In the adjacent figure is a cube having a smooth grove at an inclination of $30^{0}$ with horizontal in its vertical face. A cylinder A of mass 2 kg can slide freely inside the grove. The cube is moving with constant horizontal acceleration $a_{0}$ parallel to the shown face, so that the cylinder does not have acceleration along horizontal.

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The normal reaction acting between cylinder and cube is zero

The value of $a_{0}$ is $g \sqrt{3}$

The value of $a_{0}$ is g

Acceleration of the particle in ground frame is g

answer is A, B, D.

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Detailed Solution

The particle experiences two forces in the ground frame
i) m g (vertically down)
ii) N(which is perpendicular to the groove)
If it falls vertically in ground frame
${N=0 and a}_{PG} =g$

And the acceleration of the cube and the block must be same along the common normal.

$\Rightarrow a_{0} \cos 60^{o} = gcos 30^{o} \Rightarrow a_{0} = \sqrt{3} g$

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