A light rod carries three equal masses A, B and C as shown in figure. Find the velocity of B in vertical position of rod, if it is released from horizontal position. As shown in figure.

Options

A. `sqrt((8gl)/(7))`
B. `sqrt((4gl)/(7))`
C. `sqrt((2gl)/(7))`
D. `sqrt((10gl)/(7))`

Explanation

Applying law of conservation of mechenical energy. Loss is gravitational P.E. = Gain in rotational K.E. i.e.,
`mg.(l)/(3)+mg((2l)/(3))+mgl=(1)/(2)(m((l)/(3))^(2)+m((2l)/(3))^(2)+ml^(2))omega^(2)`
`rArr omega = sqrt((36g)/(14l))rArr V_(B)=omega l_(B)=(2l)/(3)sqrt(36g)/(14l))=sqrt(8gl)/(7))`.