A constant torque acting on a uniform circular wheel changes its angular momentum from ${1\mathrm{~A}}_0$ to $5 \mathrm{~A}_0$ to in 6 seconds, Then the magnitude of torque is -

JEE PHYSICS MCQ: SYSTEM OF PARTICLES

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Application +4 | -1

A constant torque acting on a uniform circular wheel changes its angular momentum from

{1\mathrm{~A}}_0

5 \mathrm{~A}_0

to in 6 seconds, Then the magnitude of torque is -

View answer and explanation

Correct option: A

\frac{2}{3} \mathrm{~A}_0

Option A (

\frac{2}{3} \mathrm{~A}_0

) is correct because torque is the rate of change of angular momentum, which is (5A0 - A0)/6.

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\mathrm{AB}

\mathrm{L}

\mu(x)=a+\frac{b x}{L}

x

\mathrm{A}

\mathrm{CM}

\left(\frac{7}{12}\right) \mathrm{L}

\mathrm{A}

\mathrm{b}

1.0 \mathrm{~g}

8 \mathrm{~cm}

\omega = {t^4} + 2{t^3}

t = 1\,sec

\left( {rad\,{s^{ - 2}}} \right)