Question

The length of a seconds hand in a watch is $1$ cm. The change in its velocity in $15$ s is

• A. $0cm/s$
• B. $\frac{\pi }{30\sqrt{2}}cm/s$
• C. $\frac{\pi }{30}cm/s$
• D. $\frac{\pi }{30}\sqrt{2}cm/s$

Answer 1

The correct option is D $\frac{\pi }{30}\sqrt{2}cm/s$

$\omega =\pi /30$
Hence v = $\pi /30$ cm/s
Change in velocity by 90 degrees in 15 seconds = $v\sqrt{2}=\pi /30\sqrt{2}$ cm/s