Physics - NLM Question with Solution | TestHub

Before the cut, the string simply supports the hanging block:

$${\mathrm{T}}_{\text{before }}=\mathrm{mg}$$

The two parallel springs together support the bar and the hanging block (total weight 2mg), so each spring carries mg.

Just after one spring is cut, the remaining spring has not yet changed its length, so it still pulls up with only mg while the total weight is 2mg. The system (bar + block) accelerates downward with

$$\mathrm{a}=\frac{2\mathrm{mg}-\mathrm{mg}}{2\mathrm{m}}=\frac{\mathrm{g}}{2}$$

For the hanging block:

$$\mathrm{mg}-{\mathrm{T}}_{\text{after }}=\mathrm{m}\times \frac{\mathrm{g}}{2}\Rightarrow {\mathrm{T}}_{\text{after }}=\frac{\mathrm{mg}}{2}$$

$$\Delta \mathrm{T}={\mathrm{T}}_{\text{before }}-{\mathrm{T}}_{\text{after }}=\mathrm{mg}-\frac{\mathrm{mg}}{2}=\frac{\mathrm{mg}}{2}$$

Correct option: (A).