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Answers & Solutionsforforforforfor

JEE (MAIN)-2014

Important Instructions :1. The test is of 3 hours duration.

2. The Test Booklet consists of 90 questions. The maximum marks are 360.

3. There are three parts in the question paper A, B, C consisting of Mathematics, Physics andChemistry having 30 questions in each part of equal weightage. Each question is allotted 4 (four)marks for each correct response.

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8. The CODE for this Booklet is G. Make sure that the CODE printed on Side-2 of the Answer Sheet isthe same as that on this booklet. In case of discrepancy, the candidate should immediately reportthe matter to the Invigilator for replacement of both the Test Booklet and the Answer Sheet.

(Mathematics, Physics & Chemistry)

GTest Booklet Code

(2)

1. If x = –1 and x = 2 are extreme points of

2( ) log| |f x x x x then

(1) 16, 2

(2) 12, 2

(3) 12, 2

(4) 16, 2

Answer (2)

Sol. 2( ) log| |f x x x x

( ) 2 1 0f x xx

at x = –1, 2

2 1 0 2 1 ...(i)

4 1 0

2 8 2 ...(ii)

6 3 12

2

2. The locus of the foot of perpendicular drawn fromthe centre of the ellipse x2 + 3y2 = 6 on any tangentto it is

(1) (x2 – y2)2 = 6x2 – 2y2

(2) (x2 + y2)2 = 6x2 + 2y2

(3) (x2 + y2)2 = 6x2 – 2y2

(4) (x2 – y2)2 = 6x2 + 2y2

Answer (2)

Sol. Here ellipse is 22

2 2 1yxa b

, where a2 = 6, b2 = 2

Now, equation of any variable tangent is

2 2 2y mx a m b ...(i)

where m is slope of the tangent

So, equation of perpendicular line drawn fromcentre to tangent is

xym ...(ii)

Eliminating m, we get

2 2 2 2 2 2 2( )x y a x b y

2 2 2 2 2( ) 6 2x y x y

3. Let 1( ) (sin cos )k k

kf x x xk where x R and 1k .

Then f4(x) – f6(x) equals

(1)13 (2)

14

(3)1

12(4)

16

Answer (3)

Sol. 1( ) (sin cos )k k

kf x x xk

4 6( ) ( )f x f x = 4 4 6 61 1(sin cos ) (sin cos )4 6

x x x x

= 2 2 2 21 11 2 sin cos 1 3sin cos4 6

x x x x

= 1 14 6 =

112

4. If X = {4n – 3n – 1 : n N} and Y = {9(n – 1) : n N},where N is the set of natural numbers, then X Y isequal to

(1) Y – X (2) X

(3) Y (4) N

Answer (3)

Sol. {(1 3) 3 1, }nX n n N

= 2 22 33 ( .3 ... 3 ), }n n nC C n N

= {Divisible by 9}

Y = {9(n – 1), n N}

= (All multiples of 9}

So, X Y

i.e., X Y Y

5. If A is an 3 × 3 non-singular matrix such that

AA A A and 1B A A , then BB equals

(1) I (2) B–1

(3) 1( )B (4) I + B

Answer (1)

PART–A : MATHEMATICS

(3)

Sol. 1 1' ( . ')( ( )')BB A A A A

= A–1.A.A'.(A–1)1 {as AA' = A'A}

= I(A–1A)'

= I.I = I2 = I

6. The integral

111

xxx e dx

x is equal to

(1)

1xxx e c (2)

1

( 1)x

xx e c

(3)

1xxx e c (4)

1

( 1)x

xx e c

Answer (1)

Sol. I =

1 1

211

x xx xe x e dxx

=

1

.x

xx e c

As ( '( ) ( )) ( )xf x f x dx xf x c

7. The area of the region described byA = {(x, y) : x2 + y2 1 and y2 1 – x} is

(1)

42 3 (2)

22 3

(3)

22 3 (4)

42 3

Answer (4)Sol.

x y2 2 + = 1

Shaded area

12

0

(1)2 (1 )

2x dx

13/2

0

2(1 )( 1)

2 3/2x

4 (0 ( 1))2 3

4

2 3

8. The image of the line

31 4

3 1 5yx z

in the plane 2x – y + z + 3 = 0

is the line

(1)

53 23 1 5

yx z(2)

53 2

3 1 5yx z

(3)

53 23 1 5

yx z(4)

53 2

3 1 5yx z

Answer (4)

Sol.A (1,3, 4)

( )a, b, cA

P

ˆˆ ˆ3 5i j k

ˆˆ ˆ3i j k

1 3 4

2 1 1a b c

a = 2 + 1

b = 3 –

c = 4 +

1, 3 , 4

2 2P

2( 1) 3 4 3 0

2 2

2 2 3 + 4 3 02 2

3 + 6 = 0 = – 2

a = – 3, b = 5, c = 2

So the equation of the required line is

53 2

3 1 5yx z

9. The variance of first 50 even natural numbers is

(1) 833 (2) 437

(3)4374 (4)

8334

Answer (1)

(4)

Sol. Variance =

2

2( )ix xN

22 2 22 2 4 ... 100 2 4 ... 100

50 50

=

2 2 2 2

24(1 2 3 .... 50 ) (51)50

=

250 51 1014 (51)50 6

= 3434 – 2601

2 = 833

10. If z is a complex number such that |z| 2, then

the minimum value of 12

z

(1) Lies in the interval (1, 2)

(2) Is strictly greater than 52

(3) Is strictly greater than 32 but less than

52

(4) Is equal to 52

Answer (1)Sol.

12

12

z

So, 1 1| |2 2

z z

1 122 2

z

1 32 2

z

11. Three positive numbers form an increasing G.P. Ifthe middle term in this G.P. is doubled, the newnumbers are in A.P. Then the common ratio of theG.P. is

(1) 3 2 (2) 2 3

(3) 2 3 (4) 2 3

Answer (3)

Sol. a, ar, ar2 G.P.

a, 2ar, ar2 A.P.

2 × 2ar = a + ar2

4r = 1 + r2

r2 – 4r + 1 = 0

r =

4 16 4 2 3

2

2 3r

2 3r is rejected

(r > 1)

G.P. is increasing.

12. If the coefficients of x3 and x4 in the expansion of(1 + ax + bx2) (1 – 2x)18 in powers of x are both zero,then (a, b) is equal to

(1)

25114,3 (2)

27214,3

(3)

27216,3 (4)

25116,3

Answer (3)

Sol. (1 + ax + bx2) (1 – 2x)18

(1 + ax + bx2)[18C0 – 18C1(2x) + 18C2(2x)2 –18C3(2x)3 + 18C4(2x)4 – .......]

Coeff. of x3 = –18C3.8 + a × 4.18C2 – 2b × 18 = 0

=

18 17 16 4 18 17.8 36 0

6 2a b

= –51 × 16 × 8 + a × 36 × 17 – 36b = 0

= –34 × 16 + 51a – 3b = 0

= 51a – 3b = 34 × 16 = 544

= 51a – 3b = 544 ... (i)

Only option number (3) satisfies the equationnumber (i).

(5)

13. Let a, b, c and d be non-zero numbers. If the point ofintersection of the lines 4ax + 2ay + c = 0 and5bx + 2by + d = 0 lies in the fourth quadrant and isequidistant from the two axes then

(1) 2bc + 3ad = 0 (2) 3bc – 2ad = 0

(3) 3bc + 2ad = 0 (4) 2bc – 3ad = 0

Answer (2)

Sol. Let (, -) be the point of intersection

4a – 2a + c = 0 2ca

and 5b – 2b + d = 0 3db

3bc = 2ad

3bc – 2ad = 0

Alternative method :

The point of intersection will be

2 – 2x

ad bc =–

4 – 5y

ad bc =1

8 – 10ab ab

2( – )

–2ad bcx

ab

5 – 4

–2bc ady

ab

Point of intersection is in fourth quadrant so xis positive and y is negative.

Also distance from axes is same

So x = – y ( distance from x-axis is – y as y isnegative)

( )x, y

2( – ) –(5 – 4 )

–2 –2ad bc bc ad

ab ab

2ad – 2bc = – 5bc + 4ad

3bc – 2ad = 0 ...(i)

14. If 2 [ ]a b b c c a a b c then is equal to

(1) 3 (2) 0

(3) 1 (4) 2

Answer (3)

Sol. L.H.S.

= ( ) [( ) ( )]a b b c c a

= ( ) [( ) – ( ) ]a b b c a c b c c a

= ( ) [[ ] ]a b b c a c

[ . 0]b c c

= 2[ ] ( ) [ ]a b c a b c a b c

2[ ] [ ]a b b c c a a b c

So = 1

15. Let A and B be two events such that

1 1( ) , ( )6 4

P A B P A B and 1( )4

P A , where

A stands for the complement of the event A. Thenthe events A and B are

(1) Equally likely but not independent

(2) Independent but not equally likely

(3) Independent and equally likely

(4) Mutually exclusive and independent

Answer (2)

Sol. 1 1 5( ) ( ) 1–6 6 6

P A B P A B

1 1 3( ) ( ) 1–4 4 4

P A P A

( ) ( ) ( ) – ( )P A B P A P B P A B

5 3 1( ) –6 4 4

P B

1( )3

P B

P(A) P(B) so they are not equally likely.

Also P(A) × P(B) = 3 1 14 3 4

= P(A B)

( ) ( ) ( )P A B P A P B so A & B are independent.

16. Let PS be the median of the triangle with verticesP(2, 2), Q(6, –1) and R (7, 3). The equation of the linepassing through (1, –1) and parallel to PS is

(1) 2x + 9y + 7 = 0 (2) 4x + 7y + 3 = 0

(3) 2x – 9y – 11 = 0 (4) 4x – 7y – 11 = 0

Answer (1)

(6)

Sol.P(2,2)

R(7,3)Q(6,– 1) S S is mid-point of QR

So

7 6 3 – 1,2 2

S

13 ,12

Slope of PS = 2 – 1 2–13 92 –

2Equation of line

2– (–1) – ( – 1)9

y x

9y + 9 = – 2x + 2 2x + 9y + 7 = 0

17.

2

20

sin( cos )limx

xx

is equal to

(1) 1 (2) –

(3) (4)2

Answer (3)

Sol.

2

20

sin( cos )limx

xx

2

20

sin( (1 – sin )limx

xx

2

20

( – sin )lim sinx

xx

2

20

( sin )lim sinx

xx

[ sin( ) sin ]

2 2

2 20

( sin ) sinlim sin( sin )x

x xx x

2

0

sinlim 1x

xx

18. Let and be the roots of equationpx2 + qx + r = 0, p 0. If p, q, r are in A.P. and

1 1 4 , then the value of |–| is

(1)2 17

9(2)

349

(3)2 13

9(4)

619

Answer (3)Sol.

p, q, r are in AP2q = p + r ...(i)

Also 1 1 4

4

–

4 –4

qp q rr

p...(ii)

From (i)2(–4r) = p + rp = – 9rq = – 4rr = r

Now 2| – | ( ) – 4

2– 4–

q rp p

2 – 4| |

q prp

2 216 36|–9 |r r

r

= 2 13

919. A bird is sitting on the top of a vertical pole 20 m

high and its elevation from a point O on the groundis 45°. It flies off horizontally straight away from thepoint O. After one second, the elevation of the birdfrom O is reduced to 30°. Then the speed (in m/s)of the bird is

(1) 40 3 2 (2) 20 2

(3) 20 3 1 (4) 40 2 1

Answer (3)

Sol. A B

x y

45°30°

20 20

(7)

t = 1 s

From figure 20tan 45x

and 20tan 30

x y

so, 20( 3 1)y

i.e., speed = 20( 3 1) m/s.

20. If a R and the equation–3(x – [x])2 + 2 (x – [x]) + a2 = 0(where [x] denotes the greatest integer x) has nointegral solution, then all possible values of a lie inthe interval(1) (1, 2) (2) (–2, –1)(3) (–, – 2) (2, ) (4) (–1, 0) (0, 1)

Answer (4)Sol. –3(x – [x])2 + 2[x – [x]) + a2 = 0

3 {x}2 – 2{x} – a2 = 0

a 0,

2 223 { } { }3

x x a

22 1 13 { }

3 3a x

1 1 20 { } 1 and { }3 3 3

x x

21 40 3 { }3 3

x

21 1 13 { } 13 3 3

x

For non-integral solution0 < a2 < 1 and a (–1, 0) (0, 1)Alternative–3{x}2 + 2{x} + a2 = 0Now, –3{x}2 + 2{x}

2/31

1

to have no integral roots 0 < a2 < 1 a(–1, 0) (0, 1)

21. The integral

2

0

1 4 sin 4 sin2 2x x dx equals

(1)

2 4 4 33 (2) 4 3 4

(3)

4 3 43 (4) – 4

Answer (3)

Sol.

2

0

1 4 sin 4sin2 2x x dx

0

1sin2 2

2sin 12 2 6 3

5 52 6 3

x

x xdx x

x x

/3

0 /3

1 2 sin 2 sin 12 2x xdx dx

/3

0 /34 cos 4 cos

2 2x xx x

3 34 4 0 43 2 2 3

=

4 3 43

22. If f and g are differentiable functions in [0, 1]satisfying f(0) = 2 = g(1), g(0) = 0 and f(1) = 6, thenfor some c ]0, 1[

(1) 2f (c) = 3g(c) (2) f (c) = g(c)

(3) f (c) = 2g(c) (4) 2f (c) = g(c)

Answer (3)

Sol. Using, mean value theorem

(1) (0)

'( ) 41 0

f ff c

(1) (0)

'( ) 21 0

g gg c

so, '( ) 2 '( )f c g c

(8)

23. If g is the inverse of a function f and 51'( )

1f x

x,

then g(x) is equal to

(1) 5x4 (2) 51

1 ( )g x

(3) 1 + {g(x)}5 (4) 1 + x5

Answer (3)

Sol. 51'( ) ( ( )) '( ( )) '( ) 1

1f x f g x x f g x g x

x

51'( ) 1 ( ( ))'( ( ))

g x g xf g x

24. If (10)9 + 2(11)1 (10)8 + 3(11)2 (10)7+ ... + 10(11)9 =k(10)9, then k is equal to

(1)441100 (2) 100

(3) 110 (4)12110

Answer (2)

Sol. 109 + 2(11)(10)8 + 3(11)2(10)7 +... + 10(11)9 = k(10)9

x = 109 + 2(11)(10)8 + 3(11)2(10)7+ ... +10(11)9

1110

x = 11108 + 2(11)2(10)7 +... + 9(11)9 + 1110

11110

x = 109 + 11(10)8 + 112×(10)7 +... +119 – 1110

10

9 10

11 11010 111110 110

x

10 10 10 10(11 10 ) 11 1010x

x = 1011 = k109

k = 100

25. If , 0, and f(n) = n + n and

3 1 (1) 1 (2)1 (1) 1 (2) 1 (3)1 (2) 1 (3) 1 (4)

f ff f ff f f

= K(1 – )2 (1 – )2 ( – )2, then K is equal to

(1) 1

(2) 1

(3) –1 (4)

Answer (2)

Sol.

2 2

2 2 3 3

2 2 3 3 4 4

1 1 1 1 1

1 1 1

1 1 1

2

2

2 2

11 1 11 1

1 1 11= [(1 – )(1 – )(1 – )]2

So, 1k26. The slope of the line touching both the parabolas

y2 = 4x and x2 = –32y is

(1)32

(2)18

(3)23 (4)

12

Answer (4)Sol. y2 = 4x …(1)

x2 = –32y …(2)m be slope of common tangentEquation of tangent (1)

y = mx + 1m

…(i)

Equation of tangent (2)y = mx + 8m2 …(ii)

(i) and (ii) are identical

1m

= 8m2

m3 = 18

12

m

Alternative method :

Let tangent to 2 4y x be

1

y mxm

as this is also tangent to 2 32x y

Solving 2 3232 0x mx

mSince roots are equal D = 0

2 32(32) 4 0m

3 432

m

12

m

(9)

27. The statement ~(p ~q) is(1) Equivalent to ~ p q(2) A tautology(3) A fallacy(4) Equivalent to p q

Answer (4)Sol. ~(p ~ q)

pFFTT

qFTFT

~qTFTF

p ~q

FTTF

~ ( )TFFT

p ~q

Clearly equivalent to p q28. Let the population of rabbits surviving at a time t be

governed by the differential equation

( ) 1 ( ) 200

2dp t

p tdt . If p(0) = 100, then p(t) equals

(1) 300 – 200 e–t/2 (2) 600 – 500 et/2

(3) 400 – 300 e–t/2 (4) 400 – 300 et/2

Answer (4)

Sol. ( ) 1 ( ) 200

2dp t

p tdt

( ( ))

1 ( ) 2002

d p tdt

p t

( )2 log 200

2p t

t c

2( ) 2002

tp t e k

29. Let C be the circle with centre at (1, 1) and radius = 1.If T is the circle centred at (0, y), passing throughorigin and touching the circle C externally, then theradius of T is equal to

(1) 32

(2)12

(3)14

(4)32

Answer (3)Sol.

T

C

(0, )y (1, 1)

2 2( 1) ( 1) 1C x yRadius of T = |y|T touches C externally(0 – 1)2 + (y – 1)2 = (1 + |y|)2

1 + y2 + 1 – 2y = 1 + y2 + 2|y|If y > 0,y2 + 2 – 2y = y2 + 1 + 2y 4y = 1

y = 14

If y < 0,y2 + 2 – 2y = y2 + 1 – 2y 1 = 2 (Not possible)

14

y

30. The angle between the lines whose direction cosinessatisfy the equations l + m + n = 0 andl2 = m2 + n2 is

(1)4

(2)6

(3)2

(4)3

Answer (4)Sol. l + m + n = 0

l2 = m2 + n2

Now, (–m – n)2 = m2 + n2

mn = 0m = 0 or n = 0

If m = 0 If n = 0then l = –n then l = –ml2 + m2 + n2 = 1 l2 + m2 + n2 = 1Gives 2m2 = 1

12

n 2 12

m

i.e. (l1, m1, n1) 12

m

=

1 1,0,2 2 Let

12

m

12

l

n = 0(l2, m2, n2)

=

1 1, ,02 2

1cos2

3

(10)

PART–B : PHYSICS

31. When a rubber-band is stretched by a distance x, itexerts a restoring force of magnitude F = ax + bx2

where a and b are constants. The work done instretching the unstretched rubber-band by L is

(1)

2 312 2 3

aL bL(2) aL2 + bL3

(3) 2 31 ( )2

aL bL (4) 2 3

2 3aL bL

Answer (4)

Sol. dW F dl

20 0

L LW ax dx bx dx

2 3

2 3aL bL

.

32. The coercivity of a small magnet where theferromagnet gets demagnetized is 3 103 Am–1. Thecurrent required to be passed in a solenoid oflength 10 cm and number of turns 100, so that themagnet gets demagnetized when inside thesolenoid, is(1) 6 A (2) 30 mA(3) 60 mA (4) 3 A

Answer (4)Sol. B = 0 n i

0

B ni

32

1003 1010 10

NI iL

I = 3 A.33. In a large building, there are 15 bulbs of 40 W,

5 bulbs of 100 W, 5 fans of 80 W and 1 heater of1 kW. The voltage of the electric mains is 220 V. Theminimum capacity of the main fuse of the buildingwill be(1) 14 A (2) 8 A(3) 10 A (4) 12 A

Answer (4)Sol. 15 40 + 5 100 + 5 80 + 1000 = V I

600 + 500 + 400 + 1000 = 220 I

2500

11.36220

I

I = 12 A.

34. An open glass tube is immersed in mercury in sucha way that a length of 8 cm extends above themercury level. The open end of the tube is thenclosed and sealed and the tube is raised verticallyup by additional 46 cm. What will be length of theair column above mercury in the tube now?(Atmospheric pressure = 76 cm of Hg)(1) 6 cm (2) 16 cm(3) 22 cm (4) 38 cm

Answer (2)Sol.

54 cm(54 – )x

x8 cmP

P + x = P0P = (76 – x)8 A 76 = (76 – x) A (54 – x)x = 38Length of air column = 54 – 38 = 16 cm.

35. A bob of mass m attached to an inextensible stringof length l is suspended from a vertical support.The bob rotates in a horizontal circle with anangular speed rad/s about the vertical. About thepoint of suspension(1) Angular momentum changes both in direction

and magnitude(2) Angular momentum is conserved(3) Angular momentum changes in magnitude but

not in direction(4) Angular momentum changes in direction but

not in magnitudeAnswer (4)Sol. = mg l sin . (Direction parallel to plane of

rotation of particle)

ll

m

as is perpendicular to L , direction of L changes

but magnitude remains same.

(11)

Answer (3)

Sol. By Lens maker's formula

1 1 2

1 3/2 1 1– 1 –4/3f R R

2 1 2

1 3/2 1 1– 1 –5/3f R R

1 2

1 3 1 1– 1 –2f R R

f1 = 4f & f2 = –5f

39. A parallel plate capacitor is made of two circularplates separated by a distance 5 mm and with adielectric of dielectric constant 2.2 between them.When the electric field in the dielectric is 3 × 104

V/m, the charge density of the positive plate will beclose to

(1) 6 × 104 C/m2 (2) 6 × 10–7 C/m2

(3) 3 × 10–7 C/m2 (4) 3 × 104 C/m2

Answer (2)

Sol.

0E

K

= K0 E

= 2.2 × 8.85 ×10–12 × 3 ×104 6 × 10–7 C/m2

40. In the circuit shown here, the point 'C' is keptconnected to point 'A' till the current flowing throughthe circuit becomes constant. Afterward, suddenly,point 'C' is disconnected from point 'A' and connectedto point 'B' at time t = 0. Ratio of the voltage acrossresistance and the inductor at t = L/R will be equalto

A C R

LB

(1)1 – e

e (2) 1 –e

e

(3) 1 (4) –1

Answer (4)

Sol. Applying Kirchhoff's law in closed loop, –VR – VC = 0

VR/VC = –1

Note : The sense of voltage drop has not beendefined. The answer could have been 1.

36. The current voltage relation of diode is given byI = (e1000V/T – 1) mA, where the applied V is in voltsand the temperature T is in degree kelvin. If astudent makes an error measuring ± 0.01 V whilemeasuring the current of 5 mA at 300 K, what willbe the error in the value of current in mA?(1) 0.05 mA (2) 0.2 mA(3) 0.02 mA (4) 0.5 mA

Answer (2)Sol. I = (e1000 V/T – 1)mA

When I = 5 mA, e1000 V/T = 6 mA

Also, 1000 / 1000( )V TdI e dVT

(6 mA)1000 (0.01)300

= 0.2 mA37. From a tower of height H, a particle is thrown

vertically upwards with a speed u. The time takenby the particle, to hit the ground, is n times thattaken by it to reach the highest point of its path. Therelation between H, u and n is:(1) gH = (n – 2)u2 (2) 2 gH = n2u2

(3) gH = (n – 2)2u2 (4) 2 gH = nu2 (n – 2)Answer (4)

Sol. Time taken to reach highest point is 1tug

Speed on reaching ground 2 2u gh

Now, v = u + at

2 2 –u gh u gt

H

u

u gH2 + 2

2 2

tu u gH nu

g g

22 ( – 2)gH n n u

38. A thin convex lens made from crown glass

32

has focal length f . When it is measured in two

different liquids having refractive indices 43 and

53 ,

it has the focal lengths 1f and 2f respectively. Thecorrect relation between the focal lengths is

(1) 1f and 2f both become negative

(2) 1 2f f f

(3) 1f f and 2f becomes negative

(4) 2f f and 1f becomes negative

(12)

41. Two beams, A and B, of plane polarized light withmutually perpendicular planes of polarization areseen through a polaroid. From the position when thebeam A has maximum intensity (and beam B haszero intensity), a rotation of polaroid through 30°makes the two beams appear equally bright. If theinitial intensities of the two beams are IA and IB

respectively, then A

B

II equals

(1)13 (2) 3

(3)32

(4) 1

Answer (1)Sol. By law of Malus, I = I0cos2

Now, IA = IAcos230IB = IBcos260As IA = IB

3 14 4A BI I

13

A

B

II

42. There is a circular tube in a vertical plane. Twoliquids which do not mix and of densities d1 and d2are filled in the tube. Each liquid subtends 90° angleat centre. Radius joining their interface makes an

angle with vertical. Ratio 1

2

dd is

d2

d1

(1)

1 sin1 – cos (2)

1 sin1 – sin

(3)

1 cos1 – cos (4)

1 tan1 – tan

Answer (4)Sol. Equating pressure at A

d2 R sin

R

R

( cos – sinR R )

d1

Rcos

A

(Rcos + Rsin)d2g = (Rcos – Rsin)d1g

1

2

cos sin 1 tancos – sin 1 – tan

dd

43. The pressure that has to be applied to the ends of asteel wire of length 10 cm to keep its length constantwhen its temperature is raised by 100°C is

(For steel Young's modulus is 2 × 1011 Nm–2 andcoefficient of thermal expansion is 1.1 × 10–5 K–1)

(1) 2.2 × 106 Pa (2) 2.2 × 108 Pa

(3) 2.2 × 109 Pa (4) 2.2 × 107 PaAnswer (2)

Sol. As length is constant,

Strain = LL

= Q

Now pressure = stress = Y × strain

= 2 × 1011 × 1.1 × 10–5 × 100

= 2.2 × 108 Pa44. A block of mass m is placed on a surface with a

vertical cross-section given by y = 3

.6

x If the

coefficient of friction is 0.5, the maximum heightabove the ground at which the block can be placedwithout slipping is

(1)1 m2

(2)1 m6

(3)2 m3 (4)

1 m3

Answer (2)

Sol. tan = 2

2dy xdx

At limiting equilibrium,

= tan

0.5 = 2

2x

my

x = ±1

Now, y = 16

45. Three rods of copper, brass and steel are weldedtogether to form a Y-shaped structure. Area of cross-seciton of each rod = 4 cm2. End of copper rod ismaintained at 100°C whereas ends of brass andsteel are kept at 0°C. Lengths of the copper, brass

(13)

and steel rods are 46, 13 and 12 cm respectively.The rods are thermally insulated from surroundingsexcept at ends. Thermal conductivities of copper,brass and steel are 0.92, 0.26 and 0.12 CGS unitsrespectively. Rate of heat flow through copper rod is

(1) 6.0 cal/s (2) 1.2 cal/s

(3) 2.4 cal/s (4) 4.8 cal/s

Answer (4)

Sol. 100°C

0°C0°C

SteelB

Cu

TBrass

Q = Q1 + Q2

0.92 4(100 ) 0.26 4 ( 0) 0.12 446 13 12

T T T

200 – 2T = 2T + T

T = 40°C

Q = 0.92 4 6046 = 4.8 cal/s

46.. A mass m is supported by a massless string woundaround a uniform hollow cylinder of mass m andradius R. If the string does not slip on the cylinder,with what acceleration will the mass fall on release?

m R

m

(1) g (2)23g

(3)2g

(4)56g

Answer (3)

Sol. a = R

mg – T = ma

T × R = mR2R

m amg

T

Tor T = ma

a = 2g

47. Match List-I (Electromagnetic wave type) withList - II (Its association/application) and select thecorrect option from the choices given below the lists

(a)

(b)

(c)

(d)

Infrared waves

Radio waves

X-rays

Ultraviolet rays

To treat muscular strainFor broadcasting

To detect fracture of bonesAbsorbed by the ozone layer of the atmosphere

(i)

(ii)

(iii)

(iv)

(a) (b) (c) (d)

(1) (i) (ii) (iii) (iv)

(2) (iv) (iii) (ii) (i)

(3) (i) (ii) (iv) (iii)

(4) (iii) (ii) (i) (iv)Answer (1)

Sol. (a) Infrared rays are used to treat muscular strain

(b) Radiowaves are used for broadcasting

(c) X-rays are used to detect fracture of bones(d) Ultraviolet rays are absorbed by ozone

48. The radiation corresponding to 32 transition ofhydrogen atoms falls on a metal surface to producephotoelectrons. These electrons are made to enter amagnetic field of 3 × 10–4 T. If the radius of thelargest circular path followed by these electrons is10.0 mm, the work function of the metal is close to

(1) 1.6 eV (2) 1.8 eV

(3) 1.1 eV (4) 0.8 eV

Answer (3)

Sol. mvrqB

= 2m eV

eB =

1 2mVB e

2 2

2B r eV

m = 0.8 V

For transition between 3 to 2,

1 113 64 9

E .

13 6 5

36.

= 1.88 eV

Work function = 1.88 eV – 0.8 eV

= 1.08 eV = 1.1 eV

(14)

49. During the propagation of electromagnetic waves ina medium

(1) Both electric and magnetic energy densities arezero

(2) Electric energy density is double of themagnetic energy density

(3) Electric energy density is half of the magneticenergy density

(4) Electric energy density is equal to the magneticenergy density

Answer (4)

Sol. Energy is equally divided between electric andmagnetic field

50. A green light is incident from the water to the air -water interface at the critical angle(). Select thecorrect statement

(1) The entire spectrum of visible light will comeout of the water at various angles to the normal

(2) The entire spectrum of visible light will comeout of the water at an angle of 90° to the normal

(3) The spectrum of visible light whose frequencyis less than that of green light will come out tothe air medium

(4) The spectrum of visible light whose frequencyis more than that of green light will come out tothe air medium

Answer (3)

Sol. sin 1

c

cWater

air

For greater wavelength (i.e. lesser frequency) is less

So, c would be more. So, they will not sufferreflection and come out at angles less then 90°.

51. Four particles, each of mass M and equidistant fromeach other, move along a circle of radius R underthe action of their mutual gravitational attraction.The speed of each particle is

(1) 1 (1 2 2 )2

GMR (2)

GMR

(3) 2 2 GMR (4) (1 2 2 )GM

R

Answer (1)

Sol. 2

2 2F F mvF

R

FR

FF

v

O

m m

mm

2 2 2

2 222( 2 ) 4

Gm Gm mvRR R

221 1

4 2Gm mv

R

2 44 2

vGmR

1 1 2 22

GmR

52. A particle moves with simple harmonic motion in astraight line. In first s, after starting from rest ittravels a distance a, and in next s it travels 2a insame direction then(1) Time period of oscillations is 6

(2) Amplitude of motion is 3a

(3) Time period of oscillations is 8

(4) Amplitude of motion is 4a

Answer (1)Sol. As it starts from rest, we have

x = Acost. At t = 0, x = A

when t = , x = A – a

when t = 2, x = A – 3a

A – a = Acos

A – 3a = Acos2As cos2 = 2cos2 – 1

2– 3 –2 – 1A a A aA A

2 2 2

2– 3 2 2 – 4 –A a A a Aa AA A

A2 – 3aA = A2 + 2a2 – 4Aa

a2 = 2aA

A = 2aNow, A – a = Acos

cos12

23T

T = 6

(15)

53. A conductor lies along the z-axis at –1.5 z < 1.5 mand carries a fixed current of 10.0 A in ˆ– za direction

(see figure). For a field 4 –0.2 ˆ3.0 10 x

yB e a T, findthe power required to move the conductor at constantspeed to x = 2.0 m, y = 0 m in 5 × 10–3 s. Assumeparallel motion along the x-axis

z

yB

–1.52.0

x

1.5I

(1) 29.7 W (2) 1.57 W

(3) 2.97 W (4) 14.85 W

Answer (3)

Sol. Average Power = worktime

W = 2

0Fdx

= 2 –4 –0.20

3.0 10 10 3xe dx

= 2–3 –0.20

9 10 xe dx

=

–3–0.2 29 10 – 1

0.2e

B e = 3.0 × 10 –4 –0.2x

I = 10 Al = 3 m

z

x=

–3–0.49 10 1 –

0.2e

= 9 × 10–3 × (0.33)

= 2.97 × 10–3 J

–3

–32.97 10 2.97 W

(0.2) 5 10P

54. The forward biased diode connection is

(1) –2 V +2 V

(2) +2 V –2 V

(3) –3 V –3 V

(4) 2 V 4 V

Answer (2)

Sol.

p n

For forward Bias, p-side must be at higher potentialthan n-side.

55. Hydrogen (1H1), Deuterium (1H2), singly ionisedHelium (2He4)+ and doubly ionised lithium (3Li6)++

all have one electron around the nucleus. Consideran electron transition from n = 2 to n = 1. If the wavelengths of emitted radiation are 1, 2, 3 and 4respectively then approximately which one of thefollowing is correct?

(1) 1 = 22 = 33 = 44

(2) 41 = 22 = 23 = 4

(3) 1 = 22 = 23 = 4

(4) 1 = 2 = 43 = 94

Answer (4)

Sol.

22 21

1 1 1–RZn n

21

Z for given n1 & n2

1 = 2 = 43 = 94

56. On heating water, bubbles being formed at thebottom of the vessel detatch and rise. Take thebubbles to be spheres of radius R and making acircular contact of radius r with the bottom of thevessel. If r << R, and the surface tension of water isT, value of r just before bubbles detatch is(Density of water is w)

R

2r

(1)2 3 w gRT

(2)2

3w gRT

(3)2

6w gRT

(4)2 w gR

T

Answer (No answer)

(16)

Sol. When the bubble gets detached,

Buoyant force = force due to surface tension

T dl ×

R

r

34sin3 wT dl R g

3243 wT r

r R gR

4

2 23

wrR g

2 23

wr Rg

T

57. A pipe of length 85 cm is closed from one end. Findthe number of possible natural oscillations of aircolumn in the pipe whose frequencies lie below1250 Hz. The velocity of sound in air is 340 m/s.

(1) 4 (2) 12

(3) 8 (4) 6

Answer (4)

Sol.

(2 1) 1250

4n vf

L

(2 1) 340 12500.85 4

n

2n – 1 12.5

Answer is 6.

58. Assume that an electric field 2 ˆ30E x i exists in

space. Then the potential difference VA – VO, whereVO is the potential at the origin and VA the potentialat x = 2 m is

(1) 80 J (2) 120 J

(3) –120 J (4) –80 J

Answer (4)

Sol.

dV E dx

2

2

0

30A

O

V

V

dV x dx

3 20[10 ] 80 VA OV V x

59. A student measured the length of a rod and wroteit as 3.50 cm. Which instrument did he use tomeasure it?(1) A screw gauge having 50 divisions in the

circular scale and pitch as 1 mm(2) A meter scale(3) A vernier calliper where the 10 divisions in

vernier scale matches with 9 division in mainscale and main scale has 10 divisions in 1 cm

(4) A screw gauge having 100 divisions in thecircular scale and pitch as 1 mm

Answer (3)Sol. As measured value is 3.50 cm, the least count must

be 0.01 cm = 0.1 mmFor vernier scale with 1 MSD = 1 mm and9 MSD = 10 VSD,

Least count = 1 MSD – 1 VSD= 0.1 mm

60. One mole of diatomic ideal gas undergoes a cyclicprocess ABC as shown in figure. The process BC isadiabatic. The temperatures at A, B and C are 400 K,800 K and 600 K respectively. Choose the correctstatement

P

V

A C

B800 K

600 K

400 K

(1) The change in internal energy in the process BCis – 500R

(2) The change in internal energy in whole cyclicprocess is 250R

(3) The change in internal energy in the process CAis 700R

(4) The change in internal energy in the process ABis –350R

Answer (1)

Sol. 512VRU nC T T

For BC, T = –200 K U = –500R

(17)

PART–C : CHEMISTRY

61. Which one is classified as a condensation polymer?

(1) Acrylonitrile

(2) Dacron

(3) Neoprene

(4) Teflon

Answer (2)

Sol. Dacron is polyester formed by condensationpolymerisation of terephthalic acid and ethyleneglycol.

HOOC COOH + HO—CH —CH —OH2 2

—CO CH —CH —O—2 2

nDacron

Acrylonitrile, Neoprene and Teflon are additionpolymers of acrylonitrile, isoprene and tetrafluoroethylene respectively.

62. Which one of the following properties is not shownby NO?

(1) It's bond order is 2.5

(2) It is diamagnetic in gaseous state

(3) It is a neutral oxide

(4) It combines with oxygen to form nitrogendioxide

Answer (2)

Sol. Nitric oxide is paramagnetic in the gaseous state asit has one unpaired electron in its outermost shell.The electronic configuration of NO is

2 2 1

z x y x

2 2 2 2 2* *1s 1s 2s 2s 2p 2p 2p 2p

*

However, it dimerises at low temperature to becomediamagnetic.

2 22NO N O

Its bond order is 2.5 and it combines with O2 to givenitrogen dioxide.

63. Sodium phenoxide when heated with CO2 underpressure at 125°C yields a product which onacetylation produces C.

— ONa+ CO2

125°5 Atm

BH+

Ac O2

C

The major product C would be

(1)COOH

OCOCH3

(2)COOH

OCOCH3

(3)COCH3

OH

COCH3

(4) COOCH3

OH

Answer (2)

Sol.

O OO = C = O

HC

O—

O

O—

OHCOO—

(B)

H+

OHCOOH

(CH CO) O3 2

OCOCH3

COOH

(C)

—

64. Given below are the half-cell reactions

Mn2+ + 2e– Mn; E° = — 1.18 V(Mn3+ + e– Mn2+); E° = + 1.51 V

The E° for 3 Mn2+ Mn + 2Mn3+ will be

(1) –0.33 V; the reaction will occur

(2) –2.69 V; the reaction will not occur

(3) –2.69 V; the reaction will occur

(4) –0.33 V; the reaction will not occur

(18)

Answer (2)

Sol. (1) Mn2+ + 2e Mn; E° = –1.18V ;

1G 2F( 1.18) 2.36F

(2) Mn3+ + e Mn2+ ; E° = +1.51 V;

2G F(1.51) 1.51 F

(1) – 2 × (2)

3Mn2+ Mn + 2Mn3+ ;

3 1 2G G 2 G

= [2.36 – 2(–1.51)] F

= (2.36 + 3.02) F

= 5.38 F

But 3G 2FE

5.38F = –2FE°

E° = –2.69 V

As E° value is negative reaction is non spontaneous.

65. For complete combustion of ethanol,

C2H5OH(l) + 3O2(g) 2CO2(g) + 3H2O(l),

the amount of heat produced as measured in bombcalorimeter, is 1364.47 kJ mol–1 at 25°C. Assumingideality the enthalpy of combustion, cH, for thereaction will be (R = 8.314 kJ mol–1)

(1) –1350.50 kJ mol–1

(2) –1366.95 kJ mol–1

(3) –1361.95 kJ mol–1

(4) –1460.50 kJ mol–1

Answer (2)

Sol. C2H5OH(l) + 3O2(g) 2CO2(g) + 3H2O(l)

Bomb calorimeter gives U of the reaction

So, as per question

U = –1364.47 kJ mol–1

ng = –1

H = U + ngRT

1 8.314 2981364.471000

= –1366.93 kJ mol–1

66. For the estimation of nitrogen, 1.4 g of an organiccompound was digested by Kjeldahl method and the

evolved ammonia was absorbed in 60 mL of M10

sulphuric acid. The unreacted acid required 20 mL

of M10 sodium hydroxide for complete

neutralization. The percentage of nitrogen in thecompound is

(1) 5% (2) 6%

(3) 10% (4) 3%

Answer (3)

Sol. As per question

2 4

Normality VolumeNH SO 60mL5NNaOH 20mL10

2 4 3geq H SO geq NaOH geq NH(n ) (n ) (n )

3geq NH1 60 1 20 (n )5 1000 10 1000

3geq NH6 1 (n )

500 500

3geq NH5 1(n )

500 100

3 3mol N mol NH geq NH(n ) (n ) (n ) 1

100

N14(Mass) 0.14 g100

Percentage of "N" 0.14 1001.4

= 10%

67. The major organic compound formed by thereaction of 1, 1, 1-trichloroethane with silver powderis

(1) 2-Butene (2) Acetylene

(3) Ethene (4) 2-Butyne

Answer (4)

Sol. 2Cl—C—CH3

Cl

ClCH3 3C CCH + 6AgCl

Ag

1, 1, 1-trichloroethane

(19)

68. The ratio of masses of oxygen and nitrogen in aparticular gaseous mixture is 1 : 4. The ratio ofnumber of their molecule is

(1) 3 : 16 (2) 1 : 4

(3) 7 : 32 (4) 1 : 8

Answer (3)

Sol. Let the mass of O2 = x

Mass of N2 = 4x

Number of moles of O2 x

32

Number of moles of N2 4x28 = 7

x

Ratio x x:32 7 = 7 : 32

69. The metal that cannot be obtained by electrolysis ofan aqueous solution of its salts is

(1) Cr (2) Ag

(3) Ca (4) Cu

Answer (3)

Sol. On electrolysis only in case of Ca2+ salt aqueoussolution H2 gas discharge at Cathode.

Case of Cr

At cathode : Cr3+ + 2e– Cr

So, Cr is deposited.

Case of Ag

At cathode : Ag+ + e– Ag

So, Ag is deposited.

Case of Cu

At cathode : Cu2+ + 2e– Cu

Case of Ca2+

At cathode : H2O + e– 12

H2 + OH–

70. The equivalent conductance of NaCl atconcentration C and at infinite dilution are C and, respectively. The correct relationship between Cand is given as

(Where the constant B is positive)

(1) C (B) C

(2) C (B)C

(3) C – (B)C

(4) C –(B) C

Answer (4)

Sol. According to Debye Huckle onsager equation,

C A C

Here A = B

C B C

71. The correct set of four quantum numbers for thevalence electrons of rubidium atom (Z = 37) is

(1)15, 0, 1,2

+ (2)15, 0, 0,2

+

(3)15, 1, 0,2

+ (4)15, 1, 1,2

+

Answer (2)

Sol. 37 1s22s22p63s23p63d104s24p65s1

So last electron enters 5s orbital

Hence n = 5, l = 0, ml = 0, s1m2

72. Consider separate solutions of 0.500 M C2H5OH(aq),0.100 M Mg3(PO4)2(aq), 0.250 M KBr(aq) and 0.125M Na3PO4(aq) at 25°C. Which statement is trueabout these solutions, assuming all salts to be strongelectrolytes?

(1) 0.500 M C2H5OH(aq) has the highest osmoticpressure.

(2) They all have the same osmotic pressure.

(3) 0.100 M Mg3(PO4)2(aq) has the highest osmoticpressure.

(4) 0.125 M Na3PO4(aq) has the highest osmoticpressure.

Answer (2)

Sol. i CRT

2 5C H OH 1 0.500 R T 0.5 RT

3 4 2Mg (PO ) 5 0.100 R T 0.5 RT

KBr 2 0.250 R T 0.5 RT

3 4Na PO 4 0.125 RT 0.5 RT

73. The most suitable reagent for the conversion ofR – CH2 – OH R – CHO is

(1) PCC (Pyridinium Chlorochromate)

(2) KMnO4

(3) K2Cr2O7

(4) CrO3

(20)

Answer (1)

Sol. PCC is mild oxidising agent, it will convert

2R CH OH R CHO

74. CsCl crystallises in body centred cubic lattice. If ‘a’is its edge length then which of the followingexpressions is correct?

(1) Cs Cl

r r 3a (2) Cs Cl

r r 3a

(3) Cs Cl

3ar r 2 (4) Cs Cl

3r r a2Answer (4)

Sol.

Cl—

Cl—

Cl–

ClCl—

Cs+

Cl–

Cl–

Cl—

Cl Cs2r 2r 3 a

Cl Cs3ar r2

75. In which of the following reactions H2O2 acts as areducing agent?(a) H2O2 + 2H+ + 2e– 2H2O

(b) H2O2 – 2e– O2 + 2H+

(c) H2O2 + 2e– 2OH–

(d) H2O2 + 2OH– – 2e– O2 + 2H2O

(1) (b), (d)

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

(4) (a), (c)

Answer (1)

Sol. The reducing agent oxidises itself.

(a) 1 22 2 2H O 2H 2e 2H O

(b) 0

12 2 2H O 2e O 2H

(c)

2

12 2H O 2e 2OH

(d)0

12 2 2 2H O 2OH 2e O H O

Note : Powers of 'O' are oxidation number of 'O' inthe compound.

76. For which of the following molecule significant 0?

(a)

Cl

Cl

(b)

CN

CN

(c)

OH

OH

(d)

SH

SH

(1) (c) and (d) (2) Only (a)

(3) (a) and (b) (4) Only (c)

Answer (1)

Sol. (a)

Cl

Cl = 0

(b)

CN

CN = 0

(c)

O

O

H

H

0

(d)

S

S

H

H

077. On heating an aliphatic primary amine with

chloroform and ethanolic potassium hydroxide, theorganic compound formed is

(1) An alkyl isocyanide (2) An alkanol(3) An alkanediol (4) An alkyl cyanide

Answer (1)

Sol. R – CH2 – NH2 CHCl /KOH3C H OH2 5 R– CH2 – NC

78. In SN2 reactions, the correct order of reactivity for thefollowing compounds

CH3Cl, CH3CH2Cl, (CH3)2CHCl and (CH3)3CCl is

(1) (CH3)2CHCl > CH3CH2Cl > CH3Cl > (CH3)3CCl

(2) CH3Cl > (CH3)2CHCl > CH3CH2Cl > (CH3)3CCl(3) CH3Cl > CH3CH2Cl > (CH3)2CHCl > (CH3)3CCl

(4) CH3CH2Cl > CH3Cl > (CH3)2CHCl > (CH3)3CCl

Answer (3)

(21)

Sol. Rate of SN2 reaction depends on steric crowding ofalkyl halide. So order is

CH3Cl > (CH3)CH2 – Cl > (CH3)2CH – Cl > (CH3)3CCl

79. The octahedral complex of a metal ion M3+ with fourmonodentate ligands L1, L2, L3 and L4 absorbwavelengths in the region of red, green, yellow andblue, respectively. The increasing order of ligandstrength of the four ligands is

(1) L1 < L2 < L4 < L3

(2) L4 < L3 < L2 < L1

(3) L1 < L3 < L2 < L4

(4) L3 < L2 < L4 < L1

Answer (3)Sol.

VB

G

YRO

The energy of red light is less than that of violetlight.

So energy order is

Red < Yellow < Green < Blue

The complex absorbs lower energy light lower willbe its strength. So order of ligand strength is

L1 < L3 < L2 < L4

80. The equation which is balanced and represents thecorrect product(s) is

(1) CuSO4 + 4KCN K2[Cu(CN)4] + K2SO4

(2) Li2O + 2KCl 2LiCl + K2O

(3) [CoCl(NH3)5]+ + 5H+ Co2+ + 5NH4+ + Cl—

(4) [Mg(H2O)6]2+ + (EDTA)4— excess NaOH

[Mg(EDTA)]2+ + 6H2O

Answer (3)

Sol. The complex

[CoCl(NH3)5]+ decomposes under acidic medium, so

[CoCl(NH3)5]+ + 5H+ Co2+ + 5NH4+ + Cl—.

81. In the reaction,

54 PClLiAlH Alc.KOH3CH COOH A B C,

the product C is

(1) Acetyl chloride

(2) Acetaldehyde

(3) Acetylene

(4) Ethylene

Answer (4)

Sol. Ethylene

CH COOH3

LiAlH4 CH CH OH 'A'3 2

PCl5

CH CH Cl 'B'3 2

Alc. KOH

CH = 'C'2 CH2

82. The correct statement for the molecule, CsI3, is

(1) It contains Cs+, I– and lattice I2 molecule

(2) It is a covalent molecule

(3) It contains Cs+ and –3I ions

(4) It contains Cs3+ and I– ions

Answer (3)

Sol. It contains Cs+ and I3– ions

83. For the reaction 2(g) 2(g)1SO O2

SO3(g),

if KP = KC(RT)x where the symbols have usualmeaning then the value of x is (assuming ideality)

(1) 1 (2) –1

(3) 12

(4)12

Answer (3)

Sol. 2 2 31SO (g) O (g) SO (g)2

KP = KC(RT)x

x = gn = no. of gaseous moles in product

– no. of gaseous moles in reactant

=

1 3 11 1 12 2 2

84. For the non-stoichiometre reaction 2A + B C + D,the following kinetic data were obtained in threeseparate experiments, all at 298 K.

–1 –1

–3

–3

–3

Initial Initial Initial rate of Concentration Concentration formation of C(A) (B) (mol L s )

1.2 × 100.1 M 0.1 M0.1 M 0.2 M 1.2 × 100.2 M 0.1 M 2.4 × 10

(22)

The rate law for the formation of C is

(1) dc k[A]dt

(2) dc k[A] [B]dt

(3) 2dc k[A] [B]dt

(4) 2dc k[A][B]dt

Answer (1)

Sol. 2A B C D

Rate of Reaction =

1 d[A] d[B]

2 dt dt

d[C] d[D]dt dt

Let rate of Reaction = k[A]x[B]y

Or, yxd[C]k[A] [B]

dt

Now from table,

1.2 × 10–3 = k [0.1]x[0.1]y ...(i)

1.2 × 10–3 = k [0.1]x[0.2]y ...(ii)

2.4 × 10–3 = k [0.2]x[0.1]y ...(iii)

Dividing equation (i) by (ii)

y3 x

3 yx1.2 10 k[0.1] [0.1]1.2 10 k[0.1] [0.2]

y112

y 0

Now Dividing equation (i) by (iii)

y3 x

3 yx1.2 10 k[0.1] [0.1]2.4 10 k[0.2] [0.1]

1 x1 12 2

x 1

Hence 1 0d[C] k[A] [B]dt

= .

85. Resistance of 0.2 M solution of an electrolyte is50 . The specific conductance of the solution is1.4 S m–1. The resistance of 0.5 M solution of thesame electrolyte is 280 . The molar conductivity of0.5 M solution of the electrolyte in S m2 mol–1 is

(1) 5 × 102 (2) 5 × 10–4

(3) 5 × 10–3 (4) 5 × 103

Answer (2)Sol. For 0.2 M solution

R = 50 = 1.4 S m–1 = 1.4 × 10–2 S cm–1

21 1 cm

1.4 10

Now, Ral

2R 50 1.4 10

al

For 0.5 M solutionR = 280 = ?

250 1.4 10al

Ral

1 1

R al

= 21 50 1.4 10280

= 21 70 10280

= 2.5 × 10–3 S cm–1

Now,

m1000M

= 32.5 10 1000

0.5= 5 S cm2 mol–1

= 5 × 10–4 S m2 mol–1

86. Among the following oxoacids, the correctdecreasing order of acid strength is

(1) HClO2 > HClO4 > HClO3 > HOCl

(2) HOCl > HClO2 > HClO3 > HClO4

(3) HClO4 > HOCl > HClO2 > HClO3

(4) HClO4 > HClO3 > HClO2 > HOCl

Answer (4)

(23)

Sol. 4 4HClO ClO H

3 3HClO ClO H

2 2HClO ClO H

HOCl ClO H

Resonance produced conjugate base.

(i) Cl

O

OO O–

Cl

O–

OO O

Cl

O

O

–O O

(ClO )4–

Cl

O

–OO O

(ii) Cl

O

OO–

(ClO )3–

Cl

O–

OO

Cl

O

O–O

(iii) Cl

O

O–

(ClO )2–

Cl

O–

O

(iv) ClO– is not resonance stabilized.

As per resonance stability order of conjugate base is

4 3 2ClO ClO ClO ClO

Hence acidic strength order is

4 3 2HClO HClO HClO HClO

87. Which one of the following bases is not present inDNA?

(1) Thymine (2) Quinoline

(3) Adenine (4) Cytosine

Answer (2)

Sol. DNA contains ATGC basesA – Adenine

T – Thymine

G – Guanine

C – Cytocine

So quinoline is not present.

88. Considering the basic strength of amines in aqueoussolution, which one has the smallest pKb value?

(1) C6H5NH2

(2) (CH3)2NH

(3) CH3NH2

(4) (CH3)3N

Answer (2)

Sol. Among C6H5NH2, CH3NH2, (CH3)2NH,

(CH3)3N . C6H5NH2 is least basic due to resonance.

NH2+NH2

+2NH

+2NH NH2

Out of (CH3)3N, CH3NH2, (CH3)2NH . (CH3)2NH ismost basic due to +I effect and hydrogen bonding inH2O.

N

HH C3

CH3

O

H H

+I effect

+I effect

Hydrogen bonding

89. If Z is a compressibility factor, van der Waalsequation at low pressure can be written as

(1) PbZ 1RT

(2) RTZ 1Pb

(3) aZ 1 –

VRT

(4) PbZ 1 –RT

(24)

Answer (3)

Sol. Compressibility factor PV(Z)RT

(For one mole of real gas)van der Waal equation

2a(P )(V b) RT

VAt low pressure

V b V

2

aP V RTV

aPV RTV

aPV RTV

PV a1RT VRT

So, aZ 1

VRT

90. Which series of reactions correctly representschemical reactions related to iron and itscompound?

(1) 2O ,heat CO,600ºC3 4Fe Fe O

CO,700ºCFeO Fe

(2) 2 4 2 4 2dil.H SO H SO ,O4Fe FeSO

heat2 4 3Fe (SO ) Fe

(3) 2 2 4O ,heat dil.H SOFe FeOheat

4FeSO Fe

(4) 2 heat, airCl ,heat3Fe FeCl

Zn2FeCl Fe

Answer (4)Sol. Anhydrous ferric chloride is prepared by passing

dry chlorine gas over heated iron fillings.2Fe + 3Cl2 2FeCl3FeCl3 on heating gives FeCl2 and Cl2

FeCl3 2FeCl2 + Cl2FeCl3 is reduced by Zn form Fe and ZnCl2FeCl2 + Zn Fe + ZnCl2