JEE PHYSICS MCQ: SYSTEM OF PARTICLES

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Application +4 | -1

A fly-wheel of mass

25 \mathrm{~kg}

has a radius of

0.2 \mathrm{~m}

. It is making

240 \mathrm{~rpm}

. What is the torque necessary to bring to rest in

20 \mathrm{~s}

View answer and explanation

Correct option: B

0.4 \pi \mathrm{~Nm}

Option B (

0.4 \pi \mathrm{~Nm}

) is correct because it is the torque required to bring the flywheel to rest, calculated using the moment of inertia and angular deceleration.

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)