JEE MAIN - Physics (2023 - 6th April Morning Shift - No. 1)
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 -
1
7
5
2
Explanation
$$
\begin{aligned}
& \text { Instantaneous velocity }=\omega R \\\\
& \text { Time taken }=\frac{\pi}{2 \omega} \\\\
& \text { Displacement }=R \sqrt{2} \\\\
& \text { Average velocity }=\frac{R \sqrt{2} \times 2 \omega}{\pi}=\frac{2 \sqrt{2}}{\pi} \omega R \\\\
& \Rightarrow \frac{v_{\text {ins }}}{v_{\text {avg }}}=\frac{\omega R \pi}{2 \sqrt{2} \omega R} \\\\
& \Rightarrow x=2
\end{aligned}
$$
Comments (0)
