A weightless rod of length `2l` carries two equal masses 'm', one tied at lower end `A` and the other at the middle of the rod at `B`. The rod can rotate in vertical plane about a fixed horizontal axis passing thriugh `C` . The rod of is released from rest in horizontal possion. The speed of the mass `B` at the instant rod become vertical is:

Options

A. `sqrt((3 gl)/(5))`
B. `sqrt((4 gl)/(5))`
C. `sqrt((6 gl)/(5))`
D. `sqrt((7 gl)/(5))`

Answer

C

Explanation

Let `v` be the speed of `B` at lowermost possition, the speed of `A` at lowermost position is `2v`
From conservation of energy
`(1)/(2)m (2v)^(2) + (1)/(2)m v^(2) = mg (2l)+ mg l`
Solving we get `v = sqrt((6)/(5)) gl`