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 difference in the capacitance (in F) between the metal plates at t = 0 and that at t = 500 s is (8 - n) _0, where _0 is the permittivity of free space. The value of n is:

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Correct answer: A. A

Let A be the area of each chamber. A = 1 1 = 1 m ^2. Distance between plates is d = 2 m . At t = 0, the left chamber is completely filled with liquid ( _r = 15) and right chamber is empty ( _r = 1). Initial capacitance C(0) = C_1 + C_2 = 15 _0 A d + _0 A d = 15 _0 (1) 2 + _0 (1) 2 = 8 _0. Let y_1 and y_2 be the liquid levels in the left and right chambers at time t. By conservation of volume, y_1 + y_2 = 2 m . Let h = y_1 - y_2. The velocity of efflux from the left chamber is v = 2gh . Rate of flow Q = -A dy_1 dt = a 2gh , where a = 10 10^ -4 m ^2 is the hole area. Since y_1 + y_2 = 2, we have dy_2 = -dy_1, so dh = 2dy_1 dy_1 dt = 1 2 dh dt . - 1 2 dh dt = a 2gh dh h = -2a 2g dt. Integrating from t=0 (h=2) to t=500 s (h): _ 2 ^ h h^ -1/2 dh = -2a 2g _ 0 ^ 500 dt 2 h - 2 2 = -2 ( 10 10^ -4 ) ( 20 ) (500) = - 200 10^ -4 1000 = -10 2 0.1 = - 2 . 2 h = 2 h = 1 2 h = 0.5 m . From y_1 + y_2 = 2 and y_1 - y_2 = 0.5, we get y_1 = 1.25 m and y_2 = 0.75 m . At t=500 s , each chamber acts as two capacitors in series (liquid and air). C_ left = 1 y_1 15 _0 A + 2-y_1 _0 A = 15 _0 30 - 14y_1 = 15 _0 30 - 14(1.25) = 15 _0 12.5 = 6 5 _0 = 1.2 _0. C_ right = 1 y_2 15 _0 A + 2-y_2 _0 A = 15 _0 30 - 14y_2 = 15 _0 30 - 14(0.75) = 15 _0 19.5 = 150 195 _0 = 10 13 _0. C(500) = C_ left + C_ right = 6 5 _0 + 10 13 _0 = 128 65 _0 1.97 _0. Difference in capacitance = C(0) - C(500) = 8 _0 - 128 65 _0 = (8 - 128 65 ) _0. Comparing this with the given expression (8 - n) _0, we get n = 128 65 = 1.97

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