The systeam shown is figure consists of `3` springs and two rods. If the temperature of the rod is increased by `DeltaT`, then the total energy stored in three springs is `beta xx (99)/(484)kL^(2)alpha^(2)(DeltaT)^(2)`. Datermine the value of `beta`. The soring are initially relaxed and there is no frication anywhere. For rod the coefficient of linear expansion is `alpha`

Answer

C

Explanation

Let `x_(1), x_(2)` and `x_(3)` be the compression is three springs respectively, when the temperature of the rod is increased by `DeltaT .`
Elongation in rod of length `L` is `l_(1) = LalphaDeltaT`
Elongation in rod of of length `(L)/(2)` is, `l_(2) = (LalphaDeltaT)/(2)`
Total elongation, `l_(1) + l_(2) = (3LalphaDeltaT)/(2)`
`x_(1) + x_(2) + x_(3) = (3LalphaDeltaT)/(2)` From equilibrium of rods
`kx_(1) = 2kx_(2) = 3kx_(3)`
Solving above equation we get,
`x_(1) = (9)/(11)LalphaDeltaT`
`x_(3) = (3)/(11)LalphaDeltaT`
Total energy stored in `3` springs is,
`U = (kx_(1)^(2))/(2) + (2kx_(2)^(2))/(2) + (3kx_(3)^(2))/(2)`
`= (297)/(484) xx Kalpha^(2)L^(2)(DeltaT)^(2)`
So, `beta = 3`