IITJEE Advanced 2017 Question Paper 2 with Solutions

IIT - JEE 2017 (Advanced)

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(2) Vidyalankar : IIT JEE 2017 Advanced : Question Paper & Solution

Solution to IIT JEE 2017 (Advanced) : Paper - II

PART I ? PHYSICS

SECTION 1 (Maximum Marks:21)

This section contains SEVEN questions.

Each question has FOUR options (A), (B), (C) and (D). ONLY ONE of these four

options is correct.

For each question, darken the bubble corresponding to the correct option in the ORS.

For each question, marks will be awarded in one of the following categories:

Full Marks : +3 If only the bubble corresponding to the correct option is

darkened.

Zero Marks : 0 If none of the bubbles is darkened.

Negative Marks : 1 In all other cases.

1. A photoelectric material having work-function 0 is illuminated with light of wavelength

hc 0

.

The

fastest

photoelectron

has

a

de

Broglie

wavelength

d.

A

change

in

wavelength of the incident light by results in a change d in d. Then the ratio

d/ is proportional to

(A) 3d / 2

(B) d2 / 2

(C) d/

(D) 3d /

1. (A)

KEmax

hc

d

h p

h 2mK

K

h2 2m d2

hc h2

2m d2

Differentiating both sides

hc 2

d

h2 2m 3d

dd

d 3d 2

...(i) ...(ii)

2. Three vectors P,QandR are shown in the figure. Let S be any point on the vector R . The distance between the points P and S is b| R | . The general relation among vectors P,QandS is

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IIT JEE 2017 Advanced : Question Paper & Solution (Paper ? II) (3)

(A) S(1 b)Pb2Q

(B) S(1 b2)PbQ

(C) S(1 b)PbQ

(D) S(b 1)PbQ

2. (C) | S | b| R | OS OP PS S P bR P b(Q P) S (1 b)P bQ

y P

O

P

S RQP

Q Q

x

3. A symmetric star shaped conducting wire loop is carrying a steady state current I as shown in the figure. The distance between the diametrically opposite vertices of the star is 4a. The magnitude of the magnetic field at the center of the loop is

(A) oI 6[ 31] 4a

(B) oI 3[ 31] 4a

(C) oI 6[ 31] 4a

(D) oI 3[2 3] 4a

3. (A)

Magnetic field at center of loop due to anyone section

B 0I sin sin

4a

= 30 sin = 1

a

2

= 60 sin = 3 2

Bnet =

12B

120I 4a

3 2

1 2

Bnet

=

6 0I 4a

3 1

4. Consider regular polygons with number of sides n = 3, 4, 5 ...... as shown in the figure. The center of mass of all the polygons is at height h from the ground. They roll on a horizontal surface about the leading vertex without slipping and sliding as depicted. The maximum increase in height of the locus of the center of mass for each polygon is . Then depends on n and h as

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(4) Vidyalankar : IIT JEE 2017 Advanced : Question Paper & Solution

(A)

=

h

sin

2 n

(B)

=

h

sin2

n

(C)

=

h

tan2

2n

4. (D)

1

(D)

=

h

cos

n

1

h 2

h = 2h

= h

for any regular polygon

2 = 2 n

h =

h cos

h

cos

n

h

'

h

h

1 cos

1

n

h = 2h

= 2 1 h

5. Consider an expanding sphere of instantaneous radius R whose total mass remains

constant. The expansion is such that the instantaneous density remains uniform

throughout

the

volume.

The

rate

of

fractional

change

in

density

1 p

d dt

is

constant.

The

velocity of any point on the surface of the expanding sphere is proportional to

(A) R2/3

(B) R

(C) R3

(D) 1 R

5. (B)

Total mass = v = constant

dV V d 0 dt dt

dV V d

dt

dt

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IIT JEE 2017 Advanced : Question Paper & Solution (Paper ? II) (5)

dV Vd dt dt

dV V dt

V = 4 r3 3

dV 4t2 dr 4 r3

dt

dt 3

Velocity of any point on surface

v = dr r dt

6. A person measures the depth of a well by measuring the time interval between dropping a

stone and receiving the sound of impact with the bottom of the well. The error in his

measurement of time is T = 0.01 seconds and he measures the depth of the well to be L =

20 meters. Take the acceleration due to gravity g = 10ms2 and the velocity of sound is 300 ms1. Then the fractional error in the measurement, L/L, is closest to

(A) 0.2%

(B) 5%

(C) 1%

(D) 3%

6. (C) Time taken by stone to reach the surface of well.

t1

2gh g

t1

2 20 2sec 10

............. (1)

time taken by sound to reach back

t2

h v

t2

20 300

1 15

sec

............. (2)

Neglecting time taken by sound we have total time T = 2s

L 1 gT2 2

L 2 T 2 0.01 0.01

L

T

2

percentage error

= 0.01 100

= 1%

7. A rocket is launched normal to the surface of the Earth, away from the Sun, along the line joining the Sun and the Earth. The Sun in 3 105 times heavier than the Earth and is at a distance 2.5 104 times larger than the radius of the Earth. The escape velocity from Earth's gravitational field is e = 11.2 km s1. The minimum initial velocity (s) required

for the rocket to be able to leave the SunEarth system is closest to (Ignore the rotation

and revolution of the Earth and the presence of any other planet). (A) s = 62 km s1 (B) s = 42 km s1 (C) s = 72 km s1 (D) s = 22 km s1

7. (B)

Energy conservation

1 2

mVs2

GMem R

GMsm r 2R

0

1 2

mVs2

GMem R

Gsm r

r = 2.5 104 R

Vs

Sun

Earth

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