Two identical containers are open at the top and are connected at the bottom via

Question

Two identical containers are open at the top and are connected at the bottom via a tube of negligible volume and a valve that is initially closed. Both containers are filled initially to the same height of h = 3.00 , one with oil, and the other with mercury, as the drawing below indicates. The valve is then opened. Oil and mercury have densities of 6,000 kg/m3 and 13,600 kg/m3 , respectively. The oil and mercury do not mix.

Determine the fluid level in the left container when equilibrium is established again.

Explanation / Answer

Given data Initial height in the cointainer, h = 3.0 m Density of oil, oil = 6000 kg/m3 Initial height in the cointainer, h = 3.0 m Density of mercury, Hg = 13600 kg/m3 Let the Hg move into the left contaner to a height of y The height of Hg on the right side, = (h - y) m Total pressure on the left side,                                 p1= oilgh +Hggy                                     = (6000 kg/m3)(9.8 m/s2) (3.0 m) + (13600 kg/m3)(9.8 m/s2) (y)                                     = 1.764*105 Pa + 1.3328*105 Pa y Pressure on the right side,                                p2 = Hgg(h-y)
                                    =(13600 kg/m3)(9.8 m/s2) (3.0 m - y)                                     = 3.998*105 Pa - 1.3328*105 Pa y ----------------------------------------------------------------------------------------------------------------- Equating the rightand left side pressures, and cancelling g. 1.764*105 Pa + 1.3328*105 Pa y = 3.998*105 Pa - 1.3328*105 Pa y                          2.665*105 Pa (y) = 2.234*105 Pa Then the height of mercury in left side is                                               y = 0.838 m                                Therefore the total height of the fluid level on the left side,                                             H = 3.0 +0.838
                                                 = 3.838 m Let the Hg move into the left contaner to a height of y The height of Hg on the right side, = (h - y) m Total pressure on the left side,                                 p1= oilgh +Hggy                                     = (6000 kg/m3)(9.8 m/s2) (3.0 m) + (13600 kg/m3)(9.8 m/s2) (y)                                     = 1.764*105 Pa + 1.3328*105 Pa y Pressure on the right side,                                     = (6000 kg/m3)(9.8 m/s2) (3.0 m) + (13600 kg/m3)(9.8 m/s2) (y)                                     = 1.764*105 Pa + 1.3328*105 Pa y Pressure on the right side,                                p2 = Hgg(h-y)
                                    =(13600 kg/m3)(9.8 m/s2) (3.0 m - y)                                     = 3.998*105 Pa - 1.3328*105 Pa y ----------------------------------------------------------------------------------------------------------------- ----------------------------------------------------------------------------------------------------------------- Equating the rightand left side pressures, and cancelling g. 1.764*105 Pa + 1.3328*105 Pa y = 3.998*105 Pa - 1.3328*105 Pa y                          2.665*105 Pa (y) = 2.234*105 Pa Then the height of mercury in left side is                                               y = 0.838 m                                1.764*105 Pa + 1.3328*105 Pa y = 3.998*105 Pa - 1.3328*105 Pa y                          2.665*105 Pa (y) = 2.234*105 Pa Then the height of mercury in left side is                                               y = 0.838 m                                Therefore the total height of the fluid level on the left side,                                             H = 3.0 +0.838
                                                 = 3.838 m

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