Circular MotionMediumBloom L3
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
Options
A.2p & 0 mm/s
B.$2\sqrt{\pi}$& 4.44 mm/s
C.2 & $2\sqrt{\pi}$ mm/s
D.2p &$2\sqrt{2}\pi$ mm/s
Solution
Sol. v = wr
= $\frac{2\pi}{60} \times 6 = \frac{\pi}{5}$ = 2p mm/sec
∆v = $\sqrt{2}v$
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