Question

Problem 2. (3 pts.). A simple U-shaped tube with identical sides of a diameter 25 mm (millimeters) contains mercury. We have poured 7189 of water into its left arm. (a) If adding of oil into its right arm will bring the surface of the mercury (its interface with another liquid) to the same level on each side of our tube, what is the density of the oil? (6) What is the pressure at each of the points P, and PR at h = 2.4 cm above the surface of the mercury in each arm of the tube? d=25 mm h=2.4 cm water P PR mo = 8159 oil DET NO movie CLOSmw = 718 g mercury Figure 1: Schematic graph for Problem 1.

Problem 2. (3 pts.). A simple U-shaped tube with identical sides of a diameter 25 mm (millimeters) contains mercury. We have poured 7189 of water into its left arm. (a) If adding of oil into its right arm will bring the surface of the mercury (its interface with another liquid) to the same level on each side of our tube, what is the density of the oil? (6) What is the pressure at each of the points P, and PR at h = 2.4 cm above the surface of the mercury in each arm of the tube? d=25 mm h=2.4 cm water P PR mo = 8159 oil DET NO movie CLOSmw = 718 g mercury Figure 1: Schematic graph for Problem 1.

the scratched out part says 1.58m.

Answer 1

d-25mm Oil ho 1.sem dmw-7189 T @ Surfaces of mercury on both arums becomes of same level if Pressure exerted by water and is are same. Le t.e. Pressure exerted by oil - Pressure exerted by U water ..) hox frie x g = mw xg . Area of tubes {:: P = hfg also p= force = mg } Joelo area area = 1.58 x foil = 0.718 = 4X0718 TIXdh) 2 TlXd2 - (fout d= as mm = 25 x103 m ) )Sax foil = 4x0.718. 3.14 x (25 x 103)2 - - Soil = 4x0.718 1.S@X314x (25X103)? Poil = 926.23 kg/m3

hw = 0.7188 4 6. Let height of water in left arm = ho tutt mass of water - Vol.oj water x deusity of pater 0.718 kg = tx( a)'x hus x (1000x9|m?) 3) 0.712 = 314 x/28x103) xlu x loud = 1.46 m 3.14 x (25X103)2x1000 depth of feint p from free water surface of water 5 I 7 h = 146cm - 2:4 cm 146 cm 12.40 =143.6 cm - = 1.436m Pressure at this point - P = Sinter x x h = 1000 x g. 8 X1.436 → P = 14072.8 Pa ~ 1.41X 104 Pa Depth of froint PR foom foee surface of oil- h = 150-2.4 = 155.6 cm i Te =10556m 158cm NR Pressure at this point - 12-40m PR = Sail xgx he = 926.23x9.8x1.556 = 14123.8 la ñ 1.418104 so pressure is same at both foint because they are lying in same horizontal flame.