Two springs have force constants `k_(A)` and `k_(B)` such that `k_(B)=2k_(A)`. The four ends of the springs are stretched by the same force. If energy stored in spring `A` is `E`, then energy stored in spring `B` is.

Options

A. `(E)/(2)`
B. 2 E
C. 3 E
D. 4 E

Answer

A

Explanation

Given that , `(K _(B))/(K _(A)) = (2)/(1)`
Energy stored in spring is given as `U = ( F ^(2))/( 2K )`
Here F = constant
So `( U _(B))/(U _(A)) = ( K _(A))/( K _(B)) =(1)/(2)`
We get
`U _(B) = (1)/(2) U _(A) = (E)/(2)`