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Two similar springs P and Q have spring constants $K_P$ and $K_Q$, such that $K_P > K_Q$. They are stretched, first by the same amount (case a), then by the same force (case b). The work done by the springs $W_P$ and $W_Q$ are related as in case (a) and case (b), respectively:
A
$W_P = W_Q;\ W_P > W_Q$
B
$W_P = W_Q;\ W_P = W_Q$
C
$W_P > W_Q;\ W_Q > W_P$
D
$W_P < W_Q;\ W_Q < W_P$
Detailed Solution
Given $K_P > K_Q$.
Case (a), same extension x: $W = \frac{1}{2}Kx^2 \Rightarrow W \propto K$, so $W_P > W_Q$.
Case (b), same force F: $x = \frac{F}{K}$, so $W = \frac{F^2}{2K} \Rightarrow W \propto \frac{1}{K}$, so $W_Q > W_P$.
Case (a), same extension x: $W = \frac{1}{2}Kx^2 \Rightarrow W \propto K$, so $W_P > W_Q$.
Case (b), same force F: $x = \frac{F}{K}$, so $W = \frac{F^2}{2K} \Rightarrow W \propto \frac{1}{K}$, so $W_Q > W_P$.