Given condition: same force stretches both springs

For a spring, if a force $F$ stretches it by $x$, then by Hooke’s law: $F = kx \Rightarrow x = \frac{F}{k}$

Work done in stretching from $0$ to $x$ is: $W = \frac{1}{2}kx^2$ Using $x=\frac{F}{k}$, $W = \frac{1}{2}k\left(\frac{F}{k}\right)^2 = \frac{F^2}{2k}$

Since both springs are stretched by the same force and it is given that more work is done on $S_1$ than on $S_2$, $W_1 > W_2$ $\frac{F^2}{2k_1} > \frac{F^2}{2k_2}$ $\frac{1}{k_1} > \frac{1}{k_2}$ $k_1 < k_2$

So Statement 2 is true.

Check Statement 1

Statement 1 says: if stretched by the same amount, work done on $S_1$ is more than on $S_2$.

If both are stretched by the same amount $x$, then $W = \frac{1}{2}kx^2$ Hence, $W_1 = \frac{1}{2}k_1x^2, \quad W_2 = \frac{1}{2}k_2x^2$

Since from above $k_1 < k_2$, $W_1 < W_2$

Therefore, Statement 1 is false.

Conclusion

• Statement 1: False
• Statement 2: True

Hence the correct option is: $\boxed{A}$

Comparison with stored answer

Stored correct answer: $A$

My derived answer also is $A$, so they agree.