JEE Main 17 March 2021 Shift 2 Solved Paper

Let V is the voltage across each capacitor, Now, after removing the battery the capacitor is connected.
So, using the law of conservation of charge,
2V+8V=2×10=20
⇒‌‌10V=20⇒V=2V
As we know, Q=CV=8×2=16µC
Hence, the magnitude of the charge in C2 on equilibrium condition is 16µC.