JEE Mains Ans Sol Code C 2015

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Answers & Solutionsto

JEE-Mains-2015 (Code-C)

INSTRUCTIONS

Question Paper Format

The Test Booklet consists of 90 questions.

There are Three Parts in the question paper A, B, C consisting of Mathematics, Physics and

Chemistry having 30 questions in each part of equal weightage. Each question is allotted 4 (four)

marks for each correct response.

Candidates will be awarded marks as stated above for correct response of each question.1/

4 (one fourth) marks will be deducted for indicating incorrect response of each question.

No deduction from the total score will be made if no response is indicated for an item in the answer

sheet.

Time : 3 Hrs. Max. Marks 360

PAPER-1

"The questions in the booklet are those which appeared in JEE MainsPaper 1 conducted by CBSE on 04th April 2015.

DISCLAIMER : All model answers in this Solution to JEE-Mains paper are written by Studymate Subject Matter Experts.This is not intended to be the official model solution to the question paper provided by CBSE.The purpose of this solution is to provide a guidance to students.

Answers and Solutions to JEE Mains-2015 Paper-1 (Code-C)

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PART-I : MATHEMATICS1. A complex number z is said to be unimodular if |z| = 1. Suppose z

1 and z

2 are complex numebr

such that 1 2

1 2

2

2

z z

z z

is unimodular and z

2 is not unimodular. Then the point z

1 lies on a :

(1) circle of radius 2

(2) straight line parallel to x-axis

(3) straight line parallel to y-axis

(4) circle of raidius 2.

Ans. (4)

Sol. Given

1 2

1 2

21

2

z z

z z

21 2 12 2z z z z S.B.S.

2221 2 12 2z z z z

2 2 2 21 2 1 2 1 2 1 2 1 2 1 2| | 4| | 2 2 4 | | | | 2 2z z z z z z z z z z z z

2 21 1| | 1 | | 4 0z z

1 1| | 1 or | | 2z z

2. The normal to the curve, x2 + 2xy – 3y2 = 0, at (1, 1):

(1) meets the curve agian in the fourth quadrant.

(2) does not meet the curve again.

(3) meets the curve again in the second quadrant.

(4) meets the curve again in the third quadrant.

Ans. (1)

Sol. ( )( 3 ) 0x y x y

y = x

3 0y x

Equation of normal at (1, 1) is

2 0x y with x + 3y = 0 is (3, –1)

3. The sum of first 9 terms of the series 3 3 3 3 3 31 1 2 1 2 3

... is :1 1 3 1 3 5

(1) 192 (2) 71

(3) 96 (4) 142

Ans. (3)

Sol.3 3 3 3 3 31 1 2 1 2 3

1 1 3 1 3 5

22 2

3 3 3 3

2

( 1) ( 1)1 2 3 2 41 3 5 (2 1) [2 ]

2

n

n n n nn

ann nn

21( 2 1)

4na n n

9 92

91 1

1S 2 9

4 n n

n n


-(3)-

1 9(9 1)(18 1) 2(9)(9 1)9

4 6 2

1 9 10 199 10 9

4 6

1285 99

4

38496

4

4. Let f(x) be a polynomial of degree four having extreme values at x = 1 x = 2. If

20lim 1 3,x

f x

x

then f(2) is equal to :

(1) 4 (2) –8

(3) –4 (4) 0

Ans. (4)

Sol. Let 4 3 2( )f x ax bx cx dx e

20

( )lim 1 3x

f x

x

4 3 2

20lim 1 3x

ax bx cx dx e

x

As RHS is finite so, d = e = 0

2 2

20

1 ( )lim 3x

x ax bx c

x

2

0lim 1 3x

ax bx c

c = 2

Hence, 4 3 2( ) 2f x ax bx x

3 2( ) 4 3 4f x ax bx x

As extreme value are 1 and 2.

(1) 7 3 4 0f a b ...(1)

(2) 32 12 8 0f a b ...(2)

Solve (1) and (2)

1

2a

2b

Hence 4 3 21

( ) 2 22

f x x x x

( ) 0f x

5. The negation of ~s (~r s) is equivalent to :

(1) s r (2) s ~ r

(3) s (r ~ s) (4) s r s

Ans. (1)


-(4)-

Sol.

~ ~ ~ ~ (~ ) ( ( )s r r r s s s r s v vsv rvs

T T F F F F T

T F T T F T F

F T F F T T F

F F T F T T F

which is equivalent to sr

6. If

1 2 2

2 1 2

2

A

a b

is a matrix satisfying the equation AAT = 9I, I is 3 × 3 identity matrix, then

the ordered pair (a, b) is equal to :

(1) (– 2, – 1) (2) (2, –1)

(3) (–2, 1) (4) (2, 1)

Ans. (1)

Sol.

1 2 2

A 2 1 2

2a b

T

1 2 2 1 2 9 0 0

AA 9I 2 1 2 2 1 2 0 9 0

2 2 2 0 0 9

a

a b b

2 2

9 0 4 2 9 0 0

0 9 2 2 2 0 9 0

4 2 2 2 2 4 0 0 9

a b

a b

a b a b a b

a + 2b = – 4 … (i)

a – b = – 1 … (ii)

Solving (i) and (ii)

3b = – 3

1 ; 2b a

(–2, –1)

7. The integral

32 4 41

dx

x x equals:

(1)

14 4

4

1xC

x

(2)

14 4

4

1xC

x

(3) 1

4 41x C (4) – 1

4 41x C

Ans. (1)

Sol. 2 4 3/4I

( 1)

dx

x x

3/42 3

4

1. 1

dx

x xx


-(5)-

Put 4

11 t

x

5

4dx dt

x

3/4

1

4

dt

t

34

1

4t dt

1/414 C

4t

1/4

4

11 C

x

1/44

4

1C

x

x

8. If m is the A.M. of two distinct real numbers l and n (l, n > 1) and G1, G

2 and G

3 are three

geometric means between l and n, the 4 4 41 2 32G G G equals.

(1) 4 l2m2n2 (2) 4l2mn

(3) 4lm2n (4) 4lmn2

Ans. (3)

Sol.2

l nm

and l, G

1, G

2, G

3, n are in G.P.

n = l (r)4

1

4nr

l

4 4 41 2 32G G G 4 2 4 3 4( ) 2( ) ( )lr lr lr

4 4 4 81 2.l r r r

24

2· 1 2·n n n

ll l l

32 2

2

.[ 2 ]

l nl nl n

l

= ln . (l + n)2

= ln (2m)2

= 4lnm2

9. Let y(x) be the solution of the differential equation log 2 log , 1 .dy

x x y x x xdx

Then y(e)

is equal to :

(1) 2e (2) e

(3) 0 (4) 2

Ans. (4)

Sol.1

2log

dyy

dx x x

1

logIFdx

x xe


-(6)-

1log log

dxtte e t x

. log C 2.logy x x dx

C 2 log .x x dx y . log x = C + 2 x log x – 2x … (i)

Put x = 1

y . 0 = C + 0 – 2

C 2y . log x = 2 + 2 x . log x – 2x

y (e) . 1 = 2 + 0

y (e) = 2

10. The number of integers greater than 6,000 that can be formed, using the digits 3, 5, 6, 7 and8, without repetition, is :

(1) 72 (2) 216

(3) 192 (4) 120

Ans. (3)

Sol. 4 digit number are greater than 60006

4 × 3 × 2 = 24

7

8

72Total digit greaterthan 6000

4 × 3 × 2= 24

4 × 3 × 2= 24

5 digit number greater than 6000

5 × 4 × 3 × 2 × 1 = 120Total number of digit greater than 6000 = 120 + 72

= 192

11. The number of points, haivng both co-ordinates as integers, that lie in the interior of thetriangle with vertices (0, 0), (0, 41) and (41, 0), is :

(1) 780 (2) 901

(3) 861 (4) 820

Ans. (1)

Sol. Required number of points

= 1 + 2 + .... + 39

= 39

(39 1)2

(0, 41) (2, 40)

(40, 1)

(41, 0)(0, 0)= 780

12. Let and be the roots of equation x2 – 6x – 2 = 0, If , for 1,n nna n then the value of

10 8

9

2

2

a a

a

is equal to :

(1) –3 (2) 6

(3) –6 (4) 3

Ans. (4)

Sol.1 1

1 . .n n n nn

n n n nn

a

a

. . . . . .n n n n n n

n n


-(7)-

1 1( ) ( ) ( )

( )

n n n n n n

n n

1 1( )n n

n n

a a

a a

i.e.,10 8

9 9

6 2.a a

a a

i.e.,10 8

9 9

2. 6a a

a a

i.e.,10 8

9

2 63

2 2

a a

a

13. Let 1 1 12

2 1tan tan tan , where | | .

1 3

xy x x

x

Then a value of y is :

(1)3

2

31 3

x x

x

(2)3

2

31 3

x x

x

(3)3

2

31 3x x

x

(4)3

2

31 3

x x

x

Ans. (2)

Sol. tan–1 y = tan–1 x + tan–12

2

1

x

x

tan–1 y = 3 tan–1 x3

1 12

3tan tan

1 3

x xy

x

1 1

3 3x

3

2

3

1 3

x xy

x

14. The distance of the point (1, 0, 2) from the point of intersection of the line 2 1 2

3 4 12x y z

and

the plane x – y + z = 16, is :

(1) 13 (2) 2 14

(3) 8 (4) 3 21

Ans. (1)

Sol.2 1 2 2

3 4 12

x yk

x = 3k + 2

y = 4k – 1

z = 12k + 2

x – y + z = 16

3k + 2 – 4k + 1 + 12 k + 2 = 16

11k = 11

k = 1

x = 5

y = 3

z = 14

Distance between (1, 0, 2) and (5, 3, 14)

D 16 9 144 13


-(8)-

15. The area (in sq. units) of the region described by 2, : 2 and 4 1x y y x y x is :

(1)932

(2)732

(3)564

(4)1564

Ans. (1)

Sol. {(x, y) : y2 2x and y 4x – 1}

y2 = 2x … (i)

y = 4x – 1 … (ii)

12

2

yx

X0

Y y k2 = 2

–1 41 , 1

2

12

, 1Points of intersection

2 1

2

yy

2y2 – y – 1 = 0

2y2 – 2y + y – 1 = 0

(y – 1) (2y + 1) = 0

y = 1, 1

2y

Area = 1 2

1

2

1

4 2

y ydy

12

1

2

1( 1 2 )

4y y dy

12 3

12

1 2

4 2 3

y yy

1 1 2 1 1 1 91

4 2 3 8 2 12 32

16. Let O be the vertex and Q be any point on the paraboloa, x2 = 8y. If the point P divides the linesegment OQ inernally in the ratio 1 : 3, then the locus of P is :

(1) x2 = 2y (2) x2 = y

(3) y2 = x (4) y2 = 2x

Ans. (1)

Sol. 1 10 0,

4 4

x yh k

X0

P( , )h k1

3

Q( , )x y1 1

(0, 0)x

1 = 4h, y

1 = 4k

x12 = 8y

1

(4h)2 = 8(4k)

16h2 = 32k

h2 = 2k

hence locus, x2 2y

17. The mean of the data set comprising of 16 observations is 16. If one of the observation valued16 is deleted and three new observations valued 3, 4 and 5 are addeed to the data, then themean of the resultant data, is :

(1) 14.0 (2) 16.8

(3) 16.0 (4) 15.8


-(9)-

Ans. (1)

Sol.

16

1

16

ii

xX

16

1

16 16 ii

x

16

1

256 ii

x

256 = x

1 + x

2 + .... + x

16

As 16 deleted and 3, 4, 5 add

256 – 16 + 3 + 4 + 5 = x1 + x

2 + .......+x

18

256 = x1 + x

2 +.... + x

18

18

1

18

ii

new

xX

252

18

new 14X 18. The area (in sq. units) of the quadrilateral formed by the tangents at the end points of the

latera recta to the ellipse 2 2

1,is : 9 5x y

(1) 27 (2)274

(3) 18 (4)272

Ans. (1)

Sol. Equation of tangent at L1

2

,b

aea

is

2

19 5

xae yb

a

i.e.5

13 5 3

xe y

, where a = 3; b2 = 5

0 ae

L1

3

b2

aae,

(0, 3)

9, 0

2

b2 = a2 (1 – e2)

5 = 9 – 9e2

9e2 = 4

2

3e

i.e.,2

19 3

yx

This tangent has intercepts 9

& 32

Required area 1 9

4 32 2

= 27

19. The equation of the plane contaning the line 2x – 5y + z = 3; x + y + 4z = 5, and parallel to theplane, x + 3y + 6z = 1, is :

(1) 2x + 6y + 12z = – 13 (2) 2x + 6y + 12z = 13

(3) x + 3y + 6z = – 7 (4) x + 3y + 6z = 7

Ans. (4)


-(10)-

Sol. 2x – 5y + z = 3; x + y + 4z = 5

Put z = 0 in the two given equation

2x – 5y = 3 x + y = 5

Solving we get

x = 4, y = 1

Coordinates of point is (4, 1, 0)

Hence equation of plane || to x + 3y + 6z = 1 is

1(x – 4) + 3(y – 1) + 6(z – 0) = 0

x + 3y + 6z = 7

20. The number of common tangents to the circles x2 + y2 – 4x – 6y – 12 = 0 and x2 + y2 + 6x + 18y+ 26 = 0, is :

(1) 4 (2) 1

(3) 2 (4) 3

Ans. (4)

Sol. x2 + y2 – 4x – 6y – 12 = 0 C1(2, 3)

1R 4 9 12 5

x2 + y2 + 6x + 18y + 26 = 0 C2(–3, –9),

2R 9 81 26 8

2 21 2C C (2 3) ( 9 3) 25 144 13

C1

R1C2

R2

C1C

2 = R

1 + R

2

3 common tangents

21. The set of all values of for which the system of linear equations :

2x1 – 2x

2 + x

3 = x

1

2x1 – 3x

2 + 2x

3 = x

2

–x1 – 2x

2 = x

3

has a non-trivial solution,

(1) contains more than two elements. (2) is an empty set.

(3) is a singleton. (4) contains two elements.

Ans. (4)

Sol. 1 1 2 3(2 ) 2 0x x x

1 2 32 ( 3 ) 2 0x x x

1 2 32 0x x x

D = 0

2 2 1

2 3 2 0

1 2

1 1 3( )R R R

1 0 1

2 3 2 0

1 2

(1 )

1 0 1

2 3 2 0

1 2

1 1 3( )C C C


-(11)-

(1 )

0 0 1

0 3 2 0

1 2

(1 ) (0 (3 ) ( 1)) 0

(1 ) (3 ) ( 1) 0

1, 3, 1

22. If 12 identical balls are to be placed in 3 identical boxes, then the probability that one of theboxes contains exactly 3 balls is :

(1)11

122

3

(2)11

55 23 3

(3)10

255

3

(4)12

1220

3

Ans. ()

Sol. Error in the question

23. The sum of coefficient of integral powers of x in the binomial expansion of 50

1 2 is : x

(1)501

(2 1)2

(2)501

(3 1)2

(3)501

(3 )2

(4)501

(3 1)2

Ans. (2)

Sol. 50 50(1 2 ) (1 2 )x x

50 50 50 50 50 48 50 460 2 4(2 1) (2 1) 2 C (2 ) C (2 ) C (2 )x x x x x

Put x = 1; [1 + 350] = 2{50C0 (2)50 + 50C

2(2)48 + 50C

4(2)46 + …}

i.e., Required sum = 501

(3 1)2

24. The integral 24

2 22

log

log log 36 12

xdx

x x x is equal to :

(1) 6 (2) 2

(3) 4 (4) 1

Ans. (4)

Sol.4 2

2 22

logI

log log(36 12 )

xdx

x x x

4 2

2 22

log

log log(6 )

xdx

x x

… (i)

Using ( ) ( )b b

a a

f x dx f a b x dx

4 2

2 22

log(6 )I

log(6 ) log

x

x x

… (ii)

Adding (i) and (ii)


-(12)-

4

2

2I 1dx

42

1 1I [ ] (4 2) 1

2 2x

25. If the function. 1, 0 3

2, 3 5

k x xg x

mx x

is differentiable, then the value of k + m is :

(1) 1 (2) 2

(3)165

(4)103

Ans. (2)

Sol. 1 0 3( )

2 3 5

k x xg x

mn x

Hence function is differentiable

0 3( ) 2 1

3 5

kx

g x xm x

LHD = RHD

2 1

km

x

as x = 3

4

km

k = 4m

g(x) is difference hence it’s continous

RHL = LHL = f(3)

3lim 2 1x

mx k x

3m + 2 = 2k

3m + 2 = 2(4m)

2 = 5m

2

5m

8

5k

2 8

5 5m k 2

26. Locus of the image of the point (2, 3) in the line (2x – 3y + 4)+ k(x – 2y + 3) = 0, ,k R is a:

(1) circle of radius 3 (2) straight line parallel to x-axis

(3) straight line paralle to y-axis. (4) circle of radius 2

Ans. (4)

Sol. The given family of lines pass through the intersection of 2x – 3y + 4 = 0 and 2x – 4y + 6 = 0

i.e.,

2 3 4 0

2 4 6 0

2 0

2

x y

x y

y

y

2 6 4 0x


-(13)-

2 2x

x = 1

i.e. P (1, 2)

mPM

× mAB

= –1

32

32 12 2

12

(1,2)

M

(2, 3)

( )

2 3,

2 2

P

A

B

1 31

2

2 24 3 2 2 2 2 4 3 0

Radius 1 4 3 2

27.

0

1 cos2 3 coslim

tan4x

x x

x x

is equal to :

(1)12

(2) 4

(3) 3 (4) 2

Ans. (4)

Sol.0

(1 cos 2 ) (3 cos )lim

tan 4x

x x

x x

2

20

sin (3 cos )lim2

tan44

4

x

x xxx

x

22 1 (3 1) 2 42

1 4 4

28. If the angle of elevation of the top of a tower from three collinear points A, B and C, on a lineleading to the foot of the tower, are 30°, are 30°, 45° and 60° respectively, then the ratio, AB :BC, is :

(1) 2 : 3 (2) 3 :1

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

Ans. (2)

Sol. In MNC

1cot60

3

z

h

3

hz … (i)

h

N z y x ABC

M

60° 45° 30°

In MNB

cot 45z y

h

z y h … (ii)

In MNA


-(14)-

cot 30x y z

h

3x y z h … (iii)

from (i) and (ii)

( 3 1)

3 3

h hy h

… (iv)

from (ii) and (iii)

3x h h

( 3 1)x h … (v)

AB : BC = x : y = ( 3 1) 3

1( 3 1)

3

h

h

AB : BC = 3 :1

29. Let A and B be two sets containing four and two elements respectively. Then the number ofsubsets of the set A × B, each haivng at least three elements is :

(1) 510 (2) 219

(3) 256 (4) 275

Ans. (2)

Sol. The number of subsets of A × B each having at least three elements

= 8C3 + 8C

4 + 8C

5 + 8C

6 + 8C

7 + 8C

8

= 8C0 + 8C

1 + 8C

2 + 8C

3 + 8C

4 + 8C

5 + 8C

6 + 8C

7 + 8C

8 – [8C

0 + 8C

1 + 8C

2]

= 28 – (1 + 8 + 28)

= 256 – 37

= 219

30. Let ,a b

and c be three non-zero vectors such that no two of them are collinear and

1( ) | | | .

3a b c b c a

If is the angle between vectors b

and c

, then a value of sin is :

(1)2 33

(2)

2 33

(3)2

3

(4)23

Ans. (2)

Sol.1

( ) | || |3

a b c b c a

1

( . ) ( . ) | || |3

c a b c b a b c a

i.e.,1

( . ) | || |cos | || |3

c a b b c a b c a

1( . ) | || | cos

3c a b b c a

1

cos 03

2 2

sin3


-(15)-

PART-II : PHYSICS

31. 6VP 2

9V1

3Q3

In the circuit shown, the current in the 1 resistor is:(1) 0.13 A, from P to Q (2) 1.3 A, from P to Q(3) 0 A (4) 0.13 A, from Q to P

Ans. (4)Sol. Equations are

14 6I I

16 9I I

On solving we get,

6V I1 P 2

I9V

3I

1I–I1

I1

3 Q1

42 45,

23 23I I

1

30.13

23I I

32. Distance of the centre of mass of a solid uniform cone from its vertex is z0. If the radius of itsbase is R and its height is h then z0 is equal to:

(1)23

8

h

R(2)

2

4

h

R

(3)3

4

h(4)

5

8

h

Ans. (3)33. Match List - I (Fundamental Experiment) with List - II (its conclusion) and select the correct

option from the choices given below the list:List - I List II

(A) Franck-Hertz Experiment (i) Particle nature(B) Photo-Electric Experiment (ii) Discrete energy levels of atom(C) Davison-Germer Experiment (iii) Wave nature of electron

(iv) Structure of atom(1) (A) - (iv); (B) - (iii); (C) - (ii) (2) (A) - (i); (B) - (iv); (C) - (iii)(3) (A) - (ii); (B) - (iv); (C) - (iii) (4) (A) - (ii); (B) - (i); (C) - (iii)

Ans. (4)

34. The period of oscillation of a simple pendulum is 2L

Tg

. Measured value of L is 20.0 cm

known to 1 mm accuracy and time for 100 oscillations of the pendulum is found to be 90 susing a wrist watch of 1s resolution. The accuracy in the determination of g is:(1) 5% (2) 2%(3) 3% (4) 1%

Ans. (3)Sol. g = 42L/T2

Here, T and Tt t

n n

.

T

T

t

t


-(16)-

The errors in both L and t are the least count errors.

(g/g) = (L/L) + 2(T/T)

0.1 12

20.0 90

= 0.027

Thus, the percentage error in g is

100 (g/g) = 100(L/L) + 2 × 100 (T/T)

= 3%

35. A red LED emits light at 0.1 watt uniformly around it. The amplitude of the electric field of thelight at a distance of 1 m from the diode is:

(1) 7.75 V/m (2) 1.73 V/m

(3) 2.45 V/m (4) 5.48 V/m

Ans. (3)

Sol.2

0 0 2

1

2 4

PE c

R

20 2

0

2

4

PE

R c

2 90 2 8

2 0.19 10

(1) 3 10E

0 6 2.45 V/mE

36. In the given circuit, charge Q2 on the 2µF capacitor changes as C is varied from 1 µF to 3 µF.Q2 as a function of ‘C’ is given properly by: (figures are drawn schematically and are not toscale)

C

E

1µF

2µF

(1)

Q2

1µF 3µFC

Charge

(2)

Q2

1µF 3µFC

Charge

(3)

Q2

1µF 3µFC

Charge

(4)

Q2

1µF 3µFC

Charge

Ans. (3)

Sol.3

C3eff

C

C

q = Ceff

E


-(17)-

q = 3

3

CE

C

33

1

Eq

C

As C increases the value of 3

1C

decreases, hence q increases and, the increase is non

linear.

Charge on 2F capacitor = 2

3q

37.

L

I

I

Two long current carrying thin wires, both with current I, are held by insulating threads oflength L and are in equilibrium as shown in the figure, with threads making an angle ‘’ withthe vertical. If wires have mass per unit length then the value of I is:

(g = gravitational acceleration)

(1)0

tangL

(2)

0

sincos

gL

(3)0

2sincos

gL

(4)0

2 tangL

Ans. (3)

Sol. For equilibrium2

0sin2

I dlT

r

...(1)

cos ( )T dl g ...(2)

Tcos

Fm

mgr

Tsin

T

I

I(1) / (2)

20tan

2

I dl

r dlg

2

0tan2 2 sin ·

I

L g

sin

2

r

L

2

0

4 sin · tanL gI

2

2

0

4 sin

cos

L gI

0

2sincos

LgI

38. A particle of mass m moving in the x direction with speed 2v is hit by another particle of mass2m moving in the y direction with speed v. If the collision is perfectly inelastic, the percentageloss in the energy during the collision is close to:

(1) 62% (2) 44%

(3) 50% (4) 56%


-(18)-

Ans. (4)

Sol. Using conservation of linear momentum

After collision, the combined mass will move with velocity 2 23

v

Ki = 3 mv2

24K

3f mv

Hence, percentage loss in K.E.

K56%

Ki

39.F

A B

Given in the figure are two blocks A and B of weight 20 N and 100 N, respectively. These arebeing pressed against a wall by a force F as shown. If the coefficient of friction between theblocks is 0.1 and between block B and the wall is 0.15, the frictional force applied by the wallon block B is:

(1) 150 N (2) 100 N

(3) 80 N (4) 120 N

Ans. (4)

Sol. Combined weight of A and B = 120 N

Hence, frictional force will be 120 N between wall and blocks.

40. Consider an ideal gas confined in an isolated closed chamber. As the gas undergoes an adiabaticexpansion, the average time of collision between molecules increases as Vq, where V is the

volume of the gas. The value of q is: p

v

C

C

(1)1

2

(2)

3 5

6

(3)3 5

6

(4)

1

2

Ans. (4)

Sol.0

2

1

82av

av

MT

v RTd n

0

2 82av

MVT

RTd N

0

20

82av

MVT

RTd nN

( ) 1av

V RTT

PV T

av

TT

P

As in adiabatic process 1

PV & 1

1T

V


-(19)-

12

av

VT

V

1

2avT V

1

2avT V

41. A rectangular loop of sides 10 cm and 5 cm carrying a current I of 12 A is placed in differentorientations as shown in the figures below:

(a)

B

z

I

x

y

I

I

I(b)

B

z

Ix

yI

I

I

(c)

B

z

I

x

yI

I

I (d)

B

z

Ix

yI

I

I

If there is a uniform magnetic field of 0.3 T in the positive z direction, in which orientationsthe loop would be in (i) stable equilibrium and (ii) unstable equilibrium?(1) (b) and (c), respectively (2) (a) and (b), respectively(3) (a) and (c), respectively (4) (b) and (d), respectively

Ans. (4)Sol. The angle between B and

= nIA (magnetic moment) is 0º, So, it is in stable equilibrium.B

z

Ix

yI

I

IThe angle between B and (magnetic moment) is 180º,So unstable equilibrium.

B

z

Ix

yI

I

I

42. Consider a spherical shell of radius R at temperature T. The black body radiation inside it can

be considered as an ideal gas of photons with internal energy per unit volume 4UT

Vu and

pressure 1 U

3 Vp

. If the shell now undergoes an adiabatic expansion the relation between

T and R is:

(1) 3

1T

R (2) T e– R

(3) T e–3R (4)1

TR

Ans. (4)

Sol.4 3

4

33

1

1 4 3· . .

3 3 1.

1 1 3

A

PV nRT UP

VCT R n N KT

C TT T

RR

4(As where, constant)

UCT C

V


-(20)-

43. As an electron makes a transition from an excited state to the ground state of a hydrogen -like atom/ion:

(1) kinetic energy and total energy decrease but potential energy increases

(2) its kinetic energy increases but potential energy and total energy decrease

(3) kinetic energy, potential energy and total energy decrease

(4) kinetic energy decreases, potential energy increases but total energy remains same

Ans. (2)

Sol.2

2

2

Thus as r, reducesthe PE decreases

KE increases2TE decreases

2

kePE

rke

KEr

keTE

r

From excited state to ground state, “r” reduces.

44. On a hot summer night, the refractive index of air is smallest near the ground and increaseswith height from the ground. When a light beam is directed horizontally, the Huygens’ principleleads us to conclude that as it travels, the light beam:

(1) bends upwards (2) becomes narrower

(3) goes horizontally without any defelction (4) bends downwards

Ans. (1)

Sol.

wavefrontDue to difference in RI, the rays would trave in different speeds as shown. Thus the wavefrontwould be as indicated. As wave travel perpendicular to wavefront, thus the beam would bendupwards.

45. From a solid sphere of mass M and radius R, a spherical portion of radius R

2 is removed, as

shown in the figure. Taking gravitational potential V = 0 at r = , the potential at the centreof the cavity thus formed is: (G = gravitational constant)

(1)2GM

R

(2)

GM

2R

(3)GM

R

(4)

2GM

3R

Ans. (3)

Sol. Potential due to entire sphere at center of smaller sphere2

21 3

322

GM RV R

R

1

11

8

GMV

R ...(1)

Potential at center of smaller sphere


-(21)-

2

3

8

GMV

R

1 2netV V V

11 3

8 8

GM GM

R ...(2)

GM

R

46. Monochromatic light is incident on a glass prism of angle A. If the refractive index of thematerial of the prism is µ, a ray, incident at an angle q, on the face AB would get transmittedthrough the face AC of the prism provided:

A

B C

(1)1 1 1

cos sin sinA

(2)1 1 1

sin sin sinA

(3)1 1 1

sin sin sinA

(4)1 1 1

cos sin sinA

Ans. (2)

Sol. We know critical angle at face AC is 1 1

sin

and for

r r

B C

A

transmission at face AC

1 1' sinr

r r A

1 1sinA r

1 1sinr A

and by snell’s law

sin sinr

1 1sin sin sinA

1 1 1sin sin sinA

47. Two stones are thrown up simultaneously from the edge of a cliff 240 m high with initialspeed of 10 m/s and 40 m/s respectively. Which of the following graph best represents thetime variation of relative position of the second stone with respect to the first?

(Assume stones do not rebound after hitting the ground and neglect air resistance, take g =10 m /s2)

(The figures are schematic and not drawn to scale)


-(22)-

(1)

8 12t s( )

240( – )y y m2 1

(2)

8 12t s( )

240( – )y y m2 1

t

(3)

12t s( )

240( – )y y m2 1

(4)

8 12t s( )

240( – )y y m2 1

Ans. (4)

Sol.2

1

1240 10

2y t gt

22

1240 40

2y t gt

2 1 30y y t (straight line)

but after 8 second

1 0y

and 2

2

1240 40

2y t gt

2 4x Ay

48. For a simple pendulum, a graph is plotted between its kinetic energy (KE) and potential energy(PE) against its displacement d. Which one of the following represents these correctly? (graphsare schematic and not drawn to scale)

(1)

E

PE

KE (2)

E

PE

KE

d

(3)

E

PE

KE d(4)

E

PE

KE

d

Ans. (3)

Sol.

E

PE

KE d

49. A train is moving on a straight track with speed 20 ms–1. It is blowing its whistle at thefrequency of 1000 Hz. The percentage change in the frequency heard by a person standingnear the track as the train passes him is (speed of sound = 320 ms–1) close to:


-(23)-

(1) 24% (2) 6%

(3) 12% (4) 18%

Ans. (3)

Sol. 0s

v

v v

1

3201000

320 20

2

3201000

340

100 12%

50. An LCR circuit is equivalent to a damped pendulum. In an LCR circuit the capacitor is chargedto Q0 and then connected to the L and R as shown below:

C

If a student plots graphs of the square of maximum charge (Q2Max) on the capacitor with time(t)

for two different values L1 and L2 (L1 > L2) of L then which of the following represents thisgraph correctly? (plots are schematic and not drawn to scale)

(1)Q0

Q2Max (For both L and L )1 2

t(2)

Q2Max

L 1

L2

t

(3)

Q2Max L 1

L2

t(4)

Q2Max L 1

L2

t

Ans. (2)

Sol. 0di q

L iRdt C

2

20

d q dq qL R

dt Cdt

Solving 20 sin( )

RT

Lq q e t

where 2

2

1

4

R

LC L

Now, 2max 0

RT

LQ q e

and 2 2max 0

RT

LQ q e

22max0

Rt

LdQ R

q edt L


-(24)-

So slope 1

L

As L1 > L

2

So slope of L2 should be greater than slope of 4.

51. A solid body of constant heat capacity 1 J/°C is being heated by keeping it in contact withreservoirs in two ways : -

(i) Sequentially keeping in contact with 2 reservoirs such that each reservoir suppliessame amount of heat.

(ii) Sequentially keeping in contact with 8 reservoirs such that each reservoir suppliessame amount of heat.

In both the cases body is brought from initial temperature 100°C to final temperature 200°C.Entropy change of the body in the two cases respectively is :

(1) 2ln2, 8ln2 (2) ln2, 4ln2

(3) ln2, ln2 (4) ln2, 2ln2

Ans. (3)

Sol. In the two cases, the entropy change is given by

Case I : 150 200

100 150

ln2dT dT

mc mc mcT T

Case II : 112.5 125 137.5 200

100 112.5 125 187.5

....... . ln 2dT dT dT dT

mc mc mc mc mcT T T T

Both are proportional to ln2 each.

52. An inductor (L = 0.03H) and a resistor (R = 0.15 k) are connected in series to a battery of 15VEMF in a circuit shown below. The key K

1 has been kept closed for a long time. Then at t = 0,

Ka is opened and key K

2 is closed simultaneously. At t = lms, the current in the circuit will be

15VK1

K2

0.03H 0.15 k

(1) 0.67 mA (2) 100 mA

(3) 67 mA (4) 6.7 mA

Ans. (1)

Sol.3

max 3 3

15 1500100 10 0.1

Resistance 0.15 10 15 10

EmfI I A A

After K1 is opened and K

2 is closed the decay of current is given by

0

Rt

LI I e

330.15 10

(10 )0.030.100 e

= 0.1 × e–5

5

0.1 0.10.67

150mA

e

53. A uniformly charged solid sphere of radius R has potential V0 (measured with respect to ) on

its surface. For this sphere the equipotential surfaces with potentials 0 0 03 5 3

, ,2 4 4

V V V and 0

4

V

have radius R1, R

2, R

3 and R

4 respectively. Then


-(25)-

(1) 2R < R4

(2) R1 0 and R

2 > (R

4 – R

3)

(3) R1 0 and (R

2 – R

1) > (R

4 – R

3) (4) R

1 = 0 and R

2 < (R

4 – R

3)

Ans. (4, 1)

Sol. Inside the sphere

2 2

30

(3 )

4 2

q R rV

R

2 2

0 2

3

2

R rV V

R

For 0 1

30

2V V R

For2 2

20 0 0 2

35 5

4 4 2

R RV V V V

R

22

RR

For outside

004

q RV V

r r

For 0 0 03

3 3

4 4

RV V V V

R

3

4

3

RR

For 0 0

044 4

V V RV V

R 4 4R R

So, 4 3

8

3 2

R RR R

54. A long cylindrical shell carries positive surface charge in the upper half and negativesurface charge – in the lower half. The electric field lines around the cylinder will look likefigure given in : (figures are schematic and not drawn to scale)

(1) (2)

(3) (4)

Ans. (2)

Sol. (1) is incorrect because there is discontinuity at some points in the electric field lines.

(3) is incorrect because field lines emerging from positive must end up at negative charge.

(4) is incorrect because field lines should go from positive to negative.


-(26)-

55. Assuming Human pupil to have a radiius of 0.25 cm and a comfortable viewing distance of 25cm, the minimum separation between two objects that humah eye can resolve at500 nm wavelength is :

(1) 300 µm (2) 1 µm

(3) 30 µ (4) 100 µm

Ans. (3)

Sol. Diameter of pupil (or the aperture)

= 0.25 × × 2 = 0.5 cm

Minimum angular separation () = 1.22

d

mind

r

where dmin

= minimum separation between the objects

r = viewing distance = 25 cm.

min 1.22d

r d

min

1.22d r

d

91.22 500 10 0.25

0.5

30 m

56. A signal of 5 kHz frequency is amplitude; modulated on a carrier wave of frequency, 2 MHz. Thefrequencies of the resultant; signal is/ are :

(1) 2000 kHz and 1995 kHz (2) 2 MHz only

(3) 2005 kHz, and 1995 kHz (4) 2005 kHz, 2000 kHz and 1995 kHz

Ans. (4)

Sol. The frequencies of the resultant signals are

c + m, c, c – m

c = 2 MHz = 2000 KHz

m = 5 KHz

c + m = 2000 + 5 = 2005 KHz

c – m = 2000 – 5 = 1995 KHz

Frequencies are 2005 KHz, 2000 KHz and 1995 KHz

57. Two coaxial solenoids oftiifferent radii carry current I in the same direction. Let 1F

be the

magnetic force on the inner solenoid due to the outer one and 2F

be the magnetic force onthe outer solenoid due to the inner one. Then :

(1)1F

is radially outwards and 2F

= 0

(2)1 2 0F F

(3)1F

is radially inwards and 2F

is ra radially outwards

(4)1F

is radially inwards and 2F

= 0

Ans. (2)

Sol. Force on outer solenoid due to inner one is zero as magnetic field on outer solenoid is zero.

So F2 = 0

Magnetic field due to outer solenoid in along the axis of inner solenoid.

It we take force on one turn of inner solenoid, the net force will be zero.

Hence, F1 = 0


-(27)-

B

F

dl

dl

F

58. A pendulum made of a uniform wire of cross sectional area A has time period T. When anadditional mass M is added to its bob, the time period changes to T

M. If the Young’s modulus

of the material of the wire is Y then 1

Yis equal to : (g —gravitational acceleration)

(1)

2

1M

T A

T Mg

(2)

2

1MT A

T Mg

(3)

2

1MT Mg

T A

(4)

2

1 MT A

T Mg

Ans. (2)

Sol. 2l

Tg

...(1)

with addition mass M

·Mg l

YA l

orlMg

lAY

So, nw time period

2m

l lT

g

2m

lMgT l

AY ...(2)

(2) / (1)

2

2

m

lmgl

AYgT

T l

g

2

21m

lMglT MgAY

l AYT

2

21mTMg

AY T

2

2

11mT A

Y MgT


-(28)-

59. From a solid sphere of mass M and radius R a cube of maximum possible volume is cut.Moment of inertia of cube about an .axis passing through its center and perpendicular, toone of its faces is :

(1)24

3 3

MR

(2)

2

32 2

MR

(3)2

16 2

MR

(4)

24

9 3

MR

Ans. (4)

Sol. MI of cube = 2

6

mL

3 2L R

2

3

RL

mass of cube = 3 ·L

3

3

2·

433

R Mm

R

23

3

8 2 1· ·4 63 3 33

R M R

R

22 4 1

3 63

M R

24

9 3

MR

60. When 5V potential difference is applied across a wire of length 0.1 m, the drift speed ofelectrons is 2.5 × 10–3 ms–1. If the electron density in the wire is 8 × 1028 m–3/ the resistivityof the material is close to:

(1) 1.6 × 10–5 m (2) 1.6 × 10–8 m

(3) 1.6 × 10–7 m (4) 1.6 × 10–6 m

Ans. (1)

Sol. Current in a wire

di neAv

d

VneAv

R

d

VAneAv

l

28 19 4

5

8 10 1.6 10 0.1 2.5 10d

V

ne l v

55

51.56 10

3.2 10

51.6 10 m


-(29)-

PART – II : CHEMISTRY61. The vapour pressure of acetone at 20ºC is 185 torr. When 1.2 g of a non-volatile substance was

dissolved in 100g of acetone at 20ºC, its vapour pressure was 183 torr. The molar mass (g mol–1) of the substance is

(1) 488 (2) 32

(3) 64 (4) 128

Ans.(3)

Sol. P0 = 185 torr

W2 = 1.2gram

W1 = 100gram

Ps = 183 torr

Applying the formula of relative lowering of vapour pressure

2

1 2

S nP P

P n n

2 2

1 1 2 2

/

/ /S W MP P

P W M W M

On solving m2 = 63.83gram

62. 3 g of activated charcoal was added to 50 mL of acetic acid solution (0.06N) in a flask. After anhour it was filtered and the strength of the filtrate was found to be 0.042 N. The amount ofacetic acid adsorbed (per gram of charcoal) is

(1) 54 mg (2) 18 mg

(3) 36 mg (4) 42 mg

Ans.(2)

Sol. The amount of acetic acid adsorbed per gram of charcoal is

350 0.06 0.042 10 60

3gram

= 18 mg.

63. Which of the following is the energy of a possible excited state of hydrogen ?

(1) + 6.8 eV (2) + 13.6 eV

(3) – 6.8 eV (4) – 3.4 eV

Ans.(4)

Sol. Energy of a possible excited state of hydrogen

E2 = – 13.6 × z22 21 2

1 1n n

21 1

13.6 11 2

1 113.6 1

1 4

313.6

4

= – 10.4 eVEnergy of a possible excited state of hydrogen = – 13.6 – (– 10.2)= – 3.4eV.

64. Which among the following is the most reactive ?

(1) ICl (2) Cl2

(3) Br2

(4) I2


-(30)-

Ans.(1)

Sol. Interhalogen compounds are more reactive than halogens So, ICl is the most reactive

65. Which polymer is used in the manufacture of paints and lacquers ?

(1) Poly vinyl chloride (2) Bakelite

(3) Clyptal (4) Propylene

Ans.(3)

Sol. Glyptal is used in the manufacture of paints and lacquers.

66. The molecular formula of a commercial resin used for exchanging ions in water softening isC

8H

7SO

3Na (Mol. wt. 206). What would be the maximum uptake of Ca2+ ions by the resin when

expressed in mole per gram resin ?

(1)1

412(2)

1

103

(3)1

206(4)

2

309

Ans.(1)

Sol. 2 moles of C8H

7SO

3Na uptake = 40gram Ca2+

2 × 206 gram C8H

7SO

3Na uptake = 1 mole Ca2+

1 mole of C8H

7SO

3Na uptake =

1412

gram

67. In Carius method of estimation of halogens, 250 mg of an organic compound gave 141 mg ofAgBr. The percentage of bromine in the compound is (at. mass Ag = 108; Br = 80)

(1) 60 (2) 24

(3) 36 (4) 48

Ans.(2)

Sol. 188 gram AgBr contains 80 gram Bromine, 141 milli gram AgBr 80

141180

= 60 mg

Percentage of Br in the compound is 60

100250

= 24%.

68. Assertion : Nitrogen and Oxygen are the main components in the atmosphere but these donot react to form oxides of nitrogen.

Reason : The reaction between nitrogen and oxygen requires high temperature.

(1) Both the assertion and reason are incorrect

(2) Both assertion and reason are correct, and the reason is the correct explanation for theassertion

(3) Both assertion and reason are correct, but the reason is not the correct explanation forthe assertion

(4) The assertion is incorrect, but the reason is correct

Ans.(2)

Sol. Reaction between nitrogen and oxygen require high temperature. As the reaction between N2

and O2 is endothermic in nature.

2 2 2N O NO H

69. The following reaction is performed at 298 K 2 22 ( ) 2 ( )NO g O NO g . The standard free

energy of formation of NO(g) is 86.6 kJ/mol at 298 K. What is the standard free energy offormation of NO

2(g) at 298 K? (K

p = 1.6 × 1012)

(1) 0.5 [2 × 86,600 – R (298) ln (1.6 × 1012)] (2) R (298) ln (1.6 × 1012) – 86600

(3) 8660 + R (298) ln (1.6 × 1012) (4)12ln(1.6 10 )

8660(298)R


-(31)-

Ans.(1)

Sol. G° = – RT ln Kp

0 0 lnP RG G RT Kp

–2x + 2 × 86.6 × 10 3 = RT ln 1.6 × 1012

122 86.600 ln1.6 10

2

RTx

12(298)86.600 ln1.6 10

2

R

70. Which one of the follwing alkaline earth metal sulphates has its hydration enthalpy greaterthan its lattice enthalpy ?(1) SrSO

4(2) CaSO

4

(3) BeSO4

(4) BaSO4

Ans.(3)Sol. BeSO

4 : Because of exceptionally small size of Be.

It has higher hydration entnalpy than lattice enthalpy71. The number of geometric isomers that can exist for square planar [Pt (Cl) (py) (NH

3) (NH

2OH)]+

is (py = pyridine)(1) 6 (2) 2(3) 3 (4) 4

Ans.(3)Sol. It is a [Mabcd] complex, hence has 3 isomers72. The synthesis of alkyl fluorides is best accomplished by

(1) Swarts reaction (2) Free radical fluorination(3) Sandmeyer’s reaction (4) Finkelstein reaction

Ans.(1)Sol. Swarts reaction

R – X(Cl, Br) + AgF R – F + AgX73. The intermolecular interaction that is dependent on the inverse cube of distance between

the molecules is(1) hydrogen bond (2) ion-ion intereaction(3) ion-dipole intereaction (4) London force

Ans.(3)Sol. Ion-dipole interaction74. In the context of the Hall - Heroult process for the extraction of Al, which of the following

statements is false ?(1) Na

3AlF

6serves as the electrolyte

(2) CO arid CO2 are produced in this process

(3) Al2O

3 is mixed with CaF

2 which lowers the melting point of the mixture and brings

conductivity(4) Al3+ is reduced at the cathode to form Al

Ans. (1)75. Which of the following compounds will exhibit geometrical isomerism ?

(1) 1, 1 - Diphenyl - 1 - propane(2) 1 - Phenyl - 2 - butene(3) 2 - Phenyl - 1 - butene(4) 2 - Phenyl -1 -butene

Ans. (2)

Sol. C6H5 – CH2 – CH = CH – CH3

76. The ionic radii (in Å) of N3–, O2- and F– are respectively :

(1) 1.71, 1.36 and 1.40

(2) 1.36, 1.40 and 1.71


-(32)-

(3) A.36, 1.71 and 1.40

(4) 1.71, 1.40 and 1.36

Ans. (4)

Sol. Higher the negative charge on the anion larger will be the ionic radii

77. From the following statements regarding H2O

2, choose the incorrect statement :

(1) It has to be kept away from dust

(2) It can act only as an oxidizing agent y

(3) It decomposes on exposure to light

(4) It has to be stored in plastic, or wax lined glass bottles in dark

Ans. (2)

78. Higher order (< 3) reactions are rate due to

(1) loss of active species on collision

(2) low probability of simultaneous collision of all the reacting species

(3) increase in entropy and activation energy as more molecules are involved

(4) swhifting of equilibrium towards reactants due to elastic collisions

Ans. (2)

79. Match the catalysts to the correct processes :

Catalyst Process

(A) TiCl3

(i) Wacker process

(B) PdCl2

(ii) Ziegler - Natta polymerization

(C) CuCl2

(iii) Contact process

(D) V2O

5(iv) Deacon’s process

(1) (A) - (iii), (B) - (i), (C) - (ii), (D) - (iv)

(2) (A) - (iii), (B) - (li), (C) - (iv), (D) - (i)

(3) (A) - (ii), (B) - (i), (C) - (iv), (D) - (iii)

(4) (A) - (ii), (B) - (iii), (C) - (iv), (D) - (i)

Ans. (3)

80. Which one has the highest boiling point ?

(1) Xe (2) He

(3) Ne (4) Kr

Ans. (1)

Sol. Higher the molar mass so higher the boiling point.

81. In the reaction

the product E is :NH2

CH3

2/ /20–5º

NaNO HCl CuCN KCNC

D E N the product E is

(1)

CH3

(2)

CH3

COOH

(3) H3C CH3 (4)

CH3

CN


-(33)-

Ans. (4)

Sol.

CH3

NH2

2

0 5NaNO HCl

C

CH3

N NCl+ –

/CuCN KCN

CH3

CN

+ N2

82. Which of the following compounds is not colored yellow ?

(1) Ba,Cr; O4

(2) Zn2[Fe(CN)

6]

(3) K3[Co(N0

2)6] (4) (NH

4)3 (As (Mo

3O

10)4]

Ans. (2)

83. Sodium metal crystallizes in a body centred cubic lattice with a unit cell edge of 4.29A. Theradius of sodium atom is approximately :

(1) 0.93A (2) 1.86 Å

(3) 3.22A (4) 5.72A

Ans. (2)

Sol.3

4.24 1.86 4

r Å Å

84. The standard Gibbs energy change at 300 K for the reaction 2A B + C is 2494.2 J. At agiven time, the composition of the reaction mixture is [A] = ½, [B] = 2 and [C] = ½. Thereaction proceeds in the : [R = 8.314 J/K/mol, e = 2.718]

(1) reverse direction because Q < KC

(2) forward direction because Q > KC

(3) reverse direction because Q > KC

(4) forward direction because Q < KC

Ans. (3)

Sol. 2A B C

2c

B CQ

A

2

12

2 412

cQ

G = – 2.303 RT logK

2494.2log

2.303 8.314 300K

logK = 0.43

Since KC in less than Qc So reaction will be in reverse direction

85. Which, compound would give 5 - keto - 2 - methyl hexanal upon ozonolysis ?

(1)

CH3

H3C (2) CH3

CH3

(3)

CH3

CH3

(4)

CH3

CH3

Ans. (3)


-(34)-

Sol.

CH3

CH3

+O32Zn H O

H C3

CH3

H C3

CH3

OO

H3 2 2H C C CH CH CH C H

O

CH3

OO

OO

86. Which of the following compounds is not an antacid ?

(1) Ranitidine (2) Aluminium hydroxide

(3) Cimetidine (4) Phenelzine

Ans. (4)

Sol. All are antacid acept Phenelzine which is a tranquilizer.

87. In the following sequence of reactions :

2 24

4

SOCl H /PdKMnOBaSOToluene A B C the product C is :

(1) C6H

5CHO (2) C

6H

5COOH

(3) C6H

5CH

3(4) C

6H

5CH

2OH

Ans. (1)

Sol.

NH3

4KM On

COOH

2SOCl

COCl

2

4

H /PdBaSO

CHO

88. Which of the vitamins given belb’W is water-soluble ?

(1) Vitamin K (2) Vitamin C

(3) Vitamin D (4) Vitamin E

Ans. (2)

Sol. Vitamin ‘C’ and B group are water soluble.

89. The color of KMnO4 is due to :

(1) – * transition

(2) M L charge transfer transition

(3) d – d transition

(4) L M charge transfer transition

Ans. (4)

Sol. LMCT (Ligand to Metal Charge transfer)

90. Two Faraday of electricity is passed through a solution of CuSO4. The mass of copper depositedat the cathode is : (at. mass of Cu —63.5 amu)

(1) 127 g (2) 0 g

(3) 63.5 g (4) 2g

Ans. (3)

Sol. 2 2Cu e Cu s

From the above equation

2F or 2e deposit 63.5gram of Cu

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JEE Mains Ans Sol Code C 2015

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