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QJEEAdv23Ph0936: JEE Advanced 2023 Physics - Class 12 JEEAdv23Ph0936 Solution: A container has a base of and height , as shown in the figure.…

A container has a base of 50 cm×5 cm50 \mathrm{~cm} \times 5 \mathrm{~cm} and height 50 cm50 \mathrm{~cm} , as shown in the figure. It has two parallel electrically conducting walls each of area 50 cm×50 cm50 \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 cm3 s−1250 \mathrm{~cm}^3 \mathrm{~s}^{-1} . What is the value of the capacitance of the container after 10 seconds?

[Given: Permittivity of free space ϵ0=9×10−12C2 N−1 m−2\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]

JEE Advanced 2023 Paper 1 Online Physics - Capacitor Question 2 English

Multiple Choice Options

  • A 27 pF27 ~\mathrm{pF}
  • B 63 pF63 ~\mathrm{pF}
  • C 81 pF81 ~\mathrm{pF}
  • D 135 pF135 ~\mathrm{pF}

Step-by-Step Verified Solution

JEE Advanced 2023 Paper 1 Online Physics - Capacitor Question 2 English Explanation 1

Volume of dielectric filled in 10sec10 \mathrm{sec}

amp;=250×10=2500 cm3h×50×5amp;=2500⇒hamp;=10 cm\begin{aligned}& =250 \times 10=2500 \mathrm{~cm}^3 \\\\h \times 50 \times 5 & =2500 \\\\\Rightarrow h & =10 \mathrm{~cm}\end{aligned}

JEE Advanced 2023 Paper 1 Online Physics - Capacitor Question 2 English Explanation 2
C1=ε0A1d,C2=ε0A2dKC_1=\frac{\varepsilon_0 A_1}{d}, C_2=\frac{\varepsilon_0 A_2}{d} K

A1amp;=50×(50−h)A2amp;=50×hCeq amp;=C1+C2amp;=ε050(50−h)cm25 cm+3×ε0×50×h5 cmamp;=ε0[50×405+3×50×105]×10−2amp;=7ε0=63×10−12amp;=63pF\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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