JEE PHYSICS MCQ: SYSTEM OF PARTICLES

1 Easy

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Application +4 | -1

The angular velocity of a particle is given by

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

then its angular acceleration of the particle at

t = 1\,sec

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

View answer and explanation

Correct option: C

Option C (10) is correct because the angular acceleration is the derivative of the angular velocity with respect to time, and evaluating the derivative at t=1 sec gives 10 rad/s².

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

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

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\mathrm{A}

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\mathrm{b}

1.0 \mathrm{~g}

is rotating in a circle of diameter

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. Its moment of inertia will be The angular velocity of a particle is given by

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

then its angular acceleration of the particle at

t = 1\,sec

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