Indian National Physics Olympiad 2016

Final Solutions Roll Number: 1 6 0 0 - 0 0 0 0 - 0 0 0 0

Date: 31st January 2016
Time : 09:00-12:00 (3 hours) Maximum Marks: 50

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ered for the Asian Physics Olympiad (APhO) -2016? For APhO - 2016 pre-departure
training, your presence will be required in Delhi and Hong Kong from April 25-May 10.
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Question: 1 2 3 4 5 6 Total

Marks: 6 4 10 9 9 12 50

Score:

HOMI BHABHA CENTRE FOR SCIENCE EDUCATION

Tata Institute of Fundamental Research
V. N. Purav Marg, Mankhurd, Mumbai, 400 088
Instructions:

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Table of Information

Speed of light in vacuum c = 3.00 108 ms1

Plancks constant h
Universal constant of Gravitation G
=
=

6
6.63 1034 Js
6.67 1011 Nm2 kg
2
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Magnitude of electron charge e = 1.60 1019 C
Mass of electron me = 9.11 1031 kg = 0.51 MeVc2
Stefan-Boltzmann constant = 5.67 108 Wm2 K4
Value of 1/40 = 9.00 109 Nm2 C2
Permeability constant 0 = 4 107 Hm1
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Universal Gas Constant R = 8.31 J K1 mole1

Molar mass of air = 29.0 kgkmol1
Rest mass of proton mp = 1.67 1027 kg
1 atomic mass unit 1 u = 1.661 1027 kg

HOMI BHABHA CENTRE FOR SCIENCE EDUCATION

Tata Institute of Fundamental Research
V. N. Purav Marg, Mankhurd, Mumbai, 400 088
 INPhO - 2016 - Final Solutions
Please note that alternate/equivalent methods and different way of
expressing final solutions may exist.

1. A glass rod of refractive index 1.50 of rectangular cross section {d l} is bent into a
U shape (see Fig. (A). The cross sectional view of this rod is shown in Fig. (B).

p1

R+d
p2
R

D C H G
E l
A B d F

r2 r1

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(A) (B)

Bent portion of the rod is semi-circular with inner and outer radii R and R + d re-
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spectively. Parallel monochromatic beam of light is incident normally on face ABCD.
(a) Consider two monochromatic rays r1 and r2 in Fig. (B). State whether the [1]
following statements are True or False.
Statement True/False
If r1 is total internally reflected from the semi circular section at the
HB

point p1 then r2 will necessarily be total internally reflected at the point p2 . True
If r2 is total internally reflected from the semi circular section at the
point p2 then r1 will necessarily be total internally reflected at the point p1 . False

(b) Consider the ray r1 whose point of incidence is very close to the edge BC. Assume [1]
it undergoes total internal reflection at p1 . In cross sectional view below, draw
the trajectory of this reflected ray beyond the next glass-air boundary that it
encounters.
p1
ii
i
i
O

r1
 INPhO - 2016 - Final Solutions
(c) Obtain the minimum value of the ratio R/d for which any light ray entering [2]
the glass normally through the face ABCD undergoes at least one total internal
reflection.

Solution: For total internal reflection

sin i sin c

where i is the incidence angle of the ray on bent portion of the rod (see figure
in part (b)) and c is the critical angle for glass-air boundary. For a light ray
close to edge BC
R 1

R+d nglass
For refractive index nglass = 1.5 R 2d

Minimum value of R/d = 2

(d) A glass rod with the above computed minimum ratio of R/d, is fully immersed [2]
in water of refractive index 1.33. What fraction of light flux entering the glass

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through the plane surface ABCD undergoes at least one total internal reflection?
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Solution:

O
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r3
Let the intensity of beam be I0 . Flux entering through glass slab will be dlI0 .
Assume that any light ray up to distance x from the edge BC undergoes at
least one total internal reflection. Then the flux going through at least one
total internal reflection will be (d x)lI0 . Also

R+x nwater
=
R+d nglass

For R/d = 2 and nwater = 1.33, x = 2d/3.

Fraction of light = 0.33 or 0.34.

2. A uniformly charged thin spherical shell of total charge Q and radius R is centred at [4]
the origin. There is a tiny circular hole in the shell of radius r (r  R) at z = R.
Find the electric field just outside and inside the hole, i.e., at z = R + and z = R
(  r).

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 INPhO - 2016 - Final Solutions

Solution: Electric field of a circular disc with charge density of radius r at any
point z on its axis for z  r
E~ = z
20
Given system can be considered as a spherical shell of radius R with charge density
and a disk of radius r with charge density .
~ + ) = Electric field due to shell + Electric field due to hole
E(R
R2
= 2
z z
0 (R + ) 20

z
20
Similarly
~ ) = z
E(R
20
For R  r

~ + ) = E(R
~ ) = Q
E(R z i.e. radially outward.

6
80 R2
Answers expressed up to second order approximations are also ac-
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cepted.

3. This problem is designed to illustrate the advantage that can be obtained by the use
of multiple-staged instead of single-staged rockets as launching vehicles. Suppose that
the payload (e.g., a space capsule) has mass m and is mounted on a two-stage rocket
(see figure). The total mass (both rockets fully fuelled, plus the payload) is N m.
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The mass of the second-stage rocket plus the

payload, after first-stage burnout and sepa-
1st Stage 2nd Stage Pay ration, is nm. In each stage the ratio of con-
load tainer mass to initial mass (container plus
fuel) is r, and the exhaust speed is V , con-
stant relative to the engine. Note that at the
end of each state when the fuel is completely
m
nm exhausted, the container drops off immedi-
Nm
ately without affecting the velocity of rocket.
Ignore gravity.
(a) Obtain the velocity v of the rocket gained from the first-stage burn, starting from [21/2]
rest in terms of {V, N, n, r}.

Solution: Variable mass equation gives

d~v dm
m = F~external + ~vrelative
dt dt
No gravity hence F~external = 0, |~vrelative | = v. Solving rocket equation
mi
v = V ln (1)
mf

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 INPhO - 2016 - Final Solutions

Here

initial mass mi = N m (2)

final mass mf = [N r + n(1 r)]m (3)

Answers are written in terms of speed. Answer written in terms

of velocity with proper sign (-ve) are also acceptable.

(b) Obtain a corresponding expression for the additional velocity u gained from the [1]
second stage burn.

Solution: Now mi = nm, mf = m(nr + 1 r). Equation (1) yields

n
u = V ln (4)
nr + 1 r
or
n
~u = V~ ln (5)
nr + 1 r

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(c) Adding v and u, you have the payload velocity w in terms of N , n, and r. Taking
N and r as constants, find the value of n for which w is a maximum. For this
[21/2]
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maximum condition obtain u/v.

Solution: From Eqs. (3 and 5)

Nn
w = V ln
[N r + n(1 r)][nr + 1 r]
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= V ln f (n)

Maximizing w is equivalent to maximizing f (n). Differentiating and setting

equal to zero, we obtain

u ln[ N /{r N + (1 r)}]
n= N = =1 (6)
v ln[N/{N r + N (1 r)}]

where we have used Eqs. (1 and 5).

(d) Find an expression for the payload velocity ws of a single-stage rocket with the [2]
same values of N , r, and V .

Solution: Here mi = N m and mf = m + r(N m m). Using Eq. (1)

N
ws = V ln
Nr + 1 r

(e) Suppose that it is desired to obtain a payload velocity of 10 km/s, using rockets [2]
for which V = 2.5 km/s and r = 0.1. Using the maximum condition of part (c)
obtain the value of N if the job is to be done with a two-stage rocket.

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 INPhO - 2016 - Final Solutions

Solution: Payload velocity

N
w = u + v = 2V ln
r N +1r
For the desired value of w, N = 649.4 (645 < N < 655 accepted.)

4. The realization towards the latter half of nineteenth century that blackbody radiation
in a cavity can be considered to be a gas in equilibrium became an important milestone
in the development of modern physics. The equation of state of such a gas with
pressure P and volume V is P = u/3. Here u is the internal energy per unit volume.
Also, u = T 4 , where is a constant and T is absolute temperature at which radiation
is in equilibrium with the walls of the cavity.
(a) Using the first law of thermodynamics obtain an expression for dQ in terms of [3]
volume, temperature and other related quantities.

Solution: From first law of thermodynamics we have

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dQ = dU + dW = d(uV ) + P dV

Using u = T 4 , du = 4T 3 dT
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4 4
T dV + 3T 3 V dT


dQ = (7)
3

(b) For a reversible process the entropy is defined as dS = dQ/T . Obtain an expres- [2]
HB

sion for S(V, T ) for the radiation gas.

Solution:  
dQ 4 3
= dS = d T V (8)
T 3
At T = 0, S = 0. Hence
4 3
S= T V (9)
3

(c) Imagine the universe to be a perfectly spherical adiabatic enclosure of radius r [3]
containing pure radiation. Rate of expansion of the universe is governed by a
law
dr
= r
dt
where is a constant. At t = 0, the universe has size r0 and temperature T0 .
Determine its temperature T (t) for t > 0.

Solution: As universe expands adiabatically, dQ = 0. Equation (7) gives

dV dT
+3 =0
V T

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 INPhO - 2016 - Final Solutions

Using V = 4r3 /3

dr dT
+ = 0 (10)
r T
rT = r0 T0 = constant

Using rate of expansion law

r = r0 et (11)
T = T0 et (12)

(d) Here is 72 kms1 Mpc1 where 1 Mpc= 3.26106 light years. State the [1]
dimensions of . Obtain its numerical value in SI units as per the dimensions
stated.

Solution:
[] = T1
= 2.33 1018 s1

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5. This question is about a closed electrical black box with three
terminals A, B, and C as shown. It is known that the elec-
trical elements connecting the points A, B, C inside the box A
are resistances (if any) in delta formation. A student is pro-
vided a variable power supply, an ammeter and a voltmeter. B C
Schematic symbols for these elements are given in part (a). She Black Box
HB

is allowed to connect these elements externally between only

two of the terminals (AB or BC or CA) at a time to form a
suitable circuit.
(a) Draw a suitable circuit using the above elements to measure voltage across the [11/2]
terminals A and B and the current drawn from power supply as per Ohms law.

Solution:
A Ammeter

V Voltmeter
A
Variable Power supply A
V

B C
Black Box

(b) She obtains the following readings in volt and milliampere for the three possible [6]
connections to the black box.

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 INPhO - 2016 - Final Solutions
AB BC AC
V (V) I (mA) V (V) I (mA) V (V) I (mA)
0.53 0.54 0.83 0.17 0.85 0.15
0.77 0.77 1.65 0.35 1.70 0.30
1.02 1.01 2.47 0.53 2.55 0.45
1.49 1.51 3.29 0.71 3.4 0.60
1.98 2.02 4.11 0.89 4.25 0.75
2.49 2.51 4.94 1.06 5.10 0.90
In each case plot V (on Y -axis)- I (on X-axis) on the graph papers provided.
Preferably use a pencil to plot. Calculate the values of resistances from the plots.
Show your calculations below for each plot clearly indicating graph number.

Solution: See graphs for the calculations of slopes. RAB = 0.98 k , RBC
=4.60 k, RCA =5.67 k.

(c) From your calculations above draw the arrangement of resistances inside the box [11/2]
indicating their values.

Solution:

6A
C1 0.
kW
98

B 4.60 kW C
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 INPhO - 2016 - Final Solutions

Plot for AB
2.7

2.6

2.5

2.4

2.3 (2.31,2.28)

2.2

2.1

1.9

1.8

1.7

1.6
V (V)

1.5

1.4

1.3

1.2

1.1

0.9

0.8
Slope = 0.98 k
(0.74,0.73)
0.7

0.6

0.5

0.4
0.4 0.6 0.8 1 1.2 1.4 1.6 1.8 2 2.2 2.4 2.6
I (mA)
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 INPhO - 2016 - Final Solutions

Plot for BC
5.2

4.8

4.6
(0.96,4.5)
4.4

4.2

3.8

3.6

3.4

3.2
V (V)

2.8

2.6

2.4

2.2

1.8

1.6
(0.26,1.23)
1.4
Slope = 4.6 k
1.2

0.8

0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1 1.1
I (mA)
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 INPhO - 2016 - Final Solutions

Plot for AC
5.2

4.8
(0.82,4.67)
4.6

4.4

4.2

3.8

3.6

3.4

3.2
V (V)

2.8

2.6

2.4

2.2

1.8

1.6

1.4
Slope = 5.67 k
(0.22,1.27)
1.2

0.8

0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9

I (mA)
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 INPhO - 2016 - Final Solutions
6. The Yukawa Potential and the nucleus
In 1935 the Japanese physicist Hideki Yukawa proposed that the strong attractive
central potential binds the proton and the neutron with the associated potential
energy U (r) where
er/
U (r) = g 2
r
where = ~/m c, with m , being the pion mass = 138.00 MeV/c2 and r is the
distance between nucleons. Here g 2 is the nuclear force constant. Our task is to
determine the numerical value of nuclear force constant g 2 . For simplicity we assume
that the proton and the neutron in the deuteron have equal masses, mp = mn = m =
938.00 MeV/c2 . Here subscript n,p refers to neutron and proton respectively (see
figure below). They are in circular motion under the influence of U (r) about their
centre of mass (CM) and their ground state binding energy (Eb ) is 2.22 MeV.
vp
This two mass problem can be reduced to
that of a single mass namely effective mass
CM
= mp /2 in CM frame with velocity v = n x p
vn vp where |vn | = |vp |. The total angular r/2 r/2
momentum of the n-p pair is quantized as
per the Bohr quantization formula.

6 vn
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In what follows, express numerical values of energy in MeV (1 MeV = 1.61013 J)
and length in fm (1 fm= 1015 m), mass in MeV/c2 and related quantities in terms
of these.
(a) Calculate . [1/2]
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Solution: = 1.431 fm.

(b) State the total energy E of the deuteron in terms of {, r} and relevant quantities. [1]

v 2 g 2 er/
Solution: E =
2 r

(c) Relate the centripetal force on to U (r). [11/2]

Solution:
mn vn2 d
= U (r)
rn dr
2 2 r/
 
v g e 1 1
= +
r r r

Answers written in terms of protons mass and velocity are also

accepted.

(d) State the magnitude of the total angular momentum L about the CM. [1]

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 INPhO - 2016 - Final Solutions

Solution:
L = mp vp rp + mn vn rn = vr or n~

(e) Obtain the expression for nth energy level (En ) of deuteron in terms of g 2 and [2]
rn , where rn is the radius of corresponding circular orbit.

Solution: Bohr quantization condition gives

n~
v=
r

Replace v 2 in part (b) to obtain

n2 ~2 1 g 2 exn
En = where xn = rn /
22 x2n xn

Answer can be expressed in various ways. For example in terms of

L or g 2 can also be eliminated.

6
(f) Consider the ground state of the deuteron (n = 1). Define x1 = r1 /. Here
r1 is the radius of first orbit of deuteron. Obtain a polynomial equation of x1
[2]
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involving only fundamental constants and Eb .

Solution: Using Bohr quantization condition in part (d)

g 2 2
2 ~2
(xn + xn )exn = 1 (13)
n
HB

2 xn
 2 
2 ~ 1 2g e
En = n + n (14)
22 x2n xn

Result of part (e) can be used to eliminate exn .

 2 
2 ~ 1 (1 xn )
En = n (15)
2 x2n (1 + xn )
2

For n = 1, E1 = Eb . Thus

~2
 
1 (1 x1 )
Eb = (16)
22 x21 (1 + x1 )
x31 + x21 + x1 = 0 (17)

where = ~2 /22 Eb
Answer written in different form with correct orders of polynomials
are accepted.

(g) Estimate x1 numerically. Also calculate the radius of first orbit i.e. r1 . [2]

Solution: Using = 9.1, x1 = 0.85 which gives r1 = 1.22 fm.

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 INPhO - 2016 - Final Solutions
(h) Obtain the nuclear force constant g 2 numerically in units of MeVfm. [2]

Solution: g 2 = 86.2 MeV fm

**** END OF THE QUESTION PAPER ****

6
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