Q. 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 by $\mathrm{\Delta}T$, the potential difference $V$ across the capacitance is
A) $\sqrt{\frac{2mC\mathrm{\Delta}T}{s}}$
B) $\frac{mC\mathrm{\Delta}T}{s}$
C) $\frac{ms\mathrm{\Delta}T}{C}$
D) $\sqrt{\frac{2m\ s\mathrm{\Delta}T}{C}}$
Explanation
(d) $E = \left( \frac{1}{2} \right)CV^{2}$ ...(i)
The energy stored in capacitor is lost in form of heat energy
$H = ms\ \mathrm{\Delta}T\ $...(ii)
From Eqs. (i) and (ii), we have
$ms\ \mathrm{\Delta}T = \left( \frac{1}{2} \right)CV^{2}$
$V = \ \sqrt{\frac{2\ ms\mathrm{\Delta}T}{C}}$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.