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. `sqrt((2m_1 d)/((m_1 +m_2 )g)`

Answer

B

Explanation

Let a be the common acceleration of the system. The free body diagrams of two blocks are as shown in the figure. Their equations of motion are
` T- m_1 a …..(i) `
`m_2 g - Tm_2 a …..(ii)`
from (I ) and and (ii ) we get
` a=( m_2 g)/( m_1 +m_2) .....(iii)`
Using ` ,s-ut +1/2 at^2`
`therefore d=0 xx t +1/2 (m_2 g)/(m_1 +m_2) t^2 `
` or sqrt((2d (m_1 +m_2))/(m_2 g)) `