Q 11-05-033JEE MainAIEEE 2012Medium
Two springs $S_1$ and $S_2$ of force constants $k_1$ and $k_2$ respectively, are stretched by the same force. It is found that more work is done on spring $S_1$ than on spring $S_2$.
Statement I: If stretched by the same amount, the work done on $S_1$ will be more than that on $S_2$.
Statement II: $k_1 < k_2$.
Answer: (C) Statement I is false, Statement II is true
Same force $F$: extension $x = F/k$, so $W = \dfrac{1}{2}kx^2 = \dfrac{F^2}{2k}$. More work on $S_1$ means $k_1 < k_2$. Statement II is true.
Same extension $x$: $W = \dfrac{1}{2}kx^2$, so the stiffer spring $S_2$ needs more work. Statement I is false.
Solution by Sreeraj P, M.Sc Physics