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Question Stem for Question Nos. 15 and 16
A container of height $$2\,\mathrm{m}$$, length $$2\,\mathrm{m}$$ and breadth $$1\,\mathrm{m}$$ 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 $$\sqrt{10}\,\mathrm{cm^2}$$ 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 $$10\,\mathrm{ms^{-2}}$$.]
The height (in m) of the liquid in left chamber at $$t=500\,\mathrm{s}$$ is:
Correct Answer: 1.25
Torricelli's law gives the efflux velocity as $$v = \sqrt{2g\Delta h}$$, and the rate of drainage is determined by the equation of continuity along with conservation of volume.
Given: $$H = 2\text{ m}$$, $$A_0 = 1\text{ m} \times 1\text{ m} = 1\text{ m}^2$$, $$a = \sqrt{10} \times 10^{-4}\text{ m}^2$$, $$g = 10\text{ ms}^{-2}$$
From conservation of volume: $$h_1 + h_2 = 2 \implies h_2 = 2 - h_1$$
Applying continuity equation:
$$-\frac{dh_1}{dt} = \frac{a}{A_0}\sqrt{2g(h_1 - h_2)}$$
$$-\frac{dh_1}{dt} = \frac{a}{A_0}\sqrt{2g(2h_1 - 2)} = \frac{2a\sqrt{g}}{A_0}\sqrt{h_1 - 1}$$
$$\int_{2}^{h_1} \frac{dh_1}{\sqrt{h_1 - 1}} = -\frac{2a\sqrt{g}}{A_0}\int_{0}^{500} dt$$
$$\left[2\sqrt{h_1 - 1}\right]_2^{h_1} = -\frac{2a\sqrt{g}}{A_0}(500)$$
$$2\sqrt{h_1 - 1} - 2 = -2 \cdot \frac{\sqrt{10} \times 10^{-4} \cdot \sqrt{10}}{1} \cdot 500$$
$$2\sqrt{h_1 - 1} - 2 = -1$$
$$2\sqrt{h_1 - 1} = 1 \implies \sqrt{h_1 - 1} = 0.5$$
$$h_1 - 1 = 0.25 \implies h_1 = 1.25\text{ m}$$
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