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
Concept. Spring energy can be written two ways: $U=\tfrac12kx^2$ (same stretch) or $U=\dfrac{F^2}{2k}$ (same force).
Step 1. Same force. More work on $S_1$ means $\dfrac{F^2}{2k_1}>\dfrac{F^2}{2k_2}\Rightarrow k_1\lt k_2$. Statement II is true.
Step 2. Same stretch. $U\propto k$, so $S_1$ (smaller $k$) needs less work. Statement I is false.
Answer. Statement I is false, Statement II is true (Option C).
Solution by Sreeraj P, M.Sc Physics
