A container has a base of $50 \mathrm{~cm} \times 5 \mathrm{~cm}$ and height $50 \mathrm{~cm}$, as shown in the figure. It has two parallel electrically conducting walls each of area $50 \mathrm{~cm} \times 50 \mathrm{~cm}$. The remaining walls of the container are thin and non-conducting. The container is being filled with a liquid of dielectric constant $3$ at a uniform rate of $250 \mathrm{~cm}^3 \mathrm{~s}^{-1}$. What is the value of the capacitance of the container after $10$ seconds? [Given: Permittivity of free space $\epsilon_0=9 \times 10^{-12} \mathrm{C}^2 \mathrm{~N}^{-1} \mathrm{~m}^{-2}$, the effects of the non-conducting walls on the capacitance are negligible]
IIT 2023, Medium
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$\text { In } \mathrm{t}=10 \mathrm{sec} \text { volume of liquid is }$
$\mathrm{V}=2500 \mathrm{cc}$
$\mathrm{h}=\frac{2500}{50 \times 5}=10 \mathrm{~cm}$
$\mathrm{C}_{\mathrm{d}}=\frac{\mathrm{A}_{\mathrm{d}} \varepsilon_0 \mathrm{k}}{\mathrm{d}}$
$=\frac{50 \times 10^{-2} \times 10 \times 10^{-2} \varepsilon_0 \times 3}{5 \times 10^{-2}}=3 \varepsilon_0$
$\mathrm{C}_{\mathrm{a}}=\frac{\mathrm{A}_{\mathrm{a}} \varepsilon_0}{\mathrm{~d}}=\frac{50 \times 10^{-2} \times 40 \times 10^{-2} \varepsilon_0}{5 \times 10^{-2}}=4 \varepsilon_0$
$\mathrm{C}=\mathrm{C}_{\mathrm{a}}+\mathrm{C}_{\mathrm{d}}=7 \varepsilon_0$
$=7 \times 9 \times 10^{-12}=63 \mathrm{Pf}$
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