The time taken by a block of mass m to slide | vedclass

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(C) Let $L$ be the length of the incline.For a smooth plane, the acceleration is $a_1 = g \sin	heta$. The time taken is $t_1 = \sqrt{2L/a_1}$.For a r

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$5$/$9$

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(C) Let $L$ be the length of the incline.For a smooth plane, the acceleration is $a_1 = g \sin	heta$. The time taken is $t_1 = \sqrt{2L/a_1}$.For a rough plane, the acceleration is $a_2 = g(\sin	heta - \mu \cos	heta)$. The time taken is $t_2 = \sqrt{2L/a_2}$.Given that $t_2 = 1.5 t_1$, we have $t_2^2 = 2.25 t_1^2$.Substituting the expressions for $t_1$ and $t_2$, we get $\frac{2L}{a_2} = 2.25 \frac{2L}{a_1}$, which simplifies to $a_1 = 2.25 a_2$.Substituting $a_1$ and $a_2$, we get $g \sin	heta = 2.25 g(\sin	heta - \mu \cos	heta)$.Since $	heta = 45^\circ$, $\sin	heta = \cos	heta = 1/\sqrt{2}$.Dividing by $g \sin	heta$, we get $1 = 2.25(1 - \mu)$.Thus, $1 - \mu = 1/2.25 = 4/9$.Therefore, $\mu = 1 - 4/9 = 5/9$.

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