JEE PHYSICS MCQ: SYSTEM OF PARTICLES

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Application +4 | -1

A rod of mass

m

and length

l

rotates about an axis passing through its end and perpendicular to its length. If its angular velocity it decreased from

4\omega

2\omega

in 3 seconds the torque applied on it

View answer and explanation

Correct option: C

\frac{{2n{l^2}\omega }}{9}

Option D (

\frac{{m{l^2}\omega }}{9}

) is incorrect, the correct answer is C (

\frac{{2m{l^2}\omega }}{9}

). The torque is 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)