Question

= 2.4 cm h Problem 2. (3pts.). A simple U-shaped tube with identical sides of a diameter 25 mm (millimeters) contains mercry We have poured 718 g of water into its left arm. (a) If adding 815 g 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? (b) 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 mo= 815 g water 718 g PL PR oil me= mercury Figure 1: Schematic graph for Problem 1

Answer 1

12.4 cm = 1.46 m og hw - Pegho mo areahom areah! 718 hh = 815 ho Now Pressure - & PC = mog Treyho Now 718x103 = low no hw . 718 x103 * 100w x n (25 x 103² hw hwe 718 x 15 x 4 1000xa (25 x 163)2 ho - (718x hw - 1146 718 = 1:37m V 815 a)815 x 153 = Pox xo²ho lo 2 815 8103 = 1212.516 kg/m3 X (21x103) ² x 1.37 density of oil - 1212-52 1g/m3 6) Preshore atip = Pottwg/hw-0.024) = 10 + 1000 xg (1.46-0.024). = 10s +10000 (1.436) • 114360 Pa = 1.146x105 Pa = 1.1436 atms Besare at R= Potlog (ho -0.024) -Tu +1212.516x10(1.37-0.024) = 116320 Pa or 1.1632 atms.