A cube of mass m and side length 4R has a groove made inside it in a vertical plane. The groove is a circle having radius R. A particle of mass m is moving down with a velocity $V = \sqrt{2 g R}$ and cube is at rest at that moment. Assume all surfaces are smooth. The correct statement(s) at this instant.

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Acceleration of block w.r.t ground is g

Acceleration of particle with respect to ground is $g \sqrt{2}$

Normal reaction on cube by the ground is mg

Normal reaction on particle is mg

answer is A, B, C, D.

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Detailed Solution

$N_{2} - m g = 0$

$\Rightarrow N_{2} = m g N_{1} = m a N_{1} + m a = m (\frac{V^{2}}{R}) N_{1} + m a = m 2 g \Rightarrow N_{1} = m g A n d a = g {\vec{a}}_{p . q} = {\vec{a}}_{p . c} + {\vec{a}}_{c . g} \to g ↓ = g \sqrt{2} s$

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