JEE Advance - Physics (2023 - Paper 1 Online - No. 5)
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]
[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]
$27 ~\mathrm{pF}$
$63 ~\mathrm{pF}$
$81 ~\mathrm{pF}$
$135 ~\mathrm{pF}$
Explanation
Volume of dielectric filled in $10 \mathrm{sec}$
$$ \begin{aligned} & =250 \times 10=2500 \mathrm{~cm}^3 \\\\ h \times 50 \times 5 & =2500 \\\\ \Rightarrow h & =10 \mathrm{~cm} \end{aligned} $$
$C_1=\frac{\varepsilon_0 A_1}{d}, C_2=\frac{\varepsilon_0 A_2}{d} K$
$\begin{aligned} A_1 & =50 \times(50-h) \\\\ A_2 & =50 \times h \\\\ C_{\text {eq }} & =C_1+C_2 \\\\ & =\frac{\varepsilon_0 50(50-h) \mathrm{cm}^2}{5 \mathrm{~cm}}+\frac{3 \times \varepsilon_0 \times 50 \times h}{5 \mathrm{~cm}} \\\\ & =\varepsilon_0\left[\frac{50 \times 40}{5}+\frac{3 \times 50 \times 10}{5}\right] \times 10^{-2} \\\\ & =7 \varepsilon_0=63 \times 10^{-12} \\\\ & =63 \mathrm{pF}\end{aligned}$
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