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3-2.Motion in Plane
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The second's hand of a watch has $6\, cm$ length. The speed of its tip and magnitude of difference in velocities of its at any two perpendicular positions will be respectively
A$2\pi$ and $0\, mm/s$
B$2\sqrt 2 \,\pi $ and $4.44\, mm/s$
C$2\sqrt 2 \,\pi $ and $2\pi\, mm/s$
D$2\pi$ and $2\sqrt 2 \pi\,mm/s $
Solution
$l=6 \mathrm{cm}, v=?, \omega=\frac{2 \pi}{60}=\frac{\pi}{30} \mathrm{rad} / \mathrm{s}$
So $v=\omega l=\frac{\pi}{30} \times 6=\frac{\pi}{5} \mathrm{cm} / \mathrm{s}=2 \pi \mathrm{mm} / \mathrm{s}$
Difference $=\sqrt{2} \frac{\pi}{5} \mathrm{cm} / \mathrm{s}=2 \sqrt{2} \pi \mathrm{mm} / \mathrm{s}$
So $v=\omega l=\frac{\pi}{30} \times 6=\frac{\pi}{5} \mathrm{cm} / \mathrm{s}=2 \pi \mathrm{mm} / \mathrm{s}$
Difference $=\sqrt{2} \frac{\pi}{5} \mathrm{cm} / \mathrm{s}=2 \sqrt{2} \pi \mathrm{mm} / \mathrm{s}$
Standard 11
Physics
