Indian National Physics Olympiad 2017

Date: 29th January 2017 Solutions Roll Number: 1 7 0 0 - 0 0 0 0 - 0 0 0 0

Time : 09:00-12:00 (3 hours) Maximum Marks: 75
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to a third party.
Besides the International Physics Olympiad (IPhO) 2017, do you also want to be considered for the Asian
Physics Olympiad (APhO) 2017? For APhO 2017 and its pre-departure training, your presence will be
required in Delhi and Russia from April 26 to May 10, 2017. In principle, you can participate in both
olympiads. Yes/No.
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Extra sheets attached : 0 Date Centre(e.g.Jaipur) Signature

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==================================================
Instructions
1. This booklet consists of 6 pages (excluding this sheet) and total of 6 questions.
2. This booklet is divided in two parts: Questions with Summary Answer Sheet and Detailed
Answer Sheet. Write roll number at the top wherever asked.
3. The final answer to each sub-question should be neatly written in the box provided below
each sub-question in the Questions & Summary Answer Sheet.
4. You are also required to show your detailed work for each question in a reasonably neat and coherent

5.
pages.

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way in the Detailed Answer Sheet. You must write the relevant Question Number on each of these

Marks will be awarded on the basis of what you write on both the Summary Answer Sheet and the
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Detailed Answer Sheet. Simple short answers and plots may be directly entered in the Summary
Answer Sheet. Marks may be deducted for absence of detailed work in questions involving loner
calculations. Strike out any rough work that you do not want to be evaluated.
6. Adequate space has been provided in the answersheet for you to write/calculate your answers. In case
you need extra space to write, you may request for additional blank sheets from the invigilator. Write
your roll number on the extra sheets and get them attached to your answersheet and indicate number
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of extra sheets attached at the top of this page.

7. Non-programmable scientific calculators are allowed. Mobile phones cannot be used as calculators.
8. Use blue or black pen to write answers. Pencil may be used for diagrams/graphs/sketches.
9. This entire booklet must be returned at the end of the examination.

Question Marks Score

1 5
Table of Constants
Speed of light in vacuum c 3.00 108 ms1 2 7
Plancks constant h 6.63 1034 Js
Universal constant of Gravitation G 6.67 1011 Nm2 kg
2 3 11
Magnitude of electron charge e 1.60 1019 C
Mass of electron me 9.11 1031 kg 4 14
Value of 1/40 9.00 109 Nm2 C2
Universal Gas Constant R 8.31 J K1 mole1 5 15

6 23

Total 75

HOMI BHABHA CENTRE FOR SCIENCE EDUCATION

Tata Institute of Fundamental Research
V. N. Purav Marg, Mankhurd, Mumbai, 400 088
INPhO 2017 Page 1 Questions & Summary Answers

1. A massive star of mass M is in uniform circular orbit around a supermassive black hole of mass
Mb . Initially, the radius and angular frequency of the orbit are R and respectively. According
to Einsteins theory of general relativity the space around the two objects is distorted and gravi-
tational waves are radiated. Energy is lost through this radiation and as a result the orbit of the
star shrinks gradually. One may assume, however, that the orbit remains circular throughout and
Newtonian mechanics holds.
(a) The power radiated through gravitational wave by this star is given by [1]

LG = Kcx Gy M 2 R4 6

where c is the speed of light, G is the universal gravitational constant, and K is a dimensionless
constant. Obtain x and y by dimensional analysis.

Solution: x = 5, y = 1

(b) Obtain the total mechanical energy (E) of the star in terms of M , Mb , and R. [1]

GMb M
Solution: E =
2R

(c) Derive an expression for the rate of decrease in the orbital period (dT /dt) in terms of the [3]
masses, period T and constants.

Solution: From Keplers law

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4 2 R3
T2 =
GMb
Using previous part
(GMb 2)2/3 M
E=
2T 2/3
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Also dE/dt = LG . This yields

2/3
dT 3KG5/3 Mb M (2)8/3
=
dt c5 T 5/3

Detailed answers can be found on page numbers:

2. The free surface of mercury (Hg) is a good reflecting surface. y

A tall cylinder partly filled with Hg and possessing total mo- (0,y0 )
ment of inertia I is rotated about its axis with the constant
angular velocity 0 as shown in figure. The Hg surface attains 0
a paraboloidal profile. The radius of curvature of a general
profile is given by
  Free x
 [1 + (dy/dx)2 ]3/2  surface (0,0)
=
 
d2 y/dx2

 
Hg
where the symbols have their usual meaning.
(a) Obtain the expression for of the Hg surface in terms of 0 , the distance x from the cylinder [3]
axis, and g.
INPhO 2017 Page 2 Questions & Summary Answers Roll Number:

Solution: !2 3/2
1 +
02 x
g
=
02 /g

(b) Calculate the value of at the lowest point of the Hg surface, that is (0,0), when 0 = 78 rpm [1]
(revolutions per minute).

Solution:
g
x=0 = = 14.7 cm
02

(c) Consider a point object at (0,y0 ) as shown in the figure. Obtain an expression for the image [3]
position yi in terms of given quantities. State conditions on y0 for the formation of real and
virtual images.

Solution: Using mirror equation

1 1 1
=

where f = x=0 /2. This gives

f

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yi y0
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gy0
yi =
g 202 y0
For real images y0 > g/202 . For virtual images y0 < g/202 .

Detailed answers can be found on page numbers:

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3. Two identical blocks A and B each of mass M are placed on a long inclined plane (angle of
inclination = ) with A higher up than B. The coefficients of friction between the plane and the
blocks A and B are respectively A and B with tan > B > A . The two blocks are initially
held fixed at a distance d apart. At t = 0 the two blocks are released from rest.
(a) At what time t1 will the two blocks collide? [2]
s
2d
Solution: t1 =
(B A )g cos

(b) Consider each collision to be elastic. At what time t2 and t3 will the blocks collide a second [4]
and third time respectively?

Solution: t2 = 3t1 , t3 = 5t1

(c) Draw a schematic velocity-time diagram for the two blocks from t = 0 till t = t3 . Draw below [5]
them on a single diagram and use solid line ( ) to depict block A and dashed line ( )
to depict block B.

Solution:
INPhO 2017 Page 3 Questions & Summary Answers

velocity

A
B

t1 2t1 3t1 4t1 5t1 time

Detailed answers can be found on page numbers:

4. One mole of an ideal gas (cp /cv = where sym- P

bols have their usual meanings) is subjected to an
Otto cycle (A-B-C-D) as shown in the following P - C
V diagram. Path A-B and C-D are adiabats. The 2P0
temperature at B is TB = T0 . Diagram is not to
scale.

7 P0
B D
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A
V0 2V0 V
(a) Find the temperatures at A,C, and D in terms of T0 and pressures at A and D in terms of [4]
P0 .
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T0 T0 P0 P0
Solution: TA = 1
; TC = 2T0 ; TD = 2 ; PA = ; PD = 1
2 2 2 2

(b) Find total heat absorbed (Q) by the system, the total work done (W ) and efficiency () [31/2]
of the Otto cycle in terms of and related quantities.

Solution:
1
 
Q = cv (TC TB ) + cv (TA TD ) = cv T0 1
21
W = Q
QBC 1
=1 = 1 1
QAD 2

(c) Draw below corresponding P -T and T -S(entropy) diagrams for the cycle. [61/2]

Solution:
INPhO 2017 Page 4 Questions & Summary Answers Roll Number:

P T C
C

D
B B

D A
A
T S

Detailed answers can be found on page numbers:

5. A metallic rod of mass m and length L (thick line in the figure below) can slide without friction [15]
on two perpendicular wires (thin lines in the figures). Entire arrangement is located in the hori-
zontal plane. A constant magnetic field of magnitude B exists perpendicular to this plane in the
downward direction. The wires have negligible resistance compared to the rod whose resistance
is R. Initially, the rod is along one of the wires so that one end of it is at the junction of the two
wires (see Fig. (a)).

y y y
B B

7 B
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L
L L
x x x
(a) (b) (c)

The rod is given an initial angular speed such that it slides with its two ends always in contact
with the two wires (see Fig. (b)), and just comes to rest in an aligned position with the other wire
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(see Fig. (c)). Determine . Neglect the self-inductance of the system.

Solution:

y
B

L x

An intermediate position of the rod is shown in figure. The coordinates of the centre of mass
of the rod are given by
L
xcm = cos (1)
2
L
ycm = sin (2)
2
Thus the kinetic energy T of the rod at any instant is
2 2
T
1 2 + 1 I 2 = mL
= mvcm (3)
cm
2 2 6
INPhO 2017 Page 5 Questions & Summary Answers

Magnitude of the induced emf E, is given by

1
E = BL2 cos(2) (4)
2
The power dissipated due to the current is equal to the loss of kinetic energy of the rod. We
have
dT E2
= (5)
dt R
Z
3B 2 L2 2
Z 0

= cos2 (2)d (6)
4mR 0
3B 2 L2
= = (7)
16mR

Detailed answers can be found on page numbers:

6. White light is incident at an angle i on a
prism of angle A placed in air as shown. A
i
Let D be the angular deviation (not nec- dispersed light
essarily a minimum) suffered by an emer-

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white light
gent ray of a particular wavelength.
(a) Obtain an expression for sin(D + A i) in terms of the refractive index n and trigonometric [3]
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functions of i and A only.

sin i
 
Solution: sin(D + A i) = sin A sin1

(b) Let A = 60.00 and i = 45.62 . Obtain the refractive index (n ) for a ray of wavelength [2]
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which has suffered deviation D = 49.58 .

Solution: n = 1.615

(c) A detailed microscopic theory yields the relation between the refractive index, n, of the [10]
material of the prism and the angular frequency = 2c/ of the incident light as
n2 1 N e2 1
 
=
n2 + 2 30 me 02 2
Here N is the electron density and 0 = 2c/0 the natural frequency of oscillation of the
electron of the material. The other symbols have their usual meaning. The table below lists
the refractive indices at six wavelengths.

(nm) 706.54 667.82 501.57 492.19 447.15 438.79

n 1.6087 1.6108 1.6263 1.6277 1.6358 1.6376
Re-express the above equation to get a linear relationship in terms of = (n2 + 2)/(n2 1)
and a suitable power of . Tabulate and plot so that you may obtain N and 0 . (Two graph
papers are provided with this booklet in case you make a mistake).

Solution: Linear relation:

1 1 N e2
= . Graph is plotted between vs 1/2 at the end of this booklet.
2 20 30 me (2c)2
INPhO 2017 Page 6 Questions & Summary Answers Roll Number:

1/2 (106 nm2 ) 2.0032 2.2422 3.9750 4.1280 5.0014 5.1938

2.8893 2.8813 2.8239 2.8188 2.7901 2.7839

(d) Calculate the values of N , 0 from the graph you plotted. Which part of the electromagnetic [4]
spectrum does 0 belong to?

Solution: From the drawn graph, N = 1.01 1029 m3 ; 0 = 1.78 1016 Hz. 0 belongs
to the ultraviolet part of the electromagnetic spectrum. Accepted values:
9.90 1029 N 1.10 1029 m3 and 1.68 1017 0 1.88 1017 Hz.

(e) An X-ray of energy 1.000 keV is incident on the prism. If we write n = 1 + then obtain the [3]
numerical value of for this ray.

Solution: = 7 105

(f) For the X-ray of the previous part let ic be the critical angle and c = 90 ic be the [1]
corresponding grazing angle. Obtain c .

Solution: c = 0.68

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Detailed answers can be found on page numbers:
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**** END OF THE QUESTION PAPER ****
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