JEE PHYSICS MCQ: SYSTEM OF PARTICLES

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Application +4 | -1

A wheel of moment of inertia

5 \times {10^{ - 3}}kg - {m^2}

is making 20 revolutions/sec. The torque required to stop it in 10 sec is

View answer and explanation

Correct option: A

2\pi \times {10^{ - 2}}N - m

Option A (

2\pi \times {10^{ - 2}}N - m

) is correct because it is the torque calculated using the moment of inertia and angular acceleration.

More Questions from SYSTEM OF PARTICLES

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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)