Q93 / 120 · VITEEE 2015 · Physics
A condenser of capacitance $\mathrm{C}$ is fully charged by a $200 \mathrm{~V}$ supply. It is then discharged through a small coil of resistance wire embedded in thermally insulated block of specific heat $250 \mathrm{~J} / \mathrm{kg}-\mathrm{K}$ and of mass $100 \mathrm{~g}$. If the temperature of the block rises by $0.4 \mathrm{~K}$, then the value of $\mathrm{C}$ is
- A$300 \mu \mathrm{F}$
- B$200 \mu \mathrm{F}$
- C$400 \mu \mathrm{F}$
- Correct answerD$500 \mu \mathrm{F}$
Solution
The energy stored in the capacitor
$\mathrm{U}=\frac{1}{2} \mathrm{CV}^{2}=\frac{1}{2} \mathrm{C} \times(200)^{2}=2 \mathrm{C} \times 10^{4} \mathrm{~J}$
This energy is used to heat up the block. Let $\Delta \theta$ be the rise in temperature, then heat energy
$\mathrm{Q}=\mathrm{ms} \Delta \theta=0.1 \times 250 \times 0.4=10 \mathrm{~J}$
$\begin{aligned} & \text { Now, } \\ & 2 \mathrm{C} \times 10^{4}=10 \\ \Rightarrow & \mathrm{C}=\frac{10}{2 \times 10^{4}}=5 \times 10^{-4}=500 \mu \mathrm{F} \end{aligned}$
Hence, the correct answer is option: 'D'.