Problem 145 - A Complete Resource Book in Physics for JEE Main

Question

A capacitor 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 a thermally insulated block of specific heat \(2.5 \times 10^{2} \mathrm{~J} / \mathrm{kg}-\mathrm{K}\) and of mass \(0.1 \mathrm{~kg}\). If temperature of the block rises by \(0.4 \mathrm{~K}\). Find the value of \(C\).

Step-by-Step Solution

Answer

The value of capacitance C is \(5 \times 10^{-4} F\).

Step 1: Write the equation for the energy stored in the capacitor.

The energy stored in a capacitor can be calculated using the formula: \[E_C = \frac{1}{2} \cdot C \cdot V^2\] where \(E_C\) is the energy stored in the capacitor, C is the capacitance, and V is the voltage across the capacitor. In this case, V = 200 V.

Step 2: Write the equation for the heat gained by the block.

The heat gained by the block can be calculated using the formula: \[Q = m \cdot c \cdot \Delta T\] where Q is the heat gained by the block, m is the mass of the block, c is the specific heat of the block, and \(\Delta T\) is the change in temperature. In this case, m = 0.1 kg, c = 2.5 x 10^2 J/(kg K), and \(\Delta T\) = 0.4 K.

Step 3: Equate the energy stored in the capacitor with the heat gained by the block.

Since the energy stored in the capacitor is equal to the heat gained by the block, we can equate the two formulas from Steps 1 and 2. \[\frac{1}{2} \cdot C \cdot V^2 = m \cdot c \cdot \Delta T\]

Step 4: Solve for C.

Now we can substitute the known values into the equation and solve for C. \[\frac{1}{2} \cdot C \cdot (200)^2 = (0.1) \cdot (2.5 \times 10^2) \cdot (0.4)\] Simplify the equation: \[20000 C = 10\] Now, divide both sides by 20000 to find the value of C: \[C = \frac{10}{20000} = 5 \times 10^{-4} F\] So, the value of capacitance C is \(5 \times 10^{-4} F\).