Question

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 \((M_1 \text{ and } M_2)\) 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 height (in m) of the liquid in left chamber at \(t=500\,\text{s}\) is:

Hint:

When liquid flows between two identical connected vessels, the relative height difference decreases twice as fast as the height in one vessel would if it were draining into an infinite reservoir.
Always convert areas from cm$^2$ to m$^2$ ($1\text{ cm}^2 = 10^{-4}\text{ m}^2$).
The term $(h_1 - h_2)$ in efflux velocity is equivalent to $(2h_1 - H_{total})$ in such symmetric systems.

Answer 1

Step 1: Understanding the Question:
The total length of the container is $2\text{ m}$ and it's split into two equal chambers.
So, each chamber has a base area $A = (1\text{ m} \times 1\text{ m}) = 1\text{ m}^2$.
Liquid flows from the left chamber (initially full, $H = 2\text{ m}$) to the right chamber (initially empty) through a hole of area $a = \sqrt{10}\text{ cm}^2 = \sqrt{10} \times 10^{-4}\text{ m}^2$.
The flow is driven by the height difference between the two chambers.

Step 2: Key Formula or Approach:
Torricelli's Law for efflux velocity: $v = \sqrt{2g(h_1 - h_2)}$.
Continuity equation: $A \frac{dh_1}{dt} = -a \sqrt{2g(h_1 - h_2)}$.
By conservation of volume: $h_1 + h_2 = 2 \implies h_2 = 2 - h_1$.
Substitute $h_2$: $h_1 - h_2 = h_1 - (2 - h_1) = 2h_1 - 2$.

Step 3: Detailed Explanation:

• Setting up the differential equation:
\[ \frac{dh_1}{dt} = -\frac{a}{A} \sqrt{2g(2h_1 - 2)} = -\frac{a}{A} \sqrt{4g(h_1 - 1)} = -\frac{2a\sqrt{g}}{A} \sqrt{h_1 - 1}. \]

• Integration:
\[ \int_2^h \frac{dh_1}{\sqrt{h_1 - 1}} = \int_0^t -\frac{2a\sqrt{g}}{A} dt. \]
\[ [2\sqrt{h_1 - 1}]_2^h = -\frac{2a\sqrt{g}}{A} t. \]
\[ \sqrt{h - 1} - \sqrt{2 - 1} = -\frac{a\sqrt{g}}{A} t. \]

• Substituting values ($t = 500\text{ s}$):
$a = \sqrt{10} \times 10^{-4}\text{ m}^2, g = 10\text{ ms}^{-2}, A = 1\text{ m}^2$.
\[ \sqrt{h - 1} - 1 = -\frac{\sqrt{10} \times 10^{-4} \times \sqrt{10}}{1} \times 500. \]
\[ \sqrt{h - 1} - 1 = -(10 \times 10^{-4}) \times 500 = -0.5. \]

• Solving for $h$:
\[ \sqrt{h - 1} = 1 - 0.5 = 0.5 \implies h - 1 = 0.25 \implies h = 1.25\text{ m}. \]

Step 4: Final Answer:
The height of the liquid in the left chamber at $t = 500\text{ s}$ is $1.25\text{ m}$.