A body takes n times as much time to slide do | vedclass

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(B) Let the length of the incline be $L$ and the angle of inclination be $	heta = 45^{\circ}$.For a smooth incline,the acceleration is $a_1 = g \sin

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$1-\frac{1}{n^{2}}$

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(B) Let the length of the incline be $L$ and the angle of inclination be $	heta = 45^{\circ}$.For a smooth incline,the acceleration is $a_1 = g \sin 	heta$. The time taken is $t_1 = \sqrt{\frac{2L}{g \sin 	heta}}$.For a rough incline,the acceleration is $a_2 = g(\sin 	heta - \mu \cos 	heta)$. The time taken is $t_2 = \sqrt{\frac{2L}{g(\sin 	heta - \mu \cos 	heta)}}$.Given that $t_2 = n t_1$,we have $\frac{t_2}{t_1} = n$.Squaring both sides,$\frac{t_2^2}{t_1^2} = n^2$,which implies $\frac{\sin 	heta}{\sin 	heta - \mu \cos 	heta} = n^2$.Rearranging the terms: $\sin 	heta = n^2 \sin 	heta - n^2 \mu \cos 	heta$.$n^2 \mu \cos 	heta = (n^2 - 1) \sin 	heta$.$\mu = \frac{n^2 - 1}{n^2} 	an 	heta$.Since $	heta = 45^{\circ}$,$	an 45^{\circ} = 1$.Therefore,$\mu = 1 - \frac{1}{n^2}$.

$A$ body takes $n$ times as much time to slide down a $45^{\circ}$ rough incline as it takes to slide down a smooth $45^{\circ}$ incline. The coefficient of friction between the body and the incline will be

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