`K` is the force constant of a spring. The work done in increasing its extension from `l_1` to `l_2` will be

Options

A. `- 150 J`
B. `50J`
C. `150 J`
D. None of these

Answer

C

Explanation

`U_(2) = (1)/(2)Kx_(2)^(2)`
`U_(1) = (1)/(2)Kx_(1)^(2)`
`W_(ext)=Delta U_(2) - U_(1)`
`(1)/(2)Kx_(2)^(2)- (1)/(2)Kx_(1)^(2) = (1)/(2)K(x_(2)^(2)- x_(1)^(2))`
`= (1)/(2) xx 100 xx ( ((20)/(100))^(2) - ((10)/(100))^(2)) xx 150 J`