Two springs have their force constants `K_(1)` and `K_(2)` and they are stretched to the same extension. If `K_(2) gt K_(1)` work done is

Options

A. More in spring B
B. Nothing can be said
C. More in spring A
D. Equal in both

Answer

C

Explanation

The work done in stretching a spring through a dis"tan"ce x is `(1)/(2) Kx^(2)` . But as both the springs are stretched by the same amount, `W prop K`
`therefore (W_(A))/(W_(B))=(K_(A))/(K_(B)) because K_(A) gt K_(B), W_(A) gt W_(B)`