Question

Length of the minute hand in a clock is $10\text{cm}$. Find difference between average speed and magnitude of average velocity (in $cm/min$) of tip of the minute hand from $6$ a.m. to $6:20$ a.m.

• A. $\frac{2\pi }{3}-\frac{\sqrt{3}}{2}$
• B. $\frac{2\pi }{3}-\frac{1}{2}$
• C. $\frac{\pi }{3}-\frac{\sqrt{3}}{2}$
• D. $\frac{\pi }{3}-\frac{1}{2}$

Answer 1

The correct option is C $\frac{\pi }{3}-\frac{\sqrt{3}}{2}$


Given time, $t=20min$
The distance covered by minute hand
$d=r\theta$, where $r$ is the length of the minute hand and $\theta$ in radians.

$\theta ={120}^{\circ }=\frac{\pi }{180}\times 120rad=\frac{2\pi }{3}rad$
$d=10\times \frac{2\pi }{3}cm$
Average speed:
$v=\frac{d}{t}=\frac{20\pi }{3\times 20}=\frac{\pi }{3}cm/min$

Displacement: $D=2r\sin \frac{\theta }{2}$
$D=2\times 10\times \sin \frac{120}{2}=10\sqrt{3}cm$
Magnitude of average velocity:
$V=\frac{D}{t}=\frac{10\sqrt{3}}{20}=\frac{\sqrt{3}}{2}cm/min$

$\therefore$ The difference is : $\frac{\pi }{3}-\frac{\sqrt{3}}{2}cm/min$