A rupee coin starting from rest rolls down a distance of 1 m on an inclined plane at angle of $30{^\circ}$ with the horizontal. Assuming that $g\ = \ 9.81$ $ms^{- 1}$, time taken is

Question

Dr. Abubacker · Accepted Answer

(d) Here, $l = 1m,\ 	heta = 30{^\circ},\ g = 9.81\ \ ms^{- 2},\ t = ?$
$t = \sqrt{\frac{2l(1 + K^{2}/R^{2})}{g\sin	heta}}$
For a rupee coin, $K^{2} = \frac{1}{2}R^{2}$
$t = \sqrt{\frac{2 	imes 1(1 + 1/2)}{9.81\sin{30{^\circ}}}} = \sqrt{\frac{6}{9.81}} = 0.78\ s$

Q. A rupee coin starting from rest rolls down a distance of 1 m on an inclined plane at angle of $30{^\circ}$ with the horizontal. Assuming that $g\ = \ 9.81$ $ms^{- 1}$, time taken is

Explanation

Reviewed by Dr. Abubacker

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