A container has a base of 50 text{ cm} times | vedclass

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(B) The volume of the liquid filled in $t = 10 	ext{ s}$ is $V = 250 	ext{ cm}^3 	ext{ s}^{-1} 	imes 10 	ext{ s} = 2500 	ext{ cm}^3$.The base ar

Vedclass · Accepted Answer

$63$

Vedclass · Answer

(B) The volume of the liquid filled in $t = 10 	ext{ s}$ is $V = 250 	ext{ cm}^3 	ext{ s}^{-1} 	imes 10 	ext{ s} = 2500 	ext{ cm}^3$.The base area of the container is $50 	ext{ cm} 	imes 5 	ext{ cm} = 250 	ext{ cm}^2$.The height of the liquid column is $h = \frac{V}{	ext{base area}} = \frac{2500 	ext{ cm}^3}{250 	ext{ cm}^2} = 10 	ext{ cm}$.The capacitor can be considered as two capacitors in parallel: one filled with liquid (dielectric) and one filled with air.For the liquid part: Area $A_d = 50 	ext{ cm} 	imes 10 	ext{ cm} = 500 	ext{ cm}^2 = 500 	imes 10^{-4} 	ext{ m}^2$, distance $d = 5 	ext{ cm} = 5 	imes 10^{-2} 	ext{ m}$, $k = 3$.$C_d = \frac{k \epsilon_0 A_d}{d} = \frac{3 	imes \epsilon_0 	imes 500 	imes 10^{-4}}{5 	imes 10^{-2}} = 3 \epsilon_0 	imes 10^{-2} 	ext{ F} = 3 \epsilon_0 	ext{ pF}$ (since $\epsilon_0 = 10^{-12} 	ext{ F/m}$ is not used, we use the given value).$C_d = \frac{3 	imes 9 	imes 10^{-12} 	imes 500 	imes 10^{-4}}{5 	imes 10^{-2}} = 27 	imes 10^{-12} 	ext{ F} = 27 	ext{ pF}$.For the air part: Area $A_a = 50 	ext{ cm} 	imes (50 - 10) 	ext{ cm} = 50 	ext{ cm} 	imes 40 	ext{ cm} = 2000 	ext{ cm}^2 = 2000 	imes 10^{-4} 	ext{ m}^2$.$C_a = \frac{\epsilon_0 A_a}{d} = \frac{9 	imes 10^{-12} 	imes 2000 	imes 10^{-4}}{5 	imes 10^{-2}} = 36 	imes 10^{-12} 	ext{ F} = 36 	ext{ pF}$.Total capacitance $C = C_d + C_a = 27 	ext{ pF} + 36 	ext{ pF} = 63 	ext{ pF}$.

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