A container of height 2 m, length 2 m and breadth 1 m is made of insulating vertical walls and two large area horizontal metal plates (M₁ and M₂) 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 ε_r = 15 and the right chamber is empty (ε_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 ε_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 m s^-2.]

The height (in m) of the liquid in left chamber at t = 500 s is:

The correct answer is 1.25.

Step-by-step Explanation:

1. Understanding the System Setup & Parameters:
From the given description and diagram, we have:
- Total height of the container: H = 2 m
- Length of the container: L = 2 m (partitioned into two equal chambers, so each chamber has length l = 1 m)
- Breadth of the container (into the page): b = 1 m
- Cross-sectional area of each chamber's base: A_c = l × b = 1 m × 1 m = 1 m²
- Hole area: a = $\sqrt{10}{\text{cm}}^2=\sqrt{10}\times {10}^{-4}{\text{m}}^2$
- Acceleration due to gravity: g = 10 m s^-2
- Dielectric constant of the liquid: ε_r = 15
- Dielectric constant of the region above the liquid: ε_r = 1

2. Conservation of Volume:
Initially, the left chamber is completely filled with liquid to a height of 2 m, and the right chamber is empty.
Let h₁ be the height of the liquid in the left chamber at time t, and h₂ be the height of the liquid in the right chamber at time t.
Since the total liquid volume is conserved:

$h_1+h_2=H=2$

$\Rightarrow h_2=2-h_1$

3. Applying Torricelli's Law / Speed of Efflux:
The difference in liquid levels between the left and right chambers is:
$\mathrm{\Delta h}=h_1-h_2=h_1-(2-h_1)=2h_1-2$

By Torricelli's theorem, the velocity of flow through the small hole near the bottom is:

$v=\sqrt{2g\mathrm{\Delta h}}=\sqrt{2g(2h_1-2)}$

4. Setting up the Differential Equation:
The rate of decrease of liquid volume in the left chamber is equal to the volume flow rate through the hole:

$-A_c\frac{d{h}_1}{dt}$

$-\left(1\right)\frac{d{h}_1}{dt}=a\sqrt{2g(2h_1-2)}$

$-\frac{d{h}_1}{dt}=a\sqrt{4g(h_1-1)}=2a\sqrt{g(h_1-1)}$

5. Integrating to Find h₁ at t = 500 s:
Separating variables and integrating from t = 0 (where h₁ = 2 m) to t = 500 s:

${\int }_2^{h_1}\frac{d{h}_1}{\sqrt{h_1-1}}=-2a\sqrt{g}{\int }_0^{500}dt$

${\left[2\sqrt{h_1-1}\right]}_2^{h_1}=-2a\sqrt{g}\left(500\right)$

$2\left(\sqrt{h_1-1}-\sqrt{2-1}\right)=-1000a\sqrt{g}$

Substitute g = 10 m s^-2 and a = $\sqrt{10}\times {10}^{-4}{\text{m}}^2$:

$a\sqrt{g}=\left(\sqrt{10}\times {10}^{-4}\right)\times \sqrt{10}=10\times {10}^{-4}={10}^{-3}$

Now substitute this back into the equation:

$2\left(\sqrt{h_1-1}-1\right)=-1000\times {10}^{-3}=-1$

$\sqrt{h_1-1}-1=-0.5$

$\sqrt{h_1-1}=0.5$

Squaring both sides:

$h_1-1=0.25$

$h_1=1.25\text{m}$

Thus, the height of the liquid in the left chamber at t = 500 s is 1.25 m.