A container has a base of $50 c m \times 5 c m$ and height 50 cm, as shown in the figure. It has Two parallel electrically conducting walls each of area $50 c m \times 50 c m$ . 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 $250 c m^{3} S^{- 1}$ . What is the value of the value of the capacitance of the container after 10 seconds?
[Given: Permittivity of free space $\in_{0} = 9 \times 10^{- 12} C^{2} N^{- 1} m^{- 2}$ , the effects of the non-conducting walls on the capacitance are negligible]

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27 pF

63 pF

81 pF

135 pF

answer is B.

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$h = \frac{250 \times 10}{50 \times 5} = 10 c m C_{1} = \frac{(0.40 \times 0.50) \times 9 \times 10^{- 12}}{5 \times 10^{- 2}} = 0.36 \times 10^{- 10} F C_{2} = \frac{3 \times 0.10 \times 0.5 \times 9 \times 10^{- 12}}{5 \times 10^{- 2}} C_{2} = 0.27 \times 10^{- 10} F C = C_{1} + C_{2} = 63 p F$