1

PHYSICS, CHEMISTRY AND BIOLOGY

Test date 16-04-2017

Class II Medical Excel 2017-18

Solution to NEET - 0

PART I - PHYSICS

1. Given that: ty a cos qxp

, where t represents time in second and x represents distance in metre, then

the correct choice (A) The unit of x is same as that of q (B) The unit of x is same as that of p (C) The unit of t is same as that of q (D) The unit of t is same as that of p Ans (D)

In tcos qxp

, the expression t qxp

is dimensionless.

Hence, tp

is also dimensionless.

This is possible when units of t is same as that of p. 2. The relative density of a material of a body is found by weighing it first in air and then in water. The

weight of the body in air is W1 = (8.00 0.05) N and weight in water is W2 = (6.00 0.05) N. Then the

relative density, 1r

1 2

W(W W )

, with maximum permissible error is

(A) 4.00 0.62% (B) 4.00 0.82% (C) 4.00 3.2% (D) 4.00 5.62% Ans (D)

Relative density is 1r

1 2

W(W W )

, i.e., 8.00 4.008.00 6.00

r 1 1 2

r 1 1 2

W (W W )100 100 100W W W

= 5.62%

r = 4.00 5.62% 3. A parachutist drops freely from an aeroplane for 10 s before the parachute opens out. Then he descends

with a net retardation of 25 ms2. If he bails out of the plane at a height of 2495 m and g = 10 ms2, his velocity on reaching the ground will be (A) 5 ms1 (B) 10 ms1 (C) 15 ms1 (D) 20 ms1 Ans (A) The velocity v acquired by the parachutist after 10 sec, is v = u + gt = 0 + 10 10 = 100 ms1 Let s1 be the height of fall for 10 sec.

Then, 21

1 1s ut gt 0 10 100 500 m2 2

The distance travelled by the parachutist under retardation, s2 = 2495 500 = 1995 m Let v be his velocity on reaching the ground Then, v2 v2 = 2as2 or v2 (100)2 = 2 2.5 1995 Solving, we get; v = 5 ms1

2ME1718PCBT0S 2

4. A body is thrown vertically up with a velocity u. It passes three points A, B and C in its upward journey

with velocities u u,2 3

and u4

respectively. The ratio of the separation between points A and B and

between B and C, i.e., ABBC

is

(A) 1 (B) 2 (C) 107

(D) 207

Ans (D) 2

21

uA u 2gh4

22

2uB u 2gh9

22

3uC u 2gh16

2 2u 8 3 u 5AB

2g 9 4 2g 36

2 2u 15 8 u 7BC2g 16 9 2g 144

AB 5 144 20

BC 36 7 7

5. A person walks up a stalled escalator in 90 second. When standing on the same escalator, now moving, he is carried up in 60 second. The time would it take him to walk up the moving escalator is (A) 45 s (B) 60 s (C) 54 s (D) 36 s Ans (D) Let the length of escalator be L. If v is the velocity of man (relative to escalator) and V of escalator, then according to given problem

L 90 sv …(1)

and L 60 sV …(2)

Now if the person walks up on the moving escalator his velocity relative to the ground will be (V + v). So, time taken by him to move a distance L relative to the ground will be,

L 1 V vt or

V c t L L

which in the light of equations (1) and (2) gives 1 1 1 , i.e., t 36 st 60 90

i.e., the person will take 36 s to walk up to the moving escalator. 6. A rigid rod is placed against the wall as shown in figure. When its velocity of lower end is 10 ms–1 and

its base makes an angle = 60 with horizontal, then the vertical velocity of its end B will be (in m/s):

(A) 10 3 m / s (B) 10 m / s

3

(C) 5 3m / s (D) 5 m / s3

3

Ans (B) 2 2 2x y l

dx dy2x 2y 0dt dt

But, A Bdx dyv , vdt dt

and y tan 1x

So, B10v m / s

3

OR As for a rigid rod component of velocity along rod = constant So, vA cos = vB sin

10 m / s

3

7. In projectile motion, the friction of the air causes a vertical retardation equal to 10% of the acceleration due to gravity (Take g = 10 ms2). The minimum height will be decreased by (A) 8% (B) 9% (C) 10% (D) 11% Ans (B)

Retardation due to friction of air g10

Hence, in upward motion:

Total retardation g 11gg10 10

2 2

mu sinH

2g

2 2 2 2

m mu sin 10 u sin 10H ' H11g 11 2g 112

10

% decrease in m mm

m

H H' 10H 100 1 100 9%H 11

8. If a stone is to hit at a point which is at a distance d away and at a height h above the point from where the stone starts, then the value of initial speed u if the stone is launched at an angle is

(A) g dcos 2(d tan h)

(B) d gcos 2(d tan h)

(C) 2

2

gdh cos

(D) 2gd cos

(d h)

Ans (B) 21h (u sin )t gt

2

d = (u cos) t

or dtu cos

2

2 2d 1 dh usin g

u cos 2 u cos g

2ME1718PCBT0S 4

d gucos 2(d tan h)

9. A force F pushes a block of mass 10 kg against a vertical wall as shown in the figure. The coefficient of friction between the block and wall is 0.5. The minimum value of F to start the upward motion of block is (g = 10 m/s2) (A) 500 newton (B) 400 newton (C) 250 newton (D) can’t be determined Ans (A) FBD of block From diagram

4 4FN Fcos37 F5 5

To start upward motion F sin 37 = 10 g + N

Therefore, 3 4F 100 0.5 F5 5

F = 500 newton 10. An object is attached to the bottom of two strings which are inclined at an angle of 120 with each other.

Each string can withstand a tension of 20 N. The maximum weight of the object that can be sustained without breaking the string is (A) 10 N (B) 20 N (C) 20 2N (D) 40 N Ans (B) The situation before breaking the string is shown in figure. From figure, 20 cos 60 + 20 cos 60 = mg mg = 20 N

11. In order to raise a mass of 100 kg, a man of mass 60 kg fastens a rope to it and passes the rope over a

smooth pulley. He climbs the rope with an acceleration 5g4

relative to the rope. The tension in the rope is

(g = 10 ms2) (A) 1432 N (B) 928 N (C) 1218 N (D) 642 N Ans (C) Let T be the tension in the rope and a be the acceleration of the rope.

The absolute acceleration of man is therefore 5g a4

.

Equations of motion for mass and man gives T 100 g = 100 a … (i)

5

5gT 60g 60 a4

… (ii)

Solving eqn. (i) and eqn. (ii), we get T = 1218 N

12. If the coefficient of friction between an insect and bowl is and the radius of the bowl is r, the maximum height to which the insect can crawl in the bowl is

(A) 2

r1

(B) 2

1r 11

(C) 2r 1 (D) 2r 1 1

Ans (B) The insect will crawl up the bowl till the component of its weight along the bowl is balanced by the limiting frictional force. So resolving weight perpendicular and along the bowl, we get R = mg cos … (i) fL = mg sin … (ii) Dividing eqn. (ii) by eqn. (i)

LftanR

or tan = [fL = R]

or 2 2r yy

or

2

ry1

So, 2

1h r y r 11

13. A balloon with mass m is descending down with an acceleration a (where a < g). How much mass should

be removed from it so that is starts moving up with an acceleration a?

(A) 2mag a

(B) 2mag a

(C) mag a

(D) mag a

Ans (A) Let F be the upthrust of the air. As the balloon is descending down with an acceleration a, mg F = ma … (i) Let mass m0 be removed from the balloon so that it starts moving up with an acceleration a. Then, F (m m0)g = (m m0)a F mg + m0g = ma m0a … (ii) Adding eqn. (i) and (ii), we get m0g = 2ma m0a m0g + m0a = 2ma m0(g + a) = 2ma

02mama g

14. A particle is moving with kinetic energy E, straight up an inclined plane with angle , the coefficient of friction being . The work done against friction before the particle comes to rest is

(A) E cossin cos

(B) Ecossin cos

(C) Esin cos

(D) Eg(sin cos )

Ans (A) By using work energy theorem

2ME1718PCBT0S 6

( mg cos ) d (mg sin ) d = 0 E Ed

mg(sin cos )

Work done against friction = ( mg cos ) d

E( mg cos )mg(sin cos )

E cossin cos

15. A chain is held on a frictionless table with th

1n

of its length hanging over the edge. If the chain has a

length ‘L’ and a mass ‘M’, how much work is required to pull the hanging part on to the table?

(A) 2MgL8n

(B) 2MgL4n

(C) 2MgL2n

(D) 2MgL

n

Ans (C) If m is the mass per unit length of the chain, the mass of the chain of length y will be ym and the force acting on it due to gravity will be mgy (assuming that y is the length of the chain hanging over the edge). So, the work done in pulling the dy length of the chain on the table dW = F(–dy) i.e., dW = (mgy)(–dy) So, the work done in pulling the hanging portion on the table:

0 2

2L/n

mgLW mgy dy2n

or 2MgLW2n

[as M = mL]

16. A particle of mass m is projected at an angle to the horizontal with an initial velocity u. The work done by the gravity during the time it reaches its highest point is

(A) 2 21 mu sin2

(B) 2 21 mu sin2

(C) 2 21 mu cos2

(D) 2 21 mu cos2

Ans (B) By using work energy theorem W = KE

2 2gravity 2 1

1 1W mv mv2 2

2 2 21 1mu cos mu2 2

2 21 mu (1 cos )2

2 21 mu sin2

17. A uniform solid sphere rolls on a horizontal surface at 20 ms1. It then rolls up an incline having an angle of inclination at 30 with the horizontal. If the friction losses are negligible, the value of height h above the ground where the ball stops is (A) 14.3 m (B) 28.57 m (C) 57.2 m (D) 9.8 m

7

Ans (B)

Using law of conservation of energy, 2 21 1MV I Mgh2 2

22 2

21 1 2 vMv MR Mgh2 2 5 R

Or 21 1 Mv Mgh2 5

Or 27 v 9.8 h10

27 (20)h 28.57 m10 9.8

18. A uniform equilateral triangular lamina of side a has mass m. Its moment of inertia about the axis passing through the centroid and perpendicular to the plane of the lamina is

(A) 22ma

3 (B)

2ma3

(C) 2ma

6 (D)

2ma12

Ans (D) Any uniform triangular lamina can be represented by three point masses for purpose of calculation of moment of inertia.

Required moment of inertia = 2m a3

3 2 3

2ma12

19. A uniform circular disc of radius a is taken. A circular portion of radius b

has been removed from it as shown in the figure. If the centre of hole is at a distance c from the centre of the disc, the distance x2 of the centre of mass of the remaining part from the initial centre of mass O is given by

(A) 2

2 2b

(a b )

(B) 2

2 2cb

(a b )

(C) 2

2 2c

(a b )

(D) 2

2 2a

(c b )

Ans (B) Let the circular disc of radius a be made up of the circular section of radius b and remainder. Further let the line of symmetry joining the centres O and O1 be the x-axis with O as origin. The centre of mass of the disc of radius a will be given by

1 1 2 2CM

1 2

m x m xXm m

… (i)

while YCM and ZCM will be zero (as for all points on x-axis, y and z = 0) If be the density of the material of disc, m1 = b2 and x1 = c m2 = (a2 b2) and x2 = ? M = (m1 + m2) = a2 and XCM = 0 From eqn. (i)

2 2 22

2b (c) (a b ) x0

a

2ME1718PCBT0S 8

2

2 2 2cbx

(a b )

i.e., the centre of mass of the remainder (say O2) is at a distance 2

2 2cb

(a b ) to the left of O on the line

joining the centres O and O1. 20. A rupee coin, starting from rest rolls down a distance of 1 m on a plane inclined at an angle of 30 with

the horizontal. Assuming that g = 9.81 ms2, time taken is (A) 0.32 s (B) 0.48 s (C) 0.78 s (D) 1.0 s Ans (C)

2

1 2h 1 I 2ssint 1sin g sin MR g

or 2

I2s MRt

Mgsin

For disc: 21I MR2

3M2s 3s 3 12t 0.78 secMgsin gsin 9.8 sin 30

21. The masses and radii of the earth and moon are M1, R1 and M2, R2 respectively. Their centres are at

distance d apart. The minimum speed with which a particle of mass m should be projected from a point midway between the two centres so as to escape to infinity is

(A) 1 2G(M M )d (B) 1 2G(M M )2

d (C) 1 2G(M M )

2d (D) 1 21 G(M M )

2 d

Ans (B) Potential energy of m when it is midway between M1 and M2,

1 21 2

GM GMU m(V V ) md / 2 d / 2

1 22Gm (M M )d

And as potential energy at infinity is zero, so work required to shift m from the given position to infinity,

1 22Gm(M M )W 0 Ud

As this work is provided by initial kinetic energy,

2 1 21 2Gm(M M )mv2 d

or 1 2G(M M )v 2

d

22. Figure shows elliptical orbit of a planet m about the sun S. The shaded area SCD is twice the shaded area SAB. If t1 be the time for the planet to move from C to D and t2 is the time to move from A to B, then (A) t1 = t2 (B) t1 > t2 (C) t1 = 4t2 (D) t1 = 2t2

9

Ans (D) Areal velocity = constant

So, SCD SAB

1 2

A At t

SAB SAB1 2

1 2

2A A t 2tt t

23. Four particles of equal mass M move along a circle of radius R under the action of their mutual gravitational attraction. The speed of each particle is

(A) GMR

(B) GM2 2R

(C) GM 2 2 1R

(D) GM 2 2 1

R 4

Ans (D) Gravitational force on each due to other three particles provides the necessary centripetal force.

2 2 2

2 2GM GM Mv2 cos45

(2R) R2R

Simplifying it, we get, GM 2 2 1vR 4

24. A wire of length l and radius r is welded to another wire of length 2l and radius 2r. The

free end of first wire is fixed and the free end of second wire supports a mass m. Taking the modulus of each wire as Y, the total increase in length in the combined wire is

(A) 2

2 mg3 r Y

l (B) 2

3 mg2 r Y

l

(C) mg2 r Y

l (D) 3 mg2 r Y

l

Ans (B) Here stretching force is same but stress has to be different for both wires, For first wire, Stress = Y strain

i.e., 12

mg Yr

ll

or 1 2

mgr Y

ll

For second wire

2 2 2

mg 2 1 mg(2r ) Y 2 r Y

l ll

Total increase 2

3 mg2 r Y

l

25. The length of an elastic string is l1 metre when the tension is 4 N and l2 metre when the tension is 5N. The length in metres when the tension is 9N is (A) 4l1 5l2 (B) 5l2 4l1 (C) 9l2 9l1 (D) l1 + l2 Ans (B)

TYA

ll

m

l

2l

2ME1718PCBT0S 10

1 2

1 2

T TY , YA( ) A( )

l l

l l l l and 3

3

TYA( )

l

l l

Here, T1 = 4N, T2 = 5N and T3 = 9N 1 2

1 2 2

T TA( ) A ( )

l l

l l l l

2 2 2

4 5A( ) A ( )

l l

l l l l

l = (5l1 4l2) …(i)

Again 1 3

1 3

T TA( ) A( )

l l

l l l l

1 3

4 9A( ) A( )

l l

l l l l

Using (i), we get l3 = 5l2 4l1

26. A piece of copper having an internal cavity weighs 264 g in air and 221 g in water. The volume of the cavity (Density of copper is 8.8 g/cc) is (A) 30 cc (B) 20 cc (C) 43 cc (D) 13 cc Ans (D) As Thrust = Vg, the volume of body will be

Thrust (264 221)gV 43 ccg 1 g

Now as mass of body is 264 g while the density of material of body is 8.8 g/cc, the volume of material in the body

0mass of body 264V 30 cc

density of material 8.8

So the volume of cavity = V V0 = 43 30 = 13 cc. 27. A vessel of area of cross-section A has liquid to a height H. There is a hole at the bottom of vessel

having area of cross-section a. The time taken to decrease the level from H1 to H2 will be

(A) 1 2A 2 H Ha g

(B) 2gh (C) 1 22g(H H ) (D) 1 2A g H Ha 2

Ans (A)

The average velocity of efflux, 1 22gH 2gHv

2

Let t be the time taken to empty the tank from level H1 to H2

Then, 1 21 2

2gH 2gHa t A[H H ]

2

1 2

1 2

A 2 H Hta g H H

1 2 1 2

1 2 1 2

(H H ) H HA 2a g H H H H

1 2A 2 H Ha g

11

28. Two separate air bubbles of radii 0.002 m and 0.004 m, formed of the same liquid of surface tension 0.07 N m1 come together to form a double bubble. The radius of curvature of the surface common to both bubbles is (A) 0.004 m (B) 0.003 m (C) 0.002 m (D) 0.001 m Ans(A) Let r1 and r2 be the radii of the bigger and smaller bubbles. Let r be the radius of curvature of the surface common to both bubbles.

rT4pand

rT4pp;

rT4pp

202

101

and 1212

12 r1

r1

r1

r1

r1T4

rT4ppp

m004.0)002.0(

)004.0)(002.0(rr

rrr

21

21

29. A one litre glass flask contains some mercury. It is found that at different temperatures the volume of air inside the flask remains the same. If coefficient of linear expansion of glass is 9 106 /C while of volume expansion of mercury is 1.8 104 /C, the volume of mercury in flask is (A) 50 cc (B) 100 cc (C) 150 cc (D) 200 cc Ans (C) It is given that the volume of air in the flask remains the same. This means that the expansion in volume of the vessel is exactly equal to the volume expansion of the mercury, i.e., VG = VL or VGG = VLL

6

G GL 4

L

V 1000 (3 9 10 )V 150 cc1.8 10

30. 10 g of ice at 20 C is added to 10 g of water at 50 C. Specific heat of water = 1 cal /gmC, specific heat of ice = 0.5 cal /gm-C. Latent heat of ice = 80 cal/gm. Then, resulting temperature is (A) 20 C (B) 15 C (C) 0 C (D) 50 C Ans (C) Let resulting temperature = 0 C Q1 = heat given by water if it was to cool upto 0 C = 10 1 (50 0) = 500 cal and Q2 = heat required by ice to convert totally into water at 0 C = heat required to raise the temperature of ice from 20 C + 0 C + heat required to melt 10 gm of ice at 0 C into water at 0 C = 10 0.5 20 + 10 80 = 100 + 800 = 900 cal As Q1< Q2, hence whole of the ice cannot melt. Initially, 100 cal of heat will be used up in raising the temperature of ice to 0 C and the rest 400 cal will be available for melting of ice. If it melts mgm of ice, then m 80 = 400, i.e., m = 5 gm Hence, only 5 gm of ice will melt and the remaining 5 gm of ice will remain in the mixture as ice at 0 C. The amount of water in the mixture = 10 gm + 5 m = 15 gm Final temperature of mixture = 0 C

r1 r2

p1 p2

2ME1718PCBT0S 12

31. A metal sphere of radius r and specific heat S is rotated about an axis passing through its centre at a speed of n rotations per second. It is suddenly stopped and 50% of its energy is used in increasing its temperature. Then, the rise in temperature of the sphere is

(A) 2 2 22 n r5S (B)

2 2

2n

10r S (C) 2 27 r n S

8 (D)

25( rn)14S

Ans (A)

Moment of inertia of a sphere, 22I mr5

Given: = 2n rad s1

2 2 2 2 2 21 1 2 4KE I mr (2 n) m r n2 2 5 5

Half of this energy is converted into heat

i.e., 2 2 24 1dQ m r n5 2

Specific heat, 1 dQSm dT

Or dQdTms

Or 2 2 22 m r n

5dTmS

Or 2 2 22 r ndT5S

32. Six identical conducting rods are joined as shown in figure. Points A and D are maintained at temperatures 200 C and 20 C respectively. The temperature of junction B will be

(A) 120 C (B) 100 C (C) 140 C (D) 60 C Ans (C) Equivalent electrical circuit, will be as shown in figure

Temperature difference between A and D is 180 C, which is equally distributed in all the rods. Therefore, temperature difference between A and B will be 60 C, or temperature of B should be 140 C.

33. Three moles of an ideal monoatomic gas perform a cyclic process shown in figure. The gas temperatures in different states are: T1 = 400 K, T2 = 800 K, T3 = 2400 K, T4 = 1200 K. The work done by the gas during the cycle will be (A) 20 kJ (B) 60 kJ (C) 40 kJ (D) 80 kJ

13

Ans (A) W12 = 0, W23 = PV = nRT = (3) (8.3) (2400 800) = 39840 J 40 kJ W34 = 0, W41 = PV = nRT = (3) (8.3) (400 1200) = 19920 = 20 K Total work done during the cycle = 40 kJ 20 kJ = 20 kJ

34. An ideal gas ( = 1.5) is expanded adiabatically. To reduce the root mean square velocity of molecules 2.0 times, the gas should expand by (A) 4 times (B) 16 times (C) 8 times (D) 2 times Ans (B)

rms rms3RTv or v TM

vrms is to be reduced two times, i.e., temperature of the gas will have to reduce four times or T 1T 4

During adiabatic process 1 1TV T V

1

1V TV T

1

21.5 14 4 16 V 16 V

35. If P

V

C ,C

CP being the molar specific heat of a gas at constant pressure and C its molar specific heat at

constant volume, number of degrees of freedom of a molecule of gas can be expressed as

(A) 5 ( 1)2 (B) 7 1

3 1

(C) 3 ( 1)2 (D) 2

1

Ans (D) 2 21 ff 1

36. Given diagram shows an ideal gas taken from state 1 to 2 through optional paths, A, B, C. Let Q, W and U represent the heat supplied to, the work done by, and the internal energy of the gas, respectively. Then which of the following conditions is true? (A) QA< QB< QC

(B) WA> WB> WC

(C) U1 U2 = QA WA

(D) QA WA< QB WB< QC WC

Ans (B) From first law of thermodynamics dQ = dU + dW In this case, Q = U + W

2ME1718PCBT0S 14

Here, U depends only on initial and final states, so U will be same for all curves. W represents the area surrounded by the curve or work done. so, WA> WB> WC or QA> QB> QC. Here, choice (B) is correct.

37. A Carnot engine operating between temperatures T1 and T2 has efficiency 16

.When T2 is lowered by

62 K, its efficiency increases to 13

. Then T1 and T2 are, respectively

(A) 372 K and 310 K (B) 372 K and 330 K (C) 330 K and 268 K (D) 310 K and 248 K Ans (A) The efficiency of Carnot engine.

2

1

T1T

2

1

1 T16 T

1Given,6

2 21

1

T 5 6Tor TT 6 5

... (i)

As per question, when T2 is lowered by 62 K,

then its efficiency becomes 13

.

2

1

1 T 6213 T

2

1

T 62 11T 3

[Using eqn. (i)] 2

2

5 T 62 26T 3

5T2 310 = 4T2 T2 = 310 K From equation (i)

T1 = 6 310 372 K5

38. An object of mass m is placed on a horizontal platform. The platform is performing SHM along a vertical line with an amplitude ‘A’. It is observed if the oscillation frequency is more than 0, the object leaves contact with the platform, then 0, is

(A) g2A

(B) g2A

(C) 1 g2 A

(D) Ag

Ans (C) The object will be just leaving the contact when

m2 A mg 2 gA

0

1 g2 A

15

39. On a smooth inclined plane, a body of mass m is attached between two massless springs. The other ends of the springs are fixed to firm supports. If each spring has force constant k, the period of oscillation of the body is

(A) m22k

(B) 2m2k

(C) mgsin22k

(D) 2mgsin2

k

Ans (A) keq = k1 + k2 = 2k

40. A mass M, attached to a horizontal spring, executes SHM with a amplitude A1. When the mass M passes through its mean position then a smaller mass m is placed over it and both of them move together with

amplitude A2. The ratio of 1

2

AA

is

(A) MM m

(B) M mM (C)

1/2MM m

(D)

1/2M mM

Ans (D)

1MT 2k

… (i)

When a mass m is placed on mass M, the new system is of mass = (M + m) attached to the spring. New time period of oscillation,

2(m M)T 2

k

… (ii)

Consider v1 is the velocity of mass M passing through mean position and v2 velocity of mass (m + M) passing through mean position. Using, law of conservation of linear momentum, Mv1 = (m + M)v2 M(A11) = (m + M)(A22) ( v1 = A11 and v2 = A22)

or 1 2 1

2 1 2

A (m M) m M TA M M T

1 21 2

2 2andT T

1

2

A m MA M

[Using eqn.(i) and (ii)]

41. A uniform spring has an unstretched length l and a force constant k. The spring is cut into two parts of unstretched length l1 and l2 such that l1 = l2, where is an integer. The corresponding force constants k1 and k2 are

(A) k and k( + 1) (B) k 1and k 1

(C) k 1and k 1

(D) k 1

and k 1

Ans (D) If a mass m is hung from the lower end of an unstretched length l of the spring and x is the extension produced in the spring, then

2ME1718PCBT0S 16

kx0 + mg = 0 or k = 0

mgx

Now, 0 0x or x c l l

k = mgc

l

As the spring is cut into two pieces of lengths l1and l2, so l1 + l2 = l, As l1 = l2

2 1and1 1

l ll l

Force constants k1 and k2 for pieces of lengths l1 and l2 are:

11

1mg mg 1 mg 1k kc c c

l l l and

2

2

mg mg n 1k k 1c c

l l

42. The fundamental frequency of a sonometer wire increases by 6 Hz if its tension is increased by 44% keeping the length constant. Find the change in the fundamental frequency of the sonometer wire, when the length of the wire is increased by 20% keeping the original tension in the wire. (A) 10 Hz (B) 2 Hz (C) 5 Hz (D) 1 Hz Ans (C) In case of vibration of a string, fundamental frequency is given by

1 Tf2L m

So if length of given wire is kept constant,

1/2f T

f T

and as here f = f + 6 and T = T + 0.44 T = 1.44 T

(f 6) 1.44T ,f T

i.e., f = 30 Hz

Now if keeping the original tension (T), the length of given wire is changed,

ff 1.20

l ll

[as l = l + 0.20 l = 1.20 l]

so 30f 25 Hz1.2

and hence f = ff = 25 30 = 5 Hz i.e., fundamental frequency will decrease by 5 Hz.

43. A uniform rope of length 12 m and mass 6 kg hangs vertically from a rigid support. A block of mass 2 kg is attached to the free end of the rope. A transverse pulse of wavelength 0.06 m is produced at the lower end of the rope. The wavelength of the pulse when it reaches the top of the rope is (A) 0.6 m (B) 1.2 m (C) 0.12 m (D) 0.06 m Ans (C) As the rope has a mass and a mass is also suspended from the lower end, the tension in the rope will be different at different points.

17

Now as Tvm

T T

B B

v T (6 2)g 2v T 2g

or T T

B B

f 2f

[as v = f]

Here fT = fB as frequency is the characteristic of the source producing the waves. So T = 2B = 2 0.06 = 0.12 m

44. A knife edge divides a sonometer wire into two parts which differ in length by 2 mm. The whole length of the wire is 1 metre. The two parts of the string when sounded together produce one beat per second; then the frequencies of the smaller and longer parts are (in Hz) (A) 250.5 and 249.5 (B) 249.5 and 250.5 (C) 124.5 and 125.5 (D) 125.5 and 124.5 Ans (A) L1 + L2 = 100 … (i) L1 L2 = 0.2 … (ii) Solving, we get; L1 = 50.1 cm and L2 = 49.9 cm

Now, 1 Tn2 50.1 m

or 1 Tn(50.1)2 m

… (iii)

Similarly, 1 Tn 12 49.9 m

or 1 T(n 1)(49.9)2 m

Hence, (n + 1) (49.9) = n(50.1) Solving, we get; n = 249.5 Hz and n + 1 = 250.5 Hz

45. An open organ pipe of length l is sounded together with another open organ pipe of length l + x in their fundamental mode. Speed of sound in air is v. The beat frequency heard will be (x < < l)

(A) 2

vx4l

(B) 2v

2xl (C) 2

vx2l

(D) 2vx

2l Ans (C) Beat frequency = f1 f2

v v2 2( x)

l l

1v x1 12

l l

2

v x vx1 12 2 l l l

PART II - CHEMISTRY

46. Number of hydrogen ions present in 10 millionth part of 1.33 cm3 of pure water at 25 C is (A) 6.023 million (B) 60 million (C) 8.01 million (D) 80.23 million Ans (C) Now, [H+] = 10–7 mol/L [ For water, pH = 7]

Now, 1000 mL of water contains 107 moles of H+ ions 1 mL of water contains 710

1000

moles of H+ ions

1.33 10–7 mL of water contains 1.33 10–17 moles of H+ ions [10 millionth = 107 so, 10 millionth part of 1.33 cm3 = 1.33 107 mL]

2ME1718PCBT0S 18

So, no. of H+ ions = 1.33 1017 NA = 1.33 1017 6.023 1023 = 8.01 106 = 8.01 million

47. 2.5 g of the carbonate of a metal was treated with 100 mL of 1 N H2SO4. After the completion of the reaction, the solution was boiled off to expel CO2 and was then titrated against 1 N NaOH solution. The volume of alkali that would be consumed (the equivalent mass of the metal is 20) is (A) 50 ml (B) 25 ml (C) 75 ml (D) 100 ml Ans (A) Equivalent mass of metal carbonate = 20 + 30 = 50

2.5 g of metal carbonate = 2.550

= 0.05 eq.

Number of equivalents of H2SO4 taken = 100 1 0.1 eq.1000

Number of equivalents of H2SO4 remains unreacted = 0.1 0.05 = 0.05 eq. Number of equivalents of alkali consumed = 0.05 eq.

Number of equivalents of NaOHV NormalityNaOH1000

V 10.051000

V = 50 ml 48. 1.5 litre of a solution of normality N and 2.5 litres of 2 M HCl are mixed together. The resultant solution

has a normality 5. The value of N is (A) 6 (B) 10 (C) 8 (D) 4 Ans (B) N1V1 + N2V2 = N3V3 N 1.5 + 2 2.5 = 5 4 1.5 N = 20 – 5 = 15

15N 101.5

49. The nucleus and an atom can be assumed to be spherical. The radius of the nucleus of mass number A is

given by 1.25 1013 13A cm. The atomic radius of atom is 1 Å. If the mass no. is 64, the fraction of

the atomic volume that is occupied by nucleus is (A) 1.0 103 (B) 5.0 105 (C) 2.5 102 (D) 1.25 1013 Ans (D)

3 1n 3 13 33

n n3 8 3

3a aa

4 rV r [1.25 10 (64) ]3

4V r (10 )r3

50. The 1st ionization energy of H atom is 13.6 eV, then the 2nd ionization energy of He+ ion is (A) 27.2 eV (B) 40.8 eV (C) 54.4 eV (D) 108.8 eV Ans (C)

2

n 1 H2

ZE 13.6 eV, (E ) 13.6 eVn

19

2 2

1 2He

Z 2(E ) 13.6 13.6 54.4 eVn 1

The second I.E. = E 1 He(E ) = 0 –(54.4) eV = 54.4 eV

51. The radius of Bohr’s first orbit in H-atoms is 0.053 nm. The radius of second orbit in He+ would be (A) 0.0265 nm (B) 0.0530 nm (C) 0.1060 nm (D) 0.2120 nm Ans (C)

r2 for 2nHe 0.053

z

40.053 ( n 2, z 2)2

r2 for He+ = 0.053 × 2 = 0.1060 nm 52. Statement-I: The magnetic moment of K-atom is more than Mg-atom as the former has one electron in

outermost shell. Statement-II: The magnetic moment of N-atoms is more than magnetic moment of O-atom and former has more number of unpaired electrons. (A) Statement-I and Statement-II are true and Statement-II is the correct explanation of Statement-I. (B) Statement-I and Statement-II are true, but Statement-II is not the correct explanation of Statement-I (C) Statement-I is true, but Statement-II is false (D) Statement-I is false, but Statement-II is true. Ans (B) Magnetic moment depends upon number of unpaired electrons and given by n(n 2), where n is

no. of unpaired electrons. Valence electron Unpaired electron

K 1 1 Mg 2 0 N 3 3 O 4 2

53. Among the following, the highest electron gain enthalpy is for

(A) F (B) O (C) O (D) Na Ans (C) In general, electron gain enthalpy becomes more and more negative from left to right in a period. This is due to decrease in size and increase in nuclear charge as the atomic number increases in a period. In O negatively charged electron is added to negatively charged O ion which causes repulsion and hence electron affinity value is positive.

54. The decreasing order of the first ionization energy (in kJ mol1) of He, Mg and Na is He > Mg > Na. The increasing order of the 2nd ionization energy (in kJ mol1) of these elements will be (A) Na < Mg < He (B) Mg < Na < He (C) Mg < He < Na (D) Na < He < Mg Ans (B)

2 11H1e

He : 1s 1s

2 2 6 1 2 2 61H

1eNa :1s , 2s , 2p , 3s 1s , 2s , 2p

2ME1718PCBT0S 20

2 2 6 2 2 2 6 13H1eMg : 1s , 2s , 2p , 3s 1s , 2s , 2p , 3s

He has highest value of 2nd ionisation energy due to smallest size. After removal of one electron from Na, the Na+ ion has inert gas configuration. To remove the electron from this configuration, a very high energy is required, hence Na higher 2nd ionisation energy than Mg. So, the order of 2nd ionisation energy is Mg < Na < He.

55. Maximum dipole moment value is possible in XY2 if the angle is (A) 120 (B) 90 (C) 180 (D) 150 Ans (B) Resultant dipole moment maybe calculated using vectorial method

2 2R 1 1 1 22 cos

cos = cos 120 = – 0.5 cos 90 = 0.0 cos 180 = – 1 cos 150 = – 0.87 so, if the angle between the atoms is 90, it will have highest dipole moment.

56. In 34PO ions, the formal charge on each oxygen atom and PO bond order respectively are

(A) –1.0, + 0.75 (B) –0.75, + 1.25 (C) –0.75, + 10 (D) –0.75, + 0.75 Ans (B) The structure of 3

4PO is

O

PO

O

O

3–

Here three units negative charge is shared by four atoms. Formal charge = 3 0.75

4

It has four resonating structures. Bond order of PO bond

Totalnumber of bonds between two atoms in all structuresTotalnumber of resonating strucures

2 1 1 1 5 1.254 4

57. A certain gas takes three times as long to effuse out as helium. Its molecular mass will be (A) 27 u (B) 36 u (C) 64 u (D) 9 u Ans (B) According to Graham’s law of diffusion

1 2

2 1

r M1 1rr Md M

Rate of diffusion = Volume of gas diffused (v)Time taken (t)

21

24.6

23.1

21.6

20.1

2.0 3.0 1V

(mol litre1)

PV (l

itre-

atm

mol

1)

0

[Graph not to scale]

1

1 2

2 1

2

Vt MV Mt

If same volume of two gases diffuse, then V1 = V2

2 2

1 1

t Mt M

Here t2 = 3t1, M1 = 4 u, M2 = ?

1 2 2

1

3t M M3t 4 4

22

M9 M 36 u4

58. For one mole of a van der Waal’s gas when b = 0 and T = 300 K, the PV vs. 1V

plot is shown below.

The value of the van der Waals’ constant ‘a’ (atm litre2 mol2) is

(A) 1.0 (B) 4.5 (C) 1.5 (D) 3.0 Ans (C) van der Waals’ equation for 1 mol of real gas is

2

aP [V b] RTV

Given that b = 0

2

aP (v) RTV

PV = RT aV

Following y = mx + c for the curve PV vs 1V

Slope = a

Slope = 21.6 20.1 1.52 3

a = 1.5

59. The heats of atomization of PH3(g) and P2H4(g) are 954 kJ mol1 and 1485 kJ mol1 respectively. The PP bond energy in kJ mol1 is (A) 213 (B) 426 (C) 318 (D) 1272 Ans (A) In PH3(g), energy required to break 3 P H bonds = 954 kJ mol1

Energy required to break 1 P H bond = 1954 318 kJ mol3

In P2H4(g), energy of 1 P P bond + 4 P H bonds = 1485 kJ mol1 Energy of 1 P H bond = 318 kJ mol1

24.6

23.1

21.6

20.1

2.0 3.0 1v

(mol litre1)

PV (l

itre-

atm

mol

1)

0

[Graph not to scale]

2ME1718PCBT0S 22

Energy of 4 P H bonds = 318 = 1272 kJmol1 .

Thus, the P P bond energy = 1485 1272 = 213 kJmol1

60. Equal volumes of equimolar hydrochloric acid and sulphuric acid are neutralised by dilute NaOH solution. ‘x’ kcal and ‘y’ kcal of heat are liberated respectively. The relationship between x and y is

(A) x = y (B) yx2

(C) x = 2y (D) xy2

Ans (B) By equal volumes of molar HCl and H2SO4 the ratio of H+ ions furnished = 1 : 2

Ratio of heat liberated = x 1 yxy 2 2

61. Internal energy and pressure of gas of unit volume are related as

(A) 2P U3

(B) 3P U2

(C) UP2

(D) P = 2U

Ans (A) P 1 = RT

Also internal energy, 3U RT2

3U P2

or 2P U3

62. N2O4 is dissociated to 33% and 40% at total pressure P1 and P2 atm respectively. Then the ratio 1

2

PP

is

(A) 74

(B) 73

(C) 83

(D) 85

Ans (D) (I) 2 4 2

1 0.33 0.66N O 2NO

Total moles = 1.33

2 4N O 10.67P P1.33

2NO 10.66P P1.33

2

2 4

2

1NO

P 1N O

1

0.66 PP 1.33K 0.488 P0.67P P1.33

(II) 2 4 21 0.4 0.8

N O 2NO

Total moles = 1.4

2 4N O 2

0.6P P1.4

2NO 2

0.8P P1.4

23

2

2

P 2

2

0.8 P1.4K 0.761 P0.6 P1.4

0.488 P1 = 0.761 P2

1

2

P 0.761 81.56P 0.488 5

63. Weak electrolytes are only partly dissociated in solutions. The extent to which they are dissociated in a given solution is known as the ‘Degree of dissociation’. Among the following statements the one that is true for the degree of dissociation in aqueous solution is (A) It is inversely proportional to the square root of initial concentration of the electrolyte. (B) It is directly proportional to the initial concentration of the electrolyte. (C) It is independent of the initial concentration of the electrolyte. (D) It depends on the equilibrium concentration of the electrolyte. Ans (A)

c2 = Ka aK 1orC c

.

64. In the reaction, N2 NH3. The equivalent mass of N2 and NH3 are respectively equal to

(A) 28 17,3 3

(B) 28 17,6 3

(C) 28 17,2 2

(D) 28 17,5 5

Ans (B)

0 326e (N ) 2N

2 3N NH

28 17E ; E6 3

65. Among the following processes the one that is used for the manufacture of calcium is (A) reduction of CaO with carbon (B) reduction of CaO with hydrogen (C) electrolysis of a mixture of anhydrous CaCl2 and KCl (D) electrolysis of molten Ca(OH)2

Ans (C) Calcium is obtained by electrolysis of a fused mass consisting of six parts CaCl2 and one part CaF2 at about 700 C in an electrolytic cell made of graphite which acts as anode and a water cooled cathode of iron.

22CaCl Ca 2Cl

At anode: 22Cl Cl 2e At cathode: 2Ca 2e Ca

66. Borazole is obtained by reaction of (A) NH3 + B2H6 in 2 : 1 ratio (B) NH3 + B2H6 in 1 : 2 ratio (C) NH3 + B2H6 in 1 : 4 ratio (D) NH3 + B2H6 in 4 : 1 ratio Ans (A)

523 K2 6 3 3 3 6 23B H 6NH 2B N H 12H

2ME1718PCBT0S 24

67. Among the halides 1. BCl3 2. AlCl3 3. GaCl3 4. InCl3 The order of decreasing Lewis acid character is (A) 1, 2, 3, 4 (B) 4, 3, 2, 1 (C) 3, 4, 2, 1 (D) 2, 3, 4, 1 Ans (B) The acidic character of chlorides increases down the group BCl3 is weak acid to show p p back bonding.

68. Antidote for CO poisoning is (A) carborundum (B) pure CO2 (C) carbogen (D) carbonyl chloride. Ans (C) Antidote for CO poisoning is carbogen which is a mixture of 95% O2 and 5% CO2.

69. Identify the correct statement about the following pairs of compounds.

A

OH OH

B

O

Ph

CO

Ph

D

CO2Me CO2Me

(A) A and B diastereomers, C and D diastereomers (B) A and B enantiomers, C and D diastereomers (C) A and B diastereomers, C and D enantiomers (D) A and B enantiomers, C and D enantiomers. Ans (C) C and D are mirror images of each other so, they are enantiomers. A and B have same molecular structure but do not have mirror image relationship hence, they are diastereomers.

70. Among the following tautomerism is exhibited by

OO O

PhPh

I II III (A) III only (B) Both I and III (C) Both I and II (D) Both II and III Ans (A) -Hydrogen at bridge carbon never participates in tautomerism. Thus, only (III) exhibits tautomerism.

O

Hketo - form

(III)

OH

enol - form

25

71. 29.5 mg of an organic compound containing nitrogen was digested according to Kjeldhal’s method and the evolved ammonia was absorbed in 20 mL of 0.1 M HCl solution. The excess of the acid require 15 mL of 0.1 M NaOH solution for complete neutralization. The percentage of nitrogen in the compound is (A) 29.5 (B) 59.0 (C) 47.4 (D) 23.7 Ans (D) Meq. of HCl absorbed = 20 0.1 = 2 Meq. of HCl left = 15 0.1 = 1.5 Meq of NH3 formed = 0.5

(N V) 1.4 N V% of N

w

3

1.4 0.5 23.7329.5 10

72. The structure of cis-bis (propenyl) ethene is

(A) H

H (B)

H H (C) (D)

Ans (D) The two propenyl group attached to 1, 2position of carbon in cis-form.

CH3

CH CHHC CH

CH

CH3

CH 73. The reagent for the following conversion Br Br H H is /are

(A) alc. KOH (B) alc. KOH followed by NaNH2 (C) aqueous KOH followed by NaNH2 (D) Zn/CH3OH Ans (B)

Br Br

CH2 CH2KOH alc. CH2 CHBr NaNH2 CH CH

Vinyl bromide looses HBr only by strong base.

74. Cyclohexene on ozonolysis followed by reaction with Zn dust and H2O gives E. E on further treatment with acidified KMnO4 gives

(A) CHO

(B) CHO

(C) COOH

(D) COOHCOOH

Ans (D)

O3

Zn/H2OCHOCHO

H+/KMnO4

oxidation COOHCOOH

75. Among the following the correct statement is

(A) Chloroacetic acid is less acidic than acetic acid because chlorine atom has I effect. (B) The greater the branching in a paraffin the lower is its boiling point. (C) Kjeldahl’s method is used for the estimation of azo nitrogen. (D) Duma’s method is used for estimation of sulphur.

2ME1718PCBT0S 26

Ans (B) The greater the branching, smaller is the surface area, lesser is attraction among molecules and so low boiling point.

76. Liquid hydrocarbon can be converted to a mixture of gaseous hydrocarbons by (A) oxidation (B) cracking (C) hydrolysis (D) distillation under reduced pressure Ans (B)

77. The correct bond order in the following species is (A) 2

2 2 2O O O (B) 22 2 2O O O (C) 2

2 2 2O O O (D) 22 2 2O O O

Ans (C) (a) 2O ion: Total number of electrons (16 – 1) =15.

Electronic configuration: 2 2 2 2 2 2 2 2 1 2

x y z y z 2 2 21s *1s 2s *2s 2p 2p 2p * 2p *2p O O O

b a(N N ) 10 7 3 1Bond order 12 2 2 2

(b) 2

2O

ion: Total number of electrons = (16 – 2) =14 Electronic configuration: 2 2 2 2 2

y1s *1s 2s *2s 2p 2 2z x2p 2p

b a(N N ) 10 4 6Bond order 32 2 2

(c) 2O Total number of e– = (16 +1) = 17

Bond order b a1[N N ]2

1 1[10 7] 1 1.52 2

So bond order = 22 2 2O O O

78. Molecular shapes of SF4, CF4 and XeF4 are: (A) the same with 2, 0 and 1 lone pair of electrons (B) the same with 1, 1 and 1 lone pair of electrons (C) different with 0, 1 and 2 lone pair of electrons (D) different with 1, 0 and 2 lone pair of electrons Ans (D)

SF

..F

FF

C

F

F

F

F Xe

..

..

F

F

F

F

sp3d hybrid(Trigonal bipyramidal

shape)(Contains one lone pair)

sp3d hybrid(Tetrahedral shape)

(No lone pair)

sp3d hybrid(Square planar shape

of octahedral)(Contanis two lone pair)

SF4 12

(6 + 4) = 5 sp3d 4 1

CF4 12

(4 + 4) = 4 sp3 4 0

XeF4 12

(8 + 4) = 6 sp3d2 4 2

27

79. Commercial 11.2 volume H2O2 solution has molarity of (A) 1.0 (B) 0.5 (C) 11.2 (D) 1.12 Ans (A)

2 2 2 21H O H O O2

1 mol 1 M(1L) 11.2 L O2 1 ml H2O2 11.2 ml = 1M solution.

80. In the replacement reaction C X MF+ C F MX+

The reaction will be most favourable if M happens to be (A) Na (B) K (C) Rb (D) Li Ans (C)

C X MF+ C F MX+

Since Rb is large in size, RbF has low bond dissociation enthalpy. 81. Potassium superoxide (KO2) is used in space capsules, submarines, and breathing masks, because it is

(A) diamagnetic in nature (B) high melting compound (C) helpful in removing CO2 and producing O2 (D) unstable compound Ans (C) KO2 reacts with CO2 liberating O2 4KO2 + 2CO2 2K2CO3 + 3O2

82. Variation of temperature (T) in the troposphere region with height (km) is

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

Ans (D) The temperature of atmosphere falls continuously for the first 10 km from sea level.

83. In the following carbocations, the stability order is

CH3

CH3

CH2RCH2CH2 C+ C+ +

(I) (II) (III) (IV)

(A) III > II > IV > I (B) IV > I > II > III (C) IV > III > II > I (D) III > IV > II > I Ans (A) Cyclopropylmethyl carbonation is more stable than benzyl carbonation.

2ME1718PCBT0S 28

84. The relative reactivity of 1H, 2H and 3H in bromination reaction has been found to 1 : 82 : 1600 respectively. In the reaction

CH3 CH3

CH3 CH3

Br

CH3

CH3 C CH3 CH3 CH CH2 BrCH (excess) + Br2h

(A) (B)

+

The percentage yields of the products (A) and (B) are expected to be (A) 99.4 %, 0.6% (B) 50%, 50% (C) 0.6%, 99.4% (D) 80%, 20% Ans (A)

CH3 CH3

CH3

Br

CH3

CH3

CH3 C CH3 CH3 CH CH2 .BrCH + Br2h

(A) (B)

+

1600 1 : 9 1

1600% of A 100 99.44%1609

9% of B 100 0.6%1609

85. A dye absorbs a photon of wavelength and re-emits the same energy into two photons of wavelengths 1 and 2 respectively. The wavelength is related with 1 and 2 as

(A) 1 2

1 2

(B) 1 2

1 2

(C)

2 21 2

1 2

(D) 1 2

21 2( )

Ans (B) hh+ h

1 2

hc hc hc

hc hc

1 2

1 2

1 2

1 2

1

1 2

1 2

86. The threshold frequency of metal is 1.6 1014 s1. If the kinetic energy of ejected electron is 4 eV. The frequency of incident light is (A) 1.125 1014 (B) 1.125 1015 (C) 1.125 1012 (D) 1.125 1013

Ans (B) 0 = 1.6 1014 s1, KE = 4 eV = 4 1.602 1019 J h = h0 + KE

= 6.626 1034 1.6 1014 + 4 1.602 1019

= 10.6 1020 + 6.4 1019 = 1.06 ×10–19 + 6.4 × 10–19 = 7.46 × 10–19 h = 7.46 × 10–19

19

1534

7.46 10 1.125 106.626 10

29

87. Oxidation state of nitrogen is incorrectly given for Compound Oxidation state

(A) [Co(NH3)5 Cl] Cl2 3 (B) NH2OH –1 (C) (N2H5)2 SO4 +2 (D) Mg3N2 –3 Ans (C) (N2H5)2 SO4

Let oxidation state of nitrogen be ‘x’ 4x + 10 – 2 = 0 or x = –2

88. Solubility product of silver bromide is 5.0 1018. The quantity of potassium bromide (molar mass taken as 120 g mol1) to be added to 1 L of 0.05 M solution of silver nitrate to start precipitation of AgBr is (A) 5.0 108 g L1 (B) 1.2 1010 g L1 (C) 1.2 109 g L1 (D) 6.2 g L1 Ans (C) [Ag+] [Br] = ksp 5.0 1013 = 0.05 [Br]

135.0 10[Br ] 0.05

[Br] = 1011 M Mass of KBr required = 1011 120 = 1.2 109 g L1

89. 40 ml of 0.1 M ammonia solution is mixed with 20 ml of 0.1 M HCl. The pH of the mixture is (Kb = 1.77 × 10–5) (A) 5.74 (B) 9.25 (C) 4.56 (D) 7.06 Ans (B) 20 ml 0.1 M HCl = 20 ml 0.1 M NH4OH = 20 ml 0.1 M NH4Cl Let NH4OH = 20ml of 0.1 M

[Salt] = [Base] pOH = pKb = 4.75 pH = 14 – 4.74 pH = 9.25

90. The standard free energy change for a reaction is 40 kJ. Calculate the equilibrium constant for the reaction. (A) 106 (B) 107 (C) 108 (D) 109 Ans (B) G = 2.303 RT log K 40000 = –2.303 8.314 298 log K

40000log K 2.303 8.314 298

log K = 7.0103 K = antilog (7.0103) K = 107

2ME1718PCBT0S 30

PART III – BOTANY Q.No. Version 1 Version 2 Version 3 Version 4 91. D B C D 92. C A B B 93. D D C B 94. B A B C 95. D D A D 96. B D B B 97. A B A A 98. B B A B 99. A B D B

100. D B D D 101. C A A D 102. B C A C 103. D A B B 104. D D B A 105. A B D A 106. D A D A 107. B B A B 108. B C B B 109. A C A A 110. C B D A 111. A D B D 112. A A B A 113. B C A D 114. A B B B 115. D B B C 116. D B B A 117. A A D B 118. B D A A 119. B B C C 120. A C C B 121. B A D C 122. C A A C 123. C B D A 124. B D B B 125. A D C D 126. B A C B 127. B C B D

31

Q.No. Version 1 Version 2 Version 3 Version 4 128. C D A B 129. A B B A 130. B A D D 131. D D D D 132. A A A A 133. A A A A 134. D D D D 135. C C C C

PART IV – ZOOLOGY Q.No. Version 1 Version 2 Version 3 Version 4

136. C A C A 137. A B A A 138. B D B A 139. B B A C 140. C C C D 141. C D C C 142. C C C A 143. A A A C 144. B D A A 145. D C C B 146. C B A A 147. C B A D 148. A C B B 149. D C C C 150. B C D B 151. C D C C 152. B B D B 153. A A A B 154. D A B A 155. B B B A 156. B B A C 157. D C C D 158. B A C C 159. D D B C 160. A A D C 161. B C D A 162. C C C B

2ME1718PCBT0S 32

Q.No. Version 1 Version 2 Version 3 Version 4 163. A D B B 164. A A A B 165. A A B D 166. C C C C 167. D C B B 168. A A D C 169. C A A C 170. A B C C 171. B C D A 172. C C B D 173. A A A A 174. C C B D 175. A A A B 176. C B C A 177. C B C C 178. C C C C 179. B B B B 180. C C C C

* * *