Question

contains mercury Problem 2. (3 pts.). A simple U-shaped tube with identical sides of a diameter 25 mm (millimeters) contains me We have poured 7189 of water into its left arm. (a) If adding 815g 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 PL 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 m. = 8159 water PR oil CIOSmw = 7189 P mercury Figure 1: Schematic graph for Problem 1.

Answer 1

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Sunter - materia Water Munter-Swater Atha → 718 = 1x7x(2.5) 4 he 146.269 cm at this layer URWO same, pressure must be above hence, mass in both limbs must be same this level, water * Ax hat Srg xAX ho = Mores - Mwatert Siz xaxho = Moil Ing xAX ho = (815-718) = 13.6*AX (2-5)² = 97 4 =) ho=1.45299cm , more = Sone X AX (hat ho) 815= Soie XX (2-5) 2 1 x (146.269+1-45299) 4 (146.2692, - Soil - Density of oil = 1.12393 gre / m3. Aut b) at H=2.4cm. PL = Peter Swater xgx (hotho H) = 183x9.8X1.453 P = 0.142415 Xo5 palc auge pressure Answer P = Patm t Pho = 1.013x10 to 142415 X 10 = 1.1554 x 15 Pa < Absate pressure Answer And

-anina's LD Similarly, Roo (P) = Soil & g x 19 hotho-H) gauge Pressure 2199 = 1.12393x16 49.8(1.4532199) = 16x15 Pa ) And = 1.6106 pa TRAsedute = pressure gange & Patm = 16X15 + 1:013410 = 1.7019X10 Pa And Pressure