JEE MAIN - Physics (2021 - 26th August Evening Shift - No. 3)
A particle of mass m is suspended from a ceiling through a string of length L. The particle moves in a horizontal circle of radius r such that $$r = {L \over {\sqrt 2 }}$$. The speed of particle will be :
$${\sqrt {rg} }$$
$${\sqrt {2rg} }$$
$${2\sqrt {rg} }$$
$${\sqrt {{{rg} \over 2}} }$$
Explanation
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$$r = {l \over {\sqrt 2 }}$$
$$\sin \theta = {r \over l} = {l \over {\sqrt 2 }}$$
$$\theta$$ = 45$$^\circ$$
$$T\sin \theta = {{m{v^2}} \over r}$$
$$T\cos \theta = mg$$
$$\tan \theta = {{{v^2}} \over {rg}} \Rightarrow v = \sqrt {rg} $$
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