Two springs with spring constants m ` K _(1) = 1500 N//m`and m `K_(2) = 3000 N//m` are stretched by the same force. The ratio of potential energy stored in spring will be

Options

A. `1 :4`
B. `4 : 1`
C. `2 : 1`
D. `1 : 2`

Answer

C

Explanation

`F= Kx therefore x=(F)/(K)`
and `P.E. (U)=(1)/(2)Kx^(2)=(1)/(2)K ((F^(2))/(K^(2)))=(1)/(2)(F^(2))/(K)`
`therefore (U_(1))/(U_(2))=(1)/(2)(F^(2))/(K_(1))xx(K_(2))/(2F^(2)) (because ` F is the same for both )
`=(K_(2))/(K_(1))=(3000)/(1500)=(2)/(1)`