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(D) Let $C$ be the center of mass of the block. The height of the center of mass from the base is $h/2$.For the block to topple,the torque due to the

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$\mu > \frac{a}{2h}$

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(D) Let $C$ be the center of mass of the block. The height of the center of mass from the base is $h/2$.For the block to topple,the torque due to the frictional force $f$ about the edge of the base must be greater than the torque due to the normal force $N$ about the same edge.Taking torque about the front edge of the base:$	au_{f} = f 	imes h$$	au_{N} = N 	imes \frac{a}{2}$For toppling,$	au_{f} > 	au_{N} \implies f 	imes h > N 	imes \frac{a}{2}$.Since the block is moving,$f = \mu N$.Substituting this,$\mu N h > N \frac{a}{2}$.$\mu h > \frac{a}{2} \implies \mu > \frac{a}{2h}$.Thus,the block will topple if $\mu > \frac{a}{2h}$.

$A$ rectangular block has a square base measuring $a \times a$ and its height is $h$. It moves on a horizontal surface in a direction perpendicular to one of the edges. The coefficient of friction is $\mu$. It will topple if

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