In the figure shown all the surface are smooth. All the blocks A,B and C are movable x-axis is horizontal and y-axis vertical as shown. Just after the system is relased from the position as shown.

Options

A. Acceleration of `A` relative to ground is in negative y-direction
B. Acceleration of `A` relative to B is in positive x-direction
C. The horizontal acceleration of `B` relative to ground is in negative x-direction.
D. The acceleration of `B` relative to ground directed along the inclined surface of `C` is greater than `gsintheta`.

Answer

A, B, C, D

Explanation

There is no horizontal force on block A, therefore it does not move in x-direction, whereas there is net downward force (`mg-N`) is actiong on it, making its acceleration along negative y-direction.
Block B moves downward as well as in negative x-direction. Downward acceleration of A and B will be equal due to constrain, thus w.r.t. B, A moves in positive x-direction. Due to the component of normal exterted by C on B, it moves in negative x-direction. The force acting vertically downward on block B are mg and `N_A` (normal reaction due to block A). Hence the component of net force on block B along the inclined surface of B is greater than `mgsintheta`. Therefore the acceleration of `B` relative to ground directed along the inclined surface of `C` is greater than `gsintheta`
`T=(mg)/(4)`