Jee Mains 2015 Solutions

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JJEEEE –– 22001155

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JEE MAINS2015 (Code: C)2


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PART A: MATHEMATICS 1. A complex number z is said to be unimodular if |z| = 1. Suppose z1 and z2 are complex numbers

such that 1 2

1 2

z 2z

2 z z

is unimodular and z2 is not unimodular. Then the point z1 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 radius 2 1. 4

Sol. 1 2

1 2

z 2z

2 z z

= 1

(z1 – 2z2) 1 2 1 2 1 2z 2z 2 z z 2 z z

|z1|2 + 4|z2|

2 – |z1|2 |z2|

2 – 4 = 0 4 (|z2|

2 – 1) – |z1|2 (|z2|

2 – 1) = 0 (4 – |z1|

2) (|z2|2 – 1) = 0

|z1| = 2 ( |z2| 1) 2. The normal to the curve, x2 + 2xy – 3y2 = 0, at (1, 1): (1) meets the curve again 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 2. 1 Sol. x2 + 2xy – 3y2 = 0 2x + 2(xy + y) – 6yy = 0 y (2x – 6y) = –2x – 2y

y = x y 1 1

x 3y 1 3 1

= 1

Slope of normal = –1 Equation of normal (y – 1) = –1(x – 1) x + y = 2 y = 2 – x For points of intersection of normal and the curve. x2 + 2x(2 – x) – 3(2 – x)2 = 0 x2 – 4x + 3 = 0 x = 1 x = 3 y = 1 y = –1 (1, 1) (3, –1)

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

1 1 3 1 3 5

is

(1) 192 (2) 71 (3) 96 (4) 142 3. 3

Sol. Tn = 223 2

2 2

n n 1n n 2n 1

4n 4n

Sn = Tn = n(n 1)(2n 1) 1

n(n 1) n6 4

Put n = 9 Sn = 384

4= 96



4. Let f(x) be a polynomial of degree four having extreme values at x = 1 and x = 2. If 2x 0

f(x)lim 1

x

= 3,

then f(2) is equal to (1) 4 (2) –8 (3) –4 (4) 0 4. 4

Sol. 2x 0

f(x)lim 2

x

f(x) = ax4 + bx3 + cx2 + dx + e f(0) = 0 e = 0

x 0

f (x)lim 2

2x

4ax3 + 3bx2 + 2cx + d f(0) = 0 d = 0

x 0

f (x)lim 2

2

f(0) = 4

12ax2 + 6bx + 2c = 4 2c = 4 c = 2 Now f(x) = ax4 + bx3 + 2x2 Given f(1) = f(2) = 0 4a + 3b + 4 = 0 4a + 3b = –4 ...... (1) 32a + 12b + 8 = 0 8a + 3b = –2 ..... (2)

a = 1

2; b = –2

f(x) = 4x

2– 2x3 + 2x2 f(2) = 0.

5. The negation of ~s (~r s) is equivalent to (1) s r (2) s ~r (3) s (r ~s) (4) s (r ~s) 5. 1 Sol. Let A = ~s (~r s)

r s ~r ~s ~r s ~s (~r s) ~A s r T F T F

F T T F

F T F T

T F F T

F T F F

T T F T

F F T F

F F T F

6. If A =

1 2 2

2 1 2

a 2 b

is a matrix satisfying the equation AAT = 9 I, where 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) 6. 1

Sol. A =

1 2 2

2 1 2

a 2 b

1 2 2 1 2 a

2 1 2 2 1 2 9 I

a 2 b 2 2 b



2 2

1 4 4 2 2 4 a 4 2b 9 0 0

2 2 4 4 1 4 2a 2 2b 0 9 0

0 0 9a 4 2b 2a 2 2b a 4 b

a + 2b = –4 a – b = –1 3b = –3 b = –1 a = –1 + b = –1 – 1 = –2 (a, b) = (–2, –1)

7. The integral

3 / 42 4

dx

x x 1 equals

(1)

1/ 44

4

x 1C

x

(2)

1/ 44

4

x 1C

x

(3) (x4 + 1)1/4 + C (4) –(x4 + 1)1/4 + C

7. 1

Sol.

3 / 4 3 / 42 4 5 4

dx 1dx

x x 1 x 1 x

Put 1 + x–4 = t

5

1 1dx dt

4x

=

31 1/ 444

41/ 4 43 / 4 4

1 1 1 t 1 4 1 xdt t 1 x c

34 4 4 1t x14

8. If m is the A.M. of two distinct real numbers and n (, n > 1) and G1, G2 and G3 are three geometric

means between and n, G14 + 2G2

4 + G34 equals

(1) 4 2m2n2 (2) 4 2mn (3) 4 m2n (4) 4 mn2

8. 3

Sol. 2m = + n r = 1/ 4

n

, G1, G2, G3, n

G1 = r = 1/ 4

n

(r: Common ratio of the G.P)

G14 = n 3n

n

2G24 = 2 (r2 )4 = 2

24n

= 2n22

G34 = (r3)4 =

34 3n

n

G14 + 2G2

4 + G34 = n3 + 2n22 + n3 = n (2 + 2n + n2) = n ( + n)2 = n(2m)2 = 4m2n

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

dx+ y = 2x log x, (x 1). Then y(e) is

equal to (1) 2e (2) e (3) 0 (4) 2 9. 4

Sol. Integrating factor =

1dxp dx x log xe e logx

y . log x = 2 log x dx + c y . log x = 2x (log x – 1) + c



x = 1 0 = 2 [0 – 1] + c c = 2 y . log x = 2x (log x – 1) + 2 at x = e; y = 2

10. The number of integers greater than 6,000 that can be formed, using the digits 3, 5, 6, 7 and 8, without repetition, is

(1) 72 (2) 216 (3) 192 (4) 120 10. 3 Sol. 4 digit numbers = 3 4P3 = 3 4 3 2 = 72 5 digit numbers = 5! = 120 Total = 120 + 72 = 192 11. The number of points, having both co-ordinates as integers, that lie in the interior of the triangle with

vertices (0, 0), (0, 41) and (41, 0) is (1) 780 (2) 901 (3) 861 (4) 820 11. 1

Sol. 1 140 40 40 1600 40

2 2 = 800 – 20 = 780

(41, 0)

(0, 41)

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

10 8

9

a 2a

2a

is equal to

(1) –3 (2) 6 (3) –6 (4) 3 12. 4 Sol. an = ( + ) (n – 1 – n – 1) – n – 1 + n – 1 an = 6an – 1 + 2 (an – 2) Put n = 10.

a10 = 6a9 + 2a8 10 8

9

a 2a3

2a

13. Let tan–1y = tan–1x + 12

2xtan

1 x

, where |x| <

1

3. Then a value of y is

(1) 3

2

3x x

1 3x

(2)

3

2

3x x

1 3x

(3)

3

2

3x x

1 3x

(4)

3

2

3x x

1 3x

13. 2 Sol. tan–1y = tan–1x + 2 tan–1x = 3 tan–1x

= 3

12

3x xtan

1 3x

y = 3

2

3x x

1 3x

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

3 4 12

and

the plane x – y + z = 16, is



(1) 13 (2) 2 14 (3) 8 (4) 3 21 14. 1

Sol. Put x 2 y 1 z 2

3 4 12

Any point = (3 + 2, 4 – 1, 12 + 2) It lies in the plane (3 + 2) – (4 – 1) + (12 + 2) = 16 11 = 11 = 1 Point = (5, 3, 14)

Distance of (1, 0, 2) from (5, 3, 14) = 16 9 144 13 . 15. The area (in sq. units) of the region described by {(x, y) : y2 2x and y 4x – 1} is

(1) 9

32 (2)

7

32 (3)

5

64 (4)

15

64

15. 1

Sol. A =

11 2 2 3

1/ 2 1/ 2

y 1 y 1 y y 9dy y

4 2 4 2 6 32

16. Let O be the vertex and Q be any point on the parabola, x2 = 8y. If the point P divides the line

segment OQ internally in the ratio 1 : 3, then the locus of P is (1) x2 = 2y (2) x2 = y (3) y2 = x (4) y2 = 2x 16. 1

Sol. P = 2t t

,4 32

x = t

4 16x2 = t2

y = 2t

32 32y = t2 16x2 = 32y x2 = 2y

17. The mean of the data set comprising of 16 observations is 16. If one of the observation valued 16 is

deleted and three new observations values 3, 4 and 5 are added to the data, then the mean of the resultant data, is

(1) 14.0 (2) 16.8 (3) 16.0 (4) 15.8 17. 1 Sol. Let x1, x2, ......, x16 be the data values Given mean = 16 x1 + x2 + x3 + ...... + x16 = 16 16 = 256. x1 + x2 + x3 + ...... + x14 + x15 = 256 – 16 = 240 (let x16 = 16) new data x1, x2, x3, ......, x15, 3, 4, 5

Mean = 1 2 3 15x x x x 3 4 5 240 3 4 5 25214

18 18 18

.

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 2x y

9 5 = 1 is

(1) 27 (2) 27

4 (3) 18 (4)

27

2

18. 1

Sol. L = 2b 2 5 5

ae, 3 , 2,a 3 3 3



Equation of tangent = 2x 5y

19 3 5

2x + 3y = 9 or x y

19 / 2 3

Area of quadrilateral = 4 1 9

2 2 3 = 27 sq. unit

19. The equation of the plane containing the line 2x – 5y + z = 3; x + y + 4z = 5, and parallel to the plane,

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 19. 4 Sol. (2x – 5y + z – 3) + (x + y + 4z – 5) = 0

2 5 1 4

1 3 6

6 + 3 = –5 + 2 = –11 = 11

2

(2x – 5y + z – 3) – 11

2(x + y + 4z – 5) = 0

(4x – 10y + 2z – 6) – 11x – 11y – 44z + 55 = 0 x + 3y + 6z – 7 = 0 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 20. 4 Sol. C1 = (2, 3), C2 = (–3, –9)

r1 = 4 9 12 r2 = 9 81 26 r1 = 5 r2 = 8

C1C2 = 25 144 13 C1C2 = r1 + r2 = 13 21. The set of all values of for which the system of linear equations: 2x1 – 2x2 + x3 = x1

2x1 – 3x2 + 2x3 = x2 –x1 + 2x2 = x3 has a non-trivial solution, (1) contains more than two elements (2) is an empty set (3) is a singleton (4) contains two elements 21. 4

Sol.

2 2 1

2 3 2

1 2

= 0 (2 – )((3 + ) – 4) + 2(–2 + 2) + 1(4 + (–3 – ) = 0

3 + 2 – 5 + 3 = 0 ( – 1)(2 + 2 – 3) = 0 ( – 1)( + 3)( – 1) ( – 1)2 ( + 3) = 0 = 1, 1, –3 set contains 2 elements {1, –3} 22. If 12 identical balls are to be placed in 3 identical boxes, then the probability that one of the boxes

contains exactly 3 balls is

(1) 11

122

3

(2) 11

55 2

3 3

(3) 10

255

3

(4) 12

1220

3



22.

If the balls are distinct, then the answers is 3 9

123

1 2C

3 3

; i.e., (2)

23. The sum of coefficients of integral powers of x in the binomial expansion of 501 2 x is

(1) 5012 1

2 (2) 501

3 12

(3) 5013

2 (4) 501

3 12

23. 2 Sol. Tr + 1 = 50Cr (–2 x1/2)50 – r

= 50Cr (–2)50 – r . x25 – r/2

r = 0, 2, 4, ......, 50 = 50 0

12

= 26

Sum = 50C0 (–2)50 + 50C2 (–2)48 + 50C4 (–2)46 + ...... 50C50 (2)0

= 1

2[(1 – 2)50 + (1 + 2)50] =

1

2(350 + 1)

24. The integral

4 2

2 22

logxdx

logx log 36 12x x is equal to

(1) 6 (2) 2 (3) 4 (4) 1 24. 4

Sol. I = 4 2

2 22

logxdx

logx log(6 x)

I =

24

2 22

log 6 xdx

log 6 x logx

I = 1

25. If the function g(x) = k x 1, 0 x 3

mx 2, 3 x 5

is differentiable, then the value of k + m is

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

5 (4)

10

3

25. 2 Sol. g(x) is continuous at x = 3

g(3+) = g(3–) = g(3) 3m + 2 = k 4 3m + 2 = 2k ...... (1)

g(x) = k

2 x 1 0 x 3

= m 3 x 5

g(3+) = m g(3–) = k

4

g(3+) = g(3–) m = k

4 k = 4m

3m + 2 = 8m m = 2

5 k =

8

5 k + m = 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 parallel to y-axis (4) circle of radius 2 26. 4



Sol. 2x – 3y + 4 = 0 2x – 4y + 6 = 0 y = 2, x = 1

2 2 2 2x 1 y 2 1 2 2 3

x2 + y2 – 2x – 4y + 5 = 2 x2 + y2 – 2x – 4y + 3 = 0

r = 1 4 3 2

(2, 3)

(x, y)

(1, 2) A

B

C

2x – 3y + 4 = 0 x – 2y + 3 = 0

27.

x 0

1 cos2x 3 cosxlim

x tan4x

is equal to

(1) 1

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

27. 4

Sol.

2

x 0 x 0

2 sin x 3 11 cos2x 3 cos3xlim lim

x tan4x x tan 4x

=

2

x 0 x 0

2 x 4 xlim 8 lim 2

x tan4x tan4x

28. If the angles of elevation of the top of a tower from three collinear points A, B and C on a line leading

to the foot of the tower, are 30, 45 and 60 respectively, then the ratio, AB : BC is

(1) 2 : 3 (2) 3 :1 (3) 3 : 2 (4) 1: 3 28. 2 Sol. Let height of tower be h

OC = h

3, OB = h; OA = h 3

AB OA OB h 3 h 3 1

3h 1BC OB OC h 13 3

A B C O

D

h

60 45 30

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

(1) 510 (2) 219 (3) 256 (4) 275 29. 2 Sol. Number of elements in A B = 8 Number of subsets having atleast 3 elements = 28 – 8C0 – 8C1 – 8C2 = 219

30. Let a,b and c

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

1a b c b c a

3

. If is the angle between vectors b and c

, then a value of sin is

(1) 2 3

3

(2)

2 2

3 (3)

2

3

(4)

2

3

30. 2

Sol. a b c a c b b c a

1a c b b c a b c a

3



Since a,b and c

non-collinear

1

a c 0 and b c b c3

1

b c cos b c3

cos = 1

3

sin =

2 2

3

PART B: PHYSICS 31. 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

31. 4 Sol.

(i– i1)

i1

i

19 2i i 3i 0 1 16 3(i i ) i 0

19 5i i 16 3i 4i

42

i23

1

3i

23

Ans. 0.13 A from Q to P. *32. Distance of the centre of mass of a solid uniform cone from its vertex is z0. If the radius of its base is

R and its height is h then z0 is equal to:

(1) 23h

8R (2)

2h

4R (3)

3h

4 (4)

5h

8

32. 3 Sol.

r

R

x

Z0

h

h x h

r R

rhh x

R

Rr (h x)

h

22

2

cm

Rx (h x) dx

hyM

= h2

2 22

0

R(h x 2hx)x dx

M h

2 2R h

12m



h

4

0

h 3hZ h

4 4

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 of light

(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) 33. 4

*34. The period of oscillation of a simple pendulum is T = L

2g

. 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 s using a wrist watch of 1 s resolution. The accuracy in the determination of g is:

(1) 5% (2) 2% (3) 3% (4) 1% 34. 3

Sol. L

T 2g

2

2

Lg 4

T

2logg log(4 ) logL 2logT

g T

2g L T

% in g = (% in ) + 2(% in time period)

L = 20.0 cm

= 20 0.1

100 oscillation is 90 s

1oscillation 0.9 0.01s

0.1 0.01

100 2 10020 0.9

= 3% 35. A red LED emits light at 0.1 watt uniformly around it. The amplitude of the electric field of the light 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 35. 3 Sol. Power = 0.1 w

Intensity of light waves = 1

22

0E c

Intensity at d = 1 m distance = 2

power

4 d

2

0.1

4 (1)



202

0.1 1E (c)

24 (1)

Calculating, we get E = 2.45 V/m 36. In the given circuit, charge Q2 on the 2F capacitor changes as C is varied

from 1F to 3F. Q2 as a function of ‘C’ is given properly by: (figures are drawn schematically and are not to scale)

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

36. 3 Sol. Potential difference across C and across 2f are in inverse ratio of capacitances i.e. 3 : c hence potential difference across 2F

capacitor = c

Ec 3

charge on 2F capacitor, 2Q (2 F)V

CE

2C 3

Value of 22

dQ 6E

dc (c 3)

C

2F

E

1 F

This shows that Q2 increases with value of c and the slope of the curve must be decreasing Hence it is 3. 37. Two long current carrying thin wires, both with current I, are held by insulating

threads of length L and are in equilibrium as shown in the figure, with threads making an angle ‘’ with the vertical. If wires have mass per unit length then the value of I is: (g – gravitational acceleration)

(1) 0

gLtan

(2)

0

gL

cos

(3) 0

gL2sin

cos

(4) 2

0

gLtan

37. 3 Sol. Force between the two long wires

2

0 IF

2 d

For equilibrium F = T sin mg = T cos

2

0IFtan

mg 2 mgd

Taking d = 2 L sin and m

We get

L

F

mg

I

F

d

T

0

gLI 2sin

cos



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

(1) 62% (2) 44% (3) 50% (4) 56% 38. 4 Sol. Initial momentum

iˆ ˆp 2mv i 2mv j

f iˆ ˆp p 2mv[i j]

f fp 3mv

f

2 ˆ ˆv v[ i j]3

Initial KE = 2 21 1m(2v) (2m)v

2 2

2m v

m 2v

vf

3m

= 3mv2

Final KE =

2

2f

1 1 2 2(3m)V (3m) v

2 2 3

loss of KE = Intial KE – Final KE

25mv

3

Percentage loss of KE = 2

2

5 /3mv100

3mv

56% *39. Given in the figure are two blocks A and B of weight 20 N and 100 N,

respectively. These are being pressed against a wall by a force F as shown. If the coefficient of friction between the blocks is 0.1 and between block B and the wall is 0.15, the frictional force applied by the wall on block B is:

(1) 150 N (2) 100 N (3) 80 N (4) 120 N 39. 4 Sol.

f1

A

w = 20 N

f2

B

w = 100 N

f1

A B F

For A to be in equilibrium 1f 20N

For B to be in equilibrium 1 2f w f

20 + 100 = f2 Hence f2 = 120 N.



*40. Consider an ideal gas confined in an isolated closed chamber. As the gas undergoes an adiabatic expansion, 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

40. 4 Sol.

vavg t

d

d = 2 x radius of atom

Volume swept in tiem 2avgt d v t

If no. of atoms are N and volume is v, then atoms present above volume = 2avg

Nd v t

v = no. of

collisions (Nc)

Average time per collision c avg

t V

N v

1V[TV constan t

T for adiabatic process]

1 11

2 21

V VT

1

2V

41. A rectangular loop of sides 10 cm and 5 cm carrying a current I of 12 A is placed in different

orientations as shown in the figures below

(a)

(b)

(c)

(d)

If there is a uniform magnetic field of 0.3 T in the positive z direction, in which orientations the 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 41. 4 Sol. For stable equilibrium, petential energy will be minimum.

angle between & B

will be 0. For unstable equilibrium, potential energy will be maximum.



angle between & B

will be .

*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 u = 4UT

V and pressure

1 Up

3 V

. If the shell now undergoes an adiabatic expansion the relation between T and R is

(1) 3

1T

R (2) RT e (3) 3RT e (4)

1T

R

42. 4 Sol. From 1st law of thermodynamics, for adiabatic process,

dU + dW = 0 …………………(1)

given, 4Uu T

V or 4U

TV (where is constant) …………………(2)

and 41 U T

p3 V 3

…………………(3)

substituting (2) and (3) in equation (1), 4

4 Td( VT ) dV 0

3

dV 3dT

V T

lnV 3lnT k

1

TR

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 hut 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 43. 2

Sol. As, 1

vr

, , Kinetic energy increases

As, r decreases, potential energy & total energy decreases. 44. On a hot summer night, the refractive index of air is smallest near the ground and increases with

height from the ground. When a light beam is directed horizontally, the Huygens’ principle leads us to conclude that as it travels, the light beam

(1) bends upwards (2) becomes narrower (3) goes horizontally without any deflection (4) bends downwards 44. 1 Sol. As refractive index of air increases with height, , speed of light decreases in upward direction

making an upward bend in plane wavefront.

*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 centre of the cavity thus formed is: (G = gravitational constant)

(1) 2GM

R

(2)

GM

2R

(3)

GM

R

(4)

2GM

3R

45. 3



Sol. Potential at the centre of cavity can be written as potential due to a complete solid sphere minus potential due to removed mass.

Potential due to complete solid sphere 2

2

GM 3 1 (R / 2)

R 2 2 R

11GM

8 R

Removed mass,3

3

M R MM'

2 8R

Potential due to removed mass = 3 G(M/8) 3GM

2 (R / 2) 8R

, net potential = GM

R

46. Monochromatic light is incident on a glass prism of angle A. If the refractive index

of the material of the prism is , a ray, incident at an angle , on the face AB would get transmitted through the face AC of the prism provided

(1) 1 1 1cos sin A sin

(2) 1 1 1

sin sin A sin

(3) 1 1 1sin sin A sin

(4) 1 1 1

cos sin A sin

46. 2 Sol. For limiting case i.e., TIR is just taking place at AC

r2 = ic where c

1sini

12

1r sin

For a prism A = r1 + r2

11 2

1r A r A sin

c

B C

A

r1 r2

For refraction at face AB sin sinr

Given 1 1c

1sin usin A sin

As increase, r1 will increase and r2 will decrease, hence TIR will not take place at face AC. or for c the incident ray will get transmitted

1 1 1sin usin A sin

*47. Two stones are thrown up simultaneously from the edge of a cliff 240 m high with initial speed of 10

m/s and 40 m/s respectively. Which of the following graph best represents the time 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)



(1)

(2)

(3)

(4)

47. 4 Sol. Time of flight of stone A:

S = ut + 21at

2

– 240 = 10TA – 5T2 Gives TA = 8 sec Time of flight of stone B. –240 = 20TB – 5T2 Gives TB = 12 sec First 8 seconds: both stones is motion under gravity

relative position AB wrt A = (uB – uA)t = 10t

Next 8 – 12 seconds: Only stone B will move under gravity and Stone A will be at rest. Hence the shape of the x–t curve would be identical to that of a falling object under gravity; i.e, an

inverted parabola. *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 ? (‘graphs are schematic and not drawn to scale)

(1)

(2)

(3)

(4)

48. 3 Sol. Conceptual

PE = 21Kx

2 (Parabolic)

KE = TE– PE 49. A train is moving on a straight track with speed 20 ms–1. It is blowing its whistle at the frequency of

1000 Hz. The percentage change in the frequency heard by a person standing near the track as the train passes him is (speed of sound = 320 ms–1) close to:

(1) 24% (2) 6% (3) 12% (4) 18% 49. 3



Sol. i) Train approaching fa = f0 c

c

ii) Train receding fr = f0 = c

c

Change of frequency f = fa – fr

= f0c 1 1

c c

= 02 2

2f c

c v

0

f100 %

f

change in frequency =

2 2

2c100

c

= 2 2

2 320 20100 12.5%

320 20

50. An LCR circuit is equivalent to a damped pendulum. In an LCR circuit the

capacitor is charged to Q0 and then connected to the L and R as shown

below. If a student plots graphs of the square of maximum charge 2maxQ

on the capacitor with time(t) for two different values L1 and L2 (L1 > L2) of L then which of the following represents this graph correctly ?

(plots are schematic and not drawn to scale)

(1)

(2)

(3)

(4)

50. 2 Sol. For a RLC circuit

Qmax = Rt2L

0Q e

2 2 RT /Lmax 0Q Q e

2max

dQ

dt = – 2 Rt / 2

0R

Q .eL

slope at time t = 0 = – 20

RQ

L.

Hence higher the value of L smaller the magnitude of the slope or flatter the slope. Since L1 > L2, the curve corresponding to L1 must be flatter and the shape would be exponential. *51. A solid body of constant heat capacity 1 J/°C is being heated by keeping it in contact with reservoirs

in two ways (i) Sequentially keeping in contact with 2 reservoirs such that each reservoir supplies same amount

of heat.



(ii) Sequentially keeping in contact with 8 reservoirs such that each reservoir supplies same 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

51. 3 Sol. Q = 1 (T) per contact

Case (i) S = 200

100

1 T

T

= ln 2

Case (ii) S = 200

100

1 T

T

= ln 2

52. An inductor (L = 0.03H) and a resistor (R=0.15 k) are connected in

series to a battery of 15V EMF in a circuit shown below. The key K1 has been kept closed for a long time. Then at t = 0, K1 is opened and key K2 is closed simultaneously. At t = 1 ms, the current in the circuit will be : (e5 150)

(1) 0.67 mA (2) 100 mA (3) 67 mA (4) 6.7 mA 52. 1 Sol. r = 0.15 k L = 0.03 H E = 15 V

t < 0 max 3

15i

0.15 10

= 0.1 A

t > 0 Rt /L 5max

0.1i i e 0.1 e

150 = 0.67 mA.

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 03V 5V 3V, ,

2 4 4 and 0V

4 have

radius R1, R2, R3 and R4 respectively. Then (1) 2R < R4 (2) R1 = 0 and R2 > (R4 – R3) (3) R1 0 and (R2 – R1) > (R4 – R3) (4) R1 = 0 and R2 < (R4 – R3) 53. 1 & 4

Sol. 00

Q kQv

4 R R

2 23

kQv (3R r )

2R

01

3vv R 0

2

02

5v Rv R

4 2 inside

03

3v 4Rv R

4 3 inside

04

vv R 4R

4 outside.

54. A long cylindrical shell carries positive surface charge in the upper half and negative surface

charge in the lower half. The electric field lines around the cylinder will look like figure given in



(figures are schematic and not drawn to scale)

(1)

(2)

(3)

(4)

54. 2 Sol. Field lines are comparable to dipole. 55. Assuming human pupil to have a radius of 0.25 cm and a comfortable viewing distance of 25 cm, the

minimum separation between two objects that human eye can resolve at 500 nm wavelength is (1) 300 m (2) 1 m (3) 30 m (4) 100 m 55. 3

Sol. Limit of resolution, 1.22

D

D = 0.5 cm = 500 nm

minimum seperation

25cm

minimum separation = 25 cm 1.22

D

= 30 m. 56. A signal of 5 kHz frequency is amplitude modulated on a carrier wave of frequency 2 MHz. The

frequencies 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 56. 4 Sol. Resultant signal has C C m C mf , (f f ) and (f f )

= 2000 kH, (200 + 5)kHz and (2000 – 5) kHz = 2000 kH, 2005 kHz and 1995 kHz

57. Two coaxial solenoids of different 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 on the outer solenoid due to the inner one. Then:

(1) 1F

is radially outwards and 2F

= 0

(2) 1F

= 2F

= 0

(3) 1F

is radially inwards and 2F

is radially outwards

(4) 1F

is radially inwards and 2F

= 0 57. 2 Sol. Inner Outer



F

F

F F

i

B 0

So 2F 0

1F 0

*58. A pendulum made of a uniform wire of ‘cross sectional area A has time period T. When an additional

mass M is added to its bob, the time period changes to TM. If the Young’s modulus of the material of

the wire is Y then 1

Y is equal to: (g = gravitational acceleration)

(1) 2

M

T A1

T Mg

(2)2

MT A1

T Mg

(3) 2

MT Mg1

T A

(4) 2

MT A1

T Mg

58. 2

Sol. T 2g

…………..(i)

M

'T 2

g

…………..(ii)

Where ( ' )

Y

(M/g)A

Y

' Mg

1AY

' Mg

1AY

…………..(iii)

By equation (i) and (ii)

MT '

T

1/ 2

MT Mg1

M AY

by equation (iii)

2

MTMg1

AY T

2

MTMg1

AY T

2

MT1 A1

Y Mg T

*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 to one of its faces is



(1) 24MR

3 3 (2)

2MR

32 2 (3)

2MR

16 2 (4)

24MR

9 3

59. 4

Sol. a 3 2R ……………….(1)

3m V a

3

3

M 2R4 3R3

by equation (1)

3M 8

4 3 3

m = 8M

4 3 ………………….(2)

For cube

2ma

I6

= 8M

4 3

22R

36

by equation (1) and (2)

= 8M

4 3

24R

3 6

= 24MR

9 3

60. When 5V potential difference is applied across a wire of length 0.1 m, the drift speed of electrons is

2.5 x 10–4 ms–1. If the electron density in the wire is 8 x 1028 m–3, the resistivity of the material is close to

(1) 1.6 x 10–5 m (2) 1.6 x 10–8 m (3) 1.6 x 10–7 m (4) 1.6 x 10–6 m 60. 1 Sol. j = E

ne d

1 Vv

d

V

nev

28 19 4

5

8 10 1.6 10 2.5 10 0.1

2

5

8 16 25 10

= 1.56 x 10–5 = 1.6 x 10–5 m

PART C: CHEMISTRY 61. The vapour pressure of acetone at 20C is 185 torr. When 1. 2 g of a non–volatile substance was

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

(1) 488 (2) 32 (3) 64 (4) 128 61. 3



Sol. P = 185 torr Ps = 183 torr

RLVP = x = n

N

0

s0

P P w M

m MP

185 183 1.2 100

185 m 60

m = 64g. 62. 3g of activated charcoal was added to 50 mL of acetic acid solution (0.06 N) in a flask. After an hour

it was filtered and the strength of the filterate was found to be 0.042 N. The amount of acetic acid adsorbed (per gram of charcoal) is :

(1) 54 mg (2) 18 mg (3) 36 mg (4) 42 mg 62. 2 Sol. Change in concentration for 50 mL of acetic acid = 0.06 – 0.042 = 0.018 N

Wt. of Acetic acid consumed = N GEW V

1000

= 0.018 60 50

1000

= 0.054 g

For 1 gram charcoal = 0.054

0.018g3

.

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 63. 4

Sol. Energy of e– = – 13.6 x 2

2

ZeV / atom

n

When n = 2 E = –13.6 x 2

2

13.4eV / atom

2

(it is excited state) 64. Which among the following is the most reactive ? (1) ICl (2) Cl2 (3) Br2 (4) I2 64. 1 Sol. Inter halogen compounds are more reactive than halogens (except F2). ICl = Inter halogen compound. 65. Which polymer is used in the manufacture of paints and lacquers ? (1) Poly vinyl chloride (2) Bakelite (3) Glyptal (4) Polypropene 65. 3 Sol. Polymer used in manufacture of paints and lacquers is Glyptal. 66. The molecular formula of a commercial resin used for exchanging ions in water softening is

C8H7SO3Na (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

66. 1



Sol. Molecular weight of Resin = 206. Two sodiums can replace one calcium ion. 412g of resin is required for one mole of Ca+2 ions. For 1g …………. ?

= 1

412 mole per gram of resin

67. In Carius method of estimation of halogens, 250 mg of an organic compound gave 141 mg of AgBr.

The percentage of bromine in the compound is : (at. mass Ag = 108 ; Br = 80) (1) 60 (2) 24 (3) 36 (4) 48 67. 2

Sol. % of X = At.wt.of X Wt.of AgX

100M.wt of AgX Wt.of org. compd

= 80 141

100 24188 250

.

68. Assertion : Nitrogen and Oxygen are the main components in the atmosphere but these do not 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 the assertion (3) Both assertion and reason are correct, but the reason is not the correct explanation for the

assertion. (4) The assertion is incorrect, but the reason is correct 68. 2 Sol. N2 and O2 bond energies are high, hence reaction can occur under high temperatures. Both the statements are correct and second one is the correct explanation for first one. 69. The following reaction is performed at 298 K.

2 22NO g O g 2NO g

The standard free energy of formation of NO(g) is 86.6 kJ/mol at 298 K. What is the standard free energy of formation of NO2(g) at 298 K ? (Kp = 1.6 x 1012)

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

(2) 12R 298 ln 1.6 10 86600

(3) 1286600 R 298 ln 1.6 10

(4)

12ln 1.6 1086600

R 298

69. 1 Sol. 2 22NO g O g 2NO g ;

G = – 2.303 RT log KP G = – RT ln Kp

but = G = 2 x 2

0 0NONOG 2 G

= 0 022 G NO G NO

= 022 G NO 86600

by substituting in above equation



022 g NO 86600 = R(298) ln Kp

2

0NOG 86600 = 0.5 (–R(298) ln (1.6 x 1012).

2

0 12NOG 86600 0.5 R 298 ln 1.6 10

1 is correct. 70. Which one of the following alkaline earth metal sulphates has its hydration enthalpy greater than its

lattice enthalpy ? (1) SrSO4 (2) CaSO4 (3) BeSO4 (4) BaSO4 70. 3 Sol. BeSO4 in highly souble due to high H.E of Be2+ ion.

71. The number of geometrical isomers that can exist for square planar 3 2Pt Cl py NH NH OH

is

(py = pyridine) : (1) 6 (2) 2 (3) 3 (4) 4 71. 3 Sol.

Cl

HOH2N

Py

NH3

Pt

Cl

H3N

Py

NH2OH

Pt

Cl

HOH2N

NH3

Py

Pt

72. The synthesis of alkyl fluorides is best accomplished by : (1) Swarts reaction (2) Free radical fluorination (3) Sandmeyer’s reaction (4) Finkelstein reaction 72. 1 Sol. R–X + AgF R–F + AgX (X = Cl, Br) 73. The intermolecular interaction that is dependent on the inverse cube is distance between the

molecule is : (1) hydrogen bond (2) ion–ion interaction (3) ion–dipole interaction (4) London force 73. 1 Sol. Hydrogen bonding is an example of dipole–dipole interaction

dipole–dipole interaction 3

1

r

(r = intermolecular distance) 74. In the context of the Hall–Heroult process for the extraction of Al, which of the following statements is

false ? (1) Na3AlF6 serves as the electrolyte (2) CO and CO2 are produced in this process (3) Al2O3 is mixed with CaF2 which lowers the melting point of the mixture and brings conductivity (4) Al3+ is reduced at the cathode to form Al 74. 1 Sol. In Hall Heraoult’s process for the extraction of Aluminium, the electrolyte is pure alumin a (Al2O3).

But it is a poor electrolyte & to increase its conduction, Na3AlF6 (cryolite) is added. 75. Which of the following compounds will exhibit geometrical isomerism ? (1) 1, 1–Diphenyl –1 – propane (2) 1–Phenyl –2– butene



(3) 3–phenyl – 1– butene (4) 2–Phenyl –1– butene 75. 2 Sol.

C C

H

CH3CH2

H

Ph

(cis)

C C

CH3

HCH2

H

Ph

(trans)

76. The ionc radii (in 0A ) of N3–, O2– and F– are respectively :

(1) 1.71, 1.36 and 1.40 (2) 1.36, 1.40 and 1.71 (3) 1.36, 1.71 and 1.40 (4) 1.71, 1.40 and 1.36 76. 4 Sol. N3–, O2– & F– ions have 10 electrons each. i.e., in isoelectronic species as the effective nuclear

attraction decreases ionic radius increases or vice versa decreasing order of ionic radius : N3– > O2– > F–. 77. From the following statements regarding H2O2, choose the incorrect statement : (1) It has to be kept away from dust (2) It can act only as an oxidizing agent (3) It decomposes on exposure to light (4) It has to be stored in plastic or wax lined glass bottles in dark 77. 2 Sol. H2O2 act as an oxidising agent as well as reducing agent. 78. Higher order (>3) reactions are rare 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) shifting of equilibrium towards reactants due to elastic collisions 78. 2 Sol. Higher order (>3) reactions are rare due to the rare probability of collisions between reacting

molecules sismultaneously. 79. Match the catalysts to the correct processes :

Column – I (Catalyst) Column – II (Process) (A) TiCl3 (i) Wacker process (B) PdCl2 (ii) Ziegler – Natta polymerization (C) CuCl2 (iii) Contact process (D) V2O5 (iv) Deacon’s process

(1) (A) – (iii), (B) – (i), (C) – (ii), (D) – (iv) (2) (A) – (iii), (B) – (ii), (C) – (iv), (D) – (i) (3) (A) – (ii), (B) – (i), (C) – (iv), (D) – (iii) (4) (A) – (ii), (B) – (iii), (C) – (iv), (D) – (i) 79. 1 Sol. (A) TiCl3 – Ziegler– Natta polymerization (B) PdCl2 – Wacker process (C) CuCl2 – Deacon’s process (D) V2O5 – Contact process 80. Which one has the highest boiling point ? (1) Xe (2) He (3) Ne (4) Kr 80. 1 Sol. Among the zero group elements, B.P. increases with increase in atomic weight & increase in van der

waals forces of attraction.



81. In the reaction

NH2

CH3

NaNO2 / HCl

0 - 5 °CD

CuCN/KCNE N2

The product E is :

(1)

CH3

(2)

COOH

CH3

(3) CH3 CH3

(4)

CN

CH3 81. 4 Sol.

NH2

CH3

N2+ Cl

CH3

CN

CH3

N2

CuCN/KCN

0 - 5 °C

NaNO2/HCl

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

(1) BaCrO4 (2) 2 6Zn Fe CN

(3) 3 2 6K Co NO

(4) 4 3 103 4NH As Mo O

82. 2 Sol. 1, 3, and 4 compounds are yellow ppts.

Where as 2 6Zn Fe CN

is white ppt.

83. Sodium metal crystallizes in a body centred cubic lattice with a unit cell edge of 4.29 0A . The radius

of sodium atom is approximately :

(1) 0.93 0A (2) 1.86

0A (3) 3.22

0A (4) 5.72

0A

83. 2

Sol. 3

r a4

r = 01.732 4.291.86A

4



84. The standard Gibbs energy change at 300 K for the reaction 2A B + C is 2494.2 J. At a given

time, the composition of the reaction mixture is 1

A , B 22

and 1

C2

. The reaction

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 84. 3

Sol. 2A B + C

½ 2 ½

Q = 2

12

2 41

2

G = –2.303 RT log KC

c2494.2

logK8.314 2.303 300

log Kc = –0.43 Kc = 10–0.43 Kc = 0.37 Q > Kc The reaction proceeds in reverse direction. 85. Which compound would give 5–keto – 2 – methyl hexanal upon ozonolysis ?

(A)

CH3

CH3

(B)

CH3

CH3

(C)

CH3

CH3

(D)

CH3

CH3 85. 3 Sol.

CH3

CH3

O3

Zn/H2OCH3 C

O

CH2 CH2 CH

CH3

CHO

86. Which of the following compounds is not an antacid ? (1) Ranitidine (2) Aluminium hydroxide (3) Cimetidine (4) Phenelzine 86. 4 Sol. 1, 2, 3 are antacids. Phenelzine is Tranquilizer. 87. In the following sequence of reactions :

Toluene 4 2 2

4

KMnO SOCl H /PdBaSO

A B C, the product C is :

(1) C6H5CHO (2) C6H5COOH (3) C6H5CH3 (4) C6H5CH2OH 87. 1 Sol.



CH3 COOH COCl CHO

KMnO4

(A)

SOCl2

(B)

H2/Pd

BaSO4

(C) 88. Which of the vitamins given below is water soluble ? (1) Vitamin K (2) Vitamin C (3) Vitamin D (4) Vitamin E 88. 2 Sol. Vitamin–C is water soluble, remaining are fat 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 89. 4 Sol. O2– to Mn+7 i.e., L M charge transfer transition. 90. Two Faraday of electricity is passed through a solution of CuSO4. The mass of copper deposited at

the cathode is : (at. mass of Cu = 63.5 amu). (1) 127g (2) 0g (3) 63.5g (4) 2g 90. 3 Sol. By passing 1F, the mass of Cu deposited is

1gr equivalent i.e., 63.5

31.75g2

By passing 2F, the mass of Cu deposited is 63.5g

* * *

Jee Mains 2015 Solutions

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Transcript of Jee Mains 2015 Solutions