A particle is moving with constant speed in a circular path. When the particle turns by an angle $90^{\circ}$, the ratio of instantaneous velocity to its average velocity is $\pi: x \sqrt{2}$. The value of $x$ will be -
Solution
$$
\begin{aligned}
& \text { Instantaneous velocity }=\omega R \\\\
& \text { Time taken }=\frac{\pi}{2 \omega} \\\\
& \text { Displacement }=R \sqrt{2} \\\\
& \text { Average velocity }=\frac{R \sqrt{2} \times 2 \omega}{\pi}=\frac{2 \sqrt{2}}{\pi} \omega R \\\\
& \Rightarrow \frac{v_{\text {ins }}}{v_{\text {avg }}}=\frac{\omega R \pi}{2 \sqrt{2} \omega R} \\\\
& \Rightarrow x=2
\end{aligned}
$$
About this question
Subject: Physics · Chapter: Rotational Motion · Topic: Uniform Circular Motion
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