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 value of ratio `M//m` is a) 2:3 b) 3:2 c) 4:3 d) 3:4

Options

A. `(a)2:3`
B. `(b)3:2`
C. `(c)4:3`
D. `(d)3:4`

Explanation

At angle `theta` shown in figure.
decrease in potential energy of rod
`=` increase in rotational kinetic energy of rod
`+` translational kinetic energy of blocK. ,brgt but `v=v_(p)sintheta=(lomega)sintheta`
`thereforemg((l)/(2)-(l)/(2)sintheta)=(1)/(2)((ml^(2))/(3)omega^(2))`
`+(1)/(2)M(omegal sintheta)`
From here we get
`lomega^(2)=a_(n)=(mg(1-sintheta))/(m//3+Msin^(2))theta)`
`a_(t)=lalpha=l((tau)/(I))_(0)l[(mg(l)/(2)costheta-Nlsintheta)/(ml^(2)//3)]`
`=(3)/(2)gcostheta-(3Nsintheta)/(m)` .(ii)
`N=Ma=M[a_(t)sintheta-a_(n)costheta]` ..(iii)
Now putting values of `a_(t)` and `a_(n)` from eqs. (i) and (ii) in eqs (iii) and then putting `N=0` and `theta=30^(@)` in the equation we get,
`(M)/(m)=(4)/(3)`