A block of mass `m_(1)` lies on a smooth horizontal table and is connected to another freely hanging block of mass `m_(2)` by a light inextensible string passing over a smooth fixed pulley situated at the edge o fthe table. Initially the system is at rest with `m_(1)` a distance `d` from the pulley. Then the time taken for `m_(1)` to reach the pulley is.

Options

A. `(m_(2)g)/(m_(1)+m_(2))`
B. `sqrt((2d(m_(1)+m_(2)))/(m_(2)g))`
C. `sqrt((2m_(2)d)/((m_(1)+m_(2))g)`
D. None of thess

Answer

B

Explanation

`T=m_(1)a`
`M_(2)g-T=m_(2)a`
From (1) and (2): `a=(m_(2)g)/(m_(1)+m_(2)`
Now `S=ut+(1)/(2)at^(2)`
implies `d=0xxt+(1)/(2)(m_(2)g)/(m_(1)+m_(2)t^(2))`
implies `t=sqrt(2d(m_(1)+m_(2))/(m_(2)g))` .