JEE Physics 2025 Question with Solution

Given: The first capacitor has capacitance $C_1 = 100 \, \text{pF}$ and initial voltage $V_i = 60 \, \text{V}$. The second capacitor is initially uncharged. After connection in parallel, the final common voltage is $V_f = 20 \, \text{V}$.

Find: The capacitance $C_2$ of the second capacitor.

When a charged capacitor is connected in parallel to an uncharged capacitor, charge redistributes until both capacitors have the same final voltage. Total charge is conserved.

Initial charge on the first capacitor:

$$Q_i = C_1 V_i = (100 \, \text{pF})(60 \, \text{V}) = 6000 \, \text{pC}$$

Final charge on the first capacitor:

$$Q_{f1} = C_1 V_f = (100 \, \text{pF})(20 \, \text{V}) = 2000 \, \text{pC}$$

Final charge on the second capacitor:

$$Q_{f2} = C_2 V_f = C_2 (20 \, \text{V})$$

By conservation of charge:

$$Q_i = Q_{f1} + Q_{f2}$$ $$6000 \, \text{pC} = 2000 \, \text{pC} + C_2 (20 \, \text{V})$$ $$4000 \, \text{pC} = C_2 (20 \, \text{V})$$ $$C_2 = \frac{4000 \, \text{pC}}{20 \, \text{V}} = 200 \, \text{pF}$$

Therefore, the capacitance of the second capacitor is $200 \, \text{pF}$. The correct option is B.

Given: $C_1 = 100 \, \text{pF}$, $V_i = 60 \, \text{V}$, and final voltage $V_f = 20 \, \text{V}$.

Find: $C_2$.

For a charged capacitor connected in parallel with an uncharged capacitor, the final voltage is

$$V_f = \frac{C_1 V_i}{C_1 + C_2}$$

Substitute the given values:

$$20 = \frac{(100)(60)}{100 + C_2}$$ $$20(100 + C_2) = 6000$$ $$2000 + 20C_2 = 6000$$ $$20C_2 = 4000$$ $$C_2 = \frac{4000}{20} = 200 \, \text{pF}$$

Therefore, the capacitance of the second capacitor is $200 \, \text{pF}$, so the correct option is B.