The length of a spring is increased by `2.5 `cm the elastic potential energy stored in the spring is U . If the length is increased by 5 cm its elastic potential energy become `……`

Options

A. `U/2 `
B. 4U
C. 2U
D. `U/4`

Answer

B

Explanation

The elastic potential energy stored in the spring is `U = 1/2 ky^(2)`
` :. U prop y^(2)`
` :. (U_(2))/(U_(1)) = ((y_(2))/(y_(1)))^(2) = ((5.0)/(2.5))^(2) = (2)^(2) = 4 `
` U_(2) = 4U " " [ :. U_(1) = U] `