Question

The second's hand of a watch has length 6 cm. Speed of end point and magnitude of difference of velocities at two perpendicular positions will be

• A. 6.28 and 0 mm/s
• B. 8.88 and 4.44 mm/s
• C. 8.88 and 6.28 mm/s
• D. 6.28 and 8.88 mm/s

Answer 1

The correct option is D 6.28 and 8.88 mm/s

$v=r\omega =\frac{r\times 2\pi }{T}=\frac{0.06\times 2\pi }{60}=6.28mm/s$
Magnitude of change in velocity $=|\overline{v_2}-\overline{v_1}|$
$=\sqrt{v_1^2+v_2^2}=8.88mm/s(As{v}_1=v_2=6.28mm/s)$