A rigid body rotates about a fixed axis such that its angular velocity $(\omega)$ depends on $\theta$ as $\omega=\frac{k}{\theta}$, where $k$ is a positive constant. At $t=0, \theta=\pi$, then the $\theta$ is related to time $(t)$ as

$\theta=\left(t^2+\pi^2\right)$

JEE PHYSICS MCQ: SYSTEM OF PARTICLES

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Application +4 | -1

A rigid body rotates about a fixed axis such that its angular velocity

(\omega)

depends on

\theta

\omega=\frac{k}{\theta}

, where

k

is a positive constant. At

t=0, \theta=\pi

, then the

\theta

is related to time

(t)

View answer and explanation

Correct option: B

\theta=\sqrt{\left(2 k t+\pi^2\right)}

Option B (

\theta=\sqrt{\left(2 k t+\pi^2\right)}

) is correct because it is the solution to the differential equation

\omega = \frac{{d\theta}}{{dt}} = \frac{k}{\theta}

with the initial condition

\theta(0) = \pi

Practice All Chapter MCQs →

\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)