A variable capacitor is connected to a 200 V battery. If its capacitance is changed from 2$\mathrm{\mu }$F to X $\mathrm{\mu }$F, the decrease in energy of the capacitor is 2 x 10^-2 J. The value of X is –

Energy of a capacitor $\mathrm{U}=-\frac{1}{2}{\mathrm{CV}}^2$
[Where V is voltage/potential difference between plates]
Charge in energy = Final energy of capacitor - Initial energy of capacitor
$\begin{array}{l}-2\times {10}^{-2}\mathrm{J}=\frac{1}{2}\cdot \mathrm{X}\cdot (200{)}^2-\frac{1}{2}\cdot (2\mathrm{\mu F})\cdot (200{)}^2\\ -2\times {10}^{-2}=\left(\frac{4\times {10}^4}{2}\right)\mathrm{X}-2\mathrm{\mu F}\\ -{10}^{-6}=\mathrm{X}-2\mathrm{\mu F}\\ \mathrm{X}=2\mathrm{\mu F}-{10}^{-6}\\ \mathrm{X}=2\mathrm{\mu F}-1\mathrm{\mu F}=1\mathrm{\mu F}\end{array}$