\(=54\, \mu \mathrm{F}\)
The charge will be distributed in the ratio of capacitance \(\Rightarrow \quad \frac{q_{1}}{q_{2}}=\frac{2.4}{3}=\frac{4}{5}\)
\(\therefore \quad 9 X=54\, \mu C\)
\(\therefore \quad X=6\, \mu C\)
Charge on \(4 \,\mu F\) capacitor will be \(=4 \mathrm{X}=4 \times 6\, \mu \mathrm{C}\)
\(=24\, \mu C\)