Q. The length of second’s hand in a watch is $1\ cm$. The change in velocity of its tip in 15 seconds is
A) Zero
B) $\frac{\pi}{30\sqrt{2}}\ cm/sec$
C) $\frac{\pi}{30}\ cm/sec$
D) $\frac{\pi\sqrt{2}}{30}\ cm/sec$
Explanation
(d) In 15 second’s hand rotate through $90{^\circ}$
Change in velocity $\left| \overrightarrow{\mathrm{\Delta}v} \right| = 2v\sin{(\theta/2)}$
$= 2(r\omega)\sin{(90{^\circ}/2)} = 2 \times 1 \times \frac{2\pi}{T} \times \frac{1}{\sqrt{2}}$
$= \frac{4\pi}{60\sqrt{2}} = \frac{\pi\sqrt{2}}{30}\frac{cm}{\sec}\ \ \ \lbrack As\ T = 60\ \sec\rbrack$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.