn–1(2011) + nCn (2011)0 – 1 = (2011 + 1)n – 1 = 2012n – 1


13.

(B)

Point where slope of tangent is 1 of the curve y = ex is y = ex = 1 x=0  (0,1) and for y = lnx point is (1,0)  required minimum distance is 8

14.

(B)

2

4

3

f ( x ) dx =

2

2

f ( x ) dx 

3 f ( x) dx

8

+ ...... +

7 f ( x ) dx

= 2 + 3 + 2 + 5 + 3 + 7 = 22

e cos x

1

15.

(B) I = 2012



0e

cos x

 e cos x

dx

e  cos x   cos x  e cos x 0 e 1

again I = 2012

 dx  

1

 dx

2I = 2012

I = 1006



16.

0

 1  (D) lim  2 n 4 n – 1  n

4n – r 2

n

1

1 = 2 1

=

1

 0

 r  2n 1–    2n 

2

dx

 0

  2 2  4n – n  1

2

lim 

n  r  1

4n 2 – 4

 .......... .. 

1

 = nlim  r  1

=

1

1–

x2 4

dx 4 – x2 1

 1 x   =  sin     2  0  CLASS-XII_STREAM-SB+2_PAGE # 222

17.

(C) E1 : A wins when A takes out 3 & 6 and B takes out 8 & 9 1 1 1 × = 3 3 9 E2 : A wins when A takes out 5 & 6 and B takes out any two no. 1 1 P (E2) = × 1 = 3 3

P (E1) =

Required probability = 18.

      (A) a  b  b  c  c  a      ab bc  0     ab  cb  0  a  c  b  0     a  c  b  a  c  b      (bc ) = ca a  c   c = c  a      a  c    c  a     a  c   a  c   0   (1 + ) ( a  c ) = 0    From (ii) a  c  b    abc  0 

........(i)

........(ii)



 = –1

   abc  0 centroid of ABC = 3

 n

19.

1 1 4 + = 9 3 9

2 (B)  r  r  1 r !



r 1



Tr = [(r + 1) (r – 1) – (r – 2)] r! Tr = (r – 1) (r + 1)! – (r – 2) (r!) but r = 1, 2, .............., n Tr = 1 + (n + 1)! (n – 1) 20.

1, x  A (C) If f (x , A  B) =  0, x  A Case (i) x  A, x  B then, f(x, A) + f(x, B) – f(x, A) f(x, B) = 1 + 0 – 0 = 1 Case (ii) x  A, x  B f(x, A) + f(x, B) – f(x, A) f(x, B) = 0 + 1 – 0 = 1 Case (iii) x  A, x  B f(x, A) + f(x, B) – f(x, A) f(x, B) = 0 + 0 – 0 = 0 Case (iv) x  A and x  B f(x, A) + f(x, B) – f(x, A) f(x, B) = 1 + 1 – 1 = 1 CLASS-XII_STREAM-SB+2_PAGE # 223

f(x, A B) = f(x, A) + f(x, B) – f(x, A) f(x, B)

 Aliter

1, x  A  B If f (x , A) =  0, x  A  B Option A is rejected because x  (AB) ,then f(x,A) + f(x,B) = 1+ 1 = 2 which can’t be attained by f.

Option B is rejected baecuse if x  A  B , then f(x,A) + f(x,B) –1 = 0 + 0 –1 = –1 which can’t be attained by f . Also option D is rejected because if x  A , then f(x,A) + |f(x,A) – f(x,B)| = 1 + |1 – 0| = 2 Which is also unattainble

PHYSICS 21.

(A) acceleration of cabin is g relative acceleration of A and B w.r.t. cabin is zero so both A and B continue to be exactly at rest relative to the cabin.

22.

(B)  k  k k = mg  

23.

(A) v =  A 2  r 2 from (i) and (ii)

W

v2 2 A 2



mg 

a2 4 A 2

 W = 2mg

...(i)

a = 2r

...(ii)

1

It is the equation of ellipse 24.

(A) Work done in this process is zero so that (Ui = Uf)

25.

(B) Using mole conservation 2n = n1 + n2 ...(i) PV = n1RT ...(ii) PV = n2R(2T) ...(iii) n1 = 2n2 2n = 2n2 + n2  2n 

n2 =  3   

26.

(D)

;

M2 f 2n / 3 2   M2i n 3

dQ dT  KA dt dx

K1A( T  100 ) K 2 A  (0  T )  

K1(T – 100) = K2(–T) K1T – K1 × 100 = –K2T T(K1 + K2 ) = K1 × 100  385  100   = 88°C  435 

T=  27.

(B) P1- T= constant


   T1  P     1 T   P   2  2

 233K     T 

7/5

 1

(7 / 5)1

 0.28     1 

233

T=

(0.28 )2 / 5

 466 J

So that we will required in addition an air conditionaer to cool the air injected into the cabin

28.

 v 

(C) f1 =  v  u  f0    v u

f2 =  v f0  

...(i) ...(ii)

f1  v 2   f2  v 2  u2     v2    f1 = f2  2 2  v  u  

f2 > f1 29.

(C)  = 0 (cos2)4 3  = 0 ×   4

4

 = 30% of 0

30.

 nv 

(C) f =  2    f’ =

n  v' 2

f v  v     f ' v '  1.04 v 

f’ = (1.04)(450) = 468 Hz 31.

(D)  ×  1 1  2  2

f11 = f22 f1  2 f1    2 f2 1  2f1 1 1 2  2 1 1 1  2 2

 2 = 2 1 = 2.0 mm 32.

(A) V = V0 cost CLASS-XII_STREAM-SB+2_PAGE # 225

Vrms =

V0 2

V0 = 220 2 Volt  = 50 × 2 rad/sec = 100 rad/sec. v = 220 2 Cos(100t) 33.

(A) Current through R1 is maximum so power dissipates at R1 is maximum.

34.

(B) Momentum is conserved during collision so P  P' . But some part of kinetic energy is stored in excited state So K’ < K

35.

(D) Potential difference across inductor during current growth is v = v0e–t/

36.

(A) Neutron is chargeless so graph is (1) Proton is positive charge so graph is (2) and for e– graph is (3)

37.

(D)   R 9    1



1



8

1 1   3R    R 1      2 4  4   1  1 1   5R  5R  R      3  4 9   9  4  36

1 1 1   1 2 3

38.

(A)

P  P'

AT 4 T 4 A  2

4

P 4 P'

P’ =

P 4

V i

V. i

P

 h     2 i e 

39.

(B) RH =

40.

(A) From given graph highest precision that means sharpness is maximum in graph (I)



2

i



2


CHEMISTRY 41.

(C) 2,2,2 2,2 Ni(CO)4 Ni : [Ar] 3d8 4s2 = t 2g e g

CO is a strong filed ligand

Cr (H2O)6+2 Cr+2 : [Ar] 3d4 45°

octahedral complex : d2sp3 42.

(A) Ag+ + 2CN–  [Ag(CN)2]– 2[Ag(CN)2]– + Zn+2  [Zn(CN)4]–2 + 2Ag+

43.

(A) Ideal mixture H = 0 V = 0

44.

(C)  = CRT

45.

(B) Energy distribution at two different temperatures

46.

(D)

1 0.01  2 = 0.001 = 10

The radical formed as intermediate

or

is unstable , where as in all other

cases resonance stabilised radicals are formed. 47.

(A)

The given compound is chiral and the possible stereoisomers are enatiomer to each other. 48.

(D) According to second law of thermodynamics, for spontaniety

:

SUNIV > 0


49.

(B) r1 = 30 nm

r2 = 10 nm

2  surface area  3   = 4r = 3 volume   r 4r 3

(3 / r2 ) 30 (3 / r1 ) = 10 = 3 50.

(C) The reaction is example of Pinacole – Pinacolone reaction with classical ring expanssion rearrangement.

51.

(B) H3PO2 a mild reducing agent is used to remove the diazonium group with H.

52.

(A)

0NaOAc + 0HCl – 0NaCl = 0AOAC 53.

(D)

BRAGG’S Law 2dsin = n d = , n = 1

1   = 30° 2 angle of incidence = 60° sin =

54.

(C) Zn  Zn+2 + 2e– 2mol 2mol 4mol G° = –nFE°cell = – 4 × 96500 × 1.1 = – 424.6 kJ/mol

55.

(B) 2NaBH4 + 2  2Na + B2H6 + H2 

56.

(C) [Mn(CN)6]–4 & [(Mn/Br)4]–2 Mn+2 in both Mn+2 : [Ar]3d5 4s With CN– : With Br– : Magnetic moments :

1(1  2) BM &

5 (5  2) BM


57.

(A) Half life of zero order reactions  initial concentration of the reactant.

58.

(D)

59.

(C)

60.

(C) Due to intramolecular H-bonding boiling point of (ii) is less than (iii) but greater than (i).


82.

(B) AK = I B = 0 |AB| = 0

|A|K = 1 |B| = 0

(C) Let 12 + 22 + .......+ (2n–1)2 = x and c =

n

 (4r 2 – 4r  1) = 3 (4n

2

– 1)

12  2 2  ......  (2n)2 – x < 1.01 x 2n(2n  1) ( 4n  1) < 2.01 6x n (2n  1) ( 4n  1) n 3 ( 4n 2 – 1) < 2.01 3



( 4n  1) < 2.01 (2n – 1)

 4n + 1 < (2.01) (2n – 1)  4n + 1 < 4.02 n – 2.01 3.01 < 0.02 n

301
(C) A - d , A, A + d 3A = –a A =



–a 3



– a3 a3 ab  – +a=0 27 9 3

2a3 ab a– =0 27 3

a = 0 or b 2a2 +1= 3 27

k  2h2 =  – 1 3   27 parabola whose vertex on y axis.

2x2 = 9 (y - 3) 


84.

(B) Let line y = mx + 5 x2 y2  1 16 9

ellipse

c2 = a2m2 + b2 25 = 16m2 + 9 m2 = 1 m = ±1

tangent condition

For positive slope m = 1 85.

(A) A ={R|cos2(sin) = cos2 (cos)} A = {R| sin = n ± cos} A = {R| sin ± cos = n} A = {R| sin ± cos = 0} sin ± cos[  2 , 2 ] B {R| cos (sin) sin(cos) = 0} by A B {R| cos (± cos) sin(cos) = 0} B {R|

1 sin(2cos) = 0} 2

2cos = p, p  cos =

p 2

p = 0 only cos = 0 O = 2k±

 2

so AB  86.

(B) f'(x) = 3x2 + 2ax + b f'(1) = 0  3 + 2a + b = 0  1 3 2a f '    = 0   b = 0 3 9 3    1 – 2a + 3b = 0 by (i) + (ii) we get 4 + 4b = 0  b = –1 f(2) = 8 + 4a + 2b + c = 0  8 – 4 – 2 + c = 0 c = –2  f(x) = x3 – x2 – x – 2 1

1

87.

0

.........(ii)

1

2

 f ( x ) dx  2  (  x  2) dx 1

.........(i)

 x3    =  2 3  2x  = – 14  0 3

(D) f(x) = x12 – x9 + x4 – x + 1 for x = 1 , f(1) = 1 x = 0 , f(0) = 1 for x > 1  f(x) = x9(x3–1) + x(x3–1) + 1 = positive for 0 < x < 1  f(x) = x12 – x4 (x5 –1) + (1 – x) = positive for x < 0  f(x) = x12 + x4 – x (1 + x8) + 1 = positve f(x) is always positive


88.

(A) In

1 2

= 

0

1 (1  x )n xn dx dx +  n! 1 n!

x n1 = n  1!

2

1

1 2

 (1  x )n1     n  1 !  1  0 2

n1

 1   2 In = 2 . n  1!

  1 2    2  In = 2.     2! n1  

3   1     2   .......... .......   3!  

 1 = 2 e 2  1   

89.

 1 = 2 e 3 2 

(B) |x + y| + |x – y| = 4

x2 + y2 – 4x – 6y = (x – 2)2 + (y – 3)2 – 13 A point on square ABCD which is maximum distance from (2, 3) is C(–2, –2) So, required maximum value is 28. Now,

90.

(C) Let AM and GM be ab and ba respectively AM = 10a + b GM = 10b + a

xy = 10a + b, xy = (10b + a)2 2 (x – y)2 = (x + y)2 – 4xy = 4[10b + a]2 – 4[10b + a]2 2 (x – y) = 22 × 18 (a + b) (a – b) (x – y)2 = 11 × 9 × 4 (a + b) (a – b) this should be a perfect square of an integer, only possible case when a + b = 11 a–b=1 a = 6, b = 5 x + y = 2 (10 × 6 + 5) = 130


PHYSICS 91.

92.

(B) Electric field due to uniformly charged isolated sphere (In volume) Case (a) r < R

E=

Case (b) r = R

E=

Case (c) r > R

E=

KQ R3

r

KQ R2 KQ r2

(C) Let gravitational acceleration due to mass of planet is g and due to uniformly charge on surface of planet electric field is E and which is constant because path of projectile is parabolic. All projetile have same condition I. Range for uncharged particle Range = L = II.

u2 sin 2 g

...(1)

Range for a particle of mass m and charge q L

u2 sin 2

Range = 2  qE g

...(2)

m

From eq. (1) and (2) L u2 sin 2  2  qE   g 1   mg 



L L  2 1  qE mg

 1

qE =2 mg

qE

 mg  1 III.

93.

(D)

Range for a particle of mass m and charge 2q u2 sin 2 u2 sin 2 L L    Range = 2qE   1  2 3 2 qE g  g1  m mg  


Tangential acceleration at  angle is at = g cos From energy conservation between horizontal and  position. PEi + KEi = PEf + KEf 0 + 0 = –mg sin +

1 mv2 2

v2 = 2gsin Centripetal acceleration at  angle is ac =

v2 = 2gsin 

Total acceleration at  angle is a2t  ac2  g2 cos 2   ( 2g sin )2

a= 94.

= g 3 sin2   1

(D) n = 1 mole initial temperature = T Initial volume = V Final volume = 2V Internal energy V = aV3 3 2

nCVT = aV3

 n RT = aV3

PV = nRT

From ideal gas equation

3 PV = aV3 2

P=

2aV 2 3

Work done by gas during this process is 2v

W=



2v

Pdv 

v

2a = 3



2av 2 dv 3

v

2v 2

 v dv v

2v

2a  v 3  14av 3 7 2av 2 7 = 3  3   9  3 . 3 . v  3 PV   v

W= 95.

7 RT 3

(C) VA = VC = 2m3, PA = PB = 200K Pa, PC = 500 KPa BC is isothermal process PCVC = PBVB VB =

PC VC 500  2 3  PB 200 = 5m

Total workdone by gas during one complete cycle. W = W AB + W BC + W CA = 200 × 103 (5 – 2) + 200 × 103 × 5n = 600 × 103 – 1000 × n

200 +0 500

5 = –300 kJ 2


96.

(B) For total internal reflection 1.5 sinC = 1.2 sin90° 1 .2 4  1 .5 5

sinC =

C = 53.1 maximum incident angle is 53.1 Ans. 97.

(A) Let linear charge density on wire is  and mass of the particle is m & charge on particle is q. From energy conservation between  and 2 distance. Loss in potential energy = gain in kinetic energy –2kq n

 1 2 1 2 = m( 2u)  mu 2 2 2

–2kqn2 =

1 1 m2u2  mu2 2 2

1 mu 2 2

2kqn2 =

...(i)

conservation between  and 4 distance. Loss in potential energy = gain in kinetic energy –2kq n

 1 2 1 2 = mv  mu 2 2 4e

4kqn2 =

1 1 mv 2  mu2 2 2

From eq (i) mu2 = v= 98.

1 1 mv 2  mu2 2 2 3u

(B)

As loop in pulled to right, there free downward magenetic field is decrease. So, induce emf induced as to oppose the decrement of magnetic field.

99.

Therefore directions of the induced current in the loop is clock wise and force on left side of loop is toward left and force on right side of loop is towards right (D)

V1r1  V2r2 I = r r  (r  r )R 12 2 2 Case – I

V1 = 2V

Case – II V2 = 4V from case (I and Case II)

V2 = 0V

I = 3mA

V1 = 0

I = 4mA


Case – III

 3 r1      2 r2  V2 = 10V

V1 = 10V

................. (i) I=?

10(r1  r2 ) I = r r  (r  r )R 12 1 2

................. (ii)

from (i) and (ii) I = 25 mA 100.

(A)

x2 a2



y2

=1

b2

Different w.r.t.

2x dx 2y dy  =0 ................. (i) a 2 dt b 2 dt ux = u, at (o, b), uy = 0 Different w.r.t. 2x d2 x 2

2



2  dx    a 2  dt 

2

a dt accelerat at (o, b) is

+

2 y d2 y 2

b dt

2



(from eq (i))

2  dy    =0 b 2  dt 

2xo  d2 x  2 2 2bay 2   u  (0) = 0 2  2 2 2 a  dt  a b b2

ay = –

b

u2

a2

CHEMISTRY 101.

(B) XeF6 + 3H2O  XeO3 + 6HF distorted pyramidal octahedral

102.

(C) 5e– + MnO4–

+

8H+  4H2O

C2O4–2  2CO2 + 2e– HCl  103.

1 Cl2 + H+ + e– 2

(B) 2e– + Fe+2  Fe

– 0.44 V

Fe+3 + 3e–  Fe

0.77 V

+3

Fe + 3e  Fe then 104.

–

x=

xV

2  ( 0.44 )  0.77 = – 0.037 V 3

(A)

E13 E12

1   1  2  2 3  1 8/9 32 =  1 = 1  = 3 / 4 27  2  2 2  1 CLASS-XII_STREAM-SB+2_PAGE # 235

105.

(C)

106.

(C) Due to intramolecular hydrogen bonding reaction is favored at ortho position.

107.

(B) The statement optically active (S)–– methoxy acetaldehyde is incorrect.

108.

(D) Ptotal = P°A + (P0B – P0A) XB 300 = 100 + (500 – 100) (1 – XA) 

1 – XA =

YA =

109.

1 2



XA =

1 2

1 1 2 = 6 300

100 

(B) Low spin d6 CFSE = – 4 ×

2 3 0 + 0 × 2 5 5

= – 0.4 0 High Spin d6 CFSE = –6 × 110.

2  = – 2.4 0 (ignoring pairing) 5 0

(A) Total area of droplets = 2.4 × 1018 × 12.5 × 10–16 = 3000 m2 Energy liberated = 3000 × 0.03 = 90 J


HINTS & SOLUTIONS (YEAR-2012) ANSWER KEY Ques.

1

2

3

4

5

6

7

8

9

10

11

12

13

14

15

Ans.

B

D

C

C

B

C

D

D

B

C

B

D

D

B

C

Ques.

16

17

18

19

20

21

22

23

24

25

26

27

28

29

30

Ans.

A

C

A

D

C

B

D

A

C

A

A

A

A

A

A

Ques.

31

32

33

34

35

36

37

38

39

40

41

42

43

44

45

Ans.

A

A

B

A

D

D

A

A

A

C

C

D

A

D

C

Ques.

46

47

48

49

50

51

52

53

54

55

56

57

58

59

60

Ans.

B

C

B

C

A

B

D

B

A

D

B

A

B

A

C

Ques.

61

62

63

64

65

66

67

68

69

70

71

72

73

74

75

Ans.

A

B

D

B

D

B

C

D

D

C

A

A

C

D

B

Ques.

76

77

78

79

80

81

82

83

84

85

86

87

88

89

90

Ans.

B

A

B

B

B

D

*

A

C

C

C

C

A

B

B

Ques.

91

92

93

94

95

96

97

98

99

100

101

102

103

104

105

Ans.

B

A

A

A

A

A

A

D

A

B

D

B

C

A

C

107

108

109

110

111

112

113

114

115

116

117

118

119

120

B

C

D

D

D

B

B

C

B

B

C

A

D

C

Ques. 106 Ans.

B


1.

2.

Number ways are =

720 6! = = 90 ways 8 2!2!2!

x  3  4 x 1 = x + 8 – 6 x 1 + 1 – 2 x  8  6 x 1

2 x  1 – 6 = –2

x  8 – 6 x 1

x  1 –3 = –

x  8  6 x 1 (D) Infinite many solutions.

4.

Equation of chord of T = 0 w.r.t. origin is – 3(x + 0) – p (y + 0) + 17 = 0 3x + py – 17 = 0 By Homoginization 2

 3 x  py   3 x  py   + 17   =0 x2 + y2 – (6x + 2py)   17   17 




For perpendicular coeff of x2 + coeff of y2 = 0

6.3 .9 2p2 p2 + +1– + =0 17 14 17 17 34 – 18 + 9 – p2 = 0 p2 = 25 |p| = 5 1–

  5.



6.

2x3 + ax+ b  2x3 + cx + d (a – c)x  d – b If a – c = 0 & d – b  0 Max. (a – c)2 + (b – d) = 0 + 5 Ans. e(x2 – 2 ex + e2) +  (y2 – 2y + 2) = e

x – e2  y – 2 e

 Ellipse a =

= 1 (a > b)

,b=

e

Req. PS1 + PS2 = 2  2 sin x. cos a = cot a 2 sin x sin a (D) = constant function

7.

f(x) =

8.

tan81 – tan63 – tan27 + tan9 cot9 – cot27 – tan27 – tan9 (cot9 + tan9) – (cot27 + tan 27) 2 cosec18 – 2 cosec54

 4 – 2   5 1 2 8  4

9.

4

   5  1

=4

4–

2x  5  0, 2x + 5  0 0  2x + 5  6 –5  2x  11 11 5  x  2 2

mid point is



5 11  .6 3 2 2 = = 2.2 2 2

10.

(D) at least three for every value of a

11. 

f(x) = 3x2 – 2(a1 + a2 + a3) x + a1a2 + a2a3 + a3a1  0 D0 4(a1 + a2 + a3)2 – 4.3.(a1a2 + a2a3 + a3a1)  0

a12 + a 22 + a 32 – a1a2 – a2a3 – a3a1  0




1 [(a – a2)2 + (a2 – a3)2 + (a3 – a1)2]  0 2 1 a1 = a2 = a3

12.

I1 =

/2

 (sin

x)

2

(sinx)–1 dx

0

I1 =   (cos x )(sin x ) 

I1 =

/2  2  (sin x )  0

2

/2

/2

 0

+

 (cos

2

x ) (sin x )

2 1

dx

0

/2 2 1

dx 



2

 sin x

2 1

0

 dx   

/2

I1 =

2 (I2 – I1)

Here I1 =

 (sin x )

2 1

dx

0

 2 1  1 2   = ×  ( 2  1)  2 ( 2  1)  



= 2 2



Ans.

2012

2012

 (sin x

13.

3

5

 x ) dx +

– 2012

 1. dx

– 2012

= 0 + 2012 – (– 2012) = 4024 Ans.

5

14.

 [x]{x} dx 0

2

1

3



1

0

2

1



5

 .1. {x} dx +  2 {x} dx +  3 {x} dx +  4. {x} dx

= 0.{x}dx +

=0+1

4

1

x . dx + 2 x . dx  3 



0

0

3

1

 0

4

1

xdx  4 .

 xdx 0

1

= (1 + 2 + 3 + 4)

 x . dx 0

1

 x2  = 10 .  2    0 1  = 10 .   0  2   =5


n

15.

Sn

K = k 1

nn  1 is even (n : 1, 2, ............. 49) 2

n = multi fo a 9, 8, .......... 96 Or (+) n is (multi of 4) – 1 3, 7 ............... 99 Total favourable cases 24 + 25 = 49  P[E] =

16.

49 Ans. 99

req. prob =

1 4 6   2  7 10 

=

1 40  42 41   Ans. 2 70 70

17.

—

—

—

—



HA + HB + HC = 2HO

18.

6m + 9n Unit digit of 6m is = 6 Unit digit of 9n will be = 9 or 1 For multiple of 5 unit digit of 9n must be = 9 It occur when n = odd Total number of ordered pair = 50 × 25 = 1250

19.

a2 =

a1  a 3 2

a3 =

a2  a4 2

a1 =

a 2  a 2012 2

a 2011  a1 2 Now a2 + a4 + ......... + a2012 = 3018 .................... (1) 2a2 + 2a4 + .................... + 2a2012 = 6036 a1 + a3 + a3 + a5 + .............. + a2011 + a1 = 6036 2(a1 + a3 + ................. + a2011) = 6036 a1 + a3 + ................ + a2011 = 3018 ..................... (2) By adding (1) and (2) we get a1 + a2 + a3 + ............ + a2012 = 6036 a2012 =

20.

Number of element in B = n So number of subsets of B = 2n


PHYSICS

21.

Since area of PV graph under isothermal curve is greater than area under adiabatic curve So, wi > wa > 0 22.

Q = Q0e–t/RC

Q0

= Q0e–t/RC

2 1

= e–t/RC

2

1

n

2



t RC

t  n 2 RC

t = RC

n2 2



23.

 / R

3

= 

 ML3L3  = T  2   MT  1

R 3  1/ 2

= T–1 T

=1   / R 3

2

K

R 3  K2 

K2 =

R 3  2 T 

=

R 2 RT 2 

K=

R 2 g 


24.

  mr 2   mr 2  mr 2  mr 2 2 2   m ( 2 r )  3   m ( r 2 3 )  2   m( 4r )2       P = 2  2   2   2  =

111 2 mr . 2

25.

Angular momentum conservation about S vp (a – ae) = vA(a + ae) vp vA

vp vA

26.



a  ae a  ae



1 e 1 e

max = (  0   0 )2 = 40 min = (  0   0 ) 2  0 avg = 0 + 0 = 20

27.

0 i B0 = 4 R cos  n B0 = 28.

   sin n  sin n   

0 i  tan 2 R n

Due to the negative work on rod K.E. will decrease and finally become zero.


29.

R=

R=

 A

L r 2

2R = K2rL  2 L r

2

dT dt

= K2rL

dT dt

 2 L dT = dt K 2 2 r 3

30.

RT m = 8RT m

7 = 0.73 58

So, closest ans is 1. 31.

U = mc2

KQ 2  mc 2 2R 9  10 9  (1.6  10 19 )2 = 9.1 × 10–31 × (3 × 108)2 2R

R=

9  10 9  (1.6  10 19 )2 2  9.1 10 31  (3  108 )2

= 1.4 × 10–15 32.

Probability of being found is maximum where speed is minimum. So, x = ±a

33.

Since heat capacity at high temperature is high. So for same amount of heat transfer T is more at lower temperature then at higher temperature. So, final temperature is more than 50ºC .

34.

As F = –x3 – x4 At x = 0, F = 0 Hence particle is in equilibrium.

35.

It is clear that dimension x = ML2T–2  = ML2T–2 and x=L So,

 = MLT–2

 will be dimensionless. 


36.

2gh

From conservation of momentum m 2gh  0 = mv1 + 2mv2 2gh = v1 + 2v2



...........(i)

From equation of e

v 2  v1

=



2gh  0

2gh = v2 – v1

..........(ii)

From (i) & (ii) 2gh

v1 = 

3 Hence after collision maximum height

hmax =

( 2gh / 3 )2 2g

h hmax = g 37.

Let ball is moving with speed v at anytime t hence dv = mg – Kv2 dt On the basis of equation we can rays that speed first increase and become constant when mg = kv2 Hence, Ans. (A)

Hence, m

38.

Resistance of steel =

0 .1 1 = 50  A 500 A

Resistance of copper =

0 .2 1 = 400 A 200 A

Both are connected in series hence heat current in both will be same. So

100  T  1     500 A  39.

T0 =  1     2000 A 



T = 20ºC

From N11 = N22 4 × 1020 =

n(2) n( 2) = N2 40000 20



N2 = 2 × 107


40.

 3  0    cos 2 30 º  0   2 4 2  

Final intensity will be =

3 0 8

CHEMISTRY 41.

According to MOT bond order of O2, F2, O2+ and F2– is respectively 2, 1, 2.5 and 0.5.

42.

BCl3, BeCl2 and CCl4 has zero dipole moment due to symmetric structure where as shape of NCl3 is pyramidal due to presence of lone pair.

43.

The value of n >  and m should have values from –  to +  .

44.

Option “D” Closest, if (–) is change to 

46.

KE = | T.E. |  (KE) = (T.E.) 3 × 8.3 × (T) = 1.6 × 10–19 × 6 × 1023 2

(T) =

2 9.6  10 4 × 3 8 .3

(T) = 7.6 × 103 K i.e. close to 104 K.

47.

48.

t1/2  (a)1 – n 10

( t1/ 2 )I  a1    ( t1/ 2 )II =  a2  0 .2 A0 ( t1/ 2 )II = 2 A 0

(t1/2)II = 0.4 s 49.

25

Mn2+ and 26Fe3+ both has 23 electrons.


50.

51.

52.

In metal carbonyls oxidation state of metal is equal to zero.

53.

At pH = pKa the concenteration of acid and its conjugate base are equal.

54.

It is high spin complex as Cl– is weak field effect ligand. In [CoCl4]2– oxidation state of Co is +2 in which 3 unpaired electrons are present which gives the spin-only magnetic moment equal to 3.83 BM.

55.

K = Ae –Ea / RT ln K =

–Ea + ln A RT

56.

57.

Greater is the bond order shorter will be bond length. In CO32– bond order in between 1 and 2, bond order of CO2 is 2 where as bond order for CO is in between 2 and 3.

58.

2P(g) + 3Cl2(g)

2PCl3(g)

PCl3(g) + Cl2(g)

........(i) PCl5(g)

........(ii)

2P(g) + 5Cl2(g) 2PCl5(g) ........(iii) On multipling equation (ii) by 2 and adding in (i) we obtain equation (iii) Therefore, K3 = K1K22 59.

The reagent given the question has an addition “C” within bracket. (CH3)2CH–CH2–Cl + AlCl3 (CH3)2CH–CH2–Cl - - - - AlCl3  (CH3)2CH–CH2+ + AlCl4– ,2H– shiffting (CH ) C+ (CH3)2CH–CH2+ 1 3 3   

60.

As boron is trivalent impurity it will proudce p-type semiconductor.



D1 = 4a2 – 4b2 > 0  4 (a2 – b2) > 0 a2 – b2 > 0 .......... (1) 2 2 2 D2 = 4b – 4c > 0  b – c2 > 0 ............. (2) now D = 4c2 – 4a2 = 4(c2 – b2 + b2 – a2) = – 4 (b2 – c2 + a2 – b2) = – 4 (D1 + D2) = Negative  Equation have no real root.

82.

(1+x) (1+x2)–1 (1-x)–1 

(1+x)

 r 0







x 2r ( 1)r  x r = r 0



x 2r ( 1)r

r 0

 r 0



x r +  x 2r ( 1)r r 0



 (x

r 1

)

r 0

for Coffi. of x2012 = (-1)0 + (-1)1 + (-1)2 + ....... (-1)1006) + ((-1)0 + (-1)1+ ..... +(-1)1005) =1+0 =1 No answer 83.

4x2 + 9y2 – 8x – 36y + 15 = 0 4(x2 – 2x + 1) + 9(y2 – 4y + 4) – 25 = 0 4(x – 1)2 + 9 (y – 2)2 = 25

( x  1) 2 5   2

2



( y  2) 2 5   3

2

=1

min ( x  1)2  ( y  2)2 + max. x  12  ( y  2)2





2





2

100  225 325 5 5 =   +  = = 36 36 3 2

84.

sinx +

1   cos x = sin2  x   2 4  

sinx +

1 1  cos 2 x   / 2 cosx = 2 2

1 1 1 cosx =  sin2x 2 2 2 2sinx + cosx = 1 + sin2x 2sinx + cosx = 1 + 2 sinx cosx 2sinx (1 – cosx) – 1 (1 – cosx) = 0 (1 – cosx) (2sinx – 1) = 0 sinx +

cos x = 1 or sinx =

1 2

 5 , 6 6  5 sum of roots = 0 + + = 6 6

x=0

x=


85.

P(x) = ax2 + bx + c a  0 P(x) = 2ax + b P(x) = 2a P(0) = c = 1 P(0) = b = –1  P(x) = ax2 – x + 1 P(1) = a – 1 + 1 = a  0 

87.

I=

cos 2 x

 1 a

x

dx , (a > 0)

...(1)

–

I=



cos 2 (– x )



1 a–x

–

I=



a x cos 2 x



1 ax

–

dx

dx

...(2)

(1) + (2) 2I =



cos 2 x(1  a x )



(1  a x )

–

dx

 2 2I = 2 cos x dx

 0



 1  cos 2 x   I=  2  0

1 x  cos 2 – 0  cos 0 0 2



I=

88.



=

1   1 – 1 2

=

 . 2



Consider L=

3

2012 +

3

2013 + ... +

3

3011

R=

3

2013 +

3

2014 + ... +

3

3012

3012

and I =



3 x dx .

2012

89.

P(K) = P (at least K points) x1 + x2 + x3 + ....... x10 = K Coeff xK in (x1 + x2)10 = x10(1 + x)10 coeff xk in (1 + x)10 = 10CK–10 K  10 Now

P(K) >

1 2


10

C 0  10 C1  10 C 2  ......  10 CK 10 10



1 2

2 C0 + 10C1 + ............. + 10CK–10 > 29 1 + 10 + 45 + 120 + 210 + 252 + ...... > 512 10 C0 + 10C1 + 10C2 + 10C3 + 10C4 + 10C5 > 512 K – 10 = 5 K = 15 10

90.

P = f(2) . f(f(3)) f(f(f(4)) f(f(f(5) 3 =    1

  4    5   f    f  f      2     3   5 3 5 3

 1     1    = (3) (3) (f(4)) f(f(5))

3 = (3)    1

5 =9   3

   f    

    6  f  f  f  f         4     

  f  f    

3 2 3 2

 1     1   

  6   f      4 

3    1 2  = (15)  3    1 2  = (15) (5) = 75 375 is multiple of 75.

PHYSICS 91.

H1 = AT4 × 5 = Ms1

A (2T)4 × 5 = msq2 2 2 = 81 2 = 8ºC The temperature of water would increase from 10ºC to a final temperature of 18ºC H2 = 

92.

Mechanical energy of asteroid

GMm 1 mv2 – r 2 0 Rocket will remain band to the solar system its B moving energies negative. =



GMm 1  mv 2  0 r0 2

GMm 1  mv 2 r0 2 GMm 1  m  2 v 02 r0 2 mv02 = =

1 m2v02 2

2.


93.

B = 1 + 2

n2 2n2 n2 7n2    B A 3 A 3 A B 3 =  7. A 94.

3eV  = 2.3 eV Kmax = 3 – 2.3 = 0.7 eV 5 mm  1V 1V  5mm 0.7 V  5 × 0.7 mm = 3.5 mm.

95.

V=

Kq KQ  b c

k( q  Q ) 0 c q+Q=0 q = –Q K KQ ( Q )  V b c

 1 1 KQ    V c b

Q=

bcV 4 0 . (b  c )

96.

4 (sinC) = 1 3

3 tanC =

17

=

R Hh



sinC =



R=

3 7

3 4

(H  h) .


97.

W.S. Bridge RAB = R/2. 98.

From equation 2 4 =  L



=

L = 0.6 m. 2

99.

T + k1x – mg = ma k2x + mg – T = ma T + k1x – mg = k2x + mg – T 2T = (k2 – k1)x + 2mg T=

(k 2  k 1 )x  mg 2

(k 1  k 2 )x 2m If T is 0 then a=

If T is not zero

a1 =

k 1x  mg k 1x g = m m

a2 =

k 2 x  mg k2x g = m m


100.

at = g sin ac = v2/ 

1 mv2 – mg  cos 2 aC = v2/  = 2g cos at = g sin tan = at/ac = g sin/2g cos 0+0=

=

........... (i) ........... (ii) ........... (iii)

tan  2

1  tan     = tan   2 

CHEMISTRY 101.

102.

T A 350000   0.7  G C G G=

350000 = 5,00,000 Ans. 0.7

103.

104.

Phenol on treatment with dil HNO3 gives o–nitrophenol and p-nitrophenol. o–nitrophenol has intra molecular H–bonding hence steam volatile ie P, where as “Q” is p-nitrophenol, Q has inter molecular H-bonding.

105.

Kinetic energy = h – work function KE =

1240 –1 660

KE = 0.878 eV


=

150 Å V

=

150 0.878

 = 13.07 Å = 1.3 × 10–9 m 450

a

106.

2

2

h K 

2

=

(2)2  (2)2  (1)2

=

450 = 150 pm 3

107.

H = 20 kJ/mol U = H – ng RT U = 20 × 103 – 1 × 8.314 × (273 + 60) U = 20 – 2.768 U = 17.2 kJ/mol

108.

As in Ni(CO)4 hybridisation of Ni is sp3 and CO is strong field effect ligand therefore, it is diamangetic.

109.

W = – nRT ln V 1

V2

W = – 3R × 300 ln 10 =

–900  8.314 × 2.3 = – 17.2 kJ/mol. 1000

q = – W = 17.2 kJ/mol. (For isothermal process E = 0). 110.

Rate = K . [X]p [Y]q From expt. 1 and 2, p = 2 and from expt. 1 and 3, q = 3. Therefore, over all order = 5.


(STREAM - SB/SX) : HINTS & SOLUTIONS (YEAR-2013) ANSWER KEY Ques

1

2

3

4

5

6

7

8

9

10

11

12

13

14

15

16

17

18

19

20

Ans

C

A

B

B

B

A

D

D

C

B

B

A

D

A

D

A

A

A

D

C

Ques

21

22

23

24

25

26

27

28

29

30

31

32

33

34

35

36

37

38

39

40

Ans

D

D

A

A

C

B

A

B

D

A

B

B

D

B

D

C

C

A

B

D

Ques

41

42

43

44

45

46

47

48

49

50

51

52

53

54

55

56

57

58

59

60

Ans

C

D

C

C

D

A

B

A

B

C

C

A

A

A

C

C

B

C

C

B

Ques

61

62

63

64

65

66

67

68

69

70

71

72

73

74

75

76

77

78

79

80

Ans

A

D

D

A

D

B

D

B

A

B

C

C

C

D

C

C

C

D

C

D

Ques

81

82

83

84

85

86

87

88

89

90

91

92

93

94

95

96

97

98

99

100

Ans

C

A

C

C

A

A

D

C

A

D

C

A

C

B

A

D

A

D

A

D

Ques

101

102

103

104

105

106

107

108

109

110

111

112

113

114

115

116

117

118

119

120

Ans

A

D

A

D

B

B

C

B

A

B

C

A

D

B

B

A

AC

D

C

B

PART-I : (1 Mark) MATHEMATICS 1.

f(x) = x3 + 3x2 + 3x + 3 = 0 f’(x) = 3x2 + 6x + 3 = 3(x + 1)2  0 f(x) is Increasing function Now f(– 3) = – 6 < 0 f(– 2) = 1 > 0  real root  lies between – 3 and – 2  +  + = – 3; –3<<–2 +–3<++<+–2 +–3<–3<+–2 –1<+<0

2.

log2log2log2log2log2 n < 0 < log2log2log2log2n 16 < n < 216 no. of digits in the binary expansion of 16 is 5 no. of digits in the binary expansion of 216 is 17 so no. of digits in the binary expansion of n lies in 5 to 16

3.

|a + bw + cw2| |a – c + (b – c)w| , for maximum value taking a = 1 , c = –1, b = 1 |a + bw + cw2| = |2 + 2w| = 2|w2| = 2

4.

a 9   ab = 36 4 b using AM  G.M

ab  ab  a + b  12 2


5.

2 g2  c = a 2 f2  c = b Polar coordinates of centre of circle be (rcos, rsin) g = – r cos 

and g2 – f2 =

f = – r sin 



a2  b2 4

r2 cos 2 =

B(1, 1)

a2  b2 4

P

C(19, 1)

Q

6.

P(a, 1)

 a Q  a,   9

A(0, 1) x=a

area of PQC =

1 area of ABC 2

1 a 1 1  (9  a)1     (8  1) 2 9 2 2 

(9 - a)2 = 36  a = 3

7.

Using M - N Rule (1 + 1) cot ( - B) = 1.cot B - cot C 3cotB = cot C tan B =3 tan C

8.

2sin + 3cos = 3 2 3sin + 2cos = 1 ....(2) sum of squares of equation (1) and (2) 4 + 9 + 12 sin ( + ) = 19

....(1)

1  +  = 150° or 30° 2 If  +  = 30°   = 30 -  put in equation (1) and (2) sin ( + ) =

we get 7 sin + 3 3 cos  = 6 2 7 cos  - 3 3 sin  = 2

79 6 3 = .8 < 37 2 cos < cos 30°   > 30° cos  =

  +  30°


9.

lim

x sin( e  x ) f ( x ) 

x

lim

sin( e  x ) x

x

ex

ex

f(x)

= 1 × (0) × M = 0

1

10.

 {(sin t )x

A(t) =

2

 ( 2 cos t )x  sin t } dx

0

sin t 4  cos t  sin t  sin t  cos t 3 3

A(t) =

4 cos t  sin t 3 St. I and IV are False

A’(t) =

11.

Domain of f(g(x)) is R  2 - cosx - cos2x  0 (cos x + 2)(cosx - 1) 0  - 2 cos x 1 xR Domain of g(f(x)) is [-2, 1] 2    cos  2  x  x  2 - x - x2 0 Domain of f(g(x)2) is R  2 - cos2x - cos4x 0 2 2 (cos x + 2)(cos x - 1) 0 -1 cos x  1 xR Domain of g(f3(x) is Domain of g(f(x) i.e., [-2, 1]

12.

Let r be an integer in (– 10, 10) x

Now, LHL = xlim r –

2

[t]

dt

10

8 r h  9   2[ t ] dt  2[ t ] dt  ....  2[ t ] dt  lim = h 0    9 r 1 10 







2 10  2 9  .....  2r 1 1  h = hlim 0





= 2–10 + 2–9 + ... + 2r–1

... (1)

x [t ]

lim

xr

2

dt

10

8 r h  9   2[ t ] dt  2[ t ] dt  ....  2[ t ] dt  lim = h 0    r 9 10 2 





= 2–10 + 2–9 + ..... + 2r–1



... (2)

r

f(r) =

2

[t]

dt

10

= 2–10 + 2–9 + ... + 2r–1 From (1), (2) & (3) f(x) is continous at all integers.

... (3)


n

13.

n 1 r 1

 [r ]{r } dx    r( x  r ) dx

 x2  r  rx  r 2 r 1 





r 1 r

1

n 1

n 1

 (r  1)2  r 2   r.1 2 

 r  r 1

r 1

n 1

 1

r 1

n(n  1)  2013 4



n(n - 1)  4 × 2013



n

2

1 2013  16  1  n    2 4  

1 n(n  1) 2

 r  2   2



32209 1  2 2

least n = 91

14.

1 1    2 .  required area = 2  2    2  4 15.

/4



0

4 2    2 cos x dx  =   8    

Let x = 1 favourable out comes (1, 1), (1, 2) .......(1, n) no. of favourable out comes when x = 1 n  =    1  no. of favourable out comes when x = 1 or y = 1

n  = 2 – 1  1 no. of favourable out comes when x = 2 or y = 2 but x  1, y  1 n  2   1 2 Similarly no. of favourable out comes when x = k or y = k but x, y  {1, 2, ..... k– 1} n  2   1 k  n

So probability =

n

  k   1  1.......n times k 1

n2 

2 n

2

n

n 

1

  k   n k 1


16.

µˆ 

0  x 2  x 3 .......x n 1  ( x1  x n ) µ n

ˆ 2 



1 2 y i  µˆ 2 n

1 2 0  x 22  x 32 ....( x n 1 )2  ( x1  x n )2  µ2 n



ˆ 2   2 



2x 1x n 0 n

so ˆ  

17.

In this case B, C, D are not possible. 18.

Total Integers = 999 - 99 = 900 Total Integers in which all distinct digits

9 9 8

= 648

so 900 - 648 = 252 19.

n + 1, n + 2, ...... n + 18 (A) False, if n = 19 (C) False if n = 15 16 to 33 20, 25, 30  only three multiples of 5 (D) no. of odd integers in Sn = 9 every third odd integer is multiple of 3 so maximum prime no. = 6.

20.

Sum of angles of closed polygon with 10 sides is 8. So maximum number of possible obtuse angle is 7.


PHYSICS 21.

Charge on outer most surface is zero Hence force on q is also ‘0’

22.

5 VAB 

Q=

R 3

5R R 3

V 

5 V 8

5 CV 8

25.

I - t graph is for L-R series circuit.

26.

V=

h  e Hence from graph  = 2eV

h 6  slope  e 1.6  1015 h=

27.

6  1.6  10 19 1.6  10

15

 6.0  10  34

 v  f1 = f 0  v  v  s   v  v s  ~  2v s  v  vs  fR = f1  v   f 0  v  v   1  v  f    s   = 0.990 f

28.

B due to Arc =

 0 i 4r

 0i  1 1   out of the page 8  r R 

29.

In stready state i through capacitor is zero. Hence V across 2k = V across capacitor Vacross 2k =

30.

2  6 = 4V 21

R

Net = F  2  + FR + 2FR = 3.5 FR


 tdN

1

31.

avg life =

 tdN 2

2N0

where dN1 = N e –t dt dN2 =

  – t 3 3 N  e dt



avg. life =





t(N e – t dt 

– t  N e 3 dt 3



2N0 Intergrating we got avg life = 32.

2 2.10   

For adiabatic process PV = C

PV–1 +

dp  v =0 dv

dp dv Hence n = 1

P = –V

34.

 work done = 0 Hence kinetic energy = constant

35.

i = 2r sin i  3 sin r

2cos r =

3 r = 30° & i = 60° Note : But for r = 30° TIR cannot take place at B. 36.

Check dimensionally

37.

Fr = m2r cos 45° where r = Rcos 45° Fr =

m2R 2


38.

PV 10 5  1   40 n = RT 25  300 3 N = 40 × 6.023 × 1023 ~ 24 × 1024 1

1 3 Average sep. =   ~ 1 nm N

39.

1 mv2 + 0 = 0 + 1 2 v2 = 4 or v = 2 m/s

CHEMISTRY 41.

Na+, Mg2+, F— & O2— are isoelectronic(10 electron speceis)

42.

Ct = C0 – kt ; slope = – k

43.

44.

45.

 P = KH.X(g) 1 X(g) = P × K H slope 

1 KH

46. c = 0.03 nc = 0.06 n=2 KC =

0.06 2 0.03



0.06 2 0.03

= 0.12


47.

48. Aceto acetic ester has active methylene group. 49.

BCl3  3H2O  H3BO 3  3HCl 4H3BO 3  Na 2CO3  Na 2B 4 O 7  CO 2  6H2 O

50.

XeF2 ; no. of lp on Xe = 3 XeF4 ; no. of lp on Xe = 2

51.

S =

v2 v2 nRT v 2 ln = nR ln = 2.303 nR log T v1 v1 v 1 = 2.303 × 2 × 8.314 × 1 = 38.3 J/k

52.

53.

54.

55.

U = 0, H = 0


a

56.

n = 2 sin  , dhk  =

2

h  k 2  2

a

d100 = d200 =

2

1  02  02 a 2

2  02  02

=a =

a 2

sin 1 n1 d2 sin  2 = n 2 d1 sin 30 º 2a/2 sin  2 = 1 a

or sin 2 = sin 30º or 2 = 30º

57.

Coordination number = 6 [Co(NH3)4Cl2]Cl  [Co(NH3)4Cl2]+ + Cl–

58.

Fe3+ = 3d5

is WFL so  = 5.92

While NH3 is SFL & pairing of electrons takes place so  = 1.73

59.

Carbocation stability order

So reactivity order is (CH3)3 C–Cl > CH3 – CH2 –CH2 – Cl > 60.

then formula of ionic compound is MY2


BIOLOGY 61.

HbF have more affinity for oxygen and this property help in extracting O2 from mother.

62.

In nucleolus RNA-PI is present.

63.

In GC Pairing 3 H bonds are present.

64.

Entry of T Cause formation of stop codon.

67.

Schizosaccharomyces pombe is called fission yeast.

68.

Grana bear photosystem and stroma have enzymes of dark reaction.

69.

2 alleles of a gene separate in neiosis and Non linked different genes segregate independently.

71.

UV rays damage DNA causing thymine dimerisation.

72.

In gastrulation cell migration occur forming 3 germ layers.

74.

Thinners prevent cardiovascular accidents which can cause stroke.

79.

Heterozygous population = 2pq = .48


Coefficient of highest degree term must be negative becuase if it is positive, then x , y and it is not possible, since f(x)  100. Now, graph will be like y 100

1/2

at least two real roots will be there, & if x 

x

1 , then f(x) < 100, it is not always true, as the graph can be 2

like this also 100

C

1/2

Now, let the highest coefficients, it can have is 49 49 49  1 f    49   2  ... 2 2 2   But the sum cannot be equal to 100.

then,


n

82.

 (ak

3

 bk 2  ck  d)  n 4

k 1

n

a



n

k3  b

k 1



n

k2  c

k 1

n

k

 d  n k 1

2

k 1

n4(12 - 3a) - n3(4b + 6a) - n2(6c + 6b + 3a) - n(6c + 2b + 12d) = 0 12 - 3a = 0, 4b + 6a = 0, 6c + 6b + 3a = 0,  a = 4, b = - 6, c = 4, d = - 1 |a|+|b|+|c|+|d| = 15

83.

(2x - 4)2 = 4x

6c + 2b + 12d = 0

B

C y = 2x - 4

(x - 2) 2 = x x2 - 5x + 4 = 0 x = 1, 4 C(1, -2) B(4, 4)

 AB = AC (   4 )2  16  (  1)2  4

On solving, we get  =

84.



 

9 2



Let AB  b , AC  c 



AM =  b 



AN = m c Let G divides MN in the ratio K : 1 







k c   b b  c So  k 1 3 

k 1  k 1 3

k=

 1  k 1 3

 

1 1  3   AM  GM 

1 1     2

1

2

2       3

... (1)


Now,

area of AMN  area of ABC

  1  b  c 2

= 

1  b c 2

1 1  2    0,1 maximum value of ratio = = 3  Ratio = attain when  = 1 using   3  1 3  1 derivative but  is not 1 becuase M is an interior point.

using

so 85.

4 1  ratio < 9 2

OD = R sin  AB = 2R cos 

D

A

ar( OAB) rOAB = semi  perimeter

B R

R x

O

y

1 1  OD  AB  R sin 2R cos  2 2 = = 2R  AB 2R  2R cos  2 2 rOAB =

R sin  cos  (1  cos )

drOAB (1  cos ) cos 2  sin  cos (  sin )  0 d (1  cos ) 2

at cos  =

86.

5 1 2

f(x) = 1 + x +

x2 x3 x 4   2 6 24

f’(x) = 1 + x +

x2 x3  2 6

x0

2

x >0 2 f’(x) is an increasing fn f’(x) = 0 at x = x0

f”(x) = 1 + x +

x 20 x 30  =0 2 6 x0  (-2, -1)

f’(x0) = 0  1 + x0 + f’(-2) f’(-1) < 0 f(x0) = 1 + x0 +

.... (1)

x 02 x 30 x 04 x 04    0 2 6 24 24

no solution x0


87.

AP = Rsin area of ring = (2 Rsin). Rd  2

Total area required, R =

 2R

2

sin d

0

1 = 1 – cos 2 cos =

1 2

Now, cos =

R RH

1 R  = R = H = 6400 2 R H

n 1

88.

 1

{ x} [ x ] dx  [ x] n r 1



 r 1 r

2

 1

89.



2



r 1



n



n 1

 1

{x}[ x] dx [x]

( x  r )r dx r 1

n

 r(r  1)  n  1 r 1

C 5 [1  2(3 )]

100



{x}[ x] dx  .....  [x]

r 1 r

 ( x  r )r 1     r(r  1)  r r 1 

100

3

n r 1

{ x} [ x] dx [x]

n



{x} [ x ] dx  [ x]

Suppose 1, 2, 3, 4, 5 are

C 5  5!

selected coupons.

1

=

1 20

Place of 1 is fixed Total arrangements of 5 is 2 and

1

5

arrangements of 2, 3, 4, are 2

3

1

4

5

4

2

1

3

5

4

3

1

2

5

}

3 way


90.

Let teams be T1, T2, T3, T4 & T5 Now, we can have 5 teams with the scores of 2 points each matches are (I) T1 T2 (II) T1 T3 (III) T1 T4 (IV) T1 T5 (V) T2 T3 (VII) T2T5 (VIII) T3 T4 (IX) T3 T5 (X) T4 T5

(VI) T2 T4

This score board contradicts, option D  D is not always necssarily true.

PHYSICS 91.

Apply conservation of linear momentum mv = (m + M) v0 2 2 2 mv sinR =  MR  mR   0 5  

2M  5M  R 2 h R  mv  R  0 R 5   (m + M) (h – R)0R 

2M  5M  R 2 5

0

h 10m  7M  R 5m  M

92.

 6  x 3 = (6 + x)tan 1  12  



3=

x   1.4 = x tan 1    12 



1.4 =

(6  x )(6  x ) tan  12 x 12  x  tan  12 CLASS-XII_STREAM-SB+2_PAGE # 268

30 36  x 2  14 12 x  x 2 16x2 – 360x + 36 ×14=0 x=

x=

360 



360x – 30x2 = 36 ×14 –14x2

360 2  4  36  14  16 32

360  312 48   1 .5 32 32

93.

from graph

94.

1  1  2.6 = 13.6 z2  n 2 4 2  1  1 2.6  = 22  n 2 4 2  13.6  4 2.6 1 1 = 2  13.6  4 16 n n = 3 now Energy

E=



13.6Z 2 n2

eV

13.6  4 eV = – 6eV 9

95.

V0 = Vi when no current flow through 1k \ for negative values of Vi i from D1 = 0 (always) i from D2 = 0 upto 3V So from 0 to – 3V Vi = V0 from – 3 to – 4 V V0 = – 3V For positive values of Vi i from D1 = 0 upto 1V i from D2 = 0 (always) Hence 0 to 1V Vi = V0 1 to 4 V V0 = 1 volt Correct graph is (A)

96.

mg sin – Fr = ma

97.

Fr =

2 a mr 2 2 5 r

=

5g 7R  r 

a=

5 g sin  7 R r

 T = 2

7R  r  5r

mgh =

1 1 2 2 m2 + mR2 2 2 2 5 R

mgh =

7 m 2 10

=

10 gh 7


98.

Q1 2 = w12 wTotal = w12 + w31 10 = w12 – 20 Q12 = w12 = 30 J

99.

mgH – 2 mg(d + x) – mgh = 0 h = H – 2  (d + x)

100.

kA 100  T   kA T  0   3w – 3T = T T = 75° C 3

CHEMISTRY

101.

102.

103.

Ans. (A)

104.

n PO  PS  N PO 18 18 760  750   m 108 760 18 10  6m 760 6m = 18 × 76 m = 228


105.

E0Cell = E 0 Cu2 |Cu  E 0 Zn|Zn2 E0Cell = 0.34 + 0.76 = +1.1V  G = – nFE0Cell G = – 2 × 96500 × 1.1 = – 213 V

106.

K1 =

SO3  1 O 2 2 SO 2 

K2 =

SO 2 2 O 2  SO3 2 1

Hence: K2 = K 2 1

1 K12 = K 2

107.

NH OH



4 AgNH3 2  AgNO 3  NH 4 OH  AgOH  

( excess)

Tollen' s reagent  R – CHO  R – COOH

108.

w=

Ei t 96500

w=

100  3  t 96500

m m = dV V w = 10.5 × 80 × 5 × 10–4 = 42 × 10–2 d=

42 × 10–2 =

100  3  t 96500

t = 135 sec. 109.

N1V1 = N2V2 mili eq. of hypo = 0.25 × 100 mili eq. of hypo= 25 eq. of hypo = 0.025 mole of hypo = 0.025 × 1  Vf = 1 weight of hypo = 0.025 × 248 = 6.2 g

110.


BIOLOGY 6

113.

 1 For EcoR I    10000  2.44 sites present 4 4

 1 For RsaI    10000  39.06 4 Using probability rules 114.

Early embryonic cells have yet not committed and differentiated.

115.

(Secondary Endosymbiont) Apicoplast i.e: non photosynthetic plastid is present in plasmodium (protista)

116.

Competition is harmful for both species (A is affected more)

117.

Restriction map as per data is shown above. Thus 1 kb, 3 kb, 4 kb are fragments of gene 118.

Brown fat is abundant in bibernating mammals and new born babies (mammals only)


(STREAM - SB/SX) : HINTS & SOLUTIONS (YEAR-2014)

ANSWER KEY Que s. Ans. Que s. Ans. Que s. Ans. Que s. Ans. Que s. Ans. Que s. Ans. Que s. Ans. Que s. Ans.

1 D 16 B 31 B 46 A 61 C 76 A 91 C 106 C

2 A 17 C 32 C 47 C 62 A 77 A 92 A 107 B

3 C 18 A 33 A 48 A 63 B 78 B 93 B 108 C

4 B 19 D 34 C 49 B 64 D 79 A 94 D 109 A

5 B 20 C 35 B 50 A 65 C 80 B 95 A 110 C

6 D 21 C 36 A 51 A 66 B 81 A 96 C 111 B

7 C 22 C 37 D 52 C 67 D 82 C 97 B 112 D

8 D 23 A 38 C 53 D 68 B 83 B 98 D 113 D

9 B 24 D 39 B 54 B 69 A 84 D 99 B 114 A

10 D 25 B 40 A 55 C 70 D 85 C 100 C 115 C

11 B 26 B 41 D 56 D 71 C 86 101 A 116 D

12 D 27 D 42 A 57 C 72 A 87 A 102 B 117 C

13 C 28 B 43 D 58 A 73 D 88 B 103 D 118 B

14 C 29 C 44 C 59 A 74 C 89 B 104 C 119 A

15 B 30 A 45 C 60 C 75 D 90 105 B 120 A


Let a,a1,a2 _ _ _ _ _ _ _ be the length of siden of square inscribed in circle c0,c1,c2 _ _ _ _ _ and r1, r2, r3, _ _ _ _ _ be radius of circle c1, c2, _ _ _ _ _ then, a 20  2  r12

r1 

2 

2a12 

8 



 Area  Ci  .i

a12 

 2

i0

2.

4  r22 

r2 

2 

4 8  2  2   2 2   1  3 5 ,y  2 3

(A) [ x  y ]  [ x ]  [ y ]

for x 

3 5  3  5  2  3  2  3        3  2 false

 9  10    6   1 1  

(B) [ xy ]  [ x ][ y ]

for x 

15   6   1 1  

 2  1 false

(C) [2x ]  2[7 / 2]

for x = 7/2

[27 / 2 ]  2[7 / 2]

1/2  128    23  

 x  [x]  (D)  y   [ y ] true  x  R  

Ans. (D)

3 5 ,y  2 3

 11  8 false

Ans. (A)


3.

An = (nCr) max case- n = 2k (even)  An = 2kCk  n – 1 = 2k – 1 (odd)  An–1 = 2k–1Ck–1 or 2k–1Ck

An  1.9  A 2  1.9  n –1  1.9  2  2 which is true for every n = even n = 2, 4, 6,.......... 20  10 numbers case- n = (2k – 1) (odd)

2k –1 2k – 2

10  k 4.

Ck 2 C k –1

An  An–1 = 2k–2Ck–1  1.9  A 2 n –1

An = 2k–1Ck or 2k–1Ck–1 1.9 

2k 2k – 1

Ck  2.0 ; C k –1  n  19

2k – 1  2; k  n = 19 only ; 1.9 

1.9 k  2k – 1  2k total = 11

Ans-(C)

Ans-(B)

(0,1) 1 2

,

1 2

5.

( 1 + 2,0 )

Ans - (B) 6.

AS’ = SS’ = SB  e = 1/3 a=3 Ans - (D)

7.

AB=BC Cos 

 1 2 and Sin  2 3 6

2    

cos  

8.

cos   sin

y 1  2a 6

cos 4 x 

 1  2 6

;

a  y

1 1  sin 4 x  2 cos x sin2 x

 (cos2 x  sin2 x )  1 

3 2

Ans- (C)  cos 4 x  sin4 x 

cos2 x  sin2 x sin2 x  cos2 x

1 1  2 sin x cos2 x

1    cos 2 x 1  2 2 0 x x sin  cos  

 cos2x  0

;

2x = t, t  [0,4 ]

sin2 x  cos2 x  1  t (1  t )  1

;

sin2 x (1  sin2 x )  1

 t  t2  1

 t2  t 1 0 D<0 O


9.

x  5 , x2  f (x)   x  2  1 , x  2

f(x) is discontinuous at x = 2 f(x) = 2 

10.

x 5 2 x 2

 x  5  2x  4

x9.

Ans(B)

f(x)  [x]sin x

Ans - (D)

11.

sinx = tan2x sin2x = sinx(1 – sin2x) sin x (sin2x + sin – 1) = 0

sin x  0

sin x 

x  0,  sin1

5 1 2

5 1 5 1 ,   sin1 2 2

Ans(B)

1

12.

I   f(x 2 )dx

dx 

Let x2=t ;

;

0

1

I  1 f(t) 0

13.

dt

 2 t

f'(x) = 1 +f(x)

1 1  2 2 t

1

;

1 2 t dt

dt 1

dt

 f(t) 2   f(t) 2 0

0

t

5I

Ans-(D)

f(0) = 0

f (x) 1 1 f (x)

ln (1 + f(x)) = x then f(x + y) = f(x) + f(y) f(x) f(y) Ans(C) 14.

n



0

cos(2 [ x]{xy }) 1

1

0

0

n  cos(2 [ x ]1dx )  n  | dx  n Ans(C)


15.

16.

5 1 1 5 6 P(B) = 1.  1. . .  ..........  6 6 6 6 7 Ans -(B)

x

=

i

n 3

n

x

=

2 i

n

 xi   –  n   

2



x1  x 2 x1  x 2   x 3  x 4  .........  x n 2 2 = n

y 1  y 2  y 3  ......  y n ˆ = n =

x 1  x 2  ........  x n = n

.......(1)

2

ˆ =

=

1 (2x 12  2x 22  4 x 1x 2 )  x 32  ........  x n2  4 –  n 

=

1 2 ( x1  x 22 )  x 1x 2  x 32  ......  x n2  x  2 i  2 –   n  n 



x 12  x 22  .......  x 22  x i  –  n  n  

   

2 1

n

   

2

.........(2) (B) is correct

Sol- Let side length = 2a

A

 GA=GB=a

1 3

Ans -(C) 18.

Ans -(A)

19.

Ans- (D) 4 2 3 5 4  5 2 5 2  C4     1 1 3 1 2 1 4 2 3  20 + 10×6 + 5[8 + 6] + 10 + 20 = 180 5

y

n

2 1



 cos   

20.

y



ˆ  by (1) and (2)

17.

2

 x1  x 2   x1  x 2       x 32  ......  x n2  2 2     –  n 

C2 2  5 C3

a

3 2

3 2

G

2a 3 a 2

B

D E

C

Ans - (C)


PHYSICS

2L, m

21.

J = mv J.L =

M(2L)2  12

J.L=

ML2  3

Ans.

(C)

V

 J/M

;

Cm

ke= 22.

J2 3J2  = 2J2/m 2m 2m

mgh =

1 1 mv2+ I  2 2 2

mgh =

1 1 MR 2 2 mv2 + . 2 2 2

gh =

=

3J ML

1 1 3 mv2 + mv2 = v2 2 4 4 4gh 3

V2 =

V= Vp  VR 23.

;

V=

4gh 3

2gh

4  2/3 32

Ans.

(C)

V (r) = kr U (r) = mkr

F=-

2 =

du  mk  = dr

24.

Ans. (D)

25.

A1V1 = A2V2

k r k r

;

T=

2 r k

Ans.

(A)

1 2 dv = Const. 2 d = density P1V1 = P2V2 Ans. (B) P+

26.

Ans. (B)

27.

Ans. (D)

28.

PV2 = C PV.V = K nRT.V = K VT = K CLASS-XII_STREAM-SB+2_PAGE # 277

V1T1 = V2T2 V2>V1 T2
P.

K2 C T2

P k T2 P1 T12  T12 T2 2

P1 T12  P2 T2 2

Ans.

(B)

29.

 v o  n’ = n  v  v  s  

;

 v  n” = n  v  v  s  

n' v  v s  n" v  v s

;

n' v  v s 330  30 360     1.2 n" v  v s 330  30 300

30.

(A)

31.

1 1 1   v u f

Ans.

(C)

30 cm

1 1 2   f fm fl 1 1 2 1    f  10 5 f = -5 1 1 1   v 30 5 1 1 1   v 5 30  6  1 5  30 30 v = -6cm 

Ans.

(B)

32. Ans. (D)


Vx

33. vy = ay.t eE l . m u vx = u

vy =

tan  =

34.

eEl mu2

tan 1 4 tan 2

;

 KQ  W1 = q    R 

35.

Q=

tan 1  0.1 4

Ans. (A)

+q

-q

-a/2

a/2

kq2 a

W= (C)

tan 2 =

+q

 KQ kq    ( q) W2 =  a   R

Ans.

;

C E 2

Ceq =

C.KC C2K CK   C  KC C(1  k) 1  K

;

Q=

CK E 1 k

2ck  c(k  1) 2ck  ck  c ck  c  CK C   CK C   = Ans. (B)  Q =  1 k  2 E ;  Q =  1 k  2 E = 2(1  k) 2(1  k) 2(1  k)    

36.

V = ed E=

100 20um

37.

dN  n dt

38.

N 4R 2

=

39.

40.

Ans.

N=

100 1  5v 20

;

V1 =

;

dN  dt dt

Ans.

;

6.93  10 4 = 10-4 6.9

160  6200  10 10 5  1019  10 = 4  3.14  3.24 6.62  10 34  3  108

496 5  1019  10  1019 = 1.25 ×1019 = 10 4  3.14  3.24

1 1 1 1  2  2 2 3  4 9  125 1 1 1 64 2 = 1  1  2 2 9 25 3 5

;

2 

(A)

Ans.

125 1 64

=

Ans.

(D)

16  62  10 7 20  10 34  108

(C)

Ans.

(B)

10×8 = 80 11×7.5 = 82.5 Energy required = 82.5-80 = 2.5 (A) CLASS-XII_STREAM-SB+2_PAGE # 279

CHEMISTRY 41.

+

-

-10

-3

KSP AgBr = (Ag ) (Br ) = (5×10 ) (10 ) = 5×10-13 Ksp AgBr -

-2

Concentration of Br in 10 M aq AgNO3 =

[ Ag  ]

5  10 13  5  10 11 m = (10 2 )

42.

Over bromination take place in activated Aniline.

Ans.

(A)

43.

+6 is maximum oxidation state of Cr.

Ans.

(D)

44.

Ans. (C)

45.

Al2O3 is amphoteric oxide

Ans.

(C)

46.

ln k = ln A –

Ans.

(A)

47.

Ni(CO)4 is tetrahedral with d10 configuration

Ans.

(C)

48.

Ozonolysis and intramolecular aldol

Ans.

(A)

49. 50.

Ans. (B) Ans. (A)

51.

When QC > KC Reaction move in backward direction

Ans.

(A)

52.

SN-1 Reaction, more stable carbocation is formed

Ans.

(C)

53.

Ans. (D)

54.

Wolf kishner Reduction

Ans.

(B)

55.

Compound is AsCl3

Ans.

(C)

56.

Valency factor of KMnO4 = 5 5e- + MnO4- + 8H+  Mn+2 + 4H2O

Ans.

(D)

Ea 1  R T

57.

Ans. (C)

58.

H 6000 J  S = T  213k  22J/K.mol

Ans.

(A)

59.

Ullmann reaction

Ans.

(A)

60.

Energy of t2g orbitals < eg orbitals in octahedral complexes Ans. (C)

Ans.

(D)



Ans-(A) n

82.

e

n

  k wk

2

n

1 w 

 e

R0 

 e

n

w 

 when n is multiplying of 3, then S=e, 1/e when n is not multiple of three,then s= e1/ 2 e1/ 2 Ans- (C) 83.

y = ax2 + bx + c a > 0, b < 0, c < 0 Also y(1) = a + b + c < 0 Ans-(B)

84.

Sin 

1

r1 r  2 AP1 AP2



3

AP2  AP1  r1  r2

r2 3  r1 3  r1  r2 r2 ( 3  1)  r1



3 1



r2  2 3 r1 85.

Ans- (D)

Let f(x) = 3x3 - 25x f’(x) = 9x2 - 25 5 3 f(-5/3) = 250/9  27.7 f(5/3) = -250/9  -27.7  n can take 55 values

f’(x) = 0  

86.

Ans-( )

87.

Sol- In 

 /2



0

Ans-(C)

xn cos xdx n

  In +n(n-1)In-2 =   2

.......(1)

  I  n  2 !I  In In  2    n 2 n       Also =   n! n  2 !  n 2  n  2  n!



  n    ( 2)  =    n  2  n!

   2  = e  1 2 

Ans- (A)


88.

Sol-

n



1

n

[x][ x ]dx   2dx  ............ 1

= 50

for n = 8

66

for n = 9

Ans - (B) 89.

Ans- (B), Check for n = 1, 2 , 3,......

90.

Sol- No correct Option (1,17), (2,17),(3,17),(4,17),(4,17),(6,17),(7,17)_ _ _ _ _ _ _ (16,17),(18,17) = 17 points Also (17,1), (17,2), _ _ _ _ _ _ _ _ _ _ _ _ _ (17,16), (17,18) = 17 points So there are more then 34 points. Ans-()

PHYSICS Initial

Final

w V=W'R

91. A

From cons. of anguler momentum about A

 v=

2Rw 7

92.

Ans. (A)

93.

Tmax - mg =

 W’ =

2w 7

mv 2 ....(i) R from energy cons. ki + ui = kf + uf

1 O + mg (R-R cos  ) = mv2+0 2 V2 = 2gR (1-cos  ) Put in eq. (i) Tmax = 3mg - 2mg cos 

Ans.

(C)

Tmin Tmax mg cos V mg

mg

Tmax Tmin =4 Ans.

 =60° (B)

94.

Heat cap. is define as C = Cv + C=

R 1 

3 R R R 2 1 3

Ans.

(D)

95.

Ans. (A)

96.

Ans. (C)


97.

At steady state capacitor will belave as open circuit So charge store in 1  F = 1×4 = 4  C in 2  F = 2 × 4 = 8  C Ans. B

98.

F=

2P C

;

mg =

99.

E =  + eV

;

E =  + e (.9)

;

1.1E =  + e(.9)

.1E = .3e

;

E = 3eV put

;

 2.4eV

100.

2  1.5  103 3  108

;

(MLoT-2  -4) = (ML2T-1)  (ML2T-2  -1)  (LT T-1)  Solving we get

m = 10-6kg

Ans.

(C)

Ans.

(B)

Ans.

(D)

Ans.

(B)

Ans.

(C)

Ans.

(D)

Ans.

(B)

CHEMISTRY 101.

Ans. (A)

102.

d=

PM RT M = 123

dRT P  acetic acid exist as dimer

M=

103.

 E =  H -  (PV) =  H -  ngRT

104.

Ans. (C)

105.

N2 (g) + 2H2(g)  N2H4(g)

 = 106.

 (BDE)

Reacant

  (BDE)Pr oducts

Vk+ + VCl- = 133 + 181 KCl form FCC Lattice  2(VK+ + VCl-) = a = edge length of unit cell V = a3

107. Ans.

K2Cr2O7+6FeSO4+ 7H2SO4  Cr2(SO4)3 + 3Fe2 (SO4)3+ K2SO4 + 7H2O (B)

108.

EoCell = SRP cathode - SRP anode = -0.44 - (-0.74) = 0.3V o  G = - nF E Cell (n = 6 ele )

109.

Ans. (A)

110.

XeF6 + 3H2O  XeO3 + 6HF XeO3 is SP3 hybrid and pyramidal (C)

Ans.

Ans.

(C)


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