JEE Advanced 2020 — Paper 2 — Question 13

$\frac{1}{\mathrm{C}}=\frac{1}{\mathrm{C}_{1}}+\frac{1}{\mathrm{C}_{2}}$

$\Rightarrow \frac{1}{\mathrm{C}^{2}} \Delta \mathrm{C}=\frac{1}{\mathrm{C}_{1}^{2}} \Delta \mathrm{C}_{1}+\frac{1}{\mathrm{C}_{2}^{2}} \Delta \mathrm{C}_{2}$

$\Rightarrow \frac{\Delta \mathrm{C}}{\mathrm{C}}=\mathrm{C}\left(\frac{\Delta \mathrm{C}_{1}}{\mathrm{C}_{1}^{2}}+\frac{\Delta \mathrm{C}_{2}}{\mathrm{C}_{2}^{2}}\right)==1200\left(\frac{10}{4 \times 10^{6}}+\frac{15}{9 \times 10^{6}}\right)=5 \times 10^{-3}$

Energy stored $\mathrm{U}=\frac{1}{2} \mathrm{CV}^{2} \Rightarrow \frac{\Delta \mathrm{U}}{\mathrm{U}}=\frac{\Delta \mathrm{C}}{\mathrm{C}}+2 \frac{\Delta \mathrm{~V}}{\mathrm{~V}}$

Percentage error in energy stored in the combination of capacitors.

$\frac{\Delta \mathrm{U}}{\mathrm{U}} \times 100=\left(5 \times 10^{-3}+2 \frac{0.02}{5}\right) \times 100=1.30 \%$