Q153 / 160 · AP EAMCET 2011 · Physics
A fully charged capacitor has a capacitance $C$. It is discharged through a small coil of resistance wire, embedded in a block of specific heat $s$ and mass $m$ under thermally isolated conditions. If the temperature of the block is raised by $\Delta T$, the potential difference $V$ across the capacitor initially is
- A$\left(\frac{2 m s \Delta T}{C}\right)^2$
- Correct answerB$\left(\frac{2 m s \Delta T}{C}\right)^{1 / 2}$
- C$\left(\frac{2 m s \Delta T}{C}\right)$
- D$2 m s \Delta T C$
Solution
Let $V$ be the potential across the capacitor when it is fully charged, then, energy stored in the capacitor will be
$=\frac{1}{2} C V^2$
(where $C=$ capacitance of capacitor)
When the capacitor is fully discharged, loss of energy is in the form of heat $=\Delta H$ As the system is thermally isolated so,
$\Delta H=\frac{1}{2} C V^2$
Here,
$\Delta H=m s \Delta T$
$\therefore \quad \frac{1}{2} C V^2=m s \Delta T$
or $V=\left(\frac{2 m s \Delta T}{C}\right)^{1 / 2}$
Hence, the correct answer is option: 'B'.