Two springs have their force constant as `k_(1)` and `k_(2) (k_(1) gt k_(2))`. When they are stretched by the same force.

Options

A. net work is done in case of both the same springs
B. Equal work is done in case of both the springs
C. More work is done in case of both the second springs
D. More work is done in case of both the first springs

Answer

C

Explanation

`W=(F^(2))/(2k)`
If both springs are stretched by same force then
`W=prop(1)/(2)`
As `k_(1)gt k_(2)` therefore `W_(1) lt W_(2)`
i.e, more work is done in case of second spring.