JEE Main · 2019 · 12 Apr 2019 · Shift 2 · Capacitance

Reading the network: the $1\mu \mathrm{F}$ and $5\mu \mathrm{F}$ capacitors are connected between the same pair of nodes, so they are in parallel and add to $1+5=6\mu \mathrm{F}$. This $6\mu \mathrm{F}$ is in series with the $4\mu \mathrm{F}$, giving $\frac{4\times 6}{4+6}=2.4\mu \mathrm{F}$ for that upper branch. The upper branch and the $3\mu \mathrm{F}$ branch are each joined straight across the $10\mathrm{V}$ source, so they sit in parallel and each feels the full $10\mathrm{V}$. The charge drawn by the upper branch is therefore $Q=C_{\text{branch}}V=2.4\times 10=24\mu \mathrm{C}$. Since the $4\mu \mathrm{F}$ lies in a series path within that branch, the very same $24\mu \mathrm{C}$ flows through it — every capacitor in a series chain carries an identical charge — so the charge stored on the $4\mu \mathrm{F}$ capacitor is $24\mu \mathrm{C}$.