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JEE Advanced 2022Paper 1
Exam
JEE Advanced
Year
2022
Paper
Paper 1
Subject
Physics
Chapter
Work, Power and Energy

JEE Advanced 2022 · Paper 1 · Physics · Work, Power and Energy

List I describes four systems, each with two particles A and B in...

List I describes four systems, each with two particles AandBA and B in relative motion as shown in figure. List II gives possible magnitudes of their relative velocities (in ms−1m{s}^{- 1}) at time t=π3st = \dfrac{\pi}{3}s.

List-I List-II
(I)

A and Baremovingonahorizontalcircleofradiusare moving on a horizontal circle of radius1mwithuniformangularspeedwith uniform angular speed\omega = 1rad{s}^{- 1}.Theinitialangularpositionsof. The initial angular positions ofA and Battimeat timet = 0areare\theta = 0andand\theta = \dfrac{\pi}{2}respectively.</p><p><imgsrc="https://cdn−question−pool.getmarks.app/pyq/jeeadvanced/ff049b50−ab5e−454c−8c66−cb6a46d2edb2−image.png"style="width:140px;height:128px;"/></p></td><td>(P)</td><td>respectively.</p> <p><img src="https://cdn-question-pool.Quantrex.app/pyq/jee_advanced/ff049b50-ab5e-454c-8c66-cb6a46d2edb2-image.png" style="width: 140px; height: 128px;" /></p> </td> <td>(P)</td> <td>\dfrac{\sqrt{3} + 1}{2}</td></tr><tr><td>(II)</td><td><p>Projectiles</td> </tr> <tr> <td>(II)</td> <td> <p>ProjectilesA and Barefired(inthesameverticalplane)atare fired (in the same vertical plane) att = 0 and t = 0 . 1srespectively,withthesamespeedrespectively, with the same speedv = \dfrac{5\pi}{\sqrt{2}}m{s}^{- 1}andatand at45 °fromthehorizontalplane.Theinitialseparationbetweenfrom the horizontal plane. The initial separation betweenA and Bislargeenoughsothattheydonotcollide,(g=10ms−2)is large enough so that they do not collide, \left(g = 10m{s}^{- 2}\right).

List I describes four systems, each with two particles A and B in relative motion as shown in figure. List II gives

 (Q) (3−1)2\dfrac{\left(\sqrt{3} - 1\right)}{\sqrt{2}}
(III)

Two harmonic oscillators AandBA and B moving in the xx direction according to xA=x0sintt0andxB=x0sin⁡(tt0+π2){x}_{A} = {x}_{0}sin\dfrac{t}{{t}_{0}} and {x}_{B} = {x}_{0}\sin \left(\dfrac{t}{{t}_{0}} + \dfrac{\pi}{2}\right) respectively, starting from t=0t = 0. Take x0=1m,t0=1s{x}_{0} = 1m, {t}_{0} = 1s.

List I describes four systems, each with two particles A and B in relative motion as shown in figure. List II gives

 (R) 10\sqrt{10}
(IV)

Particle AA is rotating in a horizontal circular path of radius 1m1m on the xyxy plane, with constant angular speed ω=1rads−1\omega = 1rad{s}^{- 1}. Particle BB is moving up at a constant speed 3ms−13m{s}^{- 1} in the vertical direction as shown in the figure. (Ignore gravity.)

List I describes four systems, each with two particles A and B in relative motion as shown in figure. List II gives

 (S) 2\sqrt{2}
(T) 25π2+1\sqrt{25{\pi}^{2} + 1}

Which one of the following options is correct?

Options

  1. AI→R,II→T,III→P,IV→SI \to R, II \to T, III \to P, IV \to S
  2. BI→S,II→P,III→Q,IV→RI \to S, II \to P, III \to Q, IV \to R
  3. CI→S,II→T,III→P,IV→RI \to S, II \to T, III \to P, IV \to R
  4. DI→T,II→P,III→R,IV→SI \to T, II \to P, III \to R, IV \to S

Correct answer

C. I→S,II→T,III→P,IV→RI \to S, II \to T, III \to P, IV \to R

Step-by-step solution

(I)

List I describes four systems, each with two particles A and B in relative motion as shown in figure. List II gives

Angular velocity of both the particles is same. Therefore, angle between them will remain 90°90 °.

\left|{v}_{rel}\right| =\sqrt{2}\omegaR = \sqrt{2}ms^{- 1}</p><p>Hence,</p><p>Hence,I \to S</p><p>(II)</p><p>ThevelocityofAatanytimetcanbewrittenas</p><p></p><p>(II)</p><p>The velocity of A at any time t can be written as</p><p>{v}_{A}\left(t\right) =\dfrac{5\pi}{2}\cos 45 ° \overset{^}{i} + \left[\dfrac{5\pi}{2} \times \sin 45 ° - g \times t\right]\overset{^}{j}</p><p></p><p>= \dfrac{5\pi}{2}\overset{^}{i} + \left(\dfrac{5\pi}{2} - gt\right)\overset{^}{j}</p><p>VelocityofBatanytimet</p><p></p><p>Velocity of B at any time t</p><p>{v}_{B}\left(t\right) =\dfrac{- 5\pi}{2}\overset{^}{i} + \left(\dfrac{5\pi}{2} - g\left(t - 0. 1\right)\right)\overset{^}{j}</p><p>Relative velocity at time t will be,</p><p>\overset{→}{v}rel = 5\pi\overset{^}{i} - g \times 0. 1\overset{^}{j}

\left|\overset{→}{v}rel\right| =\sqrt{25{\pi}^{2} + 1}ms^{- 1}</p><p>Hence,</p><p>Hence,II \to T</p><p>(III)</p><p>RelativepositionofAwrtBcanbewrittenas</p><p></p><p>(III) </p><p>Relative position of A wrt B can be written as</p><p>x = {x}_{A} - {x}_{B}= {x}_{0}\sin t - {x}_{0}\sin \left(t + \dfrac{\pi}{2}\right)

Using trigonometric relation

x=x =\sqrt{2}{x}_{0}sin⁡(t−π4)\sin \left(t - \dfrac{\pi}{4}\right)

Therefore,

vrel=dxdt=2x0cos⁡(t−π4)</p><p>at{v}_{rel} = \dfrac{dx}{dt} = \sqrt{2}{x}_{0}\cos \left(t - \dfrac{\pi}{4}\right)</p><p>att = π3</p><p>\dfrac{\pi}{3}</p><p>{v}_{rel} = 2\sqrt{2} \cos \left(\dfrac{\pi}{3} - \dfrac{\pi}{4}\right)=\sqrt{2} \times \dfrac{\sqrt{3} + 1}{2\sqrt{2}} = \dfrac{\sqrt{3} + 1}{2}ms−1ms^{- 1}

Hence, III→PIII \to P

(IV)

The velocity of A is in xyplaneandxyplane andit is& \left|{v}_{A}\right| =\omegar =1ms−11ms^{- 1} and the velocity of B is along zaxisand∣vB∣=zaxis and \left|{v}_{B}\right| =3ms^{- 1}.Therefore,</p><p>. Therefore,</p><p>{v}_{rel} = \sqrt{{3}^{2} + {1}^{2}} = \sqrt{10}ms−1ms^{- 1}

Hence, IV→RIV \to R

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