USE THE FOLLOWING CONSTANTS (You may not need to use all of them for this proble

Question

USE THE FOLLOWING CONSTANTS


(You may not need to use all of them for this problem)
• Unless I mention it, these experiments occur on the surface of the earth
and the atmosphere is ’air’.
• Gravity on earth is 10 meters per second squared.
• Gravity on the moon is one sixth that of earth’s gravity.
• Speed of sound in air is 350 meters per second.
• Density of water is 1000 kilograms per cubic meter.
• Density of mercury is 15000 kilograms per cubic meter.
• Density of ethyl alcohol is 750 kilograms per cubic meter.

All answers must come with explanations. Merely guessing will not give you
full credit.


4 Pascal’s Principle
You have 1 gallon of water. You place it in a ’U’ tube. The tube is of radius
1.2 centimeters. You have a pint of ethyl alcohol and pour all of it into one of
the ends of the tubes. What is the difference between the height of the ethyl
alcohol and the height of the water? Repeat the experiment by replacing the
ethyl alcohol with mercury. Your answer must include which liquid is higher
than the other and by how many cm.

Explanation / Answer

  The density of water w = 10^3 kg/m^3  The density of mercury m = 15000 kg/m^3  The density of ethyl alcohol alc = 750 kg/m^3 the length of water when we pour ethyl alcohol is L1 and when we pour mercury is L2 The volume of the water               V = 3.8*10^-3 m^3 The area of the tube A = r^2 = (1.2*10^-2m)^2 = 4.5*10^-4 m^2 Now from pascal law       the pressure on right arm is equal to the pressure on left arm        p0 + wgl = p0  + alc g(l+d)                wgl = alc g(l+d)               wgl = alc g l  + alc gd               ( w - alc) gl = alc gd Then the heigth above the water level         d1 = ( w - alc) gL1 / alc              = (1000 - 750)(9.8) L2 / 750              = 3.26 (L1) and when we use mercury m = 15000 kg/m^3       wg ( lL2+ d)  = m g L2       d = (m - w) gL2/w         = (15000 - 1000) (9.8) L2 / 1000          = 137.2 (L2) Therefore the difference          d = 137.2(L2) - 3.26(L1) here we can calculate when we know the heigths L1  and L2 The density of water w = 10^3 kg/m^3  The density of mercury m = 15000 kg/m^3  The density of ethyl alcohol alc = 750 kg/m^3 the length of water when we pour ethyl alcohol is L1 and when we pour mercury is L2 The volume of the water               V = 3.8*10^-3 m^3 The area of the tube A = r^2 = (1.2*10^-2m)^2 = 4.5*10^-4 m^2 Now from pascal law       the pressure on right arm is equal to the pressure on left arm        p0 + wgl = p0  + alc g(l+d)                wgl = alc g(l+d)               wgl = alc g l  + alc gd               ( w - alc) gl = alc gd Then the heigth above the water level         d1 = ( w - alc) gL1 / alc              = (1000 - 750)(9.8) L2 / 750              = 3.26 (L1) and when we use mercury m = 15000 kg/m^3       wg ( lL2+ d)  = m g L2       d = (m - w) gL2/w         = (15000 - 1000) (9.8) L2 / 1000          = 137.2 (L2) Therefore the difference          d = 137.2(L2) - 3.26(L1) here we can calculate when we know the heigths L1  and L2  

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