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GRAMOLY
FIZIKA Round-2
September 2021
The scoring criteria is as follows: There are no negatives for integer valued problems:
In case of Multiple choice problems, partials will be awarded as follows Marking Scheme: Full
marks, +4 if all the correct options are marked. Partial marks , +1 for each correct option provided
NO incorrect option is marked. Zero marks, if no option is marked. Negative marks, −1
The raw score of the problem is scaled based on its difficulty as follows: In case of integer valued
problems:
Marks = max (8.5 − ln( N ), 4)
Where N is the number of people who got the question correct
In case of Multiple choice questions:
8.5 · A
Marks = max
− ln( N ) , A
4
Where N is the number of people who got the question correct A is the the aggregate marks the
person has scored in the question after taking partials into account.
Selection criteria for round 2:
After every participant has been scored the following criteria will be used for selection:
Final Marks =
Your marks in round 1
Your marks in round 2
· 10 +
· 90
Average Marks scored in round 1
Average Marks scored in round 2
The top 50 from the final mark list will be selected for round 3.
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September 2021
Problem 1
A solid paraboloid of base radius R0 and height h0 is having uniform volume mass density ρs
is inverted and placed just above the surface as shown, of a liquid having volume mass density
ρ L . When paraboloid is released from rest it has been observed that when it becomes completely
ρ
submerged in the liquid the paraboloid comes to rest again. What is the ratio of ρLs for such a
paraboloid. There exist a uniform gravity of g. Assume that the liquid body is very large so that
liquid level is not changing as the paraboloid enters into liquid.
R0
ρs
h0
ρL
-Proposed by Nitin Sachan
Answer: 3
Net Force on the paraboloid:
mg − ρ L · 21 πx2 y · g = m · dv
dt
v·
dv
dy
=
dv
dt
= g−
ρL
ρS
·
x2
R2
·
yg
H
=⇒ g −
ρL
ρS
·
yR2
H · R2
· ygH
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FIZIKA Round-2
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September 2021
y
R
H
x
g·ρ
L
2
v · dv
dy = g − H 2 ρS · y
Integrating this using vi = v f = 0 and putting appropriate limits we get:
R0
RH
RH 2
gρ L
v
·
dv
=
g
·
dy
−
·
y dy
2
0
0
0
H ρ
S
Fu p
x
y
a
mg
g ρL
3 ρS
0 = gH −
·H
ρL
Hence,
=3
ρS
Also an energy conservation approach can be applied to shorten the solution.
Solution 1
Problem 2
You have a spectrometer of spectral resolving power 5 · 104 . Say n1 is the maximum value of principal quantum number for which the resulting energy spectra can be distinguished clearly from its
neighbours. Also n0 denotes the minimum value of principal quantum number for which a human
eye can see the atomic transition. Find the value of n1 − n0 . We are only working in the Balmer
transitions of hydrogen atom.
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Value of Rydberg constant is R = 1 · 107 and maximum wavelength eye can see is 700nm
Proposed by Abhiram Cherukupalli
Answer: 69
1
1
1
= R ( − 2 ), n ≥ 3
λ
4 n
dλ
∆λ
≈
∆n
dn
Note that though this is an approximate method, it can be verified by desmos, and the answers are
close.
Now using this logic we can differentiate () to get:
dλ
2R
∆λ
= − 3 λ2 ≈
dn
n
∆n
The beautiful step here is to take ∆n = 1, as it has to be distinguished from its neighbours.
This gives us:
∆λ
2R
≈ 3λ
λ
n1
The spectral power
λ
n 3 1
1
= 1 ( − 2 ) < 5 · 104
∆λ
2 4 n
=⇒ n1 < 73.69 =⇒ n1 = 73
Calculation of n0 is straight-forward:r
n0 =
4Rλ
= 3.055
Rλ − 4
Hence n0 = 4 and n1 − n0 = 69
Note that we have to use the value of R mentioned in the question not the exact value also note that
4Rλ
though the human eye data is close the value is very easy to calculate by hand as we get Rλ
−4 > 9
without need of much calculation as Rλ = 7.
Solution 2
Problem 3
We have a loop of radii 80 cm with 2 identical beads of mass 9 kg and radius 6.97 cm inserted in it,
free to move without friction. The loop starts rotating along the diameter with an angular velocity
θ1 + 3θ2
ω = 5 rad/s. If the angle lower bead makes θ1 and upper bead makes θ2 then report
. You
50
may assume that both beads are displaced to the same side of loop.
Use arcsin
6.97
80
≈ 5°,cos 5° ≈ 0.9962, cos 10° ≈ 0.9848, arccos(0.5058) ≈ 60° and arccos(0.2529) ≈
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75°
ω
g
θ2
θ1
-Proposed by Atharva Mahajan
Answer: 5
R sin
θ1
θ2 − θ1
2
=r
2
θ2
7
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September 2021
N
mω 2 R sin θ2
mω 2 R sin θ1
For Ball 1:
θ2 − θ1
2
θ1
N
θ1
mω 2 R sin θ1
mg
N cos
θ1 − θ2
= mω 2 R sin θ1 cos θ1 − mg sin θ1
2
For Ball 2:
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September 2021
θ2 − θ1
2
θ2
mg
N cos
θ2
N
mω 2 R sin θ2
θ1 − θ2
= mg sin θ2 − mω 2 R sin θ2 cos θ2
2
mω 2 R sin θ1 cos θ1 − mg sin θ1 = mg sin θ2 − mω 2 R sin θ2 cos θ2
After solving, we get θ1 = 55° and θ2 = 65°
θ1 + 3θ2
=5
50
Note that we can also directly find the position of centre of mass of the two
balls to find the angles easily.
So,
Solution 3
Problem 4
The bouncing container
There is a container containing two circular discs (of radii 0.5 mm and 1 mm respectively) which are
connected by two light rigid rods of negligible area. The space between them is filled by a unknown
ideal liquid, This container is dropped vertically on a table from a height 20cm, with which it inelastically collides. Assume that the container loses all its kinetic energy into rest. It remains at rest
for a time t seconds and then it amazingly shoots back up to the same height
Report value of 10 · t in seconds (round off to nearest integer).
Neglect the hydro-static pressure of the liquid, assume pressure everywhere outside the space between the two discs to be ambient pressure 105 Pa. Neglect all types of frictional forces i.e the discs
are free to move up and down. The mass of the disc liquid system is 15 g. Assume the container
is large enough so that the bottom piston don’t fall out and also the top piston does not reach the
conical middle. Assume the conical Portion of the container is small
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0.5mm
Large enough
1mm
Large enough
-Proposed by Abhiram Cherukupalli
Answer: 7
The assumption here was to take conical section was of negligible height.
When the impact happens the container comes to rest, but the pistons continue moving, since upper
piston has lower area the pistons moving down will cause the volume between them to increase. As
the liquid is in compressible, a vacuum will form between the pistons and the force due to pressure
difference will provide the necessary impulse for bouncing. Say ∆A is the difference in area of the
pistons and Mtotal is the total mass of the system then:
The upward force acting on the water + disc system is:F = ∆AP0 − Mtotal g
Impulse equation
F · t = 2Mtotal
p
2gh
p
2Mtotal 2gh
t=
≈ 0.7
∆AP0 − Mtotal g
Solution 4
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September 2021
Problem 5
There is a superconducting ring,made up of perfectly conducting wire of mean radius r = 2m and
cross-sectional area A and of mass 10kg which floats in a gravity free space in a region of uniform
magnetic field parallel to its induction vector ~B = 2T. Inductance L of the ring is so small that inertia
of free electrons cannot be neglected in the current building process. The free electron density in the
conductor is n, mass of an electron is me and modulus of charge on an electron is e.He found that if
the ring is turned slightly about a diameter and released, it will execute simple harmonic motion.If
calculated time period is in form of
r e
a L + bπm
ne2 A
c
Find the value of a + b − c.
~B
~B
~B
~B
~B
~B
~B
~B
-Proposed by AKIII
Answer: 7
Concept: Given Inductance L of super conducting ring is so small that inductance due to free
electrons cannot be ignored.
B
We’ll use the concept of Kinetic inductance.
Here, L0 is the kinetic inductance.
Kinetic energy of electron = 21 L0 i2 =⇒ 12 Mv2e = 21 L0 i2 Now, we know, i = neAvd =⇒ vd =
Also, by definition, M = nme A · 2πR
Hence, from our previous equation, we can write
i
neA
1
i2
1
2πRme
nme A · 2πR 2 2 2 = L0 i2 =⇒ L0 =
2
n e A
2
ne2 A
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September 2021
So, Lnet = L + L0 Now the ring is rotated by a small angle, we can calculate the torque τ due to the
magnetic field :
τ = πR2 · iB cos θ =
B2 sin θ cos θ (πR2 )2
Lnet
θ
Since its a super conducting ring, flux is constant thus, Lnet i = B sin θπR2
Hence for small angle θ, S.H.M. :
s
T = 2π
mR2
2 Lnet
B2 (πR2 )2
2
=
BR
r
mLnet
=
2
√
2mLnet
BR
Solution 5
Problem 6
A square loop of negligible resistance, side length 10 cm, mass 0.9g, total self inductance 4 · 10−8 H
is left in a vertical plane where where its bottom end touches the interface of two different uniform
horizontal magnetic fields (Bupper > Blower ) where Bupper = 1.5mT k̂ and Blower = 0.6mT k̂. There is
also a drag force, f = −kv (k = 1.8 · 10−6 ) acting on it. This square loop happens to undergo a
damped simple harmonic motion, mark the correct options:
A The damping coefficient of the resulting oscillations γ is 1.5 · 10−3 Ns/m
B The angular frequency omega of the oscillations ω is 10 rad/s
C The maximum magnetic field of the lower region for SHM to ensue is9 · 10−4 T
0.0015t
D The vertical position z(t) ≈ 0.1 1 − ecos φ (cos (5t − φ)) , φ = arctan √ γ
ω 2 − γ2
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September 2021
10 cm
Released
f rom rest
g
10 cm
1.5 m T
0.6 m T
Inter f ace
Uni f orm
-Proposed by Abhiram Cherukupalli
Answer: C
10 − x
f = −k ẋ
B1
B2
mg
x
oscillates
Net flux passing through the loop:
Φ = B1 a2 + ( B1 − B2 ) ax + LI
dφ
= ( B1 − B2 ) a ẋ − L İ = 0
dt
=⇒ mg − m ẋ − ( B1 − B2 ) aI = m ẍ
=⇒ I · R = −
Differentiating and solving we get:
ẍ + 2 ·
1η
( B1 − B2 )2 a2
ẋ +
x=g
2m
mL
1η
= 10−3
2m
r
( B1 − B2 )2 a2
Angular frequency of oscillation ω =
= 15rad/s
mL
Damping coefficient γ =
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September 2021
e0.001t
Vertical position as a function of time z(t) = 4.44 1 −
(cos(15t − φ)) cm
cos φ
where φ = arctan( √ γ
)
ω 2 − γ2
For SHM to ensue:
g
< a =⇒ | B1 − B2 | > 0.6mT =⇒ B2 < 0.9mT or B2 > 2.1mT
ω2
But since Bupper > Blower we don’t consider the second possibility as it is the best possible option.
Solution 6
Problem 7
In this problem, we will be analysing the formation of image of ring around galaxies using Corpuscluar Theory and Universal Law of Gravitation. (Note: Do not consider any aspects of Relativity
while solving this problem, i.e, solve it using only Classical Mechanics)
Consider light as a collection of particles of mass m moving with a very high speed c (c2 >> GM
b ,
where b is the impact parameter). These particles will deviate when they come under the interaction
of a star as shown in figure. Consider a simple setup with one star acting as a lens. Let the distance
of object and observer be v and u respectively. Mark the correct statements:
Image
R
b
α
θ
u
v
M
Observer
A) The radius of ring observed will be
q
2GMv(u+v)
c2 u
B) The maximum radius of star which can create such a image is
q
4GMvu
c2 ( u + v )
C) For v >> u, the value of angle subtended by ring at observer will be
D) The value of α + θ is
Object
θ, α 1
q
4GM
uc2
2GM
bc2
-Proposed by Prerak
Answers: (A), (D)
First, let’s find the deviation angle for impact parameter "b"
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September 2021
x
v x = c (almost constant)
approximately constant b
F=
GMm
b2 + x 2
M
Net impulse in y-direction, can be calculated as:
Z ∞
−∞
√
GMm
dt
x 2 + b2 x 2 + b2
b
Here, x = ct
=⇒ Net impulse in y-direction =
Z ∞
GMmb
2GMm
=
2
3/2
bc
+b )
− ∞ ( c2 t2
2GM
bc2
So angle of deviation ≈
c
δ
2GM
bc
Now,
δ
R
b
θ
v
u
α
α, β 1
By exterior angle property, it’s clear that
δ = α+θ
=⇒
2GM uv
2GM
b
b
= + =⇒ b2 = 2
= θ 2 u2
2
bc
u v
c u+v
Hence, we can write θ as
s
θ=
2GMv
+ v)
c2 u ( u
For calculating R, we can use basic trigonometry.
R = (u + v) tan θ ≈ θ (v + u)
On substituting the value of θ obtained earlier
r
R ≈ θ (v + u) =
2GMv(u + v)
uc2
Solution 7
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September 2021
Problem 8
A ball of ( R = 10m) from height h = 100m on a surface having coefficient of kinetic friction as
µk = 0.5. After the first collision with the surface, it rebounds (coefficient of restitution = 0.5) and
also moves a maximal horizontal distance of x = 1.29m. Assuming the ball to be a uniform solid
sphere, choose the correct options:
(A) It’ll slip on the surface for the complete duration of impact.
(B) It won’t slip for complete duration of impact.
(C) Maximum angular velocity being given is 0.1 rad/s.
(D) Maximum angular velocity being given is 0.2 rad/s.
-Proposed by Atharv Shivram Mahajan
Answer: BC
ω0
R
h
eV
N
v0
v
Z
Ndt = mv(1 + e)
Now first lets say it slips for all the time during collision.
Z
µ
=⇒ Time of flight before 2nd collision =
mu
R
Ndt = mv0
2eV
g
Ndt
=⇒ Range = x =
. 2eV
m
g
On substituting with values and simplifying, we get: x = 100m but x = 1.29m is given.
Hence, it can’t slip for all the time.
Z t
v0
Ndt = I ω0 −
∴ µ.R.
R
0
andµ
Z t
N.dt = mv0
0
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Using these two equations, we get:
I
R=
mv0
So,
ω0 = v
0
v0 =
1
mR2
+
IR
R
ω0 =
Also:
v0
ω0 −
R
September 2021
7v0
2R
xg
1.29 × g
= p
2eV
2gh
Using this we get:
ω = 0.1rad/s
Solution 8
Problem 9
Photon gas on a carnot engine ride
T
V
A container with a piston shown in the figure, has a photon gas characterised by state equations of
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September 2021
Internal energy U ( T, V ) and pressure P( T ) which are
U ( T, V ) = bVT 4
1 4
bT
3
where b is a dimensioned constant, V is the volume of system, T is its temperature.
Let’s take the gas through a carnot cycle, starting with V = 0 and temperature of the enclosure TH
and sequentially take it through
Process 1 : Isothermal expansion at temperature TH to a volume V1 .
Process 2 : Adiabatic expansion to a volume V2 .
Process 3 : Isothermal compression to zero volume at temperature Tc .
Process 4 : Heating back the enclosure from Tc to TH at zero volume.
P( T ) =
Mark the correct option(s):
(a) The enclosure acts as perfect reflector for photons during process-2.
(b) The change in entropy of the photon gas during process 3 is − 31 bTc3 V2 .
(c) The process equation for process-2 is PV 4/3 = constant.
(d) Ratio of number of photons in the enclosure at same volume V21 but during process-1 to process-3
is
4
TH
.
TC4
-Proposed by Janardanudu Thallaparthi
Answer: AC
Process 1-3 is isothermal
U = bVT 4 → dU = bT 4 dV = 3PdV
b
P = T 4 → dQ = dU + dW = 4PdV
3
So, W
=
1
4
3 bT ∆V
and Q
=
4
4
3 bT ∆V
Also, S
=
4
3
3 bVT
so option B is wrong.
4
For process 2, which is adiabatic “S” should not change → VT 3 = constant → PV 3 =constant, also
number of photons remain constant as walls cannot emit or absorb photons during adiabatic process.
So walls are considered perfect reflectors in process 2
Also, in adiabatic process, Number of photons N and entropy S are constant
→ N = αS, WhereN is a constant
4
3
T13
N1
S1
3 bV1 T1
=
= 4
=
as same volume
→
3
N2
S2
T23
3 bV2 T2
Solution 9
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September 2021
Problem 10
A sphere of radius r1 = 1m is placed concentrically in a thin shell of radius r2 = 3m with emmisivities
e1 and e2 respectively. The inner sphere generates a constant power P. Assume that none of the
surfaces allow transmission. Also assume that space between sphere and outer shell is a vacuum.T
is steady state temperature of inner sphere.Then Choose the correct option(s):
A.If e1 = 0.30 and e2 = 0.50, 11P = 12σπT 4
B.If e1 = 0.30 and e2 = 0.50, 91P = 3σπT 4
C.If e1 = 0.20 and e2 = 0.40,
49P = 36σπT 4
D.If e1 = 0.20 and e2 = 0.40,
92P = 3σπT 4
-Proposed by AKIII
Answer: (A), (C)
Solution-1:
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September 2021
A2 T2 , e2
A1 T1 , e1
As seen from outside, net power P should be coming out. Let T be the steady state temperature of
the inner shell and T2 be the steady state temperature of the outer shell.
∴ σe2 A2 T24 = P
Let total power absorbed by the outer shell be = k.
For steady state of the outer shell:
k = 2σe2 A2 T24
k = 2P
If the outer shell absorbs power, then it must have received ek2 power.
The radiations that outer shell receives are from some of its own radiations that fall on itself, coming
from the inner sphere.
∴It has reflected ek2 (1 − e2 ) Power.
Out of this,
A1
k
e2 (1 − e2 ). A2 e1 is absorbed by the inner shell.
So, for the inner shell:
P
k
A
A
(1 − e2 ). 1 e1 + P 1 e1 = σA1 e1 T 4
e2
A2
A2
1
( PA
A2 e1 is the direct radiation from the outer shell. )
σA1 e1 T 4 = P
4
σA1 e1 T = P
A 1 2 − e2
e1
+P
A2
e2
A 1 2 − e2
e1
+1
A2
e2
Solution-2:
X=Power falling on inner sphere
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September 2021
Y=Power coming out of the inner sphere(radiated+reflected)
=⇒ Y − X = σ
Y = X (1 − e1 ) + e1 A1 σT 4
where:
X (1 − e1 ) is the reflected part and e1 A1 σT 4 is the radiated part.
Considering the equation of the outer shell:
A1
A1
A1
A1
+P
1−
( 1 − e2 )
P
· · · + P = Ye2 + Y (1 − e2 ) 1 −
e2 + · · ·
A2
A2
A2
A2
A1
A1
Ye2 = P 2
+ e2 −
e2
A2
A2
A 1 2 − e1
+1
Y=P
A2
e2
Also,
Ye1 + P(1 − e1 ) = e1 A1 σT 4
So,
P
A1
A2
2 − e2
e2
4
e1 + e1 + 1 − e1
σA1 e1 T = P
= e1 A1 σT 4
A 1 2 − e2
e1
+1
A2
e2
∴ Energy flow from the inner sphere to the outer sphere is not considered in X and Y, we must
account it separately in ().
Solution 10
Problem 11
Consider the region y ≥ 0 in the cartesian plane filled by a certain material whose refractive index is
a function of the vertical distance (y) from the x axis.
n(y) = ay
( y > 0)
Where n(0) = n0 and a is a constant with dimension [ L]−1 A ray of light is incident to this region at
the origin making an angle θ0 with the vertical. If
y( x ) = K
n0 sin θ0 [exp{(2ax )/(n0 sin θ0 )} + 1]
a exp[( ax )/(n0 sin θ0 )]
Report the value of (2K + 1)(2K+1) as the answer. If the setup is impossible, enter 0
-Proposed by Muhammed Yaseen Nivas
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Answer: 0
As we can see n(0) = n0 , but n(0+ ) → 0. So, Total internal refraction occurs and the setup is
impossible, hence answer is 0.
But in case we assume the ray passes through to the second medium, this is how we solve it:
The angle θ, that the tangent to the graph of a function y( x ) makes with the y axis clearly satisfies
tan θ =
1
y0 ( x )
Trigonometric manipulations leads us to
sin θ = p
1
1 + ( y 0 )2
Now, coming to physics; to solve this problem, we use Snell’s law to write that
n(y) sin θ = c
Where c is determined by the initial conditions n(0) = n0 , θ (y) = θ0 . Plugging them in,
r
dy
a2 y2 − c2
ay
p
=
= c ⇐⇒
2
0
2
c
dx
1 + (y )
Z x
adx
0
c
=
Z y
0
dy
p
y2
− c2 /a2
p
y2 − c2 /a2 + y
ax
1
= ln p
c
2
y2 − c2 /a2 − y
Two cases, either the argument of the logarithm is positive, in which case we can drop the absolute
values. We will consider this first.
p
y2 − c2 /a2 + y
2ax
= p
exp
c
y2 − c2 /a2 − y
q
q
2ax
exp
( y2 − c2 /a2 − y) = y2 − c2 /a2 + y
c
q
( y2 − c2 /a2 )(exp(2ax/c) − 1) = y(exp(2ax/c) + 1)
y2 − c2 /a2 = y2 α2
Where,
α( x ) ≡
exp(2ax/c) + 1
exp(2ax/c) − 1
y2 (1 − α2 ) = c2 /a2
r
c
1
y=
a 1 − α2
n0 sin θ0
y( x ) =
a
r
1
1 − α2
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September 2021
This function is undefined, so we have to consider the other case.
p
y2 − c2 /a2 + y
2ax
exp
= p
c
− y2 − c2 /a2 + y
q
q
2ax
2
2
2
exp
(y − y − c /a ) = y2 − c2 /a2 + y
c
q
2ax
2ax
2
2
2
y(exp
− 1) = y − c /a (exp
+ 1)
c
c
y2 β2 = y2 − c2 /a2
y2 (1 − β2 ) = c2 /a2
Where β =
exp(2ax/c)−1
.
exp(2ax/c+1)
v
u
cu
1
y= t
a 1 − ( exp(2ax/c)−1 )2
exp(2ax/c)+1
c
y=
a
s
(exp(2ax/c) + 1)2
(exp(2ax/c) + 1)2 − (exp(2ax/c) − 1)2
s
c (exp(2ax/c) + 1)2
y=
a
4 exp(2ax/c)
y=
n0 sin θ0 (exp(2ax/n0 sin θ0 ) + 1)
2a exp( ax/n0 sin θ0 )
So in case we did assume ray to pass through this would be the trajectory equation.
Solution 11
Problem 12
A cube of side 2a and specific gravity s floats in a calm water. "Water line section" is the cross-section
of cube intersecting free surface of water.
If the cube is given a small angular displacement about centre of water line section and released then,
(a) Volume of cube submerged is increased due to small angular displacement.
(b) Volume of the cube submerged remains same even after small angular displacement.
(c) For the stability of rotational equilibrium of cube, the condition required is 4s2 − 4s + 1 > 0
(d) For the stability of rotational equilibrium of cube, the condition required is 6s2 − 6s + 1 > 0
-Proposed by Janardanudu Thallaparthi
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FIZIKA Round-2
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Answer: BD
Volume submerged remains same for any arbitrary shaped water line cross section. Top view of this
section is shown here:
G1
G
G2
x1
x2
For small angular displacements, G1 G moves up by x1 , θ and GG2 moves down by x2 θ, For equilibrium initially A1 x1 = A2 x2 where A1 and A2 are areas left and right of dashed horizontal line. So,
A1 x1 θ = A2 x2 θ represents equal volumes by Pappus’ theorem. Hence, volume submerged remains
unchanged. So, buoyant forces remains unchanged.
Let’s first establish an important point.
Side View
C
O
H
x-axis
H’
H and H 0 are centeres of buoyancy before and after the deflection. When a vertical line is drawn
through H 0 and it meets the line HC at M, this point ’M’ is called ’Meta center’. For stability (using
torque equation), we need HM > HC i.e. Meta center should lie above C, the center of mass.
Volume element = (δA)(δx ), its depth upon small rotation = xθ (see side-view). The vertical
displacement of center of buoyancy is of order θ 2 and to first order of θ, neglected.
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FIZIKA Round-2
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Consider Volume moments about y-axis. Let x̄ 0 and x̄ be final and initial x-positions of center of
buoyancy. We have,
V x̄ 0
= V x̄ −
Z A0
( xθ )dAx +
0
Z B0
xθ (dA) x
0
Z B0
V x̄ 0 − x̄ = θ
x2 dA
0
| {z }
A
(related to radius of gyration)
HH 0
V ( HH 0 ) = θ ( AK2 )
Top View
δA
A’
B’
x-axis
y-axis
2
2
AK
V ( HM sin θ ) =
As sin θ ≈ θ, we get HM = AK
V . For stability we get V > HC
2
So, for cube of side 2a, relative density s A = 4a2 , K2 = a3 , V = 8a3 s HC = a(1 − s), So for a stability,
θ ( AK2 )
a
> a (1 − s )
6s
. Unstable if 6s2 − 6s + 1 < 0
Solution 12
Problem 13
A particle of mass m, subjected to a central force F describes the following trajectory : r2 = a cos 2φ,
where a is a positive constant, r is distance of point from centre of force. At initial moment, r = r0 ,
speed v = V0 and makes an angle α with straight line connecting the point with centre of force!
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FIZIKA Round-2
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September 2021
α
O
F
φ
√
M
a
If the expression for force is given as
F=−
KmV02 r02 a2 sin2 α
r7
Find the value of K
-Proposed by Harshit Gupta
Answer: 3
The magnitude of central force:
F (r ) = m(r̈ − r θ̇ 2 )
Conservation of angular momentum gives us:
r2 θ̇ = constant = L
r can be converted into terms of u =
1
r
du
d 1 dt
ṙ
ṙ
=
=− 2 =−
dθ
dt r dθ
L
r θ̇
2
r̈
d u
1 dṙ dt
r̈
=−
=−
=− 2 2
dθ 2
L dt dθ
L
u
Lθ̇
Solving all the 3 equations we get the famous Binet Equation:
2 L2
d
1
1
F (r ) = − 2
+
mr
dφ2 r
r
p
Now substituting r =
a cos 2φ, L = mvo ro sin α and differentiating wrt to φ we get:
2
2
3ma ro vo sin α
3ma2 ro2 v2o sin2 α
F=−
hence k = 3
=
−
r3 cos2 φ
r7
Solution 13
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GRAMOLY
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September 2021
Problem 14
We have a tesseract cubic style arrangement with voltmeters and resistances as shown in the figure.
There are identical voltmeters to each side of smaller cube and side connecting to larger cube as
shown in fig. below. All sides have equal resistance R, and across 2 ends of bigger cube connected
with a battery of 300V. Calculate
7 n
Vi ,
99 i∑
=0
where Vi is reading of ith voltmeter
-Proposed by AKIII
Answer: 40
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FIZIKA Round-2
GRAMOLY
September 2021
0V
120V
V17
V19
V18
c
V13
d
V20
120V
P7
P8
V15
V16
180V
P5
V8
c
b
V12
V9
V14
V10
b
P4
180V
c
P3
V6
V7
P1
V11
P6
120V
V4
P2
a
V5
b
V2
V1
V3
V3
300V
180V
Applying KCL at P1:
a−300
R
+3
a−b
R
=0
⇒ 4a = 300 + 3b
Applying KCL at P7:
d
R
+3
d−c
R
=0
⇒ 4d = 3c
Applying KCL at P2:
b−180
R
+2
b−c
R
+
b− a
R
=0
⇒ 4b = a + 2c + 180
Applying KCL at P3:
c−d
R
+2
c−b
R
+
c−120
R
=0
⇒ 4c = d + 2b + 120
→
13
4 c
= 2b + 120
...from (2)
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September 2021
So,
26b = 16c + 2090
26c = 16b + 960
By solving,
a=
1380
7 V,
b=
1140
7 V,
c=
960
7 V,
d=
720
7 V
→ V1 = 300 − a, V2 = V7 = V9 = a − b, V3 = V8 = V13 = b − 180
→ V4 = V10 = V6 = V12 = V14 = V2 = b − c
→ V5 = V15 = V19 = c − 120
→ V11 = V16 = V18 = c − d, V17 = d
⇒
7
7 20
Vi =
(300 − a + 3a − 3b + 3b − 540 + 6b − 6c + 3c − 360 + 3c − 3d + d)
99 i∑
99
=1
7 3960
=
= 40
99
7
Solution 14
Problem 15
Power radiated by a point charge:
P = kµ0d qe ζ f c g
where k is a dimensionless constant. Here, µ0 is the permeability of free space, q is the charge, ζ is
the acceleration, and c is the speed of light in vacuum.
Consider a fixed positive charge Q situated at fixed point in space (vacuum). At some instant, a Li+3
ion (stable enough throughout the problem) is fired from infinity directly towards the charge Q with
a velocity v0 . The ion makes a shortest distance of approach to the fixed charge x0 ( x0 > 0). Atharva
(using his supernatural powers) extracts energy radiated away by Li+3 ion during its whole journey
and with that energy, excites an H atom from ground state to third excited state. Ignore relativistic
effects (i.e. v0 << c), and radiative losses on the motion of the particle. Neglect all kind of particle
physics interactions.
mv5
If Q 0 = ψ4 × 1025 , where m is the mass of ion, report [ψ] where [.] is the greatest integer function.
1
Take k = 6π
in the expression of power radiated by a point charge.
You may find this integral helpful :
Z ∞
ξ
x4
1
q
1
ξ
−
1
x
dx =
16
15ξ 5/2
-Proposed by Tarpan Ghosh
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FIZIKA Round-2
GRAMOLY
September 2021
Answer: 2
First, we calculate the exact expression for power radiated by a point when accelerated. Using dimensional analysis, it can be shown:
µ0 q2 ζ 2
P=k
c
This implies, the rate of work done is
µ0 q2 ζ 2
dW
=k
dt
c
Let the charge of Li+3 be q and using Coulomb’s law,
ζ=
Ω
Qq
= 2
2
4πe0 mx
x
Qq
where Ω = 4πe0 m
Therefore, the total energy radiated is (since total energy emitted is double of the energy emitted in
a single cycle)
Z
kΩ2 µ0 q2 ∞ 1
W = 2·
dx
4
c
x0 v · x
Now applying conservation of energy (no radiative losses as given in question)
v2 = v20 −
x0 =
2Ω
v20
2Ω
x
Therefore,
W = 2·
kΩ2 µ0 q2
c
Z ∞
x0
kΩ2 µ0 q2
1
dx = 2 ·
4
c
v·x
Z ∞
x0
1
x4
q
v0 −
2Ω
x
dx =
2kΩ2 µ0 q2
√
c 2Ω
Z ∞
x0
1
x4
q
1
x0
−
1
x
dx
Using the integral given in the problem, and further simplifications, we get total energy dissipated
E=
8qm 5
v
45c3 Q 0
Since Atharva is using this energy to excite the ground state single electron in H atom, we can write
8em 5
1
1
v = 13.6Ze
− 2
15c3 Q 0
n21
n2
Substituting all the values, we
Hence, [ψ] = 2 mv50
Q
= 64.5468 × 1025 . By comparing, we get ψ = 2.834449208.
Solution 15
Problem 16
A uniformly charged ring of radius Ro and total charge on the ring is Qo . There is a point P by a very
small distance of x ( x << Ro ) from the centre C in the plane of the ring as shown , The magnitude of
electric field at point P due to the stationary ring is Eo . Now the ring is rotated with constant angular
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FIZIKA Round-2
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September 2021
velocity of wo about an axis passing through its centre and perpendicular to the plane of the ring,
The magnitude of magnetic field at point P due to the rotating ring is Bo . Then the ratio of Eo /Bo at
point P is of the form
Eo
αxc2
=
Bo
ω [ βR2 + γx2 ]
Where α, β and γ are Integers and c is speed of light. Compute (α + β + γ).
C
C
P
R0
w0
P
R0
-Proposed by Nitin Sachan
Answer: 9
Calculation of the electric field:
+
Side view of the ring
+
+
+
EP
+
+
x
E=
E
KQ0 t
3/2
( R2 + t2 )
x
+
+
+
t
R0 t
+
x R0
+
t
+
EP
+
+
φ1
x
φC
t
t
2φ1 + φC = 0
φ1
2 · E · πx2 + EP · 2πx · 2t = 0
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FIZIKA Round-2
GRAMOLY
EP = −
September 2021
E·x
KQ0 t
x
= 2
·
2
3/2
2t
2t
(R + t )
EP =
KQ0 x
2R3
Direction towards the center of the ring.
Calculation of Magnetic Field:
I=
QW
2π
By+dy
By
dy
Br
By
y
x
By | axis =
P
µ0 IR2
2( R2 + y2 )3/2
Near the axis :- x R
To find Br : Using Gauss Theorem
By+dy · πx2 − By · πx2 + Br · 2πx · dy = 0
Br =
− x By+dy − By
− x dB
(
)=
·
2
dy
2 dy
Br =
− x µ0 IR2 d 2
·
· [ R + y2 ]−3/2
2
2
dy
Br = −
µ0 IR2 x −3
·
· 2y · [ R2 + y2 ]−5/2
4
2
3µ0 IR2 xy
Br =
4( R2 + y2 )5/3
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FIZIKA Round-2
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September 2021
h → 0 ⇒ Point A and B are very close to axis
B0 =
µ0 I
2R
Br B
A
B0
BP
x
O
P
Br = 0
Using Amperes Law :
Z A
|0
Z B
A
~ =
~ r · dl
B
Z x
0
~ +
~B · dl
{z }
Z B
A
~ +
~ r · dl
B
+ B0 h
Z P
|B
Z P
~ +
~ =0
~p · dl
~B · dl
B
0
{z } | {z }
−Bp h
0
3µ0 IR2 xh
3µ0 Ix2 h
3µ0 IR2 x2 h
3µ0 IR2 x2 h
=
·
dx
=
=
8R5
8R3
4( R2 + y2 )5/3
8( R2 + y2 )5/3
Rh
⇒ + B0 h − B p h +
B p = B0 +
Bp =
Bp =
3µ0 Ix2 h
=0
8R3
3µ0 Ix2
8R3
µ0 I
3x2
[1 + 2 ]
2R
4R
µ0 QW
3x2
µ0 QW
3x2
·
(1 + 2 ) =
(1 + 2 )
2R 2π
4R
4πR
4R
Ep =
kQ0 x
1
Q0 x
Q0 x
=
·
=
3
3
2R
4πe0 2R
8πe0 R3
Ep
2x2
=
Bp
w[4R2 + 3x2 ]
Solution 16
31