1. Initial energy stored in the charged capacitor

A capacitor of capacitance $C$ charged to potential difference $v$ stores energy

$U = \frac{1}{2}Cv^2.$

2. Energy converted into heat

The capacitor is discharged through the resistance wire embedded in the thermally insulated block.

Since the block is thermally insulated, all the electrical energy stored in the capacitor is converted into heat absorbed by the block.

Heat gained by the block is

$Q = ms\Delta T,$

where:

• $m$ = mass of block,
• $s$ = specific heat capacity,
• $\Delta T$ = rise in temperature.

3. Apply conservation of energy

Thus,

$\frac{1}{2}Cv^2 = ms\Delta T.$

4. Solve for $v$

$v^2 = \frac{2ms\Delta T}{C}$

$v = \sqrt{\frac{2ms\Delta T}{C}}.$

5. Match with the options

This corresponds to Option C.

Answer:

$\boxed{v = \sqrt{\frac{2ms\Delta T}{C}}}$