A pendulum bob on a 2, m string is displaced 60^{°} from vertical and then released. speed of bo

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5.Work, Energy, Power and Collision

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A pendulum bob on a $2\, m$ string is displaced $60^{°}$ from the vertical and then released. What is the speed of the bob as it passes through the lowest point in its path .......... $m/s$

A

$\sqrt 2$

B

$\sqrt {9.8}$

C

$4.43$

D

$1/\sqrt 2 $

Solution

(c)$v = \sqrt {2gl(1 – \cos \theta )} = \sqrt {2 \times 9.8 \times 2(1 – \cos 60^\circ )} $$ = 4.43\,m/s$

Standard 11

Physics

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