Physics - NLM Question with Solution | TestHub

In equilibrium each spring force balances the hanging weight: kx = mg, so

$$\mathrm{k}=\frac{\mathrm{mg}}{\mathrm{x}}$$

From the figure, the (mass, extension) pairs of the four springs are (M, s), (M/3, s/3), (2M, 2s), (4M/3, 4s/3). Hence

$${\mathrm{k}}_1=\frac{\mathrm{Mg}}{\mathrm{s}}$$

$${\mathrm{k}}_2=\frac{(\mathrm{M}/3)\mathrm{g}}{\mathrm{s}/3}=\frac{\mathrm{Mg}}{\mathrm{s}}$$

$${\mathrm{k}}_3=\frac{2\mathrm{Mg}}{2\mathrm{s}}=\frac{\mathrm{Mg}}{\mathrm{s}}$$

$${\mathrm{k}}_4=\frac{(4\mathrm{M}/3)\mathrm{g}}{4\mathrm{s}/3}=\frac{\mathrm{Mg}}{\mathrm{s}}$$

All four spring constants are equal, so statement (1) is correct and (2), (4) are wrong.

Spring force equals the supported weight: Mg, Mg/3, 2Mg, 4Mg/3. The minimum is in the second spring (Mg/3), so statement (3) is correct.

Correct options: (1), (3).