Download PART I : PHYSICS SECTION 1 (Maximum Marks : 28)

[1]
IIT JEE – 2017 / Paper-1
PART I : PHYSICS
SECTION 1 (Maximum Marks : 28)
PHYSICS

This section contains SEVEN questions.

Each question has FOUR options (A), (B), (C) and (D). ONE OR MORE THAN ONE of these four
options is(are) correct.

For each question, darken the bubble(s) corresponding to all the correct option(s) in the ORS.

For each question, marks will be awarded in one of the following categories:
Full Marks
:
+4 If only the bubble(s) corresponding to all the correct option(s) is(are)
darkened.
Partial Marks
:
+1 For darkening a bubble corresponding to each correct option, provided
NO incorrect option is darkened.
Zero Marks
:
0
If none of the bubbles is darkened.
Negative Marks
:
–2 In all other cases.

For example, if (A), (C) and (D) are all the correct options for a question, darkening all these three will get
+4 marks; darkening only (A) and (D) will get +2 marks; and darkening (A) and (B) will get –2 marks, as
a wrong option is also darkened.
1.
In the circuit, L  1H , C  1H and R  1k . They are conneted in series with an a.c. source
V  V0 sint as shown. Which of the following options is/are correct?
(A)
(B)
(C)
(D)
At  ~ 0 the circuit flowing through the circuit becomes nearly zero
The frequency at which the current will be in phase with the voltage is independent of R
The current will be in phase with the voltage if   10 4 rad. s–1
At   10 6 rad s–1 the circuit behaves like a capacitor
1
  so i  0
C
At resonance , current is in phase
with voltage and frequency
Ans. (A,B) At  ~ 0 , X C 
f 
1
2 LC
is independent of R.
"Guru-Kripa", Anasagar Circular Road, Ajmer 305 001 Tel: 0145-2631190, 2629236
[2]
IIT JEE – 2017 / Paper-1
PHYSICS
2.
For an isosceles prism of angle A and refractive index  , it is found that the angle of minimum deviation
 m  A . Which of the following options is/ are correct?
(A) For the angle of incidence i1  A , the ray inside the prism is parallel to the base of the prism
(B) At minimum deviation, the incident angle i1 and the refracting angle r1 at the first refracting surface are
related by r1  i1 / 2 
(C) For the prism, the emergent ray at the second surface will be tangential to the surface when the angle
LM
N
of incidence at the first surface is i1  sin 1 sin A 4 cos 2
A
2
 1  cos A
OP
Q
(D) For the prism, the refractive index  and the angle of prism A are related as A 
Ans. (A,B,C) At minimum deviation, i1  i2 so  m  i1  i2  A  2i  A  A  i1  A
At minimum deviation, the ray inside the prism is parallel to the base of the prism.
Further r1  r2  A / 2 and since i1  A
so
r1 
i1
2
FG A   IJ
H 2 K  sin( A)  2 cosFG A IJ
H 2K
F A I sin A
sinG J
H 2K
2
m
sin

As
i2  90 so r2   C
(1) sin i1   sin r
or
or
or
or
As
or sin i   sin A cos C  cos A sin  C
sin i   sin( A   C )
LM
MN
OP

 PQ
L
O
A
sin i  M sin A 4 cos
 1  cos AP
2
MN
PQ
L
O
F AI
i  sin M sin A 4 cos G J  1  cos AP
H 2K
MN
PQ
1
F AI
F I
  2 cosG J so A  cos G J
H 2K
H 2K
2
sin i   sin A 
2 1

cos A
2
1
2
1
"Guru-Kripa", Anasagar Circular Road, Ajmer 305 001 Tel: 0145-2631190, 2629236
1
2
cos 1
FG  IJ
H 2K
[3]
IIT JEE – 2017 / Paper-1
PHYSICS
3.
A circular insulated copper wire loop is twisted to form two loops of area A and 2A as shown in the figure.
At the point crossing the wires remain electrically insulated from each other. The entire loop lies in the

plane (of the paper). A uniform magnetic field B points into the plane of the paper. At t  0 , the loop
starts rotating about the common diameter as axis with a constant angular velocity  in the magnetic field.
Which of the following options is/are correct?
(A) The emf induced in the loop is proportional to the sum of the areas of the two loops.
(B) The rate of change of the flux is maximum when the plane of the loops is perpendicular to plane of the
paper
(C) The net emf induced due to both the loops is proportional to cost
(D) The amplitude of the maximum net emf induced due to both the loops is equal to the amplitude of
maximum emf induced in the smaller loop alone.
Ans. (B,D) The net emf induced in the loop is
  L S  
LM d  B. A cost   d  B. A cost OP
dt
N dt
Q
2
1
 B sin t  A2  A1 
 B sin t  2 A  A  AB sin t
Rate of change of flux is maximum when sint  1 i.e., plane of loop is perpendicular to the plane of paper.
4.
A flat plate is moving normal to its plane through a gas under the action of a constant force F. The gas is
kept at a very low pressure. The speed of the plane v is much less than the average speed u of the gas
molecules. Which of the following options is/are true?
(A) At a later time the external force F balance the resistive force
(B) The plate will continue to move with constant non-zero acceleration, at all times
(C) The resistive force experienced by the plate is proportional to v
(D) The pressure difference between the leading and trailing faces of the plate is proportional to uv
Ans. (A,C,D) Let  be the density of gas then force acting on leading and trailing faces are
F  A 2v  u 
2
Ft  A 2v  u 
2
2
2
Thus net resistive force FR  F  Ft  4 A  v  u    v  u   8 Avu
Thus, the resistive force FR  v and with increase in speed increases till it balances F.
Pressure difference p 
F  Ft
A
 8vu  vu
FFFl
FFtR
F
v
"Guru-Kripa", Anasagar Circular Road, Ajmer 305 001 Tel: 0145-2631190, 2629236
[4]
IIT JEE – 2017 / Paper-1
PHYSICS
5.
A block of mass M has a circular cut with a frictionless surface as shown. The block rests on the horizontal
frictionless surface of a fixed table. Initially the right edge of the block is at x = 0, in co-ordinate system
fixed to the table. A point mass m is released from rest at the topmost point of the path as shown and it
slides down. When the mass loses contact with the block, its position is x and the velocity is v. At that
instant, which of the following options is/are correct?
(A) The velocity of the point mass m is v 
2 gR
m
1
M
(B) The x component of displacement of the center of mass of the block M is 
(C) The position of the point mass is x   2
(D) The velocity of the block M is V  
m
M
mR
mR
M m
M m
2 gR
Ans. (A,B) As xcm  0 so
Mx M  m xm  0
or
Mx M  m R  xmM  0 or xmM  
xm  xmi
mR
m M
mR
mR
 x m   R  R 

m M
m M
LM
N
OP
Q
From mechanical energy conservation and momentum conservation in horizontal,
mvm  MVm  0
and
1
2
mvm2 
1
2
Mv 2M  mgR
from above equation, vm 
6.
...(1)
...(2)
2 gR
m
1
M
A block M hangs vertically at the bottom end of the uniform rope of constant mass per unit length. The top
end of the rope is attached to a fixed rigid support at O. A transverse wave pulse (Pulse 1) of the wavelength
 0 is produced at point O on the rope. The pulse takes time TOA to each point A. If the wave pulse of
wavelength  0 is produced at point A (Pulse 2) without disturbing the position of M it takes time TAO to
reach point O. Which of the following options is/are correct?
(A) The time TAO  TOA
(B) The wavelength of Pulse 1 becomes longer when it reaches point A
(C) The velocity of any pulse along the rope is independent of its frequency
and wavelength
(D) The velocities of the two pulses (Pulse 1 and Pulse 2) are the same at the
midpoint of rope.
"Guru-Kripa", Anasagar Circular Road, Ajmer 305 001 Tel: 0145-2631190, 2629236
[5]
IIT JEE – 2017 / Paper-1
PHYSICS
Ans. (A,C or A,C,D) As the variation of speed along the string is same for both pulses therefore, TAO  TOA
At mid-points, both pulses have same speeds but opposite, direction.
The velocity of pulse is only dependent on medium i.e. string not the frequency and wavelength. At A, since speed is
less so wavelength will also be less.
7.
A human body has a surface area of approximately 1 m2. The normal body temperature is 10 K above the
surrounding room temperature T0. Take the room temperature to be T0  300 K . For T0  300 K , the
value of T04  460 Wm2 (where  is the Stefan-Boltzmann constant). Which of the following options
is/are correct?
(A) If the body temperature rises significantly then the peak in the spectrum of electromagnetic radiation
emitted by the body would shift to longer wavelengths
(B) If the surrounding temperature reduces by a small amount T0  T0 , then to maintain the same body
temperature the same (living) human being needs to radiate W  4T03 T0 more energy per unit time
(C) The amount of energy radiated by the body in 1 second is close to 60 Joules
(D) Reducing the exposed surface area of the body (e.g. by curling up) allows humans to maintain the
same body temperature while reducing the energy lost by radiation.
Ans. (C,D) If the body temperature rises significantly then the peak of the spectrum of electromagnetic radiation emitted by
the body would shift to shorter wavelength. Hence option(A) is incorrect.
Amount of energy radiated by the body in one second
b
H  A T  T
4
4
o
g  AT
4
o
LMFG1  T IJ  1OP
NH T K Q
4
o
Since T  10  To
H  ATo4
FG 4T IJ  460  1  4  10  61 J
HT K
300
o
So option (C) is correct
Also on reducing the exposed surface area of the body allows humans to maintain the same body temperature while
reducing the energy lost by radiation. Hence (D) is correct.
If surrounding temperature reduces by small amount To  To , then to maintain the same body temperature the same
human body needs to radiate
W  4  To3 To more energy per unit time.
"Guru-Kripa", Anasagar Circular Road, Ajmer 305 001 Tel: 0145-2631190, 2629236
[6]
IIT JEE – 2017 / Paper-1
SECTION 2 (Maximum Marks : 15)
PHYSICS

This section contains FIVE questions.

The answer to each question is a SINGLE DIGIT INTEGER ranging from 0 to 9, both inclusive.

For each question, darken the bubble corresponding to the correct integer 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 answer is darkened.
Zero Marks
:
0
In all other cases.
8.
An electron in hydrogen atom undergoes a transition from an obrit with quantum number ni to another
with quantum number n f .Vi and V f are respectively the initial and final potential energies of the electron.
vi
If v  6.25 , then the smallest possible n f is
f
Ans. (5) Potential energy of electron in nth orbit of hydrogen atom,
Un 

Ui
Uf
nf
or ni
so
9.
27.2
n2


n 2f
ni2
eV
 6.25
5
2
n f min  5
A drope of liquid of radius R  102 m having surface tension S 
01
.
Nm1 divides itself into K identical
4
drops. In this process the total change in the surface energy U  103 J . If K  10 then the value of 
is
Ans. (6) Radius of smaller drop, r 
R
K 1/ 3
 U  U f  U i  K S 4r 2  S 4R 2
b
g b
g
10 3  S 4R 2 K 1/3  1
10 3  0.1  10 4 K 1/3  1
or K 1/ 3  1  100
 K 1/3  100 or K  10 6
so
 6
10. A stationary source emits sound of frequency f 0  492 Hz. The sound is reflected by a large car
approaching the source with a speed of 2 ms 1 . The reflected signal is received by the source and
superposed with the original. What will be the beat frequency of the resulting signal in Hz? (Given that the
speed of sound in air is 330 ms 1 and the car reflects the sound at frequency it has received).
"Guru-Kripa", Anasagar Circular Road, Ajmer 305 001 Tel: 0145-2631190, 2629236
[7]
IIT JEE – 2017 / Paper-1
Ans. (6) Here frequency of reflected single received by the source f   f o
FG V  V IJ
H V V K
car
PHYSICS
car
car
car
2 f oVcar
V  Vcar
11.
LMV  V
NV  V
f Beat  f   f o  f o

131

2  492  2
328
OP
Q
1
 6 Hz
I is an isotope of Iodine that  decays to an isotope of Xenon with a half-life of 8 days. A small amount
of a serum labelled with 131 I is injected into the blood of a person. The activity of the amount of 131 I
injected was 2.4  105 Becqueral (Bq). It is known that the injected serum will get distributed uniformly in
the blood stream in less than half an hour. After 11.5 hours, 2.5 ml of blood is drawn from the person's
body, and gives an actively of 115 Bq. The total volume of blood in the person's body, in liters is
approximately (you may use e x  1  x for x  1 and ln 2  0.7 ).
Ans. (5) Let x be the total volume of blood (in liters), then using
R  Ro e  t
115  103 x
 2.4  105 e
2.5
or


x
2.4  2.5  102
115
FG
H
2.4  2.5
 10 2 1 
115
2.4  2.5
115
e

23
24


ln 2
8  24 h
 11.5 h 
8 . 05
8  24
1
24
IJ
K
 100  5L
12. A monochromatic light is travelling in a medium of refractive index n = 1.6. It enters a stack of glass layers
from the bottom side at an angle   30º . The interfaces of the glass layers are parallel to each other. The
refractive indices of different glass layers are monotonically decreasing as nm  n  mn , where nm is the
refractive index of the mth slab and n  0.1 (see the figure). The ray is refracted out parallel to the
interface between the (m  1) th and mth slabs from the right side of the stack. What is the value of m?
Ans. (8) For refracted ray, to emerge parallel to the interface
a
f
n sin   n  m  n sin 90 o
 1.6 sin 30  1.6  m 0.1
o
m8
"Guru-Kripa", Anasagar Circular Road, Ajmer 305 001 Tel: 0145-2631190, 2629236
[8]
IIT JEE – 2017 / Paper-1
PHYSICS
SECTION 3 (Maximum Marks : 18)

This section contains SIX questions of matching type.

This section contains TWO table (each having 3 columns and 4 rows).

Based on each table, there are THREE 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.
Answer Q. 13, Q.14 and Q.15 by appropriately matching the information given in the three
columns of the following table
A charged particle (electron or proton) is introduced at the origin (x = 0, y = 0, z = 0) with a given initial


velocity v . A uniform electric field E and a uniform magnetic field B exist everywhere. The velocity v ,


electric field E and magnetic field B are given in column 1, 2 and 3, respectively. The quantities E 0 , B0
are positive in magnitude.
Column 1
Column 2
Column 3
E0

(I) Electron with v  2
x
B0

(i) E  E0 z

(P) B   B0 x
 E0
(II) Electron with v  B y
0

(III) Proton with v  0

(ii) E   E0 y

(Q) B  B0 x

(iii) E   E0 x

(R) B  B0 y
E

(IV) Proton with v  2 0 x
B0

(iv) E  E 0 x

(S) B  B0 z
13. In which case would the particle move in a stright line along the negative direction of y-axis (i.e. move along
– y )?
(A) (IV) (ii) (S)
(B) (II) (iii) (Q)
(C) (III) (ii) (R)
(D) (III) (ii) (P)


Ans. (C) To move in a straight line along the negative direction of y axis, both E & B must be along y axis and initial velocity
must be either zero or along-y axis. Hence option (c) is the only correct option.
14. In which case will the particle move in a stright line with constant velocity ?
(A) (II) (iii) (S)
(B) (III) (iii) (P)
(C) (IV) (i) (S)
(D) (III) (ii) (R)
  
Ans. (A) To move with constant velocity, net force must be zero i.e., q V  B  E  0 . Such combination is only possible
d
i
corresponding to option (A).
15. In which case will the particle describe a helical path with axis along the positive z direction?
(A) (II) (ii) (R)
(B) (III) (iii) (P)
(C) (IV) (i) (S)
(D) (IV) (ii) (R)


Ans. (C) To describe a helical path with axis along +z direction, electric field must be along z direction and V and B must be
normal to z. This is possible only for (C) in given choices.
"Guru-Kripa", Anasagar Circular Road, Ajmer 305 001 Tel: 0145-2631190, 2629236
[9]
IIT JEE – 2017 / Paper-1
PHYSICS
Answer Q. 16, Q.17 and Q.18 by appropriately matching the information given in the three
columns of the following table
An idal gas is undergoing a cyclic thermodynamic process in different ways as shown in the corresponding
P – V diagrams in column 3 of the table. Consider only the path from state 1 to state 2. W denotes the
corresponding work done on the system. The equations and plots in the table have standard notations as
used in thermodynamic processes. Here  is the ratio of heat capacities at constant pressure and constant
volume. The number of moles in the gas is n.
Column 1
Column 2
Column 3
(I) W1 2 
1
( PV
2 2  PV
1 1)
 1
(i) Isothermal
(P)
(II) W1 2   PV2  PV1
(ii) Isochoric
(Q)
(III) W1 2  0
(iii) Isobaric
(R)
(iv) Adiabatic
(S)
(IV) W1 2   nRT ln
FG V IJ
HV K
2
1
16. Which one of the following options correctly represents a thermodynamic process that is used as correction
in the determination of the speed of sound in an ideal gas?
(A) (IV) (ii) (R)
(B) (I) (ii) (Q)
(C) (I) (iv) (Q)
(D) (III) (iv) (R)
Ans. (C) Laplace made a correction that propagation of sound through air is an adiabatic process. Hence option (C) is correct.
17. Which of the following options is the only correct representation of a process in which U  Q  PV ?
(A) (II) (ii) (S)
(B) (II) (iii) (P)
(C) (III) (iii) (P)
(D) (II) (iv) (R)
Ans. (B) Here process must be isobaric and thus option (B) is correct.
18. Which one of the following options is the correct combination ?
(A) (II) (iv) (P)
(B) (III) (ii) (S)
(C) (II) (iv) (R)
(D) (IV) (ii) (S)
Ans. (B) Out of the given options (B) is correct (Isochoric combination).
"Guru-Kripa", Anasagar Circular Road, Ajmer 305 001 Tel: 0145-2631190, 2629236