Question

Find the average velocity of tip of the minute hand of an analog watch when it turn for $15\text{minutes}$.

• A. $\frac{\sqrt{2}}{4}\times {10}^{-4}\text{m/s}$
• B. $\frac{\sqrt{2}}{3}\times {10}^{-4}\text{m/s}$
• C. $\frac{\sqrt{2}}{5}\times {10}^{-4}\text{m/s}$
• D. $\frac{2\sqrt{2}}{3}\times {10}^{-4}\text{m/s}$

Answer 1

The correct option is B $\frac{\sqrt{2}}{3}\times {10}^{-4}\text{m/s}$

In $15\text{minutes}$ the position of the minute hand will be as shown

The displacement of the tip is $3\sqrt{2}\text{cm}=\frac{3\sqrt{2}}{100}\text{m}$
and time taken is $15\text{minutes}=15\times 60=900\text{sec}$
So, average velocity of the tip is
$\overline{v}=\frac{3\sqrt{2}}{100\times 900}=\frac{\sqrt{2}}{3}\times {10}^{-4}\text{m/s}$