Question

(a) adding 1.58M of oil

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-mg 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 kh=2.4 cm water mw = 7189 mo = 8159 oil Р. mercury Figure 1: Schematic graph for Problem 1.

please show all work

Answer 1

Density of water in si FS - 1000 kg/m3 diameter of tube - 25 mm radius – 25 = 12.5 mm - 0.0125m Area- r - 3.14 * (0.0125) A = 4.90x60 m2 let the height of water coloumn = a. .. volume of water coloumn V = Areaxx V = (4.90x10xx i mass of that water - S.V od 1000 x (4.90x10") . According to Question 718 = 0.49 x x x = 1.46 m 1000 The height of water coloumn = 1.46m

@ Let the density of oil is Poil height of oil coloumn - 1.58m. volume of oil coloumn - Axh = 1.58x (4.9x104) m3 mass of oil - Soil & (1.58 x 4.9x104) m3 Now, as they exert some pressure on mercury (as mercury leves in both Coloumn if game). (ing) water (mg) oil in water a moil 718 700g a son x 1.58 x 4.9x104 Son = 927,4 kg/m3 Pressure formula at h depth (6) 2.4 cm à 0.024m. water coloum height = 1:46m So at 0.024 in height from .Pa heg me rcury the pressure PL - Score & g x depth (1.96- 0-624) = 1000 x 9.8 x 1.436 = 14072,8 N = 14.07KN/m2 water 23

PR = Sox qx depth (1.58-0:024) = 927:4 x 9.8 x 1.556 = 14141.7 = 14.14 KN/m2