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In the system shown in the figure the mass m moves in a circular path of angular amplitude $60^{\circ}$ , and mass 4m is at rest.

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The minimum value of coefficient of friction between the mass 4 m and the surface of table is 0.5

The work done by gravitation force on the block m is positive when it moves from A to B

The power delivered by the tension when m moves from A to B is non-zero

The kinetic energy of m in position B equals the work done by gravitational force on the block when the moves from position A to position B.

answer is A, B, D.

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Detailed Solution

$T = \frac{M_{A} g}{2}$
$\frac{M_{A} g l}{2} = \frac{1}{2} M_{A} V^{2}$
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$\Rightarrow \frac{M_{A} V^{2}}{l} = M_{A} g$

$\Rightarrow T = 2 M_{A} g$ (maximum tension)
$\Rightarrow f = T \Rightarrow f_{\max} = 2 m_{A} g \Rightarrow 2 m g \leq μ (4 M g) \Rightarrow μ \geq \frac{1}{2}$

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