Q 12-02-156JEE MainJEE Main 2020 (2 Sep, Shift 2)Easy
A $10\ \mu\text{F}$ capacitor is fully charged to a potential difference of $50\ \text{V}$. After removing the source voltage it is connected to an uncharged capacitor in parallel. Now the potential difference across them becomes $20\ \text{V}$. The capacitance of the second capacitor is:
Answer: (A) $15\ \mu\text{F}$
Charge is conserved: $10\times50 = (10 + C)\times20 \Rightarrow C = 15\ \mu\text{F}$.
Solution by Sreeraj P, M.Sc Physics