Passage: 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 (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 verti

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Passage: 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 (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 10 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 _r = 15 and the right chamber is empty ( _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 _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 ms^ -2 .] The height (in m) of the liquid in left chamber at t = 500 s is:

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Correct answer: A. A

Let A be the cross-sectional area of each chamber and a be the cross-sectional area of the hole. From the given dimensions, the area of each chamber is A = 1 m 1 m = 1 m ^2. The area of the hole is a = 10 cm ^2 = 10 10^ -4 m ^2. Let h_1 and h_2 be the heights of the liquid in the left and right chambers at time t, respectively. By conservation of volume, A h_1 + A h_2 = A H, where H = 2 m is the initial height of the liquid in the left chamber. h_1 + h_2 = 2 The velocity of efflux through the hole is given by Torricelli's law: v = 2g(h_1 - h_2) The rate of flow of the liquid is: -A dh_1 dt = a 2g(h_1 - h_2) Substitute h_2 = 2 - h_1 into the equation: -A dh_1 dt = a 2g(2h_1 - 2) Let y = 2h_1 - 2. Then dy = 2 dh_1, which gives dh_1 = dy 2 . The differential equation becomes: - A 2 dy dt = a 2gy dy y = - 2a 2g A dt Integrating both sides from t = 0 to t = 500 s. At t = 0, h_1 = 2 m, so y(0) = 2(2) - 2 = 2 m. _ 2 ^ y y^ -1/2 dy = - 2a 2g A _ 0 ^ 500 dt [2 y ]_ 2 ^ y = - 2 ( 10 10^ -4 ) 20 1 500 2 y - 2 2 = -1000 10^ -4 200 2 y - 2 2 = -0.1 10 2 2 y - 2 2 = - 2 2 y = 2 y = 2 2 Squaring both sides, we get: y = 0.5 Since y = 2h_1 - 2, we have: 2h_1 - 2 = 0.5 2h_1 = 2.5 h_1 = 1.25 m Answer: 1.25

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