In the given arrangement shown in figure, surface is frictionless, string is inextensible and massless, pulley `B` is frictionless. Length of the string is `l`. At `t=0`, `AB=l`. Find out the velocity of block `C`, when distance between blocks is minimum, it was released from rest at `t=0`.

Options

A. `sqrt(2gl)`
B. `sqrt(gl)`
C. `sqrt((gl)/(2))`
D. `sqrt((gl)/(4))`

Answer

C

Explanation

Distance between blocks `s^(2)=x^(2)+(l-x)^(2)`
for minimum distance `(ds)/(dx)=0`
`rArr2x-2(l-x)=0rArrx=(l)/(2)`
Acceleration of system `=(mg)/(2m)=(g)/(2)`
Velocity of `C` at that instant`=sqrt(2xx(g)/(2)xx(l)/(2))=sqrt((gl)/(2))`