Question

The length of second hand in a watch is $1cm$. The magnitude of change in velocity of its tip in $15s$ is :

• A. zero
• B. $\left(\frac{\pi }{15\sqrt{2}}\right)cm{s}^{-1}$
• C. $\left(\frac{\pi }{15}\right)cm{s}^{-1}$
• D. $\left(\frac{\pi \sqrt{2}}{15}\right)cm{s}^{-1}$

Answer 1

The correct option is B $\left(\frac{\pi }{15\sqrt{2}}\right)cm{s}^{-1}$

In $15$ seconds the angle travelled by second hand will be ${90}^0$
$\Delta v=2v\sin \theta /2=2\times \frac{2\pi }{60}\times \frac{1}{\sqrt{2}}$ = $\frac{\pi }{15\sqrt{2}}$ cm/s