please answer A vertical cylinder of 2m diameter carries water upto a height of
ID: 1920186 • Letter: P
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A vertical cylinder of 2m diameter carries water upto a height of 2.5 m and then oil upto a height of 3m. If specific gravity of oil is 0.8, determine the force to which the bottom of the cylinder will be subjected. Also determine the intensity of pressure at the interface of oil and water. A closed tank is filled partially with air, oil and water as shown in below. Water is raised to a height of 10m in a pipe fitted at the bottom of the tank. Determine the pressure recorded by the pressure gauge fitted to the top of the tank. Determine the pressure of a liquid of specific gravity 1.01 contained in a pipe line to which a simple manometer is connected . Mercury is used as the manometric liquid . Tie height of the free surface of mercury in the right hand limb of the U tube from the centre of the pipeline is 20 cm and the common surface of contact between the pipe liquid in f mercury in the left hand limb of the U tube is 50 cm before the free surface of mercury in the right hand limb of the U tubeExplanation / Answer
Pressure at a depth h = rho*g*h
rho water = 1000 Kg/m^3
1 atm = 10e5 Pa
Answer may vary because of choice of value of 1 atm but the method id correct
8)at 2.5 m in oil, P = 0.8*1000*10*2.5 = 20 kPa + 1 atm = 120 kPa
at 2 m + 2.5 m inside water, P = 120e3 + 1000*10*2.5 = 145 kPa
9) P(air) + p(oil) + p(2.5 m water) = 1 atm + p(10 m water)
p(air) + 0.8*1000*10*1 + 1000*10*2.5 = 100000 + 1000*10*10
p(air) = 16.7 kPa
10) P + P(50-20 cm of liquid) = 1atm + P(50cm mercury)
P + 1.01*1000*10*(50-20)*10^-2 = 100000 + 13.6*1000*10*50*10^-2
P = 168000 - 3003 = 13797 Pa
11) The pressure at point C and D will be same
At point C, Pressure = P(container) + P (50mm water)
At point D, Pressure = 1atm + P (40 mm mercury)
P (container) + 1000*10*0.05 = 100000 + 13.6*1000*10*0.04
P (container) = 104940 Pa
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