`A` and `B` are smooth light hinges equidistant from `C`, which can slides on `ABC`. The spring of force constant `K` is fixed at its one end `C` and conncented to light rods `AD` and `BD` at point `D`. A block of mass `m` is suspended at `D`. Find the velocity of the block, when `/_ CAD` changes from `30^@` to `45^@. AD = BD = L`.
.

Options

A. `[gL-(KL^(2))/(2m) (sqrt(2) -1)^(2)]^((1)/(2))`
B. `[gL sqrt(2) -(KL^(2))/(2m) (sqrt(2) -1)^(2)]^((1)/(2))`
C. `[gL(sqrt(2) -1)-(KL^(2))/(4m) (sqrt(2)-1)^(2)]^((1)/(2))`
D. `[gL -(KL^(2))/(2m)]^((1)/(2))`

Answer

C

Explanation

Initially, `CD = L sin 30^@ = (L)/(2)`
Finally, `CD = L sin 45^@ = (L)/(sqrt(2))`
Increase in elastic `PE` = Increase
in `PE` = Decrease in `PE`
`(1)/(2)K((L)/(sqrt(2))-(L)/(2))^(2) +(1)/(2)mv^(2)=mg ((L)/(sqrt(2))-(L)/(2))`
On solving, `v =[gL (sqrt(2)-1)-(KL^(2))/(4m)(sqrt(2) -1)^(2)]^((1)/(2))`.