A narrow, U-shaped glass tube with open ends is filled with 25.0 cm of oil (of specific gravity 0.80) and 25.0 cm of water on opposite sides, with a barrier separating the liquids (Fig. P12.58). (a) Assume that the two liquids do not mix, and find the final heights of the columns of liquid in each side of the tube after the barrier is removed. (b) For the following cases, arrive at your answer by simple physical reasoning, not by calculations: (i) What would be the height on each side if the oil and water had equal densities? (ii) What would the heights be if the oil’s density were much less than that of water?

Solution 58P Introduction We have to calculate the final height in both side after the barrier is removed. Step 1 Since the density of the water is greater than the density of the oil, the water side will come down and push the oil upward. Let us consider that the water will come down by h, that is the oil will go up by h from its initial position. Now the height of the water in right hand side is 25.0 cm h Then the height of the water in the right hand side is h And the height of the oil in the right hand side is 25.0 cm. Hence the pressure from the left hand side is P l (25.0 cm h)g Considering is the density of water. Now the specific gravity of oil is 0.8, Hence the density of oil is 0.8 Hence the pressure from the right hand side si P r hg + (25.0 cm)(0.8)g = hg + (20 cm)g For equilibrium, this two pressure will be equal, hence we can write that (25.0 cm h)g = hg + (20 cm)g (25.0 cm)g hg = hg + (20 cm)g 2hg = (5.00 cm)g h = (52gcm)= 2.5 cm Hence the height in the left hand side is 25.0 cm 2.5 cm = 22.5 cm And height of the right hand side si 25.0 cm + 2.5 cm = 27.5 cm.