In the diagram shown below, the rod is uniform having mas M and length L. When it is making an angle `= 60^(@)` the rod is released from the state of rest (and spring in its natural lenth) The rod comes to state of rest when the angle theta reduces to `30^(@)`. What should be the initial compression in the spring (when `theta = 60^(@))` so that the whole system can remain in the state of rest? (there is no friction between any surfaces)

Options

A. `((sqrt(3)-1))/(2sqrt(3)`
B. `3(sqrt(3)-1)/(2)l`
C. `((sqrt(3)-1))/(4sqrt(3))l`
D. `(2sqrt(3-1))/sqrt(3)l`

Answer

C

Explanation

`mg,(1)/(2)(sin60-sin30)=(1)/(2)kx^(2)`
`x=lcos30-lcos60`
`=mg,(1)/(2)(sqrt(3)/(2)-(1)/(2))=(1)/(2)kl^(2)(sqrt(3)/(2)-(1)/(2))^(2)`
`= (2mg)/(l(sqrt(3-1)))=k`

`tau_(A)=0implies mg ,(l)/(2)costheta=N_(1)lsintheta`
`N_(1)=(mgcottheta)/(2)`
`N_(1) = kx implies x=(mgcottheta)/(2k)`
`x=(mg cot 60^(@))/(2,(2mg)/(1.(sqrt(3-1))))=(sqrt(3-1))/(4sqrt(3))l`