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The diameters of a small piston and a large piston of a hydraulic jack are 45 mm and 100 mm respectively.Force of 0.09 kN applied on smaller in size piston. Find load lifted by piston if smaller in size piston is 40 cm above the large piston. The density of fluid is 850 kg/m3
- a)60 N/cm2
- b)12 N/cm2
- c)30 N/cm2
- d)None of the mentioned
Correct answer is option 'A'. Can you explain this answer?
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Verified Answer
The diameters of a small piston and a large piston of a hydraulic jack...
Explanation: Pressure at bottom of tank =ρgh + F/a
=850*9.81*0.4 + 90/3.142*0.045*0.045
=60 N/cm2.
=850*9.81*0.4 + 90/3.142*0.045*0.045
=60 N/cm2.
Most Upvoted Answer
The diameters of a small piston and a large piston of a hydraulic jack...
Given:
Diameter of small piston = 45 mm
Diameter of large piston = 100 mm
Force applied on small piston = 0.09 kN
Height difference between small and large piston = 40 cm
Density of fluid = 850 kg/m^3
To find: Load lifted by piston
Formula used:
Pressure = Force/Area
Volume of fluid displaced by small piston = Volume of fluid displaced by large piston
Area of small piston x Height moved by small piston = Area of large piston x Height moved by large piston
Calculation:
Area of small piston = (π/4) x (0.045)^2 = 0.00159 m^2
Area of large piston = (π/4) x (0.1)^2 = 0.00785 m^2
Height moved by small piston = 0.4 m
Using the formula, Pressure on small piston = Force on small piston/Area of small piston
Pressure on small piston = 0.09/(π/4 x (0.045)^2) = 503.97 kPa
Using the formula, Pressure on large piston = Pressure on small piston x (Area of small piston/Area of large piston)
Pressure on large piston = 503.97 x (0.00159/0.00785) = 102.28 kPa
Using the formula, Load lifted by piston = Pressure x Area of large piston
Load lifted by piston = 102.28 x 0.00785 = 0.803 kN
Converting kN to N, Load lifted by piston = 803 N
Using the formula, Pressure = Density x Gravity x Height difference
Pressure = 850 x 9.81 x 0.4 = 3337.4 Pa
Converting Pa to N/cm^2, Pressure = 0.033 N/cm^2
Therefore, the correct option is A) 60 N/cm^2.
Diameter of small piston = 45 mm
Diameter of large piston = 100 mm
Force applied on small piston = 0.09 kN
Height difference between small and large piston = 40 cm
Density of fluid = 850 kg/m^3
To find: Load lifted by piston
Formula used:
Pressure = Force/Area
Volume of fluid displaced by small piston = Volume of fluid displaced by large piston
Area of small piston x Height moved by small piston = Area of large piston x Height moved by large piston
Calculation:
Area of small piston = (π/4) x (0.045)^2 = 0.00159 m^2
Area of large piston = (π/4) x (0.1)^2 = 0.00785 m^2
Height moved by small piston = 0.4 m
Using the formula, Pressure on small piston = Force on small piston/Area of small piston
Pressure on small piston = 0.09/(π/4 x (0.045)^2) = 503.97 kPa
Using the formula, Pressure on large piston = Pressure on small piston x (Area of small piston/Area of large piston)
Pressure on large piston = 503.97 x (0.00159/0.00785) = 102.28 kPa
Using the formula, Load lifted by piston = Pressure x Area of large piston
Load lifted by piston = 102.28 x 0.00785 = 0.803 kN
Converting kN to N, Load lifted by piston = 803 N
Using the formula, Pressure = Density x Gravity x Height difference
Pressure = 850 x 9.81 x 0.4 = 3337.4 Pa
Converting Pa to N/cm^2, Pressure = 0.033 N/cm^2
Therefore, the correct option is A) 60 N/cm^2.
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The diameters of a small piston and a large piston of a hydraulic jack are 45 mm and 100 mm respectively.Force of 0.09 kN applied on smaller in size piston. Find load lifted by piston if smaller in size piston is 40 cm above the large piston. The density of fluid is 850 kg/m3a)60 N/cm2b)12 N/cm2c)30 N/cm2d)None of the mentionedCorrect answer is option 'A'. Can you explain this answer?
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