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JEE ADVANCED 2021 P1
PHYSICS SOLUTIONS
SECTION 1 (maximum marks: 12)
This section contains FOUR (4) questions
Each question has FOUR options. ONLY ONE of these four options is the correct answer
For each question, choose the option corresponding to the correct answer
Answer to each question will be evaluated according to the following marking scheme
Full marks: +3 if only the correct option is chosen
Zero marks: 0 if none of the options is chosen
Negative marks: −1 in all other cases
Kalyan’s Physics Challenge / JEE ADVANCED 2020 P1
1. The smallest division on the main scale of a Vernier calipers is 0.1 cm. Ten divisions of the Vernier scale correspond to nine divisions of the main scale. The figure below on the left shows the reading of this calipers with no gap between its two jaws. The figure on the right shows the reading with a solid sphere held between the jaws. The correct diameter of the sphere is
(A) 3.07 cm (B) 3.11 cm (C) 3.15 cm (D) 3.17 cm
Solution:
Smallest division on the main scale of Vernier Calipers: 0.1 cm
10 divisions on Vernier scale correspond to 9 divisions on main scale
Least count of the Vernier calipers: LC = 1 MSD – 1 VSD = 1 MSD –9
10MSD = 1 −
9
10MSD =
1
10x 0.1 = 0.01 cm
With no gap between the jaws, 6th division of Vernier scale coincides with one division on main scale
As 0 of Vernier scale is to the left of 0 of mains scale, the zero error is negative:
Zero error: − 10 − 6 𝐿𝐶 = −4 x 0.01 = −0.04 cm
Reading of the Vernier calipers: MSR + VC (LC) = 3.1 + 1 (0.01) = 3.1 + 0.01 = 3.11 cm
Correct diameter of the sphere: d = 3.11 + zero correction
d = 3.11 + 0.04 = 3.15 cm
Kalyan’s Physics Challenge / JEE ADVANCED 2021 P1
2. An ideal gas undergoes a four step cycle as shown in P – V diagram below. During this cycle, heat is absorbed by the gas in
(A) steps 1 and 2 (B) steps 1 and 3 (C) steps 1 and 4 (D) steps 2 and 4
Solution:
Step 1: isobaric expansion: dQ = dW + dU (work done and change in internal energy are positive)
Heat is supplied to the gas
Step 2: isochoric process: dW = 0; dU is negative as the temperature decreases: dQ is negative
Heat is rejected by the gas
Step 3: isobaric compression: work done is negative and internal energy decreases as the temperature decreases
Heat is rejected by the gas
Step 4: isochoric process: dW = 0; dU is positive as the temperature increases: dQ is positive
Heat is absorbed by the gas
Kalyan’s Physics Challenge / JEE ADVANCED 2021 P1
3. An extended object is placed at point O, 10 cm in front of a convex lens L1 and a concave lens L2 is placed 10 cm behind it, as shown in figure. The radii of curvature of all the curved surfaces in both the lenses are 20 cm. The refractive index of both lenses is 1.5. The total magnification of this lens system is
(A) 0.4 (B) 0.8 (C) 1.3 (D) 1.6
Solution:
Focal length of convex lens: 1
𝑓1= (n – 1)
2
𝑅= (1.5 – 1)
2
20= 0.05 → 𝑓1 = 20 cm
Focal length of concave lens: 1
𝑓2= (n – 1) −
2
𝑅= (1.5 – 1) −
2
20= −0.05 → 𝑓2 = −20 cm
For convex lens: 1
𝑣−
1
𝑢=
1
𝑓1→
1
𝑣−
1
−10=
1
20→
1
𝑣= −
1
20→ 𝑣 = −20 cm
Magnitude of magnification provided by convex lens: m1 = 𝑣
𝑢= 20
10= 2
For concave lens: 1
𝑣−
1
𝑢=
1
𝑓2→
1
𝑣−
1
−30=
1
−20→
1
𝑣= −
5
60= −
1
12→ 𝑣 = −12 cm
Magnitude of magnification provided by concave lens: m2 = 𝑣
𝑢= 12
30= 2
5
Total magnification of the lens system: m = m1m2 = 2 x 2
5= 4
5= 0.8
Kalyan’s Physics Challenge / JEE ADVANCED 2021 P1
4. A heavy nucleus Q of half life 20 min undergoes alpha decay with probability of 60% and beta decay with probability of 40%. Initially, the number of Q nuclei is 1000. The number of alpha decays of Q in the first one hour is
(A) 50 (B) 75 (C) 350 (D) 525
Solution:
Half life of the radioactive nucleus: T = 20 min
Probability of alpha decay: 60%
Probability of beta decay: 40%
Initial number of nuclei: N0 = 1000
60% of 1000 = 600 radioactive nuclei may undergo alpha decay
Number of nuceli remaining after 1 hour: Nα = 𝑁0(𝛼)1
2
𝑡/𝑇= 600
1
2
60/20= 600
8= 75
Number of alpha decays of Q in the first 1 hour: 600 – 75 = 525
Kalyan’s Physics Challenge / JEE ADVANCED 2021 P1
SECTION 2 (maximum marks: 12)
This section contains THREE (3) question stems
There are TWO (2) questions corresponding to each question stem
The answer to each equation is a Numerical Value
For each question, enter the correct numerical value corresponding to the answer in the designated place
If the numerical value has more than two decimal places, truncate/round-off the value to Two decimal places
Answer to each question will be evaluated according to the following marking scheme
full marks: +2 if only the correct numerical value is entered at the designated place
zero marks: 0 in all other cases
Kalyan’s Physics Challenge / JEE ADVANCED 2021 P1
Question Stem for Question Numbers: 5 & 6
A projectile is thrown from a point O on the ground at an angle 450 from the vertical and with a speed 5 2 m/s. The projectile at the highest point of its trajectory splits into two equal parts. One part falls vertically down to the ground, 0.5 sec after thesplitting. The other part, t seconds after the splitting, falls to the ground at a distance 𝑥 m from the point O. The acceleration due to gravity g = 10 m/s2
5. Th value of t is 6. The value of 𝑥 is
Solution:
Maximum height attained by the projectile: H = 𝑢2 𝑠𝑖𝑛2 𝜃
2𝑔= 50 × 𝑠𝑖𝑛2 (450)
2(10)= 2.5 x
1
2= 1.25 m
Time taken by the freely falling part to reach ground: t = 2𝐻
𝑔=
2(1.25)
10= 0.5 sec
Range of the original projectile: R = 𝑢2 sin 2𝜃
𝑔= 50 × sin 90
10= 5 m
Consrvation of momentum at the highest point of trajectory: mu cos θ = 𝑚
2v1 +
𝑚
2v2 → 5 2 cos 450 =
1
20 +
1
2v2 → v2 = 10 m/s
Range of horizontal projectile (second part): r = v2
2𝐻
𝑔= 10
2(1.25)
10= 10 x 0.5 = 5 m
Distance of the second part from point of projection: 𝑥 = 𝑅
2+ 𝑟 = 2.5 + 5 = 7.5 m
Kalyan’s Physics Challenge / JEE ADVANCED 2021 P1
Question stem for Question Numbers: 7 & 8
In the circuit shown below, the switch S is connected to position P for a long time so that the charge on the capacitor becomes q1 𝜇C. Then S is switched to position Q. After a long time, the charge on the capacitor is q2 𝜇C
7. The magnitude of q1 is 8. The magnitude of q2 is
Solution:
When the switch is connected to position P, after a long time the capacitor gets fully charged.
A fully charged capacitor behaves as open circuit
Apply KVL to the loop: – 1(i) – 1 + 2 – 2 (i) = 0 → i = 1
3A
Apply KVL to the loop: VA – 1 (i) – 1 = VB → VA – VB = 1 + 1 1
3= 4
3volt
Potential difference across the capacitor: VAB = 4
3volt
Charge on the capacitor: q1 = CVAB = 1 x 4
3= 1.33 𝝁C
Kalyan’s Physics Challenge / JEE ADVANCED 2021 P1
Solution:
When the switch is connected to position Q, after a long time the capacitor gets fully charged.
A fully charged capacitor behaves as open circuit
Apply KVL to the loop: – 1(i) + 2 – 2 (i) = 0 → i = 2
3A
Apply KVL to the loop: VA – 1 (i) = VB → VA – VB = 1 x 2
3= 2
3volt
Potential difference across the capacitor: VAB = 2
3volt
Charge on the capacitor: q1 = CVAB = 1 x 2
3= 0.67 𝝁C
Kalyan’s Physics Challenge / JEE ADVANCED 2021 P1
Question stem for Question Numbers: 9 & 10
Two point charges −Q and +𝑄
3are placed in XY plane at the origin (0, 0) and a point (2, 0) respectively, as shown in the figure.
This results in an equipotential circle of radius R and potential V = 0 in the XY plane with its centre at (b, 0). All the lengths are measured in meters
9. The value of R is10. The value of b is
Solution:
Consider a point P (x, y) in XY plane which is at a distance r1 from A and r2 from B as shown in figure
Potential at point P: V = −𝑘𝑄
𝑟1+
𝑘𝑄
3𝑟2= 𝑘𝑄 −
1
𝑥2+𝑦2+
1
3
1
(𝑥−2)2+𝑦2
0 = −1
𝑥2+𝑦2+
1
3
1
(𝑥−2)2+𝑦2→
1
𝑥2+𝑦2=
1
3
1
(𝑥−2)2+𝑦2→
1
𝑥2+𝑦2= 1
3
1
((𝑥−2)2+𝑦2)→ 𝑥2 + 𝑦2 = 3 ((𝑥 − 2)2+𝑦2)
𝑥2 + 𝑦2 = 3 𝑥2 + 22 − 4𝑥 + 𝑦2 → 2𝑥2 + 2𝑦2 − 12𝑥 + 12 = 0 → 𝑥2 + 𝑦2 − 6𝑥 + 6 = 0
𝑥2 − 2 3 𝑥 + 6 + 3 − 3 + 𝑦2 = 0 → (𝑥 − 3)2 + 𝑦2 = 32
Radius of the circle: R = 3 = 1.73 m and centre of the circle: (b, 0) = (3, 0)
Kalyan’s Physics Challenge / JEE ADVANCED 2021 P1
SECTION 3 (maximum marks: 24)
This section contains SIX (6) questions
Each question has FOUR options. One or More than One of these four option(s) is(are) correct answer(s)
For each question, choose the option(s) corresponding to all the correct answer(s)
Answer to each question will be evaluated according to the following marking scheme
Full marks: +4 if only all the correct options(s) is(are) chosen
Partial marks: +3 if all the four options are correct but only three options are chosen
Partial marks: +2 if three or more options are correct but only two options are chosen, both of which are correct
Partial marks: +1 if two or more options are correct but only one option is chosen and it is a correct option
Zero marks: 0 if unanswered
Negative marks: −2 in all other cases
Kalyan’s Physics Challenge / JEE ADVANCED 2021 P1
11. A horizontal force F is applied at the centre of mass of a cylindrical object of mass ‘m’ and radius R, perpendicular to its axis as shown in figure. The coefficient of friction between the object and the ground is μ. The center of mass of the object has an acceleration ‘a’. The acceleration due to gravity is g. Given that the object rolls without slipping, which of the following statement(s) is(are) correct?
(A) For the same F, the value of ‘a’ does not depend on whether the cylinder is solid or hollow(B) For a solid cylinder, the maximum possible value of ‘a’ is 2μg(C) The magnitude of the frictional force on the object due to the ground is always μmg
(D) For a thin walled hollow cylinder, a = 𝑭
𝟐𝒎
Solution:
For translation: Fnet = ma → F – f = ma ---- (1)
For rotation: 𝜏𝑛𝑒𝑡 = Iα → fR = Iα → f = 𝐼𝛼
𝑅---- (2)
Condition for accelerated pure rolling: a = Rα ---- (3)
From the above equations: F −𝐼𝛼
𝑅= ma → F −
𝐼𝑎
𝑅2= ma → F = a 𝑚 +
𝐼
𝑅2→ a =
𝐹
𝑚 +𝐼
𝑅2
(D) For thin walled hollow cylinder: I = mR2
Acceleration of center of mass: a = 𝐹
𝑚 +𝑚𝑅2
𝑅2
= 𝑭
𝟐𝒎
Kalyan’s Physics Challenge / JEE ADVANCED 2021 P1
Solution:
(B) For a solid cylinder: I = 1
2mR2
Acceleration of center of mass: a = 𝐹
𝑚 +
12𝑚𝑅2
𝑅2
= 2
3
𝐹
𝑚
From equation (1): F – f = ma →3𝑚𝑎
2− 𝜇𝑁 = 𝑚𝑎 →
1
2𝑚𝑎 = 𝜇𝑚𝑔 → 𝑎 = 2𝜇𝑔
Maximum possible acceleration of the solid cylinder: 𝟐𝝁𝒈
Kalyan’s Physics Challenge / JEE ADVANCED 2021 P1
12. A wide slab consisting of two media of refractive indices n1 and n2 is placed in air as shown in the figure. A ray of light is
incident from medium n1 to n2 at an angle θ, where sin θ is slightly larger than 1
𝑛1.
Take refractive index of air as 1. Which of the following statement(s) is(are) correct?
(A) The light ray enters air if n2 = n1
(B) The light ray is finally reflected back into the medium of refractive index n1 if n2 < n1
(C) The light ray is finally reflected back into the medium of refractive index n1 if n2 > n1
(D) The light ray is reflected back into the medium of refractive index n1 if n2 = 1
Solution:
(D) When the angle of incidence is greater than critical angle, the light rays reflect back into denser medium (TIR)
Condition for TIR between n1 and n2: n1 sin θ = n2 sin 90 → n1 sin θ = 1 x 1 → θ = 𝑠𝑖𝑛−11
𝑛1
Since the angle of incidence is greater than critical angle [sin θ is slightly larger than 1/𝑛1], the light ray gets back to n1
Kalyan’s Physics Challenge / JEE ADVANCED 2021 P1
Solution:
(C) If n2 > n1; light ray travels from rarer medium to denser medium.
The angle of refraction in n2 is less than the angle of incidence in n1: n1 sin θ = n2 sin r → sin r = 𝑛1
𝑛2sin θ
Since sin θ is slightly larger than 1
𝑛1, r is slightly greater than the crictical angle for n2 and air: 𝑠𝑖𝑛−1
1
𝑛2
Angle of incidence at n2 – air interface: r
Since r is greater than critical angle, the light ray undergoes TIR at n2 – air interface and travels back to n1
Kalyan’s Physics Challenge / JEE ADVANCED 2021 P1
13. A particle of mass M = 0.2 kg is initially at rest in the XY plane at a point (𝑥 = −𝑙, 𝑦 = −ℎ), where 𝑙 = 10 m and ℎ = 1 m. The particle is accelerated at time t = 0 with a constant acceleration a = 10 m/s2 along the +ve X direction. Its angular momentum
and torque with respect to the origin, in SI units, are represented by ത𝐿 and ҧ𝜏, respectively. Ƹ𝑖, Ƹ𝑗 and 𝑘 are unit vectors along the +ve
X, Y and Z directions, respectively. If 𝑘 = Ƹ𝑖 x Ƹ𝑗 then which of the following statement(s) is(are) correct?
(A) The particle arrives at the point (𝒙 = 𝒍, 𝒚 = −𝒉) at time t = 2 sec
(B) ത𝝉 = 2 𝒌 when the particle passes through the point (𝒙 = 𝒍, 𝒚 = −𝒉)
(C) ത𝑳 = 4 𝒌 when the particle passes through the point (𝒙 = 𝒍, 𝒚 = −𝒉)
(D) ҧ𝜏 = 𝑘 when the particle passes through the point (𝑥 = 0, 𝑦 = −ℎ)
Solution:
(A) Mass of the particle: M = 0.2 kg
Initial position of the particle: (x, y) = −𝑙,−ℎ = −10,−1
Acceleration of the particle: a = 10 m/s2 (along +ve X direction)
Displacement of the particle: x = 1
2at2 = 0.5 x 10 x 4 = 20 m
Final position of the particle (after 2 sec): (x, y) = (10 m, −1 m)
Kalyan’s Physics Challenge / JEE ADVANCED 2021 P1
Solution:
(B) Force acting on the particle: ത𝐹 = Mത𝑎 = 0.2 x 10 Ƹ𝑖 = 2 Ƹ𝑖 N
Position vector of the particle: ҧ𝑟 = 10 Ƹ𝑖 − Ƹ𝑗
Torque acting on the particle: ҧ𝜏 = ҧ𝑟 x ത𝐹 = Ƹ𝑖 Ƹ𝑗 𝑘
10 −1 02 0 0
= Ƹ𝑖 (0 – 0) − Ƹ𝑗 (0 – 0) + 𝑘 (0 + 2) = 2 𝒌
(C) Mass of the particle: M = 0.2 kg
Velocity of the particle: ҧ𝑣 = ത𝑎t = 10 Ƹ𝑖 (2) = 20 Ƹ𝑖
Position vector of the particle: ҧ𝑟 = 10 Ƹ𝑖 − Ƹ𝑗
Angular momentum of the particle: ത𝐿 = m ҧ𝑟 × ҧ𝑣 = 0.2 Ƹ𝑖 Ƹ𝑗 𝑘
10 −1 020 0 0
= 0.2 Ƹ𝑖 (0 – 0) − Ƹ𝑗 (0 – 0) + 𝑘 (0 + 20) = 4 𝒌
Kalyan’s Physics Challenge / JEE ADVANCED 2021 P1
14. Which of the following statement(s) is(are) correct about the spectrum of hydrogen atom?
(A) The ratio of the longest wavelength to the shortest wavelength in Balmer series is 𝟗
𝟓(B) There is an overlap between the wavelength ranges of Balmer and Paschen series
(C) The wavelengths of Lyman series are given by 1 +1
𝑚2 λ0 where λ0 is the shortest wavelength of Lyman series and m is an integer
(D) The wavelength ranges of Lyman and Balmer series do not overlap
Solution:
Wavenumber of the emitted photon during electron transition: 1
λ= R
1
𝑛𝑖2 −
1
𝑛𝑓2
Longest wavelength in Balmer series (n = 3 to n = 2): 1
λ𝑙= R
1
4−
1
9= 5𝑅
36
Shortest wavelength in Balmer series (n = ∞ to n = 2): 1
λ𝑠= R
1
4−
1
∞= 𝑅
4
Ratio of longest wavelength to shortest wavelength: λ𝑙
λ𝑠= 36
5𝑅x 𝑅
4= 𝟗
𝟓
Longest wavelength of Lyman series (n = 2 to n = 1): 1
λ𝐿= R 1 −
1
4= 3𝑅
4→ λ𝐿 =
4
3𝑅
Shortest wavelength of Balmer series (n = ∞ to n = 2): 1
λ𝐵= R
1
4−
1
∞= 𝑅
4→ λ𝐵 =
4
𝑅
So, Lyman and Balmer series do not overlap
Kalyan’s Physics Challenge / JEE ADVANCED 2021 P1
15. A long straight wire carries a current I = 2 A. A semi circular conducting rod is placed beside it on two conducting parallel rails of negligible resistance. Both the rails are parallel to the wire. The wire, the rod and the rails lie in the same horizontal plane, as shown in figure. Two ends of the semi – circular rod are at distances 1 cm and 4 cm from the wire. At time t = 0, the rod starts moving on the rails with a speed v = 3 m/s (see the figure). A resistor R = 1.4 Ω and a capacitor C0 = 5.0 μF are connected in series between the rails. At time t = 0, C0 is uncharged. Which of the following statement(s) is(are) correct?[μ0 = 4𝜋 x 10−7 SI units, take ln 2 = 0.7]
(A) maximum current through R is 1.2 x 𝟏𝟎−𝟔 A (B) maximum current through R is 3.8 x 10−6 A(C) maximum charge on capacitor C0 is 8.4 x 𝟏𝟎−𝟏𝟐 C (D) maximum charge on capacitor C0 is 2.4 x 10−12 C
Solution:
Magnetic field due to infinitely long current carrying straight conductor: B = 𝜇0
2𝜋
𝐼
𝑥
Consider an element of thickness ‘dx’ in the semi-circular rod at distance ‘x’ from the straight wire
Emf (motional) induced in the element: de = B (dx)v = 𝜇0
2𝜋
𝐼
𝑥v (dx) =
𝜇0𝐼𝑣
2𝜋
1
𝑥𝑑𝑥
Emf induced in the semi-circular rod: e = 𝑑𝑒 = 𝜇0𝐼𝑣
2𝜋14 1
𝑥𝑑𝑥 =
𝜇0𝐼𝑣
2𝜋[ln x] [limits: 1 to 4]
Emf induced in the semi-circular rod: e = 𝜇0𝐼𝑣
2𝜋[ln 4 – ln 1] =
𝜇0𝐼𝑣
2𝜋(2 ln 2) = 2 × 10−7 x 2 x 3 (2 x 0.7) = 16.8 x 10−7 volt
Maximum current through the resistor: imax = 𝑒
𝑅= 16.8 × 10−7
1.4= 1.2 x 𝟏𝟎−𝟔 A
Maximum charge on the capacitor: qmax = eC0 = 16.8 x 10−7 x 5 x 10−6 = 8.4 x 𝟏𝟎−𝟏𝟐 C
Kalyan’s Physics Challenge / JEE ADVANCED 2021 P1
16. A cylindrical tube, with its base as shown in the figure, is filled with water. It is moving down with a constant acceleration ‘a’ along a fixed inclined plane with angle θ = 450. P1 and P2 are pressure at points 1 and 2, respectively, located at the base of the
tube. Let β = 𝑃1−𝑃2
𝜌𝑔𝑑, where ρ is density of water, d is the inner diameter of the tube and g is the acceleration due to gravity.
Which of the following statement(s) is(are) correct?
(A) β = 0 when a = g/ 𝟐 (B) β > 0 when a = g/ 2 (C) β = 𝟐−𝟏
𝟐when a = g/2 (D) β =
1
2when a = g/2
Solution:
Acceleration in horizontal direction: ax = a cos 450 = 𝑎
2
Acceleration in vertical direction: ay = g – a sin 450 = g −𝑎
2
Consider a point 3 (vertically above 1) at the same horizontal level of point 2 as shown in figure
Pressure difference between 1 & 3: P1 – P3 = ρayh = ρ 𝑔 −𝑎
2
𝑑
tan 45= ρ 𝑔 −
𝑎
2d
Pressure difference between 2 & 3: P2 – P3 = ρax(d) = ρ𝑎
2d (pressure decreases in the direction of acceleration)
Pressure difference between 1 & 2: P1 – P2 = ρ 𝑔 −𝑎
2d − ρ
𝑎
2d = ρd 𝑔 −
𝑎
2−
𝑎
2= ρd 𝑔 −
2𝑎
2= ρgd 1 −
2𝑎
𝑔
Given: β = 𝑃1−𝑃2
𝜌𝑔𝑑= 1 −
2𝑎
𝑔= 0 → a =
𝒈
𝟐
For a = 𝑔
2; β = 1 −
2
𝑔
𝑔
2= 1 −
1
2=
𝟐−𝟏
𝟐
Kalyan’s Physics Challenge / JEE ADVANCED 2021 P1
SECTION 4 (maximum marks: 12)
This section contains THREE (3) questions
The answer to each question is a non negative integer
For each question, enter the correct integer corresponding to the answer
Answer to each question will be evaluated according to the following marking scheme
Full marks: +4 if only the correct integer is entered
Zero marks: 0 in all other cases
Kalyan’s Physics Challenge / JEE ADVANCED 2021 P1
17. An α particle (mass 4 amu) and a singly charged sulfur ion (mass 32 amu) are initially at rest. They are accelerated through a potential V and then allowed to pass into a region of uniform magnetic field which is normal to the velocities of the particles. Within this region, the α particle and the sulfur ion move in circular orbits of radii 𝑟𝛼 and 𝑟𝑆,
respectively. The ratio 𝑟𝑆
𝑟𝛼is _____
Solution:
Charge on alpha particle: 𝑞𝛼 = 2 units
Mass of alpha particle: 𝑚𝛼 = 4 amu
Charge on singly charged sulfur ion: 𝑞𝑆 = 1 unit
Mass of singly charged sulfur ion: 𝑚𝑆 = 32 amu
Radius of the circular path followed by the charged particle in uniform magnetic field:
r = 𝑚𝑣
𝐵𝑞=
𝑃
𝐵𝑞=
2𝑚𝑘
𝐵𝑞=
2𝑚𝑞𝑉
𝐵𝑞=
2𝑉
𝐵
𝑚
𝑞
Both the charged particles are accelerated by the same potential difference and both are moving in same magnetic field
Ratio of radii: 𝑟𝑆
𝑟𝛼=
𝑚𝑆
𝑚𝛼
𝑞𝛼
𝑞𝑆=
32
4×
2
1= 4
Kalyan’s Physics Challenge / JEE ADVANCED 2021 P1
18. A thin rod of mass M and length ‘a’ is free to rotate in horizontal plane about a fixed vertical axis passing through point O. A thin circular disc of mass M and of radius ‘a/4’ is pivoted on this rod with its center at a distance a/4 from the free end so that it can rotate freely about its vertical axis, as shown in figure. Assume that both the rod and the disc have uniform density andthey remain horizontal during the motion. An outside stationary observer finds the rod rotating with an angular velocity Ω and the disc rotating about its vertical axis with angular velocity 4Ω. The total angular momentum of the system about the point O is 𝑀𝑎2Ω
48𝑛. The value of 𝑛 is _____
Solution:
Angular momentum of the rod about O: L1 = Iω = 𝑀𝑎2
3Ω
Angular momentum of the disc about O: L2 = Iω = M 𝑎 −𝑎
4
2Ω =
9𝑀𝑎2
16Ω [for rotation with the rod]
Angular momentum of the disc about its axis: L3 = Iω = 𝑀
𝑎
4
2
24Ω =
𝑀𝑎2
8Ω [for rotation about own axis]
Angular momentum of the system: L = L1 + L2 + L3 = 𝑀𝑎2Ω1
3+
9
16+
1
8=
𝑀𝑎2Ω
4849
The value of n is 49
Kalyan’s Physics Challenge / JEE ADVANCED 2021 P1
19. A small object is placed at the centre of a large evacuated hollow spherical container. Assume that the container is maintained at 0 K. At time t = 0, the temperature of the object is 200 K. The temperature of the object becomes 100 K at t = t1 and 50 K at t = t2. Assume the object and the container to be ideal black bodies. The heat capacity of the object
does not depend on temperature. The ratio𝑡2
𝑡1is _____
Solution:
Power radiated by the black body: P = σAT4 ---- (1)
Rate of colling of the black body: 𝑑𝑄
𝑑𝑡= −𝑚𝑠
𝑑𝑇
𝑑𝑡---- (2)
From (1) & (2): σAT4 = −𝑚𝑠𝑑𝑇
𝑑𝑡→
1
𝑇4𝑑𝑇 = −
𝜎𝐴
𝑚𝑠𝑑𝑡 →
1
𝑇4𝑑𝑇 = −k 𝑑𝑡
First time interval: 200100 1
𝑇4𝑑𝑇 = −k 0
𝑡1 𝑑𝑡 →1
3𝑇3(limits: 200 to 100) = kt1 →
1
3
1
1003−
1
2003= kt1 ---- (3)
Second time interval: 20050 1
𝑇4𝑑𝑇 = −k 0
𝑡2 𝑑𝑡 →1
3𝑇3(limits: 200 to 50) = kt2 →
1
3
1
503−
1
2003= kt2 ---- (4)
(4)/(3): 𝑡2
𝑡1=
1
503−
1
2003
1
1003−
1
2003
= 1 −
1
431
23−
1
43
= 1 −
1
641
8−
1
64
= 63/64
7/64= 63
7= 9
Kalyan’s Physics Challenge / JEE ADVANCED 2021 P1