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

• A

  $\frac{\pi}{2}$
• B

  $\frac{3}{\pi}$
• C

  $\frac{2}{\sqrt{3} \pi}$
• Correct answer
  D

  $\frac{\pi}{2 \sqrt{2}}$

Solution

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.

Hence, the correct answer is option: 'D'.