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(C) In the first case,the block is on the verge of moving upwards,so the friction force acts downwards. The force balance equation is: $F = mg \sin 30

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$\frac{\sqrt{3} - 1}{\sqrt{3} + 1}$

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(C) In the first case,the block is on the verge of moving upwards,so the friction force acts downwards. The force balance equation is: $F = mg \sin 30^{\circ} + \mu mg \cos 30^{\circ} \dots (1)$In the second case,the block is on the verge of sliding downwards,so the friction force acts upwards. The force balance equation is: $F + \mu mg \cos 60^{\circ} = mg \sin 60^{\circ} \implies F = mg \sin 60^{\circ} - \mu mg \cos 60^{\circ} \dots (2)$Since the magnitude of $F$ is the same in both cases,equate $(1)$ and $(2)$:$mg \sin 30^{\circ} + \mu mg \cos 30^{\circ} = mg \sin 60^{\circ} - \mu mg \cos 60^{\circ}$$\mu (\cos 30^{\circ} + \cos 60^{\circ}) = \sin 60^{\circ} - \sin 30^{\circ}$$\mu (\frac{\sqrt{3}}{2} + \frac{1}{2}) = \frac{\sqrt{3}}{2} - \frac{1}{2}$$\mu = \frac{\sqrt{3} - 1}{\sqrt{3} + 1}$

Column-$I$	Column-$II$
$(1)$ Static friction	$(a)$ Limiting friction
$(2)$ Rolling friction	$(b)$ Ball bearing
	$(c)$ Object moving on a road

Similar Questions

Consider the following statements $A$ and $B$ and identify the correct answer given below:
$(A)$ $A$ body initially at rest is acted upon by a constant force. The rate of change of its kinetic energy varies linearly with time.
$(B)$ When a body is at rest, it must be in equilibrium.

$A$ body of mass $m = 10^{-2} \ kg$ is moving in a medium and experiences a frictional force $F = -kv^2$. Its initial speed is $v_0 = 10 \ ms^{-1}$. If,after $10 \ s$,its energy is $\frac{1}{8} mv_0^2$,the value of $k$ will be

Match Column-$I$ with Column-$II$ correctly.

Column-$I$ Column-$II$

$(1)$ Static friction $(a)$ Limiting friction

$(2)$ Rolling friction $(b)$ Ball bearing

$(c)$ Object moving on a road

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