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JEE Main 202125 JulyMorning Shift
Exam
JEE Main
Year
2021
Date
25 July
Shift
Morning Shift
Subject
Physics
Chapter
Rotational Motion

JEE Main 2021 · 25 July · Morning Shift · Physics · Rotational Motion

A particle of mass m is moving in time t on a trajectory given by,...

A particle of mass mm is moving in time tt on a trajectory given by,

\overset{\to}{r} = 10\alpha{t}^{2}\overset{^}{i} + 5\beta\left(t - 5\right)\overset{^}{j}

where α\alpha and β\beta are dimensional constants.

The angular momentum of the particle becomes the same as it was for t=0t = 0 at time t=t =_____ seconds.

Correct answer

0

Step-by-step solution

\overset{\to}{r} = 10\alpha{t}^{2}\overset{^}{i} + 5\beta\left(t - 5\right)\overset{^}{j}

\overset{\to}{v} = 20\alpha + \overset{^}{i} + 5\beta\overset{^}{j}

L→=m(r→×v→)\overset{\to}{L} = m\left(\overset{\to}{r} \times \overset{\to}{v}\right)

= m\left[10\alpha{t}^{2}\overset{^}{i} + 5\beta\left(t - 5\right)\overset{^}{j}\right] \times \left[20\alpha\overset{^}{i} + 5\beta\overset{^}{j}\right]

\overset{\to}{L} = m\left[50\alpha\beta{t}^{2}\overset{^}{k} - 100\alpha\beta\left({t}^{2} - 5t\right)\overset{^}{k}\right]

At t=0,L→=0→t = 0, \overset{\to}{L} = \overset{\to}{0}

50αβt2−100αβ(t2−5t)=050\alpha\beta{t}^{2} - 100\alpha\beta\left({t}^{2} - 5t\right) = 0

t−2(t−5)=0t - 2\left(t - 5\right) = 0

t=10sect = 10sec

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