Physics · Capacitors and R-C Circuits

JEE Advanced 2023 — Paper 1 — Question 21

A container has a base of 50 cm×5 cm50 \mathrm{~cm} \times 5 \mathrm{~cm} and height 50 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−1\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\varepsilon_{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

Question figure
  1. Option A:

    27 pF

  2. Option B:

    63 pF

    Correct
  3. Option C:

    81 pF

  4. Option D:

    135 pF

Answer: B

Step-by-step solution

Let container is filled upto height x in 10 sec

250×10=50×5×x250 \times 10=50 \times 5 \times x

X=10 cm\mathrm{X}=10 \mathrm{~cm}

C=C1+C2\mathrm{C}=\mathrm{C}_{1}+\mathrm{C}_{2}

C=A1∈0d+KA2∈0dC=\frac{A_{1} \in_{0}}{d}+\frac{K A_{2} \in_{0}}{d}

C=ϵ0 d[ A1+KA2]\mathrm{C}=\frac{\epsilon_{0}}{\mathrm{~d}}\left[\mathrm{~A}_{1}+\mathrm{KA}_{2}\right] C=9×10−125×10−2[40×50×10−4+3×50×10×10−4]\mathrm{C}=\frac{9 \times 10^{-12}}{5 \times 10^{-2}}\left[40 \times 50 \times 10^{-4}+3 \times 50 \times 10 \times 10^{-4}\right]

=63×10−12 F=63 \times 10^{-12} \mathrm{~F}

C=63pF\mathrm{C}=63 \mathrm{pF}

Solution figure

Answer key and solution verified before publishing.

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Exam
JEE Advanced 2023
Paper
Paper 1
Subject
Physics
Chapter
Capacitors and R-C Circuits
Topic
Effect of Dielectrics
A container has a base of 50 cm × 5 cm and height 50 cm , as shown in… | JEE Advanced 2023 PYQ with Solution · DhiX AI