Q 12-02-209JEE MainJEE Main 2025 (3 Apr, Shift 2)Easy
Using a battery, a $100\ \text{pF}$ capacitor is charged to $60\ \text{V}$ and then the battery is removed. After that, a second uncharged capacitor is connected to the first capacitor in parallel. If the final voltage across the second capacitor is $20\ \text{V}$, its capacitance is (in pF):
Answer: (B) $200$
Charge is conserved once the battery is removed: $Q = 100\times60 = 6000\ \text{pC}$.
In parallel both capacitors share the common voltage:
$$(100 + C)\times20 = 6000 \Rightarrow C = 200\ \text{pF}$$
Solution by Sreeraj P, M.Sc Physics