In the figure the block on the smooth table is set into motion in a circular orbit of radius $r$ around the centre hole. The hanging mass is identical to the mass on the table and remains in equilibrium. Neglect friction. The string connecting the two blocks is massless and inextensible. Select the correct answer.

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Kinetic energy as function of $r$ is given by $\frac{m g r}{2}$

The block on table is in equilibrium.

Angular momentum about the hole is conserved even if hanging block is pulled down

The angular speed $ω$ of the block in its circular motion is $\sqrt{\frac{g}{r}}$

answer is A, B, C.

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Detailed Solution

For (A) $T = m g$

For (B) $T = \frac{m v^{2}}{r} = m ω^{2} r$

$m ω^{2} r = m g ω = \sqrt{\frac{g}{r}} . . . (1)$

K.E. of system $= \frac{1}{2} m_{A} v_{A}^{2}$

$= \frac{1}{2} (m) {(\sqrt{\frac{g}{r}})}^{2} r^{2} = \frac{m g r}{2}$

For (3), tension is central force on $A,$ so angular momentum remains conserved

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