Question

A condenser of capacitance \( C \) is fully charged by a 200V supply. It is then discharged through a small coil of resistance \( r \) embedded in thermally insulated block of specific heat 250 J/K-g and mass 100 g. If the temperature of the block rises by 0.4 K, then the value of \( C \) is

• A. 300 \(\mu\)F
• B. 200 \(\mu\)F
• C. 400 \(\mu\)F
• D. 500 \(\mu\)F

Hint:

Energy dissipated by the capacitor equals the heat absorbed by the block, allowing us to find the capacitance.

Answer 1

The Correct Option is C

Using the energy dissipated in the resistor and equating it to the thermal energy absorbed by the block, we can solve for \( C \). The energy dissipated by the capacitor is given by \( E = \frac{1}{2} C V^2 \), and the thermal energy is \( Q = m c \Delta T \).