Rod `AO_(3)` of length `L` can rotate abput `A`. Initially rod was at position `AO_(2)`, when spring `OB` of force constant `K`, attached to block `B` of mass `m` was at position `OA` with unstretched length `L`. The smooth block `B` can slide on rod when pulled by the block `D` of mass `m` through a massless spring and smooth pulley at `O_(1)`.
Find the velocity of the block `B`, when the rod and spring at `B` make an angle of `30^@` with their respective initial positions : (`B` is the middle point of the block)
.

Options

A. `[(10mgL-KL^(2)(2- sqrt(3))^(2))/(8m)]^((1)/(2))`
B. `[(2mgL-KL^(2)(sqrt(2)-1))/(4m)]^((1)/(2))`
C. `[(5mgL-KL^(2)(sqrt(2)-1))/(4m)]^((1)/(2))`
D. `[(6mgL-KL^(2)(sqrt(2)-1))/(4m)]^((1)/(2))`

Answer

A

Explanation

`/_ ABO = 90^@`
(since, `/_ AOB = 30^@` and `/_OAB = 60^@`)
`OB =L cos 30^@ = (L sqrt(3))/(2) ,AB = (L)/(2)`
`BC = BA sin 30^@ = (L)/(4)`
Distance by which `B ahs` gone down `= BC = (L)/(4)`
Distance by which `D` has gone down `= AB = L sin 30^@ = L`
Decrease in `PE` = Increase in `KE` + Increase in elastic `PE`
``mgL +mg `(L)/(4) =(1)/(2)xx 2m xxv^(2) +(1)/(2) L(L-L(sqrt(3))/(2))^(2)`
on solving, `v =[(10mgL-KL^(2)(2- sqrt(3))^(2)]/(8m)]^((1)/(2))`.