2017-Jee-Advanced

Question Paper-1_Key & Solutions PART-1:PHYSICS SECTION -1 (Maximum Marks: 28)

   



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 options(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 options(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. **************************************************************************************************

Q1.

In the circuit shown,

L  1 H , C  1 Fand R  1k  . They are connected in series with an a.c. source

V  V0 sin t as shown. Which of the following options is/are correct? L  1 H

C  1 F R  1 K 

V0 sin t

A) At  0 the current flowing through the circuit becomes nearly zero B) The frequency at which the current will be phase with the voltage is independent of R C) The current will be in phase with the voltage if

  104 rad . s 1

  106 rad .s 1 , the circuit behaves like a capacitor

Key:

D) At A, B

Sol:

 ~ O  V0 sin A  so no current and option A is correct (A) current in phase with voltage implies resonance so,  L 

1 or 0  C

1  frequency depends on LC

capacitance and inductance not on resistance. (B) is correct

1 6

10 10

6

  10  X L  X C 6

 106 rad01 so (C) is not correct. or circuit behaves as inductive so (D) is not correct.

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  0 

2017-Jee-Advanced Q2.

For an isosceles prism of angle A and refractive index

Question Paper-1_Key & Solutions  , it is found that the angle of minimum deviation

 m  A . Which of the following options is/are correct?

 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  A) For the angle of incidence i1

C) For this prism, the emergent ray at the second surface will be tangential to the surface when the angle of incidence at the first surface is i1  sin D) For this prism the refractive index

Sol:

  2 A  1  cos A sin A 4cos 2  

1   and the angle of prism A are related as A  cos 1   2 2 



A, B, C

 A  m  sin   sin A 2  m  A &     sin A / 2 sin A / 2 2sin A / 2cos A / 2 or    2cos A / 2 sin A / 2 1 or cos A / 2    / 2  and  A / 2   cos   / 2 

  / 2 (D) is not correct.  m   i1  i2   A   i1  i2   2 Aand i1  i2  A

or A  2cos

1

So option A is correct.

sin i1 sin r1 sin A 2sin A / 2cos A / 2 2cos A / 2   sin r1  sin r1 2cos A / 2  r1  A / 2  ri / 2 option (B) is correct.

r1  r2 and  

Emergent ray tangential to the surface means i2  90  r2  C (critical angle)

r1   A  C 



sin i1 sin r1

sin i1   sin r1  2cos A / 2 sin  A  C 

 2cos A / 2  sin A cos C  cos A sin C 

 1  2cos A / 2 sin A 1  sin 2 C  cos A     1 1  2cos A / 2 sin A 1  2  cos A       1 1  2cos A / 2 sin A 1   cos A  2 4cos A / 2 2cos A / 2  

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Key:

1

2017-Jee-Advanced

Question Paper-1_Key & Solutions

 sin A 4cos 2 A / 2  1  1  2cos A / 2   cos A  2cos A / 2 2cos A / 2    sin A 4cos2 A / 2  1  cos A   1  1 2 i1  sin sin A 4cos A / 2  1  cos A option C is correct   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 of 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?

B Area A

Area 2A

 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 loops is proportional to cos t D) The amplitude of the maximum net emf induced due to the both loops is equal to the amplitude of maximum

emf induced in the smaller loop alone

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Q3.

2017-Jee-Advanced

Question Paper-1_Key & Solutions

Key:

B,D

Sol:

emf induced in the loop 1 

d  BA cos t  dt d emf induced in the loop 2   B 2 A cos t  dt But these two are in opposite in sense as Area vectors are in opposite direction.  net emf induced  difference of area so option A is wrong

d d  BA cos t  or  B  2 A cos t  dt dt Will be max when t  90 so option B is correct and option D is currect Rate of change of flux

Key: Sol:

A flat plate is moving normal to its plane through as a gas under the action of a constant force F . The gas is kept at a very low pressure. The speed of the plate 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 resistance force B) The plate will continue to the 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 ACD S  area of plate F

V

2

1

u  v  P f 1

p1  p2 

2

 2

2

 p2

u  v  

2

2

u  v   u  v   2

2

 



 4Uv 2 p1  p2  2uv Resistance form   p1  p2  s  2uvS   2US V  KV K  2US (constant) Resistance force  v p1  p2 

After long time, resistance can balance external force.

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Q4.

2017-Jee-Advanced Q5.

Question Paper-1_Key & Solutions

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 a 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?

R y

m R

x

M

x0 A) The velocity of the point mass m is: v 

2 gR m 1 M

B) The x component of displacement of the centre 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   Key:

A,B

Sol:

1) mV  MV 1

 v and v

1

mR M m

mR M m

m M

2 gR

are velocities of m and M 

2 1 1 mV 2  M V 1  2 2 2 1 1 m  2 mgR  mV  M  V  2 2 M 

2) mgR 

2 gR V m 1 M Option (B) m  R  x   Mx  0 Option (A)

x

mR M m

But in negative x direction hence B is correct.

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2017-Jee-Advanced Q6.

Question Paper-1_Key & Solutions

A block M hangs vertically at the bottom end of a 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 wavelength 0 is produced

TOA to reach point A. If 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 at point O on the rope. The pulse takes time following options is/are correct?

O

Pulse1

Pulse 2 A M A) The time

Key: Sol:

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 velocity of the two pulses (Pulse 1 and pulse 2) are the same at the midpoint of rope AC option A is correct as variation of velocity is same w.r.t “y”

T

Velocity  Q7.

TAO  TOA

 2

A human body has a surface area of approximately 1m . 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  460Wm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 T0 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 BCD or D (A) is clearly wrong from win’s displacement law. 3



(B) correct:- W1   A T  T0



4

W2   A T 4  T0  T0 

4

4





W2  W1  4 AT03T0 (C) net heat radiated is 60J per second. (D) True, by reducing are a net power emitted is reduced while his body will same heat inside so will be easy to maintain temperature.

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Key: Sol:

2017-Jee-Advanced

Question Paper-1_Key & Solutions SECTION -2 (Maximum Marks: 15)

   

Q8.

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. Negative Marks : -2 In all other cases. **************************************************************************************************

An electron in a hydrogen atom undergoes a transition from an orbit 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 electrons. If

Vi  6.25 , then the smallest possible n f is Vf Key:

5

Sol:

Potential energy 

27.2  ineV  n2 27.2 27.2 and V f  Vi  2 ni n 2f 2 Vi n f   6.25 (given) V f ni2

nf ni

 2.5

2 n f  5 (smallest possible value)

for ni

0.1 Nm1 divides itself into K identical drops. 4 3  In this process the total change the surface energy U  10 J . If K  10 then the value of  is 2

Q9.

A drop of liquid of radius R  10 m having surface tension S 

Key:

6

Sol:

Surface energy U1  4 R S (for big drop) 2

U 2  K 4 r 2 S ( for small drop) R R3  Kr 3  r  1/3 K 2 U  K 4 r S  4 R2 S 4 R 2 S K  4 R 2 S  4 R 2 S  K 1/3  1 2/3 K 2 0.1 1/3  4 R 2 SK 1/3  4 102  K  103 4 5 1/3 3  10 K  10 K 1/3  102 10 /3  102  6

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2017-Jee-Advanced Q10.

Question Paper-1_Key & Solutions

A stationary surface emits sound of frequency

f0  492 Hz . The sound is reflected by a large car approaching

1

the source with a speed of 2ms . 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

330ms 1 and the car reflects the sound at the frequency it has received). Key: Sol:

6 frequency of sound received by car is

f1 

V 2  492 V

V  330us  1

Reflected sound from the car will have a frequency f 2 



V V  2  492 V  2  V

V  2    492 V  2  Beat frequency  f1  f

Q11.

f

V f1 V 2

 492

V  2    492  492 V  2  V  2   V  2  492  V 2 4  492  6 328 131 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 131 131 serum laelled with I is injected into the blood of a person. The activity of the amount of I injected was 2.4 105 Becquerel (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 activity of 115 Bq. The total volume of blood in the person’s body, in liters is approximately (you may use e  1  x for x  1 and ln 2  0.7 ). x

5

I  Xe dN   N dt

131

 2.4 105

i.e. initially.

N0  N

2.4  105



2.4 105



e

 t



2.4 105



e



11.50.693 824

For 2.5 ml we get 115 Bq So for “V” vol. we get

115 V  N 2.5

115V 2.4 105  t  e 2.5  V  0.0521105   0.95  5lit In litres.

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Key: Sol:

2017-Jee-Advanced Q12.

Question Paper-1_Key & Solutions

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   30 . 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 0

th

index of the m slab and

 m  1 m m 1

th

and m

th

 n  0.1 (see the figure). The ray is refracted out parallel to the interface between the

slabs from the right side of the stack. What is the value of m?

n  mn n   m  1 n



 n3 n n2 n n n n

3 2 1

 Key: 8

n sin 300   n  mn  sin 900

" m  8" ( but n for a material i.e  n  mn  can’t be less than “1” ).

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age-11

Sol:

2017-Jee-Advanced

Question Paper-1_Key & Solutions SECTION -3 (Maximum Marks: 18)

     

This section Contains SIX questions of matching type. This section Contains TWO tables (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. **************************************************************************************************

A charged particle (electron or proton) is introduced at the origin (x=0,y=0,z=0) with a given initial velocity  . A uniform electric field E and a uniform magnetic field B exist everywhere. The velocity  , electric field E and magnetic field B are given columns 1,2 and 3, respectively. The quantities

E0 , B0 are positive in magnitude.

Column 1

Column 2

(I) Electron with   2 (II) Electron with   (III) Proton with 

E0 x B0

E0 y B0

0

(IV) Proton with   2

E0 x B0

Column 3

(i) E  E0 z

(P) B   B0 x

(ii) E   E0 y

(Q) B  B0 x

(iii) E   E0 x

(R) B  B0 y

(iv) E  E0 x

(S) B  B0 z

Q13. In which case would the particle move in a straight line along the negative direction of y-axis (i.e., move along- y )? A) (IV) (ii) (S) C) (III) (ii) (R) Key: C Sol:

B) (II) (iii) (Q) D) (III) (ii) (P)

Y By ton pro

e eas l e r

X Ey

Proton released from rest at origin

F magnetic  q(V y  B y ) qE y t j m

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Vy 

2017-Jee-Advanced

Question Paper-1_Key & Solutions

V y is antiparallel to B y

F magnetic  0 Q14.

Key: Sol:

Because of Electrostatic force, if the proton moves along –ve y-direction. In which case will the particle move in a straight line with constant velocity? A) (II) (iii) (S) B) (III) (iii) (P) C) (IV) (i) (S) D) (III) (ii) (R) A Y

V

FM

E

X

FE

When F m  F E  0, the electron moves with constant velocity along y-axis.

qVy B0  qE0 E0 B0  qE0 B0  Fnet  0 q

V y remains constant. 15.

Key:

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) C

Y

E

V

X

B

The proton describes helical path with increasing pitch, with axis of helix along z-axis and plane of helix is

  X , Y 

An ideal 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 ration 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  PV   1 2 2 1 1

(i) Isothermal

(P) P

1

2

V

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Sol:

Z

2017-Jee-Advanced (II)

Question Paper-1_Key & Solutions

W12   PV 2 2  PV 1 1

(ii) Isochoric

(Q) P 1

2 V

(III)

W12  0

P

(iii) Isobaric

1 2

V

(R) (IV) W12  nRT ln(

P

(iv) Adiabatic

V2 ) V1

1

2

(S) Q16. `

V

Key:

Which one of the following options correctly represents a thermodynamic process that is used as a 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) C

Sol:

Vsound 

 RT M0

as the sound wave propagates, the air in a chamber undergoes compression & rare fraction, &

Undergo a adiabatic process. So curves are steeper than isothermal

 dP  P          1  dV  Adi V   dP  P           2  dV  Iso V  Graph ' Q ' satisfies eqn(1) Q17.

Key: Sol:

Which of the following options is the only correct representation of a process in which U  Q  PV ? A) (II) (iii) (S) B) (II) (iii) (P) C) (III) (iii) (P) D) (II) (iv) (R) B

U  Q  PV U  PV  Q As U  0 W  0 Q  0 The Process represents, Isobaric process

Wgas   P  V    P  v2  v1    Pv2  Pv1

Key: Sol:

graph ‘p’ satisfies isobaricprocess 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) B Work done in isochoric process is zero

W12  0 as V  0

Graph ‘S’ represents isochoric processes

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Q18.