JEE Advanced 2023 — Paper 1 — PHY-II solution

JEE Advanced2023Paper 1PHY-II

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: Permittivity of free space $ε_{0} = 9 \times 1 0^{- 12} C^{2} N^{- 1} m^{- 2}$ , the effects of the non-conducting walls on the capacitance are negligible]

A
$27$ pF
B
$63$ pF
C
$81$ pF
D
$135$ pF

Solution

Liquid height after 10 s. Volume filled $= 250 \times 10 = 2500 cm^{3}$ . Base area $= 50 \times 5 = 250 cm^{2}$ , so liquid height $h = 10$ cm. Remaining air column has height $40$ cm.

Geometry of plates. Each plate has horizontal width $50$ cm and vertical extent $50$ cm. Plate separation $d = 5 cm = 5 \times 1 0^{- 2}$ m. The plates are vertical, so the dielectric region (lower $10$ cm) and air region (upper $40$ cm) form two capacitors in parallel.

Dielectric portion. Area $A_{d} = 0.50 \times 0.10 = 5 \times 1 0^{- 2} m^{2}$ . $C_{d} = \frac{K ε _{0} A _{d}}{d} = \frac{3 ε _{0} \cdot 5 \times 1 0 ^{- 2}}{5 \times 1 0 ^{- 2}} = 3 ε_{0}$ .

Air portion. Area $A_{a} = 0.50 \times 0.40 = 0.20 m^{2}$ . $C_{a} = \frac{ε _{0} A _{a}}{d} = \frac{ε _{0} \cdot 0.20}{5 \times 1 0 ^{- 2}} = 4 ε_{0}$ .

Total. $C = C_{d} + C_{a} = 7 ε_{0} = 7 \times 9 \times 1 0^{- 12} F = 63$ pF.

Answer: (B).

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