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A ball moves one-fourth $\left(\frac{1^{\text {th }}}{4}\right)$ of a circle of radius $R$ in time $T$. Let $v_1$ and $v_2$ be the magnitudes of mean speed and mean velocity vector. The ratio $\frac{v_1}{v_2}$ will be
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2414 Upvotes
Verified Answer
The correct answer is:
$\frac{\pi}{2 \sqrt{2}}$
Key idea Mean speed of a moving body,
$$
v_{\mathrm{rms}}=\frac{\text { total distance }}{\text { total time taken }}
$$
Mean velocity vector for a moving body,
$$
v_{w v}=\frac{\text { total displacement }}{\text { total time taken }}
$$
A ball moving in a circular arc is shown in the figure below,
Here, $v_{\mathrm{rms}}=\frac{2 \pi R}{4} \times \frac{1}{T}=v_1$
Similarly, $v_{\mathrm{vv}}=\frac{\sqrt{2} R}{T}=v_2$
Hence, $\frac{v_1}{v_2}=\frac{\pi}{2 \sqrt{2}}$
Option (4) is correct.
$$
v_{\mathrm{rms}}=\frac{\text { total distance }}{\text { total time taken }}
$$
Mean velocity vector for a moving body,
$$
v_{w v}=\frac{\text { total displacement }}{\text { total time taken }}
$$
A ball moving in a circular arc is shown in the figure below,
Here, $v_{\mathrm{rms}}=\frac{2 \pi R}{4} \times \frac{1}{T}=v_1$
Similarly, $v_{\mathrm{vv}}=\frac{\sqrt{2} R}{T}=v_2$
Hence, $\frac{v_1}{v_2}=\frac{\pi}{2 \sqrt{2}}$
Option (4) is correct.
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