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A spring with spring constant K when str...

A spring with spring constant `K` when stretched through `1 cm`, the potential energy is `U`. If it stretched by `4 cm`, the potential energy will be

A

`4U`

B

`8U`

C

`16U`

D

`2U`

Text Solution

Verified by Experts

The correct Answer is:
`(3)`
`U = 1/2 Kx^(2) Rightarrow U_(1)/U_(2) = x_(1)^(2)/x_(2)^(2)`
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