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CBSE 12th

Non-Medical

The Aspire Scholarship

Test 2016 (Maths + Physics + Chemistry)

(Solved)

General Instructions:

The question paper contains 90 objective multiple choice questions.

There are three parts in the question paper consisting of

Section-A MATHEMATICS (1 to 30)

Section-B PHYSICS (31 to 60),

Section-C CHEMISTRY (61 to 90).

Each right answer carries 4 marks and Minus 1 for every wrong answer.

The paper consists of 90 questions. The maximum marks are 360.

Maximum Time 3Hrs.

Give your response in the OMR Sheet provided with the Question Paper.




Section – I {Mathematics}

1. The domain of the function, f(x) = 1

2

sin x 3is

9 x

(a) [2, 3] (b) [2, 3) (c) [1, 2] (d) [1, 2)

Ans. (b)

Solution:

Given function 1

2

sin x 3f x

9 xwill be defined, if

1 x 3 1 2 x 4 …(i)

and 9 – x2> 0 –3 < x < 3 …(ii)

From Eqs. (i) and (ii), we get

2 x 3

Hence, domain of the given function is [2, 3)

2. A function f from the set of natural numbers to integers defined by

n 1,when n is odd

2f n is

n, when n is even

2

(a) one-one but not onto (b) onto but not one-one

(c) one-one and onto both (d) neither one-one nor onto

Ans. (c)

Solution:

Given that,

n 1, when n is odd

2f n

n, when n is even

2

and f : N I, where N is the set of natural numbers and I is the set of integers.

Let x, y N both are even.

Then,f(x) = f(y)

x yx y

2 2

Again, x, y N and both are odd.




Then, f(x) = f(y)

x 1 y 1x y

2 2

So, mapping is one-one.

Since, each negative integer is an image of even natural number and positive integer is an

image of odd natural number. So, mapping is onto. Hence, mapping is one-one onto.

3. If the system of linear equations

x + 2ay + az = 0

x + 3by + bz = 0

and x + 4cy + cz = 0

has a non-zero solution, then a, b, c

(a) are in AP (b) are in GP

(c) are in HP (d) satisfy a + 2b + 3c = 0

Ans. (c)

Solution :

Since, the system of linear equations has a non-zero solutions, then

1 2a a

1 3b b 0

1 4c c

Applying 2 2 1 3 3 1R R R ,R R R

1 2a a

0 3b 2a b a 0

0 4c 2a c a

3b 2a c a 4c 2a b a 0

3bc – 3ba – 2ac + 2a2

= 4bc – 2ab – 4ac + 2a2

2ac = bc + ab

On dividing by abc both sides, we get

2 1 1

b a c

a, b, c are in HP.




4. Let A =

1 1 1 4 2 2

2 1 3 and 10 B 5 0

1 1 1 1 2 3

If B is the inverse of matrix A, then is

(a) – 2 (b) 1 (c) 2 (d) 5

Ans.(d)

Solution:

Since, B is the inverse of matrix A, ie, B= A–1.

1

4 2 2

10A 5 0

1 2 3

1

4 2 2

10A A 5 0 A

1 2 3

4 2 2 1 1 1

10I 5 0 2 1 3

1 2 3 1 1 1

1A A I

10 0 0 10 0 0

0 10 0 5 5 5

0 0 10 0 0 10

5 0 and 5 10

5

5. Let a, b, c be such that (b + c) 0.

If n 2 n 1 n

a a 1 a 1 a 1 b 1 c 1

b b 1 b 1 a 1 b 1 c 1 0

c c 1 c 1 1 a 1 b 1 c

Then the value of ‘n’ is

(a) zero (b) any even integer (c) any odd integer (d) any integer

Ans. (c)

Solution:




1

n

a a 1 a 1 a 1 b 1 c 1

b b 1 b 1 1 a 1 b 1 c 1

c c 1 c 1 a b c

= n

a a 1 a 1 a 1 a 1 a

b b 1 b 1 1 b 1 b 1 b

c c 1 c 1 c 1 c 1 c

= n 1

2 3

a a 1 a 1 a 1 a a 1

b b 1 b 1 1 b 1 b b 1 C C

c c 1 c 1 c 1 c c 1

= n 2

a a 1 a 1

1 1 b b 1 b 1

c c 1 c 1

This is equal to zero only, if n + 2 is odd ie, n is an odd integer.

6. The number of 3 × 3 non-singular matrices, with four entries as 1 and all other entries as 0, is

(a) less than 4 (b) 5 (c) 6 (d) at least 7

Ans. (d)

Solution:

Consider

1 * *

* 1 *

* * 1

. By placing a1 in any one of the 6* position and 0 elsewhere. We get 6 non-

singular matrices.

Similarly,

* * 1

* 1 *

1 * *

gives at least one non-singular.

7. The normal to a curve at P(x, y) meets the x-axis at G. If the distance of G from the origin is

twice the abscissa of P, then the curve is a

(a) ellipse (b) parabola (c) circle (d) hyperbola

Ans. (d)

Solution:

Let the equation of normal is

dxY y X x

dy




It meets the x-axis at G. Therefore, coordinates of G are dy

x y ,0 .dx

According to given condition,

dyx y 2x

dx

dyy x

dx

y dy x dx

On integrating, we get

2 2y xc

2 2

2 2x y 2c

8. Let y be an implicit function of x defined by x2x – 2xx cot y – 1 = 0. Then, y’(1) equals

(a) – 1 (b) 1 (c) log 2 (d) – log 2

Ans. (a)

Solution :

2x xx 2x cot y 1 0 …(i)

Now, x = 1,

1 – 2 cot y – 1 = 0

cot y = 0

y2

On differentiating Eq. (i), w. r. t. ‘x’, we get

2x x 2 xdy2x 1 log x 2[x cosec y cot y x 1 log x ] 0

dx

At 1, ,2

1,2

dy2 1 log 1 2 1 1 0 0

dx




1,2

dy2 2 0

dx

1,2

dy1

dx

9. The values of p and q for which the function

2

3/2

sin p 1 x sinx,x 0

x

f x q ,x 0

x x x,x 0

x

is continuous for all x in R, are

(a) 5 1

p , q2 2

(b) 3 1

p , q2 2

(c) 1 3

p ,q2 2

(d) 1 3

p ,q2 2

Ans. (b)

Solution:

Here,

2

3

2

sin p 1 x sin x, x 0

x

f x q , x 0 Since,

x x x, x 0

x

f(x) is continuous for x R.

Continuous at x = 0.

RHL at x= 0.

2

3h 0 h 02

h h h h { h 1 1}lim lim

h hh

h 0

h 1 1 h 1 1lim

h h 1 1

= h 0

h 1 1lim

h{ h 1 1}

h 0

1 1lim

2h 1 1




LHL at x = 0

h 0

sin p 1 h sin hlim

h

= h 0

sin p 1 h sin hlim

h h

p 1 1 P 2 ..(ii)

f(0) = q ..(iii)

From Eqs. (i), (ii) and (iii)

1q p 2

2

3 1p ,q

2 2

10. How many real solutions does the equation x7 + 14x5 + 16x3 + 30x – 560 = 0 have ?

(a) 5 (b) 7 (c) 1 (d) 3

Ans. (c)

Solution :

Let f(x) = x7 + 14x5 + 16x3 + 30x – 560

f’(x) = 7x6 + 70x4 + 48x2 + 30 > 0, x R

f(x) is increasing.

f(x) = 0 has only one solution.

11. Let and be the distinct roots of ax2 + bx + c = 0, then 2

2x

1 cos ax bx clim

x is equal to

(a) 21

2 (b)

22a

2 (c) 0 (d)

22a

2

Ans. (d)

Solution:

Now, 2

2x a

1 cos ax bx clim

x

=

22

2x

ax bx c2 sin

2lim

x




=

22

2

2x2 2

a2sin x x

a2lim x

2ax x

2

= 2

2

x

alim x

2

x 0

sin xlim 1

x

= 2

2a

2

12. The normal to the curve x = a cos sin , y a sin cos at any point is such that

(a) it is at a constant distance from the origin (b) it passes through a

, a2

(c) it makes angle 2

with the x-axis (d) it passes through the origin

Ans. (a)

Solution:

Given that, x = a cos sin

and y a sin cos

On differentiating w. r. t. respectively, we get

dxa sin sin cos

d

dxa cos

d ..(i)

and dy

a cos sind

dya sin

d ..(ii)

On dividing Eqs. (ii) by (i), we get

dytan

dx

Since, slope of normal

= dx

cotdy

So, equation of normal is




y a sin a cos

= cos

x a cos a sinsin

2y sin a sin a cos sin

= x cos y sin a

It is always at a constant distance ‘a’ from origin.

13. Let f(x) be a polynomial function of second degree. If f(1) = f(–1) and a, b, c are in AP, then

f’(a), f’(b) and f’(c) are in

(a) AP (b) GP

(c) HP (d) Arithmetico-Geometric Progression

Ans. (a)

Solution:

Let f(x) = Ax2 + Bx + C

f(1) = A + B + C

and f(–1) = A – B + C

f(1) = f(–1) (given)

A + B + C = A – B + C

2B = 0 B = 0

f(x) = Ax2 + C

f’(x) = 2 Ax

f’(a) = 2 Aa, f’(b) = 2Ab

and f’(c) = 2Ac

Also, a, b, c are in AP.

2Aa, 2Ab, 2Ac are in AP.

f’(a), f’(b), f’(c) are in AP.




14. x 0

1 cos 2xlim is

2 x

(a) (b) –1 (c) zero (d) does not exist

Ans. (d)

Solution:

Now, x 0

1 cos 2xlim

2 x

= x 0 x 0

2 sin x sin xlim lim

x2 x

Let f(x) = sin x

x

Now, h 0

sin 0 hLHS lim

0 h

= h 0

sin hlim 1

h

and h 0

sin 0 hRHL lim

0 h

= h 0

sin hlim 1

h

LHL RHL

Hence, x 0

sin xlim

xdoes not exist.

15. A lizard, at an initial distance of 21 cm behind an insect, moves from rest with an acceleration of 2

cm/s2 and pursues the insect which is crawling uniformly along a straight line at a speed of 20 cm/s.

Then the lizard will catch the insect after

(a) 24 s (b) 21 s (c) 1 s (d) 20 s

Ans. (b)

Solution :

Let lizard catch the insect C and distance covered by insect = S time taken by insect, S

t20

Distance covered by lizard = 21 + S

2121 S 2 .t

2 ..(ii)




21 + 20t – 21 = 0 [from Eq. (i)]

t2 – 20t – 21 = 0

t2 – 21t + t – 21 = 0

t(t – 21) + 1(t – 21) = 0

(t + 1) (t – 21) = 0

t = 21s

16. If 0

xf (sin x)dx = A /2

0f sin x dx, then A is equal to

(a) 0 (b) (c) 4

(d) 2

Ans. (b)

Solution:

Let 0

I xf sin x dx

0I x f[sin x ]dx

0I x f sin x dx

On adding Eqs. (i) and (ii), we get

02I x x f sin x dx

02I f sin x dx

/2

02I 2 f sin x dx

/2

0I f sin x dx

/2 /2

0 0A f sin x dx f sin x dx

/2

0I A f sin x dx given

A

17. The area bounded by the curve y = 2x – y2 and the straight line y = – x is given by

(a) 9

sq unit2

(b)43

6 sq unit (c)

35

6sq unit (d) None of these

Ans. (a)

Solution :




The equations of given curve and a line are

y = 2x – x2 ..(i)

and y = – x .(ii)

On solving Eqs. (i) and (ii), we get the points of intersection of curves which are (0, 0) and (3, – 3).

Required area = 3

2

0[ 2x x x ]dx

= 3

2

03x x dx

=

32 3

0

3x x 27 27 9sq unit

2 3 2 3 2

18.

2

2

log x 1

1 log xdx is equal to

(a)2

xc

log x 1 (b)

x

2

xec

1 x (c)

2

xc

x 1 (d)

2

log xc

log x 1

Ans. (b)

Solution:

2

2

log x 1

1 log xdx

=

2

2 2

log x 1 2 log xdx

[ log x 1]

=

2

2 2

1log x 1 2x log x.

xdx

[ log x 1]




= 2

d xf dx

dx log x 1

= 2

xc

log x 1

19. 0

[cot x] dx,[.] denotes the greatest integer function, is equal to

(a) 2

(b) 1 (c) –1 (d) 2

Ans. (d)

Solution:

Let 0

I [cot x]dx ..(i)

0I [cot x ]dx

= 0

[ cot x]dx ..(ii)

On adding Eqs. (i) and (ii),

0 02I [cot x]dx [ cot x]dx

= 0

1 dx

[ [x] [ x] 1, if x z 0, if x z]

= 0[ x]

I2

20. The solution of the differential equation (1 + y2) + tan 1 dyx e 0, is

dx

(a) 2 1x 2 ce tan y (b) 1tan y 2 1 y2xe e tan c

(c) 1tan y 1 yxe tan c (d)

1 12tan y tan yxe e c

Ans. (b)

Solution:

Given differential equation can be rewritten as

12 tan ydx1 y x e

dy




or

1tan

2 2

dx 1 e yx

dy 1 y 1 y

12

1dyP dy tan yx yIF e e e

Therefore, required solution is

12 1y

tan y12

e tanxe dy c

1 y

1 1tan y 2tan y1

1xe e c

2

1 1tan y 2tan y2xe e c

21. If the straight lines x = 1 + s, y= – 3 – t

s, z 1 s and x ,y 1 t, z 2 t2

with parameters s

and t respectively, are coplanar, then equals

(a) –2 (b) – 1 (c)1

2 (d) 0

Ans. (a)

Solution:

The given straight line, can be rewritten as

x 1 y 3 z 1

s1

and x 0 y 1 z 2

t1 2 2

These two lines are coplanar, if

1 0 3 1 1 2

1 0

1 2 2

1 4 1

1 0

1 2 2

1 2 2 4 2 1 2 0

8 4 2 0

10 5 2




22. Let the linex 2 y 1 z 2

3 5 2 lies in the plane x + 3y – z 0. Then, , equals

(a) (6, – 17) (b) (–6, 7) (c) (5, – 15) (d) (–5, 15)

Ans. (b)

Solution:

Dr’s of given line are (3, –5, 2).

Dr’s of normal to the plane = 1,3,

Line is perpendicular to the normal

3 1 5 3 2 0

3 15 2 0

2 12

6

Also, point (2, 1, –2) lies on the plane

2 + 3 + 6(–2) + 0

7

, 6,7

23. The distance of the point (1, –5, 9) from the plane x – y + z = 5 measured along a straight line

x = y = z is

(a) 3 5 (b) 10 3 (c) 5 3 (d) 3 10

Ans. (b)

Solution:

Equation of PQ, is

x 1 y 5 z 9

1 1 1

x 1, y 5, z 9 lies on the plane x – y + z = 5




1 5 9 5

10

Coordinate of Q, is

Q(–9, – 15, – 1)

2 2 2PQ 10 10 10

= 10 3

24. The vectors AB 3i 4k, and AC 5i 2j 4k are the sides of a triangle ABC. The length of the

median through A is

(a) 18 (b) 72 (c) 33 (d) 288

Ans. (c)

Solution :

Let D be the mid point of BC.

AB ACAD

2

= 3i 4k 5i 2j 4k

2

= 4i j 4k

Length of median,

22 2AD 4 1 4

= 16 1 16 33

25. The distance between the line r 2i 2j 3k i j 4k and the plane r. i 5j k 5 is

(a) 10

3 (b)

3

10 (c)

10

3 3 (d)

10

9




Ans. (c)

Solution:

Line is parallel to plane as

i j 4k . i 5j k 1 5 4 0.

General point on the line is 2, 2,4 3 . For 0 a point on this line is (2, –2, 3) and

distance from

r. i 5j k 5 or x 5y z 5 is

2 5 2 3 5d

1 25 1

10 10d

3 3 3 3

26. Let a i j k, b i j 2k c xi x 2 j k. and If the vector c lies in the plane of a and b , then x

equals

(a) 0 (b) 1 (c) – 4 (d) – 2

Ans. (d)

Solution:

Since, given vectors a, b and c are coplanar.

1 1 1

1 1 2 0

x x 2 1

x{1 2 x 2 } 1 1 2x 1 x 2 x 0

1 –2x + 4 + 1 + 2x + 2x – 2 = 0

2x = – 4 x = – 2

27. Let a j k and c i j k. Then the vector b satisfying a b c 0 and a. b 3, is

(a) i j 2k (b) 2i j 2k (c) i j 2k (d) i j 2k

Ans. (a)

Solution:

We have, a b c 0




a a b a c 0

a.b a a.a b a c 0

3a 2b a c 0

2b 3a a c

2b 3j 3k 2i j k 2i 2j 4k

b i j 2k

28. If 1 1

a 3i k and b 2i 3j 6k ,710

then the value of 2a b . [ a b a 2b ] is

(a) –3 (b) 5 (c) 3 (d) –5

Ans. (d)

Solution:

1 1a 3i k and b 2i 3j 6k

710

2a b .{ a b a 2b }

2a b .{ a b a a b 2b}

2a b .{ a . a b b a 2 a . b b 2 b . b a}

2a b .{1 b 0 a 2 0 b 2 1 a}

{as a. b 0 and a. a b. b 1}

2a b b 2a

2 2

4( a 4a. b b )

{4 0 1} 5

29. 1 1cot ( cos ) tan ( cos ) x, then sin x is equal to

(a) 2tan2

(b) 2cot2

(c) tan (d) cot2

Ans. (a)

Solution:

Given that,




1 1cos ( cos ) tan ( cos ) x ..(i)

We know that,

1 1cos ( cos ) tan ( cos )2

…(ii)

On adding Eqs. (i) and (ii), we get

12cot ( cos ) x2

xcos cot

4 2

xcot 1

2cosx

1 cot2

x xcos sin

2 2cosx x

cos sin2 2

On squaring both sides, we get

1 sin xcos

1 sin x

2

2

1 tan1 sin x21 sin x1 tan

2

Applying componendo and dividend rule, we get

2sin x tan2

30. If 1 1 ycos x cos ,

2 then 2 24x 4xy cos y is equal to

(a) 24sin (b) 24sin (c) 4 (d) 2sin2 Ans. (b)

Solution:

Given that, 1 1 ycos x cos

2

21 2xy y

cos 1 x 12 4




22xy y

1 x 1 cos2 4

22 y

2 1 x 1 2 cos xy2

On squaring both sides, we get 2 2

2 2 24 1 x 4 y

4 cos x y 4xy cos4

2 2 2 2 2 2 24 4x y x y 4 cos x y 4xy cos

2 2 24x 4xy cos y sin




Section – II {Physics}

31. A Partical of mass m and charge + q is midway between two fixed charge particles, each having a charge

+q and at a distance 2L apart. The middle charge is displaced slightly along the line joining the fixed

charges and released. The time period of oscillation is proportional to.

(a) L1/2 (b) L (c) L3/2 (d) L2

Ans. (c)

Solution:

If the middle charge is displaced by a distance x, the net force acting it, when it is released, is

2 2

2 2

0 0

1 q 1 qF

4 4L x L x

= 2

22 2

0

4q Lx

4 L x

For 2

30

q xx L, F kx

L

where 2

30

qk

L

Now m

T 2k

So, the correct choice is (c).

32. Two isolated metal spheres of radii R and 2R are charged such that both have the same surface charge

density . The spheres are located far away from each other. When they are connected by a thin

conducting wire, the new surface charge density on the bigger sphere will be

(a) 2

3 (b)

3

5 (c)

5

6 (d)

2

Ans. (c)

Solution:

Initial charge on sphere of radius R = 21Q 4 R and on sphere of radius 2R is Q2 = 24 2R

= 216 R

Total initial charge is Q = Q1 + Q2 = 20 2R

Initial potential of the sphere of radius R is

11

0

QV

4 R




and of the sphere of radius 2R is

22

0

QV

4 2R

When the spheres are connected by a thin wire, charge will flow from one sphere to the other until their

potentials become equal . Let Q’1 and Q’2 be the new charges, then the potential of each sphere will be

1 2

0 0

Q' Q'V

4 R 4 2R

which gives 21

Q'Q'

2

From conservation of charge, we have

Q1 + Q2 = Q’1 + Q’2

2 2 22

Q' 3Q'20 R Q'

2 2

22

40Q' R

3

New surface charge density on the sphere of radius 2R is

2

22 2

40R

Q' 53'16 R 64 2R

33. Two equal points charges are fixed at x = –a and x = + a on the x-axis. Another point charge Q is placed

at the origin. The charge in the electrical potential energy of Q, when it is displaced by a small distance x

along the x-axis, is approximately proportional to

(a) x (b) x2 (c) x3 (d) 1/x

Ans. (b)

Solution:

Potential energy of the system when charge Q is at O is

0

qQ qQ 2qQU

a a a

When charge Q is shifted to position O’, the potential energy will be see fig.




2 2

qQ 2aqQ qQU

a x a x a x

=

12

2

2qQ x1

a a

= 2

2

2qQ x1

a a x a

2

0 2

2qQ x 2qQU U U 1

a a a

= 2

3

2qQx

a

Hence 2U x which is choice (b).

34. Find the equivalent capacitance between A and B in the circuit shown in fig. Given C = 5 F.

(a) 3 F (b) 5 F (c) 7 F (d) 9 F

Ans. (c)

Solution:

Let us imagine a battery of emf E connected between A and B. Let the positive terminal send a charge

+Q and let negative terminal send a charge – Q. The distribution of charges on the capacitor plates is

shown in the figure using the principle of change conservation. Then




A B A P P BV V V V V V

11Q QQ

EC 2C

1 1 12CE 2Q Q Q Q Q (1)

Also A B A P P R R BV V V V V V V V

11 12Q QQ Q

EC C C

1 1 1 1CE Q 2Q Q Q 4Q Q (2)

Eliminating Q1 from (1) and (2), we get

7CEQ

5

eq

Q 7C 7 5 FC 7 F

E 5 5

[ C 5 F given ]

35. A rigid insulated wire frame in the form of a right-angled triangle ABC, is set in a vertical plane as

shown in fig. Two beads of equal masses m each and carrying charges q1 and q2 are connected by a

cord of length l and can slide without friction on the wires. The beads are stationary.

Then the value of angle is

(a) 300 (b) 450 (c) 600 (d) 750

Ans. (c)

Solution:

Let us consider forces acting on bead P as shown in fig. These forces are :

(i) Weight mg vertically downwards

(ii) Tension T in the string




(iii) Electric force between P and Q given by

F = 1 22

0

q q1.

4 l

(iv) Normal reaction N1.

The net force along the string is (T – F). Bead P will be in equilibrium , if the net force acting on it is

zero. Resolving forces mg and (T – F) parallel and perpendicular to plane AB, we get, when the bead P is

in equilibrium,

mg cos 600 = (T – F) cos (1)

and 0

1N mg cos 30 T F sin (2)

For the bead at Q, we have

mg sin 600 = (T – F) sin (3)

and 02N mg cos 60 T F cos (4)

Dividing Eq. (3) by Eq. (1), we get

tan 0 0tan 60 or 60 , which is choice (c).

36. Calculate the steady state current in the 3 resistor in the circuit shown in fig. The internal resistance

of the cell is negligible.

(a) 0.2A (b) 0.4A (c) 0.6A (d) 0.8A

Ans. (c)

Solution:

The resistance of the parallel combination of 2 and 3 is

2 3R' 1.2

2 3

In the steady state, no current flows through the capacitor and hence through the 5 resistor.

Therefore, the circuit reduces to that shown in fig.




Current 6

I 1.5A1.2 2.8

Potential differenceacross R’ is

V’ = IR’ = 1.5 × 1.2 = 1.8V

This is also the p.d. across the 3 resistor. Hence the current through the resistor is 1.8

0.6A.3

37. In the circuit shown in fig, the equivalent resistance between points A and B is

(a) 3R

4 (b)

5R

6 (c)

7R

10 (d)

11R

9

Discarded

38. An RC circuit consists of a resistance R = 5 M and a capacitance C = 1.0 C connected in series with a

battery. In how many time will the potential difference across the capacitor become 8 times that across

the resistor ? (Given loge(3) = 1.1)

(a) 5.5 (b) 11s (c) 44s (d) 88s

Ans. (b)

Solution:

At instant of time t, the charge on the capacitor is given by

t /RC0q q 1 e

and the potential drop across the capacitor is given by V q/C

t /RCC 0V V 1 e

where V0 is the voltage of the battery. The potential drop across the resistor is




t /RC t /RCR 0 C 0 0 0V V V V V 1 e V e

t /RCt /RCC

t /RCR

V 1 ee 1

V e

Given C

R

V8.

V Therefore,

t /RC8 e 1

or 2t /RCe 9 3

or e

t2 log 3

RC

or t = RC × 2 loge (3)

= (5 × 106) × (1 × 10–6) × 2 × 1.1

= 11s

Hence the correct choice is (b).

39. A wire of length L and 3 identical cells of negligible internal resistances are connected in series. Due to

the current, the temperature of the wire is raised by T in a time t. A number N of similar cells is now

connected in series with a wire of the same material and cross section but the length 2L. The

temperature of the wire is raised by the same amount T in the same time t. The value of N is

(a) 4 (b) 6 (c) 8 (d) 9

Ans. (b)

Solution:

The heat energy generated in time t is Q = V2t/R, where V is the terminal voltage of a single cell. Here Q

= ms T where m is the mass of the wire, s its specific heat and R its resistance.

In the first case:

23V t

ms TR

(1)

Now m length and R length. Hence, in the second case,

2NV t

2m s T2R

(2)

Dividing (2) by (1), we have

22N /2

2 or N 36 or N 6,9

which is choice (b).




40. The walls of a closed cubical box of edge 50 cm are made of a material of thickness 1 mm and thermal

conductivity 1.68 × 10–1 W m–1 K–1. The interior of the box is maintained at 1000C above the outside

temperature by a heater placed inside the box and connected to a 420 V d.c. source. What must be the

resistance of the heater element?

(a)5 (b)6 (c) 7 (d) 8

Ans. (c)

Solution:

The heat transmitted per second through the walls of the closed box is given by

2 1K AQ

t d

where A is the total surface area of the box.

Since a cube has 6 faces, the total surface area is

A = 6 × surface area of one face

= 6 × (0.5 m × 0.5m) = 1.5 m2

1

3

1.68 10 1.5 100Q

t 1 10

= 2.52 × 104 J s–1

In order to maintain the temperature difference between the inside and the outside, the heat lost must

be compensated by the outside, the heat lost must be compensated by the production of heat through

electric current in the coil. The heat produced per second is given by

24V

H 2.52 10R

or 4

4.20 420R 7 ,

2.52 10 which is choice (c).

41. Figure shows a wire loop ABCDEA carrying a current I as shown. Given AE = ED = a and AB = CD =

a/2.Find the magnitude of the magnetic field at point F where BF = CF = a/2.




(a) 0I 11

a 2 2 (b) 0I 1

12 a 2 2

(c) 0I 1

a 2 2 (d) 0I 1

2 a 2 2

Ans. (a)

Solution:

Magnetic field at F is

B = BAB + BBC + BCD + BDE + BEA

Since point F lies in line with current elements AB and CD, BAB = BCD = 0

Also 0 00 0DE EA

I IB B sin sin45

4 a 4 2 a

directed out of the page and towards the reader.

0 00BC

IB (sin 45 sin45 )

4 BC/2

directed into the page and away from the reader. Now

2 2

2 2 a a aBC BF FC

2 2 2

0 0BC

I I1 1B

a a2 24

2 2

directed into the page.

Now BBC> BDE + BEA the net field, B is directed into the page and has magnitude

0 0 0I I I 1B 1

a a2 2 a 2 2

42. A uniform wire is bent into the shape of an equilateral triangle of side a. It is suspended from a vertex at

place where a uniform magnetic field B exists parallel to its plane. Find the magnitude of the torque

acting on the coil when a current I is passed through it.




(a) Ia2B (b) 23Ia B

4 (c) 21

Ia B2

(d) zero

Ans. (b)

Solution:

Area of the coil is (AB = a, BD = a/2)

A = 2 × area of triangle ABD

= 2 × 1

AD BD2

= 1 3 a

2 a2 2 2

= 23

a4

Magnetic moment of the loop is

23M IA I a

4

Since the current is clockwise the direction of vector M is perpendicular to the plane of the coil directed

inwards as shown in fig. Here 090 . The magnitude of the torque acting on the coil is

2 0 23 3MB sin Ia B sin 90 Ia B

4 4

43. A particle of charge q moves with a velocity v = a i in a magnetic field B = b j ck where a, b and c are

constants. The magnitude of force experienced by the particle is

(a) zero (b) qa(b + c) (c) qa(b2– c2)1/2 (d) qa(b2 + c2)1/2




Ans. (d)

Solution:

F q v B q{ai bj ck }

= q ab i j aci k

= q abk acj qa bk cj

Magnitude of F = 2 2 1/2[ qab q ac ]

= 1/2

2 2qa b c

44. Two particles A and B of masses mA and mB respectively and having the same charge are moving in a

plane. A uniform magnetic field exists perpendicular to this plane. The seeds of the particles are

A Band respectively and the trajectories are as shown in fig. Then

(a) A A B Bm m (b) A A B Bm m

(c) A B A Bm m and (d) A B A Bm m and

Ans. (b)

Solution :

The radius r of the circular path of a particle of mass m and charge q moving with velocity

perpendicular to a magnetic field B is given by

2mq B

r

or m qrB. Hence A B A B B Bm qr B and m q r B.

A A

B B B

m A r

m r

It follows from the figure that rA> rB. Hence A A B Bm m . Thus the correct choice is (b).

45. A bar magnet is suspended at a place where it is acted upon by two magnetic fields which are inclined

to each other at an angle of 750. One of the fields has a magnitude 22 10 T. The magnet attains stable




equilibrium at an angle of 300 with this field. Find the magnitude of the other field.

(a) 10–1 T (b) 10–2 T (c) 10–3 T (d) 10–4 T

Ans. (b)

Solution:

Magnetic field B1 exerts anticlockwise torque 1 to orient M along itself and magnetic field B2 exerts a

Clockwise torque 2 to orient M along itself fig. For equilibrium,

1 2

1 1 2 2MB sin MB sin

2 01 1 2

2 02

B sin 2 10 sin30B 10 T

sin sin 45

46. A current carrying wire AB is placed near another CD as shown in fig. Wire CD is fixed while wire AB is

free to move. When a current is passed through wire AB, it will have

(a) only translational motion

(b) only rotational motion

(c) both translational as well as rotational motions

(d) neither translational nor rotational motion.

Ans. (c)

Solution:




Since the magnetic field due to wire CD is non-uniform, wire AB will experience as force as well as a

torque. Hence it will have both translational as well as rotational motions.

47. A metal rod PQ of length l slides with a velocity on two parallel rails AB and CD parallel to a long

straight wire XY carrying a current I as shown in fig. A resistance R is connected between the rails as

shown. The velocity of rod PQ is kept constant by applying force.

then the expression for the current induced in resistance R.

(a) 0Iv a lln

R a (b) 0Iv a

lnR a l

(c) 0Iv a lln

2 R a (d) 0Iv a

ln2 R a l

Ans. (c)

Solution:

In this case the induced emf is due to change in magnetic flux which is due to the change in the area of

ACPQ with time.

Magnetic flux through an infinitesimal area element of with dx at a distance x from PQ is

0Ird BdA Brdx dx

2 x

where r is the position of PQ at an instant of time t.

Magnetic flux through ACPQ is

a l

0

a

Ird dx

2 x

=

a l

0 0

a

Ir Irdx a lln

2 x 2 a




Induced emf e = 0Id a l drln

dt 2 a dt

= 0I a lln

2 a

dr

dt

(a) Current induced in R is

0Ie a li ln

R 2 R a

48. Two circular coils A and B of radii a and b respectively (with b > a) have their plane perpendicular to

the plane of the page. They are separated co-axially by a distance x = 3 b as shown in fig. A transient

current I flows through coil B for a very short time interval. If the resistance of coil A is R obtain the

expression for the charge that flows through coil A during the short time interval.

(a) 2

0I a

bR (b)

20I a

8bR (c)

20I a

16bR (d)

20I a

4bR

Ans. (c)

Solution :

Magnetic field at the centre of coil A due to current I in coil B is

20 0

AB 3/22 2

Ib IB

16b2 b x x 3 b

Since the magnetic field is along the axis of coil A, it is perpendicular to the plane of A, hence 00 .

Therefore, magnetic flux through A is

20 0

AB

I aB area of coil A cos 0

16b

Induced emf isd

edt

dIR

dt




IRdt d

1

Idt dR R

Or 2

0I aQ

R 16bR

49. A thin non-conducting disc of radius R and mass M is held horizontally and is capable of rotation about

an axis passing through its centre and perpendicular to its plane. A charge Q is distributed uniformly

over the surface of the disc.

A time-varying magnetic field B = kt (where k is a constant and t is the time) directed perpendicular to

the plane of the disc is applied to it. If the disc is stationary initially (i.e. at t = 0).

Find the torque acting on the disc.

(a) 2kQR (b) 2kQR

2 (c)

2kQR

4 (d) 22kQR

Ans. (c)

Solution :

Area of disc = 2R

Charge per unit area = 2

Q

R

Area of small elements of width dx at a distance x from the centre of the disc = Therefore 2 xdx. , charge

of the element is

2

Qdq 2 xdx

R

A time-varying magnetic field gives rise to an electric field E. Since

dVE

dl

dV Edl E 2 x




V E 2 x

where V is emf induced in the element, which is given by

d dV BA

dt dt

= 2 2dkt x kx

dt (2)

From (1) and (2) we get

2E 2 x kx

kxE

2 (3)

Force acting on the element is

2

Q kxdF dq E 2 xdx

R 2

= 2

2

kQx dx

R

Torque acting on the disc is

R R 23

2

0 0

kQ kQRxdF x dx

R 4

50. An equilateral triangular loop PQR of side a is at the edge of a uniform magnetic field B at t = 0 as shown

in fig. It is pulled to the right with a constant velocity and its edges R leaves the region of magnetic

field at t = t0. Which of the graphs shown in fig. Represents the variation of induced current

i with time t?




Ans. (b)

Solution:

Refer to fig. In time t, the loop moves to the right a distance CD = t Let RD = b.

Now

RC CT

RD QD

RD CD CT

RD QD

b t CT

b a/2

a b tCT

2b

aST 2CT b t

b




Induced emf a

e B ST B a tb

If R is the resistance of the loop, the induced current is

e B ai a t

R R b

2B a B ai t

Rb R

Thus the graph of i against t has a positive intercept B a

Rand a negative slope

2B a.

R b

Hence the correct choice is (b).

51. In the circuit shown in fig, I2 = 3A in the steady state. The potential difference across the 4 resistor is

(a) 12 V (b) 18 V (c) 20 V (d) 24 V

Ans. (c)

Solution:

In the steady state, the resistance of the inductor is zero, i.e. it behaves as a short circuit. Hence the

circuit can be drawn as follows. Fig

I2 = 3 A. Therefore, p.d. across 1 + 3 = 4 resistance = 3 × 4 = 12 V. Hence 1

12I 2A.

6

Therefore I = I1 + I2 = 2 + 3 = 5A, which is the current flowing through 4 resistance. So p.d. across

4 resistance = 5 × 4 = 20 V.




52. Two thin equi-convex lenses each of focal length f and made of glass g

3

2are placed in contact. The

space between them is filled with water g

4

3. The focal length of the combination is

(a) f

2 (b)

2f

3 (c)

3f

5 (d)

4f

7

Ans. (d)

Solution :

1 3 2 11

f 2 R R 1 2R R R

which gives f = R. When the space between the lenses is filled with water, we have a concave water lens

of 4/3 surrounded by a medium of g 3/2. Therefore, for the water lens,

3 4

1 22 3

f ' 4/3 R1 2R R and R R

= 1 1

4R 4f

The focal length of the combination of the three lenses is given by

1 1 1 1

F f f f '

= 2 1 7

f 4f 4f

4fF ,

7which is choice (d).

53. What is the relation between refractive indices 1 2, and if the behaviour of light rays is as shown in

fig.

(a) 2 1 (b) 2 1 (c) 2 1; (d) 2 1 2;

Ans. (c)




Solution :

The ray does not suffer any deviation on entering the lens. Hence 1 . The ray leaves the second

surface of the lens bending towards the normal. Hence 2 . Thus the correct choice is (c).

54. Figure shows a lens having radii of curvature R1 and R2 and 1 2 3 . If the thickness of the lens is

negligible and R1 = R2 = R, the focal length of the lens will be

(a) 3

3 1

Rf (b) 2

3 1

Rf (c) 1

3 2

Rf (d) 2 1

3 1

Rf

Ans. (a)

Refraction at the first surface : u = , and R = + R1. We have

2 1 2 1

1R

or 2 2 1

1R (i)

Refraction at the second surface: u = t t is negligible ,

1 2and R R . We have

3 3 22

1 2R (ii)

Since the incident ray is parallel to the principal axis, 1 f , the focal length of the lens and using 1 f




(i) in (ii), we get

3 3 22 1

1 2f R R

or 3 22 1

3 1 3 2

1 1 1. .

f R R (iii)

This is the expression for the focal length. If R1= R2 we get [put R1 = R2 = R in (iii)]

3 1

3

1 1

f R

So the correct choice is (a).

55. Two identical thin isosceles prisms of refracting angle (in radian)and refractive index are placed

with their bases touching each other as shown in fig. A ray of light is incident on the prism at an small

height h. The focal length of this crude converging lens is

(a) h

f (b) h

f1

(c) h

f (d) h

f1

Ans. (d)

Two isosceles prisms ABC and DBC are placed with their bases BC touching each other as shown in fig,

shows the path of parallel incident rays which come to a focus at point F. Distance OF = f is the focal

length of the system. It follows from figure that

htan

f




Since the prism is very thin and if h is small, i.e. the incident rays are close to the axis PQ of the system,

then will be very small and we can replace tan by where is measured in radian. Thus

h

f

Now, for a prism having a small refracting angle and for a small h, the deviation produced by a prism

is given by (here angle is expressed in radian)

1

Thus h h

1 or ff 1

So the correct choice is (d).

56. Parallel rays from a distant object fall on a solid transparent sphere of radius R and refractive index

. The distance of the image from the sphere is

(a) R 2

2 1 (b)

R

1 (c)

R

1 (d) R 1

Ans. (a)

Solution :

Refer to fig. For refraction at face I,

1 1

1 1

u R

Since 1u , we have

1

1

R

or 1

R

1 (1)




For refraction at face II, 2 1 1u 2R 2R .

Using Eq. (1), we get

2

R 2Ru 2R

1 1 (2)

The image distance 2 is given by

2 2

1 1

u R (3)

Using Eq. (2) in Eq. (3) and simplifying, we get

2

R 2.

2 1

The correct choice is (a).

57. In Young’s double slit experiment, the intensity at a point P on the screen is half the maximum intensity

in the interference pattern. If the wavelength of light used is and d is the distance between the slits,

the angular separation between point P and the centre of the screen is

(a) 1sind

(b) 1sin2d

(c) 1sin3d

(d) 1sin4d

Ans. (d)

Solution :

If is the phase difference between the interfering waves at point P, then the intensity at point P is

given by (see fig.)

2maxI I cos

2

Given maxII .Hence

2

2 1cos , which gives

2 2 2 4




or 2

The angular separation between points P and O is given by tan y/D. Since is very small,

tan sin . Hence

ysin

D (1)

If is the fringe width, then

y /2 1

2 4 (2)

This is so because the phase difference between two consecutive maxima is 2 . Now D

.d

Using this in Eq. (2), we get

yd 1

D 4

or y 1

D 4d (3)

Using Eq. (1) in Eq. (3), we have

1sin or sin ,4d 4d

which is choice (d).

58. Photoelectric effect experiments are performed using three different metal plates p, q and r having

work functions p q2.0 eV. 2.5 eV and r = 3.0 eV, respectively. A light beam containing

wavelengths of 550 nm, 450 nm and 350 nm with equal intensities illuminates each of the plates. The

correct I-V graph for the experiment is (see fig. ) [Take hc = 1240 eV nm]




Ans.(a)

Solution:

For photoelectric emission, the wavelength of the incident radiation must be less than the cut-off

wavelength of the metal given by 0 0hc/ . For metals p, q and r, the cut-off wavelengths are

p

p

hc 1240eV nm620 nm

2.0 eV

q

1240496 nm

2.5

r

1240413.3 nm

3.0

Hence metal plate p emits photoelectrons for all the three given radiations, metal plate q units

photoelectrons for radiation of wavelengths 450 nm and 350 nm and metal plate r emits

photoelectrons only for wavelength 350 nm. Therefore, photoelectric current is maximum for metal p

and minimum for r, i.e. Ip> Iq> Ir. So the correct choice is (a).

59. Which one of the following statements is Wrong in the context of X-rays generated from a X-ray tube ?

(a) Wavelength of characteristic X-ray decreases when the atomic number of the target increases.

(b) Cut-off wavelength of the continuous X-rays depends on the atomic number of the target

(c) Intensity of the characteristic X-rays depends on the electric power given to the X-rays tube

(d) Cut-off wavelength of the continuous X-rays depends on the energy of the electrons in the X-ray

tube.

Ans. (a)




Solution :

min

hc,

eVwhich is independent. Hence choice (a) is wrong.

60. A particle at rest disintegrates into two fragments of masses in the ratio of 1 : 2 having non-zero

velocities. The ratio of the de Broglie wavelength of the lighter particle to that of the heavier particle is

(a) 1

2 (b) 1 (c) 2 (d)

1

4

Ans. (b)

Solution :

From the conservation of linear momentum, the two fragments will have equal and opposite momenta.

Now de Broglie wavelength h/p. Hence

1 2

2 1

p

p

Since 11 2

2

p p , 1.

Section – III {Chemistry}

61. The activity of a sample has 30% as much radioactivity as present originally. If its half-life is 20 y, the

life of the sample undergoing disintegration is about

(a) 34.8 y (b) 49.2 y (c) 25 y (d) 55 y

Ans. (a)

Solution :

100% activity t 1/2 50% activity

t 1/2 25% activity

The given activity of 30% is a littler larger than 25% activity, hence time required to achieve this

activity with slightly less than two half-lives.

Alternatively,




0

[A] 30 0.693In k t ln t

[A] 100 20y

or 20y 20 2.303 0.477

t 2.303 log 3 y 34.75y0.693 0.693

62. In the following cleavage reactions of ether :

The products are

(a) CH3I + (CH3)3COH in both the reactions

(b) CH3OH + (CH3)3CI in both the reactions

(c) CH3OH + (CH3)3CI in Reaction (i) and CH3I + (CH3)3COH in Reaction (ii)

(d) CH3I + (CH3)3COH in Reaction (i) and CH3OH + (CH3)3CI in Reaction (ii)

Ans. (d)

Solution :

The high polarity of the solvent (H2O) in reaction (ii) favours SN1 mechanism. The low polarity of

solvent (ether) in reaction (i) favours SN2 mechanism.

Reaction (ii) N

H|

S 1 I3 3 3 3 33 3 3

CH O C CH CH OH CH C CH Cl

63. The volume required to prepare 2.0 L of 3.0 M H2SO4 solution from 90% H2SO4 solution (density 2.0 g

mL–1) would be

(a) 250.5 mL (b) 372.2 mL (c) 326.7 mL (d) 402.2 mL

Ans. (c)

Solution :

Amount of H2SO4 required to prepare 2.0 L of 3.0 M solution is

n = VM = (2.0 L) (3.0 mol L–1) = 6.0 mol

Mass of H2SO4 required, nMm = (6.0 mol) (98 g mol–1) = 588 g

Mass of 90% H2SO4 required = 588 g × 100/90 = 653.3 g




Volume of 90% H2SO4 required , m/ = 1

653.2g326.7 mL

2g mL

64. Which of the following conditions must hold good for the disproportion reaction of Cu+(aq) under

standard conditions?

(a) 2

o o

Cu |Cu Cu , Cu |PtE E with positive values of both the potentials.

(b) 2

o o

Cu |Cu Cu , Cu |PtE E with positive values of both the potentials.

(c) 2

0 o

Cu |Cu Cu ,Cu |PtE E with negative value of 0

Cu |CuE and positive value of 2

0

Cu ,Cu |PtE

(d) 2

o o

Cu |Cu Cu ,Cu |PtE E with negative values of both the potentials.

Ans. (a)

Solution:

The disproportion reaction is 22Cu Cu Cu

The reaction is obtained as

Subtract 2

2

o

Cu |Cu

2 o

Cu , Cu |Pt

2 o o o

Cu |Cu Cu , Cu |Pt

Cu e Cu E

Cu e Cu E

2Cu Cu Cu E E E

Since the disproportion reaction is spontaneous under standard conditions, we will have E0> 0.

This condition is satisfied by the choice (a) only.

65. Which of the following pairs is expected to involve sp3d2 hybridization?

(a) 3 32 36 6

[Ti H O ] and [Cr NH ] (b) 2 23 36 6

[Ni NH ] and [Cu NH ]

(c) 4 366

[Mn CN ] and [MnF ] (d) 236 6

[Fe CN ] and [Fe NH ]

Ans. (b)

Solution :

The electronic configurations of 22Ti3+, 24Cr3+, 25Mn2+, 26Fe2+, 28Ni2+ and 29Cu2+ are as follows.




F– is a weak ligand. It has no effect in pairing of 3d electrons.

H2O, NH3 are CN– are strong ligands, It may cause pairing of 3d electrons.

66. The reactivity of acid derivative towards a nucleophile : Nu is

(a) acid chloride > ester > amide (b) acid chloride > amide > ester

(c) ester > acid chloride > amide (d) ester > amide > acid chloride

Ans. (a)

Solution :

The order of reactivity decreases as the base strength of the leaving group increases which in the

present case is 2Cl RCOO H N

67. Which of the following statements regarding cleavage of ethers by HI is correct ?

(a) Cleavage follows SN1 mechanism in high-polarity solvent and SN2 mechanism in low-polarity

solvent.

(b) Cleavage follows SN2 mechanism in high-polarity solvent and SN1 mechanism in low-polarity

solvent.

(c) Cleavage follows SN1 mechanism in both high-polarity and low-polarity solvents.

(d) Cleavage follows SN2 mechanism in both high-polarity and low-polarity solvents.

Ans. (a)

Solution:

The low-polarity of solvent favours SN2 mechanism and high-polarity favour SN1 mechanism.

68. For a reaction A B, the activation energy for the forward reaction is 65 kJ mol–1. If enthalpy of

reaction is 42 kJ mol–1, the activation energy for the reaction B A would be

(a) 23 kJ mol–1 (b) 65 kJ mol–1 (c) 107 kJ mol–1 (d) 32 kJ mol–1




Ans. (c)

Solution:

The activation energy of backward reaction will be (65 + 42) kJ mol–1 = 107 kJ mol–1 as shown in the

following diagram.

69. In the bis-(dimethylglyoximato)nickel(II)chealated complex, the number of five and six membered

rings, respectively, are

(a) 1, 9 (b) 2, 2 (c) 3, 3 (d) 1, 2

Ans. (b)

Solution:

The structure is

70. The oxidation of methylnaphthalene with CrO3/HOAc mainly produces

Ans. (c)

Solution:

In methylnaphthalene the positions being more reactive are easily oxidized than the methyl

group.




71. The mass of silver (molar mass = 107.2 g mol–1) deposited on a serving tray by electrolysis from a

solution of silver nitrate when a current of 8.1 ampere is passed for 10 hrs would be about

(a) 234 g (b) 324 g (c) 412 g (d) 482 g

Ans. (b)

Solution:

e

It Mm

F V=

1

8.1A 10 60 60s 107.2g/mol324g

96500Cmol 1

Ve is valency of silver. i.e.+ 1

72. The coagulation of 10 mL of a colloidal gold solution is prevented by the addition of 0.036 g of a

compound followed by 1 mL of 10% solution of sodium chloride. The gold number of the compound is

(a) 0.036 (b) 0.36 (c) 3.6 (d) 36

Ans. (d)

Solution:

Gold number is the minimum mass in mg of the dry protective colloid which will check the coagulation

of 10 mL of a standard gold sol on adding 1 mL of 10% sodium chloride solution.

73. Polymer obtained by using Ziegler-Natta catalyst is

(a) short-chain branching polymer

(b) long-chain branching polymer

(c) linear polymer

(d) a mixture of linear and non-linear polymers

Ans. (c)

Solution:

Linear Polymer is obtained.

74. The formation of oximes and other ammonia derivatives of a keto group is carried out in a medium

having pH round about

(a) 1.0 (b) 3.5 (c) 7.0 (d) 8.0

Ans. (b)

Solution :

The reaction is carried out when the pH is round about 3.5. In more acidic conditions, the unshared pair

of electrons of N is protonated to give electrophilic 3H NG species which react with the keto group. In

basic medium, there is no protonation of C = O group.




75. Ammonium chloride crystallizes in a body-centred cubic lattice with edge length of unit cell equal to

390 pm. If the size of Cl– ion is 180 pm, the size of 4NH ion will be

(a) 210 pm (b) 105 pm (c) 157.7 pm (d) 96 pm

Ans. (c)

Solution:

In body-centred ionic lattice 2 r r 3 a

Therefore 1.732 390 pm

r 3 a /2 r 180 pm 157.7 pm2

76. Methyl ethyl ketone (CH3COCH2CH3) forms two enols :

Which of the following facts is correct ?

(a) Both A and B are equally stable (b) Both A and B are equally unstable

(c) A is more stable than B (d) A is less stable than B

Ans. (d)

Solution:

B is more stable than A because it has more substituted double bond

77. Which of the following alcohols does not give iodoform test?

(a) (CH3)2CH(OH)CH3 (b) PhCH(OH)CH2CH3

(c) 1-methylcyclohexanol (d) CH3CH2CH(OH)CH3

Ans. (b)

Solution:

PhCH(OH)CH2CH3 does not contain–COCH3 or –CH(OH)CH3.

78. Water can be added across a double bond through the following two methods.

Method – I Oxymercuration and demercuration

Method –II Hydroboration-oxidation

Which of the following statements is correct regarding the addition of water ?

(a) Markovnikov’s rule in both the methods

(b) anti-Markovnikov’s rule in both the methods

(c) Markovnikov’s rule in the method-I and anti-Markovnikov’s rule in the method-II

(d) anti-Markovnikov’s rule in the method-I and Markovnikov’s rule in the method-II.




Ans. (c)

Solution :

Oxymercuration - demercuration follows Markovnikov’s addition while hydroboration – oxidation

follows anti-Markovnikov’s addition.

79. The van’t Hoff factor for 0.1 molal Ba(NO3)2 solution is 2.74. Its degree of dissociation is

(a) 0.74 (b) 0.87 (c) 0.91 (d) 1.0

Ans. (b)

Solution:

We have i 1 2.74 1 1.74

0.87.v 1 3 1 2

Here is the number of species produced on ionization of the

salt.

80. The IUPAC name of the compound

(a) 1-Amino-1-phenyl-2-methylpropane

(b) 1-Amino-2-methyl-1-phenylpropane

(c) 2-Methyl-1-amino-1-phenylpropane

(d) 2-Methyl-1-phenylpropane

Ans. (b)

Solution:

The parent skeleton is propane. Its name is 1-amino-2-methyl-1-phenyl propane.

81. Which of the following orders is correct regarding the base strength of substituted aniline ?

(a) p-Nitroaniline < p–aminobenzaldehyde < p-bromoaniline

(b) p-Nitroaniline < p-bromoaniline < p-aminobenzaldehyde

(c) p-Bromoaniline < p-nitroaniline < p-aminobenzaldehyde

(d) p-Bromoaniline < p-aminobenzaldehyde < p-nitroaniline

Ans. (a)

Solution:

Electron-releasing group increasing basicity of aniline while electron-withdrawing group decrease the

basicity.

82. Mineral that does not contain Al is

(a) feldspar (b) fluorspar (c) cryolite (d) mica




Ans. (b)

Solution:

Fluorspar is CaF2. Feldspar is KAl Si3O8; mics is KAlSi3O10(OH)2 and cryolite is Na3AlF6

83. Ferrimagnetism is found when the magnetic moments are aligned

(a) in parallel and antiparallel directions in equal numbers

(b) in parallel and antiparallel directions in unequal numbers

(c) all in one direction pointing upward

(d) all in one direction pointing downward

Ans. (b)

Solution:

For ferrimagnetism, the magnetic moments are aligned parallel and antiparallel directions in unequal

numbers.

84. An involatile solute A dimerizes in a solvent as 2 A A2. If m is the molality of solute in the solution

then the equilibrium constant of the reaction is given as

(a) b b b

2

b b

K K m T

2 T K m (b) b b b

2

b b

K K m T

2 T K m (c) b b b

2

b b

K K m T

2 T K m (d) b b b

2

b b

K K m T

2 T K m

Ans. (b)

Solution:

2m 1 m /2

2A A Total molality = m 1 m /2 m 1 /2

We know that , b bT K m

b

2 b b bbbb b eq 2 2 2

b b b2 b

b

Tm 1

A K K m TK mTT K m 1 /2 2 1 ; K

K m 2 T K mA Tm 1 2 1

K m

85. The treatment of 3 2 2 2 3CH CH C CCH CH CH with KMnO4 under neutral conditions at room temperature

gives

(a) 3 2 2 2 3| |OH OH

CH CH C C HCH CH CH

(b) 3 2 2 2 3| ||

OH O

CH CH CH C CH CH CH

(c) 3 2 2 2 3|| ||O O

CH CH C C CH CH CH




(d) CH3CH2COOH + CH3CH2CH2COOH

Ans. (c)

Solution:

Under neutral conditions at room temperature, the product formed is CH3CH2COCOCH2CH2CH3.

86. For a zero-order reaction, the plot of [A]t versus t for a reaction A B is linear with a

(a) positive slope and zero intercept (b) positive slope and nonzero intercept

(c) negative slope and zero intercept (d) negative slope and nonzero intercept

Ans. (d)

Solution:

For a zero order reaction A B , we have –d[A]/dt = k, which on integration gives t 0[A] [A] kt.

Slope is –k and intercept is [A]0.

87. Which one of the following will be most readily dehydrated in acidic condition ?

(a) CH2COCH2CH(OH) CH3 (b) CH2CH(OH)CH2CH2CH3

(c) CH2COCH(OH)CH2CH3 (d) CH3COCH2CH2CH(OH)CH3

Ans. (a)

Solution:

The ketohydroxy compound is most readily dehydrated.

88. Which of the following orders of ligands towards d-orbitals splitting of the central metal in a complex

ion is correct ?

(a) Cl–< OH–< CN–< NH3 (b) Cl–< OH–< NH3< CN–

(c) Cl–< NH3< OH–< CN– (d) Cl–< NH3< CN–< OH–

Ans. (b)

Solution:

The spectrochemical series is 3 2 3 2I Br Cl NO F OH H O NH NO CN

89. The compound formed on heating chlorobenzene with chloral in the presence of concentrated

sulphuric acid is

(a) gammexene (b) DDT (c) freon (d) hexachloroethane




Ans. (b)

Solution:

The reaction is

Therefore, the choice is b is correct.

90. Reaction of cyclohexanone with dim-ethylamine in the presence of catalytic amount of an acid forms a

compound if water during the reaction is continuously removed. The compound is generally known as

(a) an amine (b) an amine (c) a Schiff’s base (d) an enamine

Ans. (d)

Solution:

the best way to success iit cbse 12th · 2016-01-09 · cbse 12th non-medical the aspire...

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