Question

A fully charged capacitor has a capacitance ‘C’. It is discharged through a small coil of resistance wire embedded in a thermally insulated block of specific heat capacity ‘s’ and mass ‘m’. If the temperature of the block is raised ‘$\Delta T$’, the potential difference ‘V’ across the capacitance is:
$
  {\text{A}}{\text{. }}\dfrac{{mC\Delta T}}{s} \\
  {\text{B}}{\text{. }}\sqrt {\dfrac{{2mC\Delta T}}{s}} \\
  {\text{C}}{\text{. }}\sqrt {\dfrac{{2ms\Delta T}}{C}} \\
  {\text{D}}{\text{. }}\dfrac{{ms\Delta T}}{C} \\
$

Answer 1

Hint: Use the conservation of energy principle. Compare the energy stored by the capacitor and the heat energy stored in the block to get the value of potential difference.

Formula Used:
${E_c} = \dfrac{{C{V^2}}}{2}$, energy stored in a capacitor.
$H = ms\Delta T$, heat energy stored in the block, raised by temperature $\Delta T$.

Complete Step-by-Step solution:
As energy is transformed,
So, H = E
$
  ms\Delta T = \dfrac{{C{V^2}}}{2} \\
  V = \sqrt {\dfrac{{2ms\Delta T}}{C}} \\
$
The correct option is (C).

Note: Since, the capacitor is connected to the block by a wire, it implies that the energy stored in the capacitor is transferred to the block and it gets stored as the heat energy in the block. So, simple conservation of energy.