A uniform rod of mass `m` and length `l` is applied pivoted at point `O`. The rod is initially in vertical position and touching a block of mass `M` which is at rest on a horizontal surface. The rod is given a slight jerk and it starts rotating about point `O` this causes the block to move forward as shown The rod loses contact with the block at `theta=30^(@)` all surfaces are smooth now answer the following questions.
Q. The velocity of block when the rod loses contact with the block is

Options

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

Explanation

`(M)/(m)=(4)/(3)`
`M=(4)/(3)m`
`omegalcos60^(@)=(omegal)/(2)`
`(omegal)/(2)=vthereforeomega=(2v)/(l)`
decrease in potential energy of rod
`=` increase in rotaional kinetic energy of rod
`+` translational kinetic energy of block
`thereforemg((l)/(2)-(l)/(2)sin30^(@))=(1)/(2)((ml)/(3))^(2)((2v)/(l))^(2)`
`+(1)/(2)((4m)/(3)(v^(2))`
Solving this equation we get
`v=(sqrt(3gl))/(4)`