Kvpy 2010 Class XII

8/8/2019 Kvpy 2010 Class XII

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QUESTIONS & SOLUTIONS OF

KISHORE VAIGYANIK PROTSAHAN YOJANA (KVPY) 2010

TEST PAPER CLASS - XII

PART-I

One-Mark Questions

MATHEMATICS

1. Let denote the matrix

0i

i0, where i2 = 1, and let I denote the identity matrix

10

01. Then

I + A + A2 + .............+ A2010 is

(A)

00

00(B)

0i

i0(C)

1i

i1(D)

10

01

Ans. (C)

Sol. I =

10

01 + A + A2 + A3 + ............+ A2010

A =

0i

i0( + A + A2 + A3) + A4 (+ A + A2 + A3) ..... + A2008 (+ A + A2)

A2 =

10

01

A3 =

0i

i0= 0 +

0i

i0

A4 = A4n =

+ A + A2 + A3 = 0

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

(A)

2

51,0 (B)

2

52,

2

51(C)

2

51,

2

51(D)

,

2

52

Ans. (A)

(TEST PAPER HELD ON 31-10-2010)


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Sol. Let a, ar, ar2 are sides.

Case 1 : r > 1

a + ar > ar2

r2 r 1 < 0

r

2

51,

2

51 r

2

51,1

Case 2 : r = 1 equilateral triangle

Case 3 : r < 1

Final solution

ar2 + ar > a r

2

15,1

2

15,0

r2 + r 1 > 0 r

2

51,0

3. The number of rectangles that can be obtained by joining four of the twelve vertices of a 12-sided regular

polygon is

(A) 66 (B) 30 (C) 24 (D) 15

Ans. (D)

Sol. Number of diagonals passing through centre = 6

Number of rectangles = 6C2= 15

4. Let 1, and 2

be the cube roots of unity. The least possible degree of a polynomial, with real coefficients,having 22, 3 + 4, 3 + 42 and 5 2 as roots is

(A) 4 (B) 5 (C) 6 (D) 8

Ans. (B)

Sol. 22, 3 + 4, 3 + 42, 5 2

1 3 i , 1 + 32 i , 1 32 i, 6

where =2

i31 , 2 =

2

i31

1 + 32 i , 1 32 i are conjugate of each other..

Least possible degree = 5

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

contact of the circle and the directrix is

(A) 2 (B) 2 (C) 22 (D) 4

Ans. (A)


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Sol. 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

024 = 2

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

(A) 54 (B) 108

(C) 48 (D) not determinable with the above data

Ans. (B)

Sol. BKN = 6

9

1

APB

BKN

3

1

AB

BN

1

2

BN

AN

APB =1

9 6

=1

54

ABC = 108

7. Let P be an arbitrary point on the ellipse1

b

y

a

x

2

2

2

2

, a > b > = 0. Suppose F1 and F2 are the foci of the

ellipse. The locus of the centroid of the triangle PF1F

2as P moves on the ellipse is

(A) a circle (B) an ellipse (C) a parabola (D) a hyperbola

Ans. (B)


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Sol. 2

2

2

2

b

y

a

x = 1 , a > b > 0

h =3

x, k =

3

y

2

2

2

2

b)k3(

a)h3( = 1 which is ellipse

8. 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

Ans. (A)

Sol. 1

6

1

7 sin1cos

LH.S is less than or equal 1while R.H.S is greater or equal to 1

possible when

cos = 1 and sin = 0

= 0, 2

9. The product

(1 + tan 1) (1 + tan 2) (1 + tan 3) ... (1 + tan 45)

equals

(A) 221 (B) 222 (C) 223 (D) 224

Ans. (C)

Sol. (1 + tan 1) (1+ tan 2) .........(1+ tan 45)

(1 + tan A) (1 + tan B) = 2 if A + B = 45

Ans = 223

10. 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 (B) 2 (C) 1 (D) 0

Ans. (B)

Sol. 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,


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11. 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

Ans. (B)

Sol. 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. The figure shown below is the graph of the derivative of some function y = (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

Ans. (C)

Sol.

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


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13. The following figure shows the graph of continuous function y = (x) on the interval [1, 3]. The points A, B,

C have coordinates (1, 1), (3, 2), (2, 3) respectively, and the lines 1and

2are parallel, with

1being tangent

to the curve at C. If the area under the graph of y = (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

Ans. (A)

Sol. 3

i

4dx)x(f

shaded area = Trapezium AreaArea under curve

=2

)13(2

7

2

5

4 = 2

14. Let e

1

nn )x(log dx, where n is a non-negative integer..

Then I2011

+ 2011 I2010

is equal to

(A) 1000

+ 999 998

(B) 889

+ 890 891

(C) 100

+ 100 99

(D) 53

+ 54 52

Ans. (C)


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18. 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

Ans. (D)

Sol.

Number of ways = 4P2 4P

2 10 = 1440

19. The number of natural number n in the interval [1005, 2010] for which the polynomial

1 + x + x2 + x3 +.......+ xn1 divides the polynomial 1 + x2 + x4 + x6 + .........+ x2010 is

(A) 0 (B) 100 (C) 503 (D) 1006

Ans. (C)

Sol. 1 + x2 + x4 + .........+ x2010

=2

10062

x1

)x(1

=

)x1)(x1(

)x(121006

=)x1(

)x1(

)x1(

)x1( 10061006

= (1 + x1006))x1(

)x1(

)x1(

)x1( 503503

= (1 + x1006) (1 + x + x2 + ...............+ x502) (1 x + x2 x3 + ............ + x502)

n 1 = 502 n = 503

20. Let a0= 0 and a

n= 3 a

n1+ 1 for n 1. Then the remainder obtained on dividing a

2010by 11 is

(A) 0 (B) 7 (C) 3 (D) 4

Ans. (A)

Sol. an

= 3an1

+ 1

= 3(3an2

+ 1) + 1

= 32 an2

+ 3 + 1

= 32 (3an3

+ 1) + 3 + 1

= 33 an3

+ 32 + 3 + 1

aN

= 3n a0

+ 3N1 + 3N2 + .... 1


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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

=2

1)1242(

2

1)243(

2

13

2

13 40240240252010

= 11 k (k I)

Hence remainder is 0

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 :(A) (m + M) (

1+

2)g (B) (m

1+ M

2)g

(C) (m 1+ (m + M)

2)g (D) m(

1+

2)g

Ans. (A)

Sol.

F 1mg 2(m + M)g = M. mmg1

F = 1mg +

2(M + m)g +

1Mg

= 1g(M + m) +

2g(m + M)

= (1g +

2g) (M + m)

22. Two masses m1

and m2

connected by a spring of spring constant k rest on a frictionless surface. If the

masses are pulled apart and let go, the time period of oscillation is :

(A) T =

21

21

mm

mm

k

12 (B) T =

21

21

mm

mmk2 (C) T =

k

m2 1 (D) T =

k

m2 2

Ans. (A)

Sol. T =k

1

mm

mm2

21

21

we can use concept of reduce mass (A).


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23. A bead of mass m is attached to the midpoint 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 ofoscillation is :

(A)m

T2

2

1

(B)

m

T4

2

1

(C)

m

T4

2

1

(D)

m

T2

2

1

Ans. (B)

Sol. 2T cos = ma ; 22

x4

x

m

T2

= a

If x < < < R

2from the center of the shell. Then the electric

field in the hollow cavity.

(A) depends on both +q and +Q (B) is zero

(C) is only that due to +Q (D) is only that due to +q

Ans. (D)

Sol. There is no net effect of outside charge.

So, (D).

28. The following travelling electromagnetic wave

Ex

= 0, Ey

= E0sin(kx + t), E

z= 2E

0sin(kx + t) is :

(A) elliptically polarized (B) linearly polarized

(C) circularly polarized (D) unpolarized

Ans. (C)


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29. 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 :

(A)1

h

(B) 1

h

2 (C)

1

h2

(D)1

h

2

Ans. (B)

Sol. tanC =h

r

from snells law

sinC = 1

r =1

h

2

30. 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

Ans. (B)

Sol. 2

> 3

1sin i

1=

2 sin r

1

< 2

From condition TIR

2

> 3

and 1

> 3

So, 2

> 1

> 3

.

31. A nucleus has a halflife of 30 minute. At 3PM its decay rate was measured as 120,000 counts/sec. What

will be the decay rate at 5 PM ?

(A) 120,000 counts/sec. (B) 60,000 counts/sec.

(C) 30,000 counts/sec. (D) 7,500 counts/sec.

Ans. (D)

Sol. A = T/t0

2

AA = 42

120000=

2222

10100120

= 7500 (D).


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32. 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 contact with the

shelf ? (Assume that g = 10 m/s2)

(A) 20 Hz (B) 3.18 Hz (C) 125.6 Hz (D) 10 Hz

Ans. (D)

Sol. 2 A = g

A

g

f =21

A

g

= 2105.2

8.9

2

1

= 3.18

33. A vander Waal's gas obeys the equation of state

2

2

V

anP (V nb) = nRT. Its internal energy is given by

U = CT V

an2. The equation of a quasistatic adiabat for this gas is given by :

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

(C) TC/nR (V nb) = constant (D) P(C+nR)/nR (V nb) = constant

Ans. (A)

Sol. dQ = 0

du = dw

nCvdt = Pdv

du = C dT +2

2

v

andv

cdT + dv

v

an

2

2

= Pdv

P = 2

2

v

an

nbv

nRT

cdT + dvv

an2

2

= dvv

an

nbv

nRT2

2

nR

C

nbv

dv

T

dT


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nR

CnT = n (v nb) + C

n nR/CT n (v nb) = C

nbv

T nR/C

= ec = K

nR/CT = Kv Knb

Kv nR/CT = Knb = constant

v nR/CT = constant.

34. 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 T1and T

2are the maximum and minimum temperatures reached during the cycle, then the efficiency is

given by 1 1

2

T

T

(D) The cycle can only be carried out in the reverse of the direction shown in the figure

Ans. (B)

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 ms1, 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 passenger

Ans. (B)

Sol.


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Page # 15

v0

= 39.6 kmph = 39.6 18

5= 2.2 5 = 111

t1

=330

d

t2 = 330

3011d = 330

d

330

3011

t = t1 t

2= 1

time interval = 30 1 = 29 second

Ans. (B)

36. Velocity of sound measured at a given temperature in oxygen and hydrogen is in the ratio :

(A) 1 : 4 (B) 4 : 1 (C) 1 : 1 (D) 32 : 1

Ans. (A)

Sol. v =M

RT

4

1

32

2

M

M

v

v

0

H

H

0

37. 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 (B) 2.0 mm (C) 0.5 mm (D) 1.5 mm

Ans. (A)

Sol. d = 0.1 mm

D = 1m

= 600 nm

given IR

= 75% of maximum = 75% of 4 = 30

R

= 3 = + + 2

cos

=3

x =

D

dy=

2

3

=

6

y =d6

D= 3

9

101.06

106001

y = 1 meter


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Page # 16

38. Two masses m1and m

2are 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 v0is suddenly imparted to the first particle. At a later

time t, the center of mass of the two masses is at :

(A) x =21

2

mm

m

(B) x =21

02

21

1

mm

tvm

mm

m

(C) x =21

02

21

2

mm

tvm

mm

m

(D) x =21

01

21

2

mm

tvm

mm

m

Ans. (D)

Sol.

vcom

=21

01

mm

0vm

x

i=

21

21

mm

)(m)0(m

xcm = xi + vcom t

xcm

=21

01

21

2

mm

tvm

mm

)(m

39. A charged particle of charge q and mass m, gets deflected through an angle upon passing 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) cotam

qB(B) tana

m

qB(C) 2cota

m

qB(D) 2tana

m

qB

Ans.

Sol. According to given information

sin =R

a


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Page # 17

CHEMISTRY

41. The number of isomers of Co(diethylene triamine) Cl3is.

(A) 2 (B) 3 (C) 4 (D) 5

Ans. (B)

Sol. [Co(dien)Cl3]

dien = NH2 CH

2 CH

2 NH CH

2 CH

2 NH

2

[Co(dien)Cl3] have two geometrical isomers cis and trans.

cis form is optically active but trans form is optical inactive.

42. Among the following, the -acid ligand is :

(A) F (B) NH3

(C) CN (D) I

Ans. (C)

Sol. The ligands which allow back bonding to sufficient extent are called acid acceptors.

Example CO, CN, NO etc.

43. The bond order in O22 is

(A) 2 (B) 3 (C) 1.5 (D) 1

Ans. (D)

Sol. O2

2 : Total electron = 18 ; Bond order = 1.

Peroxide (O2

2) : (1s)2 (*1s)2 (2s)2 (*2s)2 (2pz)2 (2p2

x= 2p2

y) (*2p

x2 = *2p2

y)

Bond order =

2

NN AB=

2

810= 1.


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44. The energy of a photon of wavelength = 1 meter is (Plank's constant = 6.626 1034 Js, speed of light = 3 108

ms1)

(A) 1.988 1025 J (B) 1.988 1030 J (C) 1.988 1028 (D) 1.988 1031

Ans. (A)

Sol. E =

hc =1

10310262.6 834 = 1.988 1025 J.

45. 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

(A) zero order reaction (B) second order reaction

(C) first order reaction (D) third order reaction

Ans. (B)

Sol. t1/2

(half-life) is independent of initial concentration so reaction is first order.

46. The shape of the molecule CIF3is :

(A) trigonal planar (B) T-shaped (C) pyramidal (D) Y-shaped

Ans. (C)

Sol. ClF3

F

Cl

F

87.5

87.5

F

F

..

..

F nearly 'T' shaped.

47. FriedalCrafts 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

Ans. (D)

Sol. 3AlCl

O||

ClCR

Freidal craft acylation


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Page # 19

48. The order of acidity of compounds IIV, 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

Ans. (A)

Sol. > > >

Stability conjugate base order

> > >

49. The most stable conformation for n-butane is

(A) (B) (C) (D)

Ans. (A)

Sol. anti conformation is most stable

50. In the nuclear reaction

Th23490 Pa23491 + X

X is :

(A) e01 (B) e01 (C) H

11 (D) H

21

Ans. (A)

Sol. ePaTh0

1

234

91

234

90


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Page # 20

51. 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.

Ans. (A)

Sol. Entropy change is positive, but enthalpy change is negative.

52. 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

Ans. (B)

Sol. On increasing temperature, change in activation energy is not significant.

53. In metallic solids, the number of atoms for the face centred and the body-centered cubic unit cells, are,

respectively.

(A) 2,4 (B) 2,2 (C) 4,2 (D) 4,4

Ans. (C)

Sol. For FCC unit cell number of atom per unit cell

Z = 8[corner] 81 + 6 [face center]

21 = 4.

For body center unit cell number of atom per unit cell

Z = 8[corner] 8

1+ 1 [body center] 1 = 2.

54. From equation 1 and 2

CO2

CO +2

1O

2[

1cK = 9.1 1012 at 1000C] (eq 1)

H2O H

2+

2

1O

2[

2cK = 7.1 1012 at 1000C] (eq 2)

(A) 0.78 (B) 2.0 (C) 16.2 (D) 1.28

Ans. (D)


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Page # 21

Sol. CO2

CO +2

1O

2 1CK = 9.1 1012 .....(i)

H2O H

2+

2

1O

2 2CK = 7.1 1012 .....(ii)

Target eq.

CO2 + H2 CO + H2O KC = ?Target eq.

[Eq. (i) Eq(ii)] KC

=2

1

C

C

K

K

=12

12

101.7

101.9

= 1.28.

55. For a first order reaction R P, the rate constant is k. If the initial concentration of R is [R0], the concentra-

tion of R at any time 't' is given by the expression.

(A) [R0] ekt (B) [R

0] (ekt) (C) [R

0] ekt (D) [R

0] (ekt)

Ans. (B)

Sol. R P

t=0 R0

0

t=t (R0 x) x

Rate = dt

]R[d= K[R]1

=dt

dx= K[R

0 x] =

x

0

0

dx.)xR(

1=

t

0

dt.K

= ln

xR

R

0

0= Kt

[R] = (R0 x] = R

0eKt

56. The correct structure of PCl3F

2is

(A) (B) (C) (D)

Ans. (C)

Sol. PCl3F

2:


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Page # 22

57. 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

Ans. (B)

Sol. (I) & (IV) are enantiomeric pair

I IV

58. 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 = NO

2(B) X = O

2, Y = NO, Z = NO

2

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

2(D) X = O

2, Y= NO

2, Z = NO

Ans. (A)


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Page # 23

Sol. 2NO2(g) 2NO (g) + O

2(g)

NO2is reactant so its concentration is decrease with time while NO and O

2are product so their concentration

increases with time.

Formation of moles of NO is double than O2.

So X = NO, Y = O2

and Z = NO2.

59. The aromatic carbocation among the following is

(A) (B) (C) (D)

Ans. (C)

Sol. Cyclic

Planar

Complete conjugation

(4n + 2) pe

Tropylium carbocation

60. Cyclohexene is reacted with bromine in CCl4in the dark. The product of the reaction is

(A) (B) (C) (D)

Ans. (A)

Sol. Anti addition takes place.

BIOLOGY

61. Ribouncleic Acid (RNA) that catalyze enzymatic reactions are called ribozymes. Which one of the following

acts as a ribozyme?

(A) Ribosome (B) Amylase (C) tRNA (D) Riboflavin

Ans. (B)

62. 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

Ans. (A)


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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

Ans. (B)

64. A horse has 64 chromosomes and a donkey has 62. Mules result from crossing a horse and donkey. State

which of the following in 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

Ans. (C)

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?

(A) Shining green light

(B) Shining sunlight

(C) Shining blue light

(D) Shining ultraviolet light

Ans. (B)

66. 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

Ans. (C)

67. 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

Ans. (A)


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68. If you compare adults of two herbivore species of different sizes, but form 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 possible to predict which would be more

Ans. (A)

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 is fruit ripening.

Ans. (C)

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

Ans. (D)

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

Ans. (A)

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

Ans. (A)

73. A couple went to a doctor and reported that both of them are "carries" 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%

Ans. (C)


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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

Ans. (D)

75. Conversion of the Bt protoxin produced by Bacillus thuringinesis 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

Ans. (A)

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 the heat?

(A) Imbibition

(B) Exothermic reaction between water and seed coats

(C) Friction amount seeds due to swelling

(D) Respiration

Ans. (D)

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

Ans. (B)

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

Ans. (A)

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

Ans. (A)

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

Ans. (D)


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Page # 27

Two-Marks Questions

PART - II

MATHEMATICS

81. Arrange the expansion of

n

4/1

2/1

x2

1x

in decreasing powers of x. Suppose the coefficients of the first

three terms form an arithmetic progression. Then the number of terms in the expansion having integer

powers of x is

(A) 1 (B) 2 (C) 3 (D) more than 3

Ans. (C)

Sol.

n

4/1

2/1

x2

1x

=

n

4/3

2/n

x2

11x

= xn/2

.........

x2

1c....

x2

1c

x2

1cc

4/3rn

2

4/32n

4/31n

0n

nc0

, 22

n1

n

2

c,

2

care in AP

1,2

n,

8

)1n(nare in APAP

n = 1 + 8)1n(n

8(n 1) n (n 1) = 0

(n 1) (8 n) = 0 n = 1 or n = 8

but n = 1 i.e. not posible

n = 8

Now

8

4/1

2/1

x2

1x

=

r

4/1

8

0r

r82/1r

8

x2

1)x(c

=4r34r8

0r

r8 x

2

1c

required number of terms = 3 (r = 0, 4, 8) ans (C)


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Page # 28

82. Let r be a real number and n N be such that the polynomial 2x2 + 2x +1 divides the polynomial (x + 1)nr.

Then (n, r) can be

(A) (4000, 41000) (B)

10004

1,4000 (C)

1000

1000

4

1,4 (D)

4000

1,4000

Ans. (B)

Sol. Roots of 2x2 + 2x = 1 = 0 are ,2

i1

It will satisfies the (x + 1)n r = 0

n

12

i1

= r

n

12

i1

= r,, r R

n

4

in

e2

1

= r

n

4

i

e

will be real only if n is multiple of 4.

For n = 4000

r =4000

2

1

= 10004

1

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 (B) Fig.2 (C) Fig.3 (D) Fig.4

Ans. (B)

Sol. y = ax + b and 1b

y

a

x 22

fig. 1 and 3 incorrect (a < 0 b the equation of line y = ax + b)

Now for a > 0 and b < 0 fig. 2 is correct fig. 2


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Page # 29

84. Among all cyclic quadrilaterals inscribed in a circle of radius R with one of its angles equal to 120, consider

the one with maximum possible area. Its area is

(A) 2R2 (B) 2 R2 (C) 2R3 (D) 2R32

Ans. (C)

Sol. 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 60 = 3 R.

maximum Area of ABC = 2)R3(4

3=

2R4

33

maximum area of ADC =2

1(AC) (DM)

= 21

60cot2

R3)R3(

=4

R3 2

area of ABCD =222 R3R

4

3R

4

33

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

0).

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


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Page # 30

(A) Fig.1 (B) Fig.2 (C) Fig.3 (D) Fig.4

Ans. (B)

Sol.

f(x) is monotonic increasing in [a, )

f(x) > 0 x [a,)

f(x) is monotonic decreasing in (, b] f(x) < 0 x (, b]

Now g(x) =x

)x(f

g(x) = 2x

)x(f)x(xf

h(x) = xf(x) f(x)

h(x) = f(x) + xf(x) f(x)

h(x) = xf < 0 { x > 0 and f(x) < 0 concavity}

h(x) is M.D.

So, at x = a it may be M.I. followed by M.D. till x = b

or

M.D. through out a

But cant be M.I. followed by M.D.

So (B) is correct answer.


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Page # 31

86. Let V1be the volume of a given right circular cone with O as the centre of the base and A as its apex. Let V

2

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)25

3(B)

9

4(C)

27

4(D)

27

8

Ans. (C)

Sol.

Let radius of base of given cone is r and length

as h V1

=3

1r2h ........(1)

Let radius of base 1 and R1 and height H. For the cane with apex O

V2

=3

1R2H ..........(2)

NowHh

h

R

r i

r

R

h

H

H = h

r

R1

V =

r

R1hR

3

1 2

r

R2R2h

3

1

dR

dv 2

= 0 R =3

r2

For maximum volume

R 3

r2 V

2=

r

R1hR

3

1 2

=

3

21

9

r4h

3

1 2

= h81

4 2 .......(2)

27

4

1V

V2


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Page # 32

87. Let : R R be a continuous function satisfying x

0

dt)t(x)x( , for all x R. Then the number of

elements in the set S = {x R (x) = 0 } is

(A) 1 (B) 2 (C) 3 (D) 4

Ans. (A)

Sol. f(x) = 1 + f(x) dx

dy= 1 + y

n (1 + y) = x + c

y = ex + c 1

y = ex 1

f(x) = ex 1 ..... (1)

Now f(x) = x + x

0

dt)t(f

ex 1 = x + et (t)0

x

ex 1 = x + ex x

= 1

f(x) = ex 1

for f(x) = 0 ex 1 = 0

x = 0

88. The value of

2

0

1 )x(coscos|,x|min dx is

(A)4

2(B)

2

2(C)

8

2(D) 2

Ans. (B)

Sol.

A =2

1.

2

+

2

1.

2

2

2


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Page # 33

89. Let ABC be a triangle and P be a point inside ABC such that 0PC3PB2PA . The ratio of the area of

triangle ABC to that of APC is

(A) 2 (B)2

3(C)

3

5(D) 3

Ans. (D)

Sol. Let P is origin and p.v. of A, B, C are a, b, c respectively

Given 0PC3PB2PA

0c3b2a

...... (i)

Area of ABC =2

1|a

b

+ b

c

+ c

a

|

= 2

1

|a

2

ac3

+

2

ac3

c

+ c

a

|

=2

1|2

3( c

a

) 0 0 +2

1( c

a

) + c

a

|

=2

3|c

a

| ...... (ii)

Area of APC =2

1|c

a

|

APC

ABC

=1

3

90. Suppose m, n are positive integers such that 6m + 2m+n 3n + 2n = 332. The value of the expression m2 + mn +

n2 is

(A) 7 (B) 13 (C) 19 (D) 21

Ans. (C)

Sol. 6m + 2m+n. 3m + 2n = 332 = 4 83

2m2. 3m + 2m+n2 3m + 2n2 = 83 .......(1)

83 is prime number so that this is possible only ifm = 2

by (1) 32 + 2n 32 + 2n2 = 83

2n. 9 + 2n2 = 74

2n2 (36 + 1) = 74

2n2 = 21

n = 3

Hence. m2 + mn + n2

= 4 + 6 + 9 = 19


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Page # 34

PHYSICS91. 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 theball up to the 10th impact ?

(A)r1

r1h2

10

(B)

2

20

r1

r1h

(C) h

r1

r1h2

2

22

(D) h

r1

r1h2

2

20

Ans. (D)

Sol. h10upto........g

Vr

g

Vr

g

V th20

420

220

=g

V20[1 + r2 + r4 + .............upto 10th] h

= hr1

)r(1h2

2

102

= hr1

r1h2

2

20

.

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

(A) 2GT4

3

f1

f4

(B) 2GT4

3

f1

f4

(C) 2GT4

3

f1

f34

(D) 2GT4

3

f1

f24

Ans. (A)

Sol. 2r

GMm=

r

mv2 v =

r

GM

T =V

r2T =

GM

r2

3

T = rGM

r2=

GM

r2

3

geff

= g2 Re cos2

f(g 2 Re cos260) = (g2Re cos20)

fg g = 2R

1

4

f

g[f 1] =4

R2[f 4] R =

)4f(

)1f(g42

= 2R

Gm(f 1) =

4

R2(f 4)

)4f(

)1f(GM42

= R3

)4f(4

)1f(GM4T2

2

= R3

)4f(

)1f(

= R3 =3

R3

4

M

=)1f(GT4

)4f(32


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Page # 35

93. Three equal charges +q are placed at the three vertices of an equilateral triangle centred at the origin. Theyare held in equilibrium by a restoring force of magnitude F(r) = kr directed towards the origin, where k is aconstant. What is the distance of the three charges from the origin ?

(A)

3/12

0 k

q

6

1

(B)

3/12

0 k

q

12

3

(C)

3/22

0 k

q

6

1

(D)

3/22

0 k

q

4

3

Ans. (B)

Sol. a = 2r 2

32

2

q

Kq2

2

q

Kq2cos30

= r3

kr = 2

2

q

kq2 cos30

k.r.3r2 =04

2

.q2

2

3

r =

3/1

0

2

k12

3q

94. Consider the infinite ladder circuit shown below :

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

(A)2

LC(B)

LC

2(C)

LC

1(D)

C

L2

Ans. (B)

Sol.

Let the equivalent impedence of circuit be ZSo, Z = wL + Z

Z = L +C

C

XZ

ZX

= L +

C

1Z

C

1Z

On solving we get,


36/45

Page # 36

Z =C2

LC4CLLC 222

For Z to be purely inductive 0LC4CL 222 = 0

=LC

2Ans.

95. 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))1(2

)2(R

(B))1(2

)2(R

(C))1(2

)2(R

(D))1(2

)2(R

Ans. (A)

Sol.

1V1

=R

1 R1

v1

1Rv1

)VR2(V

1

1f

=R

1

1

RR2

R

1

V

1

f

)RR2R2(

)1(

R

1

=

21

R

1

)2(

)2(

R

)1(=

2

2

R

1 v

f=

)1(2

)2(R

96. A particle of mass m undergoes oscillations about x = 0 in a potential given by V(x) =2

1kx2 V

0cos

a

x,

where V0,k, a are constants. If the amplitude of oscillation is much smaller than a, the time period is given by

(A)0

2

2

Vka

ma2

(B)

k

m2 (C)

0

2

V

ma2 (D)

02

2

Vka

ma2

Ans. (A)

Sol. V(x)

=2

1kx2 v

0cos

a

x

G =dx

dv

= kx + v0

sin

a

x.a

1

Since, x


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Page # 37

EQ

= kx +a

x.

a

V0= x

a

VK

2

0

=

2

0

a

VK = 2

02

a

VKa K = m2

T =

2= 02

2

vka

ma

2 .

97. An ideal gas with heat capacity at constant volume CV

undergoes a quasistatic process described by PV ina PV diagram, where is a constant. The heat capacity of the gas during this process is given by :

(A) CV

(B) CV

+1

nR(C) C

V+ nR (D) C

V+ 21

nR

Ans. (B)

Sol. CV

PV

Direct formulae C = Cv

+

1

nR.

98. An ideal gas with constant heat capacity CV

=2

3nR is made to carry out

a cycle that is depicted by a triangle in the figure given below :The following statement is true about the cycle :

(A) The efficiency is given by22

11

VP

VP1

(B) The efficiency is given by22

11

VP

VP

2

11

(C) Net heat absorbed in the cycle is (P2 P

1) (V

2 V

1)

(D) Heat absorbed in part AC is given by

2(P2V

2 P

1V

1) +

2

1(P

1V

2 P

2V

1)

Ans. (D)

Sol. CV

=2

3nR, C

P= C

v+ R =

2

nR5

f = 3

w =2

1(V

2 V

1) (P

2 P

1)

A B Q = nCpT = n

2

5RT =

2

5(P

1V

1 P

2V

2)

A C, QAC

=2

1(P

1+ P

2) (V

2 V

1) + nC

VT

QAC

=2

1(P

1 P

2) (V

2 1) +

2

3(P

2V

2 P

1V

1)

=2

3nR (T

2 T

1) +

2

VPVP)TT(nR 122112

= 2nR (T2 T1) + 2

VPVP 1221


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Page # 38

=)VPVP(

2

3)VV)(PP(

2

1

)PP)(VV(2

1

11221221

1212

=112222221121

1212

VP3VP3)VPVPVPVP(

)PP)(VV(

=12211122

11212122

VPVPVP4VP3PVPVVPVP

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

(A) 20

2

mv8

q

(B) 2

0

2

mv4

q

(C) 2

0

2

mv2

q

(D) 0

Ans. (B)

Sol. V V

0mv2

12 2

= 0 + 0

d

Kqq

d =

2

2

0 mv

q.

4

1

= 20

2

mv4

q

100. At time t = 0, a container has N0radioactive 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 att = 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)

Ans. (C)

Sol. N0 Initial nucleon

at t = 0 N0N

Add. at a constan rate (N C) =dt

dN

k

0

dt =

N

N0CN

dN

t =NN0

)CN(n1


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Page # 39

t =

CN

CNn

1

0

et = CNCN

0

N C = elt (N0 C)

N =

teC(N

0 C)

= C

+ N0et C

.e(t)

=C

[1 e(t)] + N0

edt .

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 (B) 0.08 N, 3.60 (C) 0.04 N, 3.60 (D) 0.02 N, 10.08

Ans. (.....)

Sol. H2C

2O

4.2H

2O (GMM = 126)

Molarity (M) =100126

100052.2

= 0.2M

M1V

1= M

2V

2

0.2 10 = M2 500

M2

= 4 103.

valence factor of H2C2O4.2H2O = 2.Normality = [M] V.F. = 2 [4 103] = 8 103 N

Amount of oxalic acid (mg/mL) =3

33

10

10126104

mL

mg= 0.504.

Note : Options are not match with solution.

102. Two isomeric compounds I and II are heated with HBr.

I II

The products obtained are


40/45

Page # 40

(A) (B)

(C) (D)

Ans. (A)

Sol. H

Br

HBr + CH3OH

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

(A) 0, 1 (B) 1 , 0 (C) 0, 2 (D) 1, 1

Ans. (A)

Sol.


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Page # 41

104. For the reaction A B, H = 7.5 kJ mol1 and S = 25 J mol1, the value of G and the temperature, at

which the reaction reaches equilibrium are, respectively,

(A) 0 kJ mol1 and 400 K (B) 2.5 kJ mol1 and 400 K

(C) 2.5 kJ mol1 and 200 K (D) 0 kJ mol1 and 300 K

Ans. (D)

Sol. A B

H = 7.5 kJ/mole ; S = 25 J/mole.

G = H TS at equilibrium G = 0

0 = 7.5 103 T(25)

T =25

105.7 3= 300 K

105. The solubility product of Mg(OH)2is 1.0 1012. 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 (B) 7.8 (C) 8.0 (D) 9.0

Ans. (D)

Sol. [Mg2+] = 102 M

)2)OH(Mg(SPK = [Mg

2+] [OH]2 = 1.0 1012

= [OH]2 =2

12

10

100.1

= 1010

[OH] = 1 105

pOH = 5 ; pH = 9

106. 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 (B) 6.40 107 (C) 2.19 107 (D) 9.20 107

Ans. (C)

Sol. For FCC unit cell

4r = a2

a =

2

4.1414= 400 pm

Volume of unit cell a3 = (400)3 = 6.4 107 pm3.


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Page # 42

107. Identify the cyclic silicate ion given in the figure below

(A) [Si6O

24]24 (B) [Si

6O

18]18 (C) [Si

6O

18]12 (D) [Si

6O

24]12

Ans. (B)

Sol. Formula of cyclic silicate : [SinO

3n]2n

Cyclic silicates : (SiO32)

nor (SiO

3)

n2n

108. Diborane is formed the elements as shown in equation (1)

2B (s) + 3H2(g) B

2H

6(g) ... (1)

Given that

H2O () H2O (g) H10 = 44 kJ2B + 3/2 O

2(g) B

2O

3(s) H

20 = 1273 kJ

B2H

6(g) + 3 O

2(g) B

2O

3(s) + 3 H

2O (g) H

30 = 2035 kJ

H2(g) + 1/2 O

2(g) H

2O () H

40 = 286 kJ

the H0 for the reaction (1) is :(A) 36 kJ (B) 520 kJ (C) 509 kJ (D) 3550 kJ

Ans. (A)

Sol. Target eq.

2B (s) + 3H2(g) B

2H

6(g)

Given thatH2O () H

2O (g) H

10 = 44 kJ .....(i)

2B + 3/2 O2

(g) B2O

3(s) H

20 = 1273 kJ .....(ii)

B2H

6(g) + 3 O

2(g) B

2O

3(s) + 3 H

2O (g) H

30 = 2035 kJ .....(iii)

H2(g) + 1/2 O

2(g) H

2O () H

40 = 286 kJ .....(iv)

Target eq. :

Eq.(ii) Eq.(iii) + 3[Eq.(iv)] + 3 [Eq.(i)]

= 1273 [ 2035] + 3[286] + 3[44]

= 36 KJ/mole.


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Page # 43

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 (B) 2.0 and 3 BM

(C) 0.4 and 24 BM (D) 2.4 and 0 BM

Ans. (B)

Sol. K3[Fe(CN)

6]

3 + x 6 = 0 x = + 3

26Fe = 3d6 4s2

26Fe3+ d5 with S.L. t

2g2,2,1, eg0,0 CFSE = 2.0

0+ 2P ~ 2.0

0

Number of unpaired electron = 1.

So, = )2n(n = 3 B.M

110. 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= 1.85C mol1 then the molar mass of nicotine is

:

(A) 16 (B) 80 (C) 320 (D) 160

Ans. (D)

Sol. Tf= K

f molality

= 0.925 = 1.85

92M

10008

M = 173.9 ~ 174

Closest answer = 160.

BIOLOGY

111. A hot 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% (B) two-fold (C) ten-fold (D) hundred-fold

Ans. (B)

112. In a diploid organism, there are three different alleles for a particular gene. Of these three alleles one isrecessive and the other two alleles exhibit co-dominance. How many phenotypes are possible with this set

of alleles?

(A) 3 (B) 6 (C) 4 (D) 2

Ans. (C)


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Page # 44

113. 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). 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

Ans. (D)

114. The amino acid sequences of a bacterial protein and 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% identi-

cal. 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.

Ans. (B)

115. The following DNA sequence (5 3) specifies part of a protein coding sequence, starting form 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

Ans. (D)

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

(A) P1 + S P2 + E (B) E + S P1 + P2

(C) P1 + P2 + E S (D) E + S P1 + P2 + E

Ans. (D)117. Four species of birds have different egg colors: [1] while with no markings, [2] pale brown with ano markings,

[3] gray-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 (B) 2 (C) 3 (D) 4

Ans. (C)

118. 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

Ans. (C)


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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?

(A) (B)

(C) (D)

Ans. (A)

120. 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 to have grown more apart than the two sports 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.

Ans. (C)

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Kvpy 2010 Class XII