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JEE ADVANCED 2021 P1
PHYSICS SOLUTIONS
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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: โ1 in all other cases
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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 โ 1 VSD = 1 MSD โ9
10MSD = 1 โ
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: โ 10 โ 6 ๐ฟ๐ถ = โ4 x 0.01 = โ0.04 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
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2. An ideal gas undergoes a four step cycle as shown in P โ 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
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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
๐1= (n โ 1)
2
๐ = (1.5 โ 1)
2
20= 0.05 โ ๐1 = 20 cm
Focal length of concave lens: 1
๐2= (n โ 1) โ
2
๐ = (1.5 โ 1) โ
2
20= โ0.05 โ ๐2 = โ20 cm
For convex lens: 1
๐ฃโ
1
๐ข=
1
๐1โ
1
๐ฃโ
1
โ10=
1
20โ
1
๐ฃ= โ
1
20โ ๐ฃ = โ20 cm
Magnitude of magnification provided by convex lens: m1 = ๐ฃ
๐ข= 20
10= 2
For concave lens: 1
๐ฃโ
1
๐ข=
1
๐2โ
1
๐ฃโ
1
โ30=
1
โ20โ
1
๐ฃ= โ
5
60= โ
1
12โ ๐ฃ = โ12 cm
Magnitude of magnification provided by concave lens: m2 = ๐ฃ
๐ข= 12
30= 2
5
Total magnification of the lens system: m = m1m2 = 2 x 2
5= 4
5= 0.8
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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: Nฮฑ = ๐0(๐ผ)1
2
๐ก/๐= 600
1
2
60/20= 600
8= 75
Number of alpha decays of Q in the first 1 hour: 600 โ 75 = 525
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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
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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 ๐ฅ m from the point O. The acceleration due to gravity g = 10 m/s2
5. Th value of t is 6. The value of ๐ฅ is
Solution:
Maximum height attained by the projectile: H = ๐ข2 ๐ ๐๐2 ๐
2๐= 50 ร ๐ ๐๐2 (450)
2(10)= 2.5 x
1
2= 1.25 m
Time taken by the freely falling part to reach ground: t = 2๐ป
๐=
2(1.25)
10= 0.5 sec
Range of the original projectile: R = ๐ข2 sin 2๐
๐= 50 ร sin 90
10= 5 m
Consrvation of momentum at the highest point of trajectory: mu cos ฮธ = ๐
2v1 +
๐
2v2 โ 5 2 cos 450 =
1
20 +
1
2v2 โ v2 = 10 m/s
Range of horizontal projectile (second part): r = v2
2๐ป
๐= 10
2(1.25)
10= 10 x 0.5 = 5 m
Distance of the second part from point of projection: ๐ฅ = ๐
2+ ๐ = 2.5 + 5 = 7.5 m
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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 ๐C. Then S is switched to position Q. After a long time, the charge on the capacitor is q2 ๐C
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: โ 1(i) โ 1 + 2 โ 2 (i) = 0 โ i = 1
3A
Apply KVL to the loop: VA โ 1 (i) โ 1 = VB โ VA โ 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 ๐C
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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: โ 1(i) + 2 โ 2 (i) = 0 โ i = 2
3A
Apply KVL to the loop: VA โ 1 (i) = VB โ VA โ 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 ๐C
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Question stem for Question Numbers: 9 & 10
Two point charges โQ and +๐
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 = โ๐๐
๐1+
๐๐
3๐2= ๐๐ โ
1
๐ฅ2+๐ฆ2+
1
3
1
(๐ฅโ2)2+๐ฆ2
0 = โ1
๐ฅ2+๐ฆ2+
1
3
1
(๐ฅโ2)2+๐ฆ2โ
1
๐ฅ2+๐ฆ2=
1
3
1
(๐ฅโ2)2+๐ฆ2โ
1
๐ฅ2+๐ฆ2= 1
3
1
((๐ฅโ2)2+๐ฆ2)โ ๐ฅ2 + ๐ฆ2 = 3 ((๐ฅ โ 2)2+๐ฆ2)
๐ฅ2 + ๐ฆ2 = 3 ๐ฅ2 + 22 โ 4๐ฅ + ๐ฆ2 โ 2๐ฅ2 + 2๐ฆ2 โ 12๐ฅ + 12 = 0 โ ๐ฅ2 + ๐ฆ2 โ 6๐ฅ + 6 = 0
๐ฅ2 โ 2 3 ๐ฅ + 6 + 3 โ 3 + ๐ฆ2 = 0 โ (๐ฅ โ 3)2 + ๐ฆ2 = 32
Radius of the circle: R = 3 = 1.73 m and centre of the circle: (b, 0) = (3, 0)
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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: โ2 in all other cases
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11. A horizontal force F is applied at the centre of mass of a cylindrical object of mass โmโ and radius R, perpendicular to its axis as shown in figure. The coefficient of friction between the object and the ground is ฮผ. The center of mass of the object has an acceleration โaโ. 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 โaโ does not depend on whether the cylinder is solid or hollow(B) For a solid cylinder, the maximum possible value of โaโ is 2ฮผg(C) The magnitude of the frictional force on the object due to the ground is always ฮผmg
(D) For a thin walled hollow cylinder, a = ๐ญ
๐๐
Solution:
For translation: Fnet = ma โ F โ f = ma ---- (1)
For rotation: ๐๐๐๐ก = Iฮฑ โ fR = Iฮฑ โ f = ๐ผ๐ผ
๐ ---- (2)
Condition for accelerated pure rolling: a = Rฮฑ ---- (3)
From the above equations: F โ๐ผ๐ผ
๐ = ma โ F โ
๐ผ๐
๐ 2= ma โ F = a ๐ +
๐ผ
๐ 2โ a =
๐น
๐ +๐ผ
๐ 2
(D) For thin walled hollow cylinder: I = mR2
Acceleration of center of mass: a = ๐น
๐ +๐๐ 2
๐ 2
= ๐ญ
๐๐
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Solution:
(B) For a solid cylinder: I = 1
2mR2
Acceleration of center of mass: a = ๐น
๐ +
12๐๐ 2
๐ 2
= 2
3
๐น
๐
From equation (1): F โ f = ma โ3๐๐
2โ ๐๐ = ๐๐ โ
1
2๐๐ = ๐๐๐ โ ๐ = 2๐๐
Maximum possible acceleration of the solid cylinder: ๐๐๐
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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 ฮธ, where sin ฮธ is slightly larger than 1
๐1.
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 ฮธ = n2 sin 90 โ n1 sin ฮธ = 1 x 1 โ ฮธ = ๐ ๐๐โ11
๐1
Since the angle of incidence is greater than critical angle [sin ฮธ is slightly larger than 1/๐1], the light ray gets back to n1
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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 ฮธ = n2 sin r โ sin r = ๐1
๐2sin ฮธ
Since sin ฮธ is slightly larger than 1
๐1, r is slightly greater than the crictical angle for n2 and air: ๐ ๐๐โ1
1
๐2
Angle of incidence at n2 โ air interface: r
Since r is greater than critical angle, the light ray undergoes TIR at n2 โ air interface and travels back to n1
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13. A particle of mass M = 0.2 kg is initially at rest in the XY plane at a point (๐ฅ = โ๐, ๐ฆ = โโ), where ๐ = 10 m and โ = 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 เดค๐ฟ and าง๐, respectively. ฦธ๐, ฦธ๐ and ๐ are unit vectors along the +ve
X, Y and Z directions, respectively. If ๐ = ฦธ๐ x ฦธ๐ then which of the following statement(s) is(are) correct?
(A) The particle arrives at the point (๐ = ๐, ๐ = โ๐) at time t = 2 sec
(B) เดค๐ = 2 ๐ when the particle passes through the point (๐ = ๐, ๐ = โ๐)
(C) เดค๐ณ = 4 ๐ when the particle passes through the point (๐ = ๐, ๐ = โ๐)
(D) าง๐ = ๐ when the particle passes through the point (๐ฅ = 0, ๐ฆ = โโ)
Solution:
(A) Mass of the particle: M = 0.2 kg
Initial position of the particle: (x, y) = โ๐,โโ = โ10,โ1
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, โ1 m)
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Solution:
(B) Force acting on the particle: เดค๐น = Mเดค๐ = 0.2 x 10 ฦธ๐ = 2 ฦธ๐ N
Position vector of the particle: าง๐ = 10 ฦธ๐ โ ฦธ๐
Torque acting on the particle: าง๐ = าง๐ x เดค๐น = ฦธ๐ ฦธ๐ ๐
10 โ1 02 0 0
= ฦธ๐ (0 โ 0) โ ฦธ๐ (0 โ 0) + ๐ (0 + 2) = 2 ๐
(C) Mass of the particle: M = 0.2 kg
Velocity of the particle: าง๐ฃ = เดค๐t = 10 ฦธ๐ (2) = 20 ฦธ๐
Position vector of the particle: าง๐ = 10 ฦธ๐ โ ฦธ๐
Angular momentum of the particle: เดค๐ฟ = m าง๐ ร าง๐ฃ = 0.2 ฦธ๐ ฦธ๐ ๐
10 โ1 020 0 0
= 0.2 ฦธ๐ (0 โ 0) โ ฦธ๐ (0 โ 0) + ๐ (0 + 20) = 4 ๐
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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 ๐
๐(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
๐2 ฮป0 where ฮป0 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
ฮป= R
1
๐๐2 โ
1
๐๐2
Longest wavelength in Balmer series (n = 3 to n = 2): 1
ฮป๐= R
1
4โ
1
9= 5๐
36
Shortest wavelength in Balmer series (n = โ to n = 2): 1
ฮป๐ = R
1
4โ
1
โ= ๐
4
Ratio of longest wavelength to shortest wavelength: ฮป๐
ฮป๐ = 36
5๐ x ๐
4= ๐
๐
Longest wavelength of Lyman series (n = 2 to n = 1): 1
ฮป๐ฟ= R 1 โ
1
4= 3๐
4โ ฮป๐ฟ =
4
3๐
Shortest wavelength of Balmer series (n = โ to n = 2): 1
ฮป๐ต= R
1
4โ
1
โ= ๐
4โ ฮป๐ต =
4
๐
So, Lyman and Balmer series do not overlap
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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 โ 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 ฮฉ and a capacitor C0 = 5.0 ฮผF are connected in series between the rails. At time t = 0, C0 is uncharged. Which of the following statement(s) is(are) correct?[ฮผ0 = 4๐ x 10โ7 SI units, take ln 2 = 0.7]
(A) maximum current through R is 1.2 x ๐๐โ๐ A (B) maximum current through R is 3.8 x 10โ6 A(C) maximum charge on capacitor C0 is 8.4 x ๐๐โ๐๐ C (D) maximum charge on capacitor C0 is 2.4 x 10โ12 C
Solution:
Magnetic field due to infinitely long current carrying straight conductor: B = ๐0
2๐
๐ผ
๐ฅ
Consider an element of thickness โdxโ in the semi-circular rod at distance โxโ from the straight wire
Emf (motional) induced in the element: de = B (dx)v = ๐0
2๐
๐ผ
๐ฅv (dx) =
๐0๐ผ๐ฃ
2๐
1
๐ฅ๐๐ฅ
Emf induced in the semi-circular rod: e = ๐๐ = ๐0๐ผ๐ฃ
2๐14 1
๐ฅ๐๐ฅ =
๐0๐ผ๐ฃ
2๐[ln x] [limits: 1 to 4]
Emf induced in the semi-circular rod: e = ๐0๐ผ๐ฃ
2๐[ln 4 โ ln 1] =
๐0๐ผ๐ฃ
2๐(2 ln 2) = 2 ร 10โ7 x 2 x 3 (2 x 0.7) = 16.8 x 10โ7 volt
Maximum current through the resistor: imax = ๐
๐ = 16.8 ร 10โ7
1.4= 1.2 x ๐๐โ๐ A
Maximum charge on the capacitor: qmax = eC0 = 16.8 x 10โ7 x 5 x 10โ6 = 8.4 x ๐๐โ๐๐ C
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16. A cylindrical tube, with its base as shown in the figure, is filled with water. It is moving down with a constant acceleration โaโ along a fixed inclined plane with angle ฮธ = 450. P1 and P2 are pressure at points 1 and 2, respectively, located at the base of the
tube. Let ฮฒ = ๐1โ๐2
๐๐๐, where ฯ 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) ฮฒ = 0 when a = g/ ๐ (B) ฮฒ > 0 when a = g/ 2 (C) ฮฒ = ๐โ๐
๐when a = g/2 (D) ฮฒ =
1
2when a = g/2
Solution:
Acceleration in horizontal direction: ax = a cos 450 = ๐
2
Acceleration in vertical direction: ay = g โ a sin 450 = g โ๐
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 โ P3 = ฯayh = ฯ ๐ โ๐
2
๐
tan 45= ฯ ๐ โ
๐
2d
Pressure difference between 2 & 3: P2 โ P3 = ฯax(d) = ฯ๐
2d (pressure decreases in the direction of acceleration)
Pressure difference between 1 & 2: P1 โ P2 = ฯ ๐ โ๐
2d โ ฯ
๐
2d = ฯd ๐ โ
๐
2โ
๐
2= ฯd ๐ โ
2๐
2= ฯgd 1 โ
2๐
๐
Given: ฮฒ = ๐1โ๐2
๐๐๐= 1 โ
2๐
๐= 0 โ a =
๐
๐
For a = ๐
2; ฮฒ = 1 โ
2
๐
๐
2= 1 โ
1
2=
๐โ๐
๐
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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
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17. An ฮฑ 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 ฮฑ particle and the sulfur ion move in circular orbits of radii ๐๐ผ and ๐๐,
respectively. The ratio ๐๐
๐๐ผis _____
Solution:
Charge on alpha particle: ๐๐ผ = 2 units
Mass of alpha particle: ๐๐ผ = 4 amu
Charge on singly charged sulfur ion: ๐๐ = 1 unit
Mass of singly charged sulfur ion: ๐๐ = 32 amu
Radius of the circular path followed by the charged particle in uniform magnetic field:
r = ๐๐ฃ
๐ต๐=
๐
๐ต๐=
2๐๐
๐ต๐=
2๐๐๐
๐ต๐=
2๐
๐ต
๐
๐
Both the charged particles are accelerated by the same potential difference and both are moving in same magnetic field
Ratio of radii: ๐๐
๐๐ผ=
๐๐
๐๐ผ
๐๐ผ
๐๐=
32
4ร
2
1= 4
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18. A thin rod of mass M and length โaโ 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 โa/4โ 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 ฮฉ and the disc rotating about its vertical axis with angular velocity 4ฮฉ. The total angular momentum of the system about the point O is ๐๐2ฮฉ
48๐. The value of ๐ is _____
Solution:
Angular momentum of the rod about O: L1 = Iฯ = ๐๐2
3ฮฉ
Angular momentum of the disc about O: L2 = Iฯ = M ๐ โ๐
4
2ฮฉ =
9๐๐2
16ฮฉ [for rotation with the rod]
Angular momentum of the disc about its axis: L3 = Iฯ = ๐
๐
4
2
24ฮฉ =
๐๐2
8ฮฉ [for rotation about own axis]
Angular momentum of the system: L = L1 + L2 + L3 = ๐๐2ฮฉ1
3+
9
16+
1
8=
๐๐2ฮฉ
4849
The value of n is 49
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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 ratio๐ก2
๐ก1is _____
Solution:
Power radiated by the black body: P = ฯAT4 ---- (1)
Rate of colling of the black body: ๐๐
๐๐ก= โ๐๐
๐๐
๐๐ก---- (2)
From (1) & (2): ฯAT4 = โ๐๐ ๐๐
๐๐กโ
1
๐4๐๐ = โ
๐๐ด
๐๐ ๐๐ก โ
1
๐4๐๐ = โk ๐๐ก
First time interval: 200100 1
๐4๐๐ = โk 0
๐ก1 ๐๐ก โ1
3๐3(limits: 200 to 100) = kt1 โ
1
3
1
1003โ
1
2003= kt1 ---- (3)
Second time interval: 20050 1
๐4๐๐ = โk 0
๐ก2 ๐๐ก โ1
3๐3(limits: 200 to 50) = kt2 โ
1
3
1
503โ
1
2003= kt2 ---- (4)
(4)/(3): ๐ก2
๐ก1=
1
503โ
1
2003
1
1003โ
1
2003
= 1 โ
1
431
23โ
1
43
= 1 โ
1
641
8โ
1
64
= 63/64
7/64= 63
7= 9
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