Question 16

NUMERICALHARD

Liquid Efflux and Variable Dielectric Capacitance

A container of height $2 \text{ m}$ , length $2 \text{ m}$ and breadth $1 \text{ m}$ is made of insulating vertical walls and two large area horizontal metal plates ( $M_1$ and $M_2$ ) which extend far beyond the vertical walls in all directions. The container is partitioned into two equal chambers with a thin insulating vertical wall. The partition wall contains a small hole of cross-sectional area $\sqrt{10} \text{ cm}^2$ near its bottom edge. Initially the hole is closed and the left chamber of the container is completely filled with a liquid of dielectric constant $\epsilon_r = 15$ and the right chamber is empty ( $\epsilon_r = 1$ ). At time $t = 0$ , the hole is opened and the liquid flows from the left chamber to the right chamber. In both the chambers, the space above the liquid has $\epsilon_r = 1$ and is maintained at atmospheric pressure. The schematic of the container at a time $t > 0$ is shown in the figure. [Given: acceleration due to gravity is $10 \text{ ms}^{-2}$ ]

The difference in the capacitance (in F) between the metal plates at $t = 0$ and that at $t = 500 \text{ s}$ is $(8 - n)\epsilon_0$ , where $\epsilon_0$ is the permittivity of free space. The value of $n$ is:

Correct Answer: 1.97

Detailed Solution

The total capacitance $C$ is the sum of the capacitances of the left and right chambers ( $C = C_L + C_R$ ) as they are in parallel.

In each chamber, the liquid and air act as two capacitors in series. For a chamber with liquid height $h$ :

$C_{\text{chamber}} = \frac{\epsilon_0 A}{\frac{h}{\epsilon_r} + \frac{2-h}{1}} = \frac{\epsilon_0 (1)}{\frac{h}{15} + 2 - h} = \frac{15\epsilon_0}{30 - 14h}$

At $t = 0$ , $h_1 = 2$ and $h_2 = 0$ :

$C(0) = \frac{15\epsilon_0}{30 - 28} + \frac{15\epsilon_0}{30 - 0} = 7.5\epsilon_0 + 0.5\epsilon_0 = 8\epsilon_0$

At $t = 500 \text{ s}$ , $h_1 = 1.25 \text{ m}$ and $h_2 = 0.75 \text{ m}$ :

$C_L(500) = \frac{15\epsilon_0}{30 - 14(1.25)} = \frac{15\epsilon_0}{12.5} = 1.2\epsilon_0$

$C_R(500) = \frac{15\epsilon_0}{30 - 14(0.75)} = \frac{15\epsilon_0}{19.5} = \frac{10}{13}\epsilon_0 \approx 0.769\epsilon_0$

$C(500) = (1.2 + 10/13)\epsilon_0 = \frac{128}{65}\epsilon_0 \approx 1.969\epsilon_0$

The difference is $C(0) - C(500) = 8\epsilon_0 - 1.969\epsilon_0 = (8 - 1.969)\epsilon_0$ .

Comparing with $(8-n)\epsilon_0$ , we get $n = 128/65 \approx 1.97$ .

Question 16

Liquid Efflux and Variable Dielectric Capacitance

Detailed Solution

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