JEE MAIN - Physics (2023 - 29th January Evening Shift - No. 9)
The time taken by an object to slide down 45$$^\circ$$ rough inclined plane is n times as it takes to slide down a perfectly smooth 45$$^\circ$$ incline plane. The coefficient of kinetic friction between the object and the incline plane is :
$$1 - {1 \over {{n^2}}}$$
$$1 + {1 \over {{n^2}}}$$
$$\sqrt {1 - {1 \over {{n^2}}}} $$
$$\sqrt {{1 \over {1 - {n^2}}}} $$
Explanation
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Smooth case:
$$a = g\sin 45^\circ = {9 \over {\sqrt 2 }}$$
$${t_1} = \sqrt {{{2L} \over a}} = \sqrt {{{2L} \over {g/\sqrt 2 }}} = \sqrt {{{2\sqrt 2 L} \over g}} $$ ..... (1)
Rough case:
$$a = g\sin 45^\circ - \mu g\cos 45^\circ $$
$$ = {g \over {\sqrt 2 }}(1 - \mu )$$
$${t_2} = \sqrt {{{2L} \over a}} = \sqrt {{{2\sqrt 2 L} \over {g(1 - \mu )}}} $$ ..... (2)
From (1) to (2) and $${t_1} = {{{t_2}} \over n}$$ we have
$$\sqrt {{{2\sqrt 2 L} \over g}} = {1 \over n}\sqrt {{{2\sqrt 2 L} \over {g(1 - \mu )}}} \Rightarrow \mu = 1 - {1 \over {{n^2}}}$$
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