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

Options

A. No work is done in case of both the springs
B. No work is done in case of both the springs
C. No work is done in case of both the springs
D. More work is done in case of first spring

Answer

C

Explanation

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