JEE Advanced 2023 Paper 1 Q05 Physics Electrostatics & Circuits Capacitors+Dielectric Medium

JEE Advanced 2023 Paper 1 · Q05 · Capacitors+Dielectric

A container has a base of $50$ cm $\times$ $5$ cm and height $50$ cm, as shown in the figure. It has two parallel electrically conducting walls each of area $50$ cm $\times$ $50$ 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 $250$ cm$^3$ s$^{-1}$. What is the value of the capacitance of the container after $10$ seconds? [Given: $\varepsilon_0 = 9 \times 10^{-12}$ C$^2$N$^{-1}$m$^{-2}$]

  1. A. $27$ pF
  2. B. $63$ pF
  3. C. $81$ pF
  4. D. $135$ pF
Reveal answer + step-by-step solution

Correct answer:B

Solution

Volume filled in $10$ s = $2500$ cm$^3$. Base area = $50 \times 5 = 250$ cm$^2$, so liquid height $h = 10$ cm. The capacitor has plate separation $d = 5$ cm = $0.05$ m. Wet portion (height $0.10$ m, width $0.50$ m) is dielectric-filled; dry portion (height $0.40$ m) is air. Areas $A_1 = 0.5\cdot 0.1 = 0.05$ m$^2$, $A_2 = 0.5\cdot 0.4 = 0.20$ m$^2$. Capacitance $C = (\varepsilon_0/d)[K_1 A_1 + K_2 A_2] = (9\times 10^{-12}/0.05)[3\cdot 0.05 + 1\cdot 0.20] = (1.8\times 10^{-10})(0.35) = 6.3\times 10^{-11}$ F = $63$ pF.

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