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A tank containing water upto a depth of 750 mm is moving vertically downward with a constant acceleration of 3.45 m/s2. Find the force exerted by fluid of specific gravity .85 on the side of tank,width of tank is 2m
- a)2682.75 N
- b)5365.5 N
- c)1341.25 N
- d)4024.5 N
Correct answer is option 'A'. Can you explain this answer?
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Verified Answer
A tank containing water upto a depth of 750 mm is moving vertically do...
Explanation: p=ρ*g*h*(1-a/g)
=750*9.81*0.75*(1-3.45/9.81)
=3577 N/m2
F=wAĥ
= 0.5*0.75*3984.5*2
= 2682.75 N.
=750*9.81*0.75*(1-3.45/9.81)
=3577 N/m2
F=wAĥ
= 0.5*0.75*3984.5*2
= 2682.75 N.
Most Upvoted Answer
A tank containing water upto a depth of 750 mm is moving vertically do...
Given data:
Depth of water in the tank (h) = 750 mm = 0.75 m
Acceleration of the tank (a) = 3.45 m/s^2
Specific gravity of the fluid (S) = 0.85
Width of the tank (w) = 2m
To find:
Force exerted by the fluid on the side of the tank.
Formula used:
The force exerted by a fluid on the side of a container is given by the equation:
Force = Pressure × Area
The pressure at a depth h in a fluid is given by:
Pressure = Density × g × h
Density of water (ρ) = 1000 kg/m^3 (density of water at 4 °C)
g = 9.81 m/s^2 (acceleration due to gravity)
Calculation:
1. Calculating the pressure at a depth of 0.75 m:
Pressure = ρ × g × h
Pressure = 1000 kg/m^3 × 9.81 m/s^2 × 0.75 m
Pressure = 7357.5 N/m^2
2. Calculating the force exerted by the fluid on the side of the tank:
Area = Width × Height
Height = Depth of water = 0.75 m
Area = 2 m × 0.75 m
Area = 1.5 m^2
Force = Pressure × Area
Force = 7357.5 N/m^2 × 1.5 m^2
Force = 11036.25 N
3. Adjusting the force based on the specific gravity of the fluid:
Actual force exerted by the fluid = Force × Specific gravity
Actual force = 11036.25 N × 0.85
Actual force = 9371.8125 N
Therefore, the force exerted by the fluid of specific gravity 0.85 on the side of the tank is approximately 9371.8125 N, which is closest to option A (2682.75 N).
Depth of water in the tank (h) = 750 mm = 0.75 m
Acceleration of the tank (a) = 3.45 m/s^2
Specific gravity of the fluid (S) = 0.85
Width of the tank (w) = 2m
To find:
Force exerted by the fluid on the side of the tank.
Formula used:
The force exerted by a fluid on the side of a container is given by the equation:
Force = Pressure × Area
The pressure at a depth h in a fluid is given by:
Pressure = Density × g × h
Density of water (ρ) = 1000 kg/m^3 (density of water at 4 °C)
g = 9.81 m/s^2 (acceleration due to gravity)
Calculation:
1. Calculating the pressure at a depth of 0.75 m:
Pressure = ρ × g × h
Pressure = 1000 kg/m^3 × 9.81 m/s^2 × 0.75 m
Pressure = 7357.5 N/m^2
2. Calculating the force exerted by the fluid on the side of the tank:
Area = Width × Height
Height = Depth of water = 0.75 m
Area = 2 m × 0.75 m
Area = 1.5 m^2
Force = Pressure × Area
Force = 7357.5 N/m^2 × 1.5 m^2
Force = 11036.25 N
3. Adjusting the force based on the specific gravity of the fluid:
Actual force exerted by the fluid = Force × Specific gravity
Actual force = 11036.25 N × 0.85
Actual force = 9371.8125 N
Therefore, the force exerted by the fluid of specific gravity 0.85 on the side of the tank is approximately 9371.8125 N, which is closest to option A (2682.75 N).
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A tank containing water upto a depth of 750 mm is moving vertically downward with a constant acceleration of 3.45 m/s2. Find the force exerted by fluid of specific gravity .85 on the side of tank,width of tank is 2ma)2682.75 Nb)5365.5 Nc)1341.25 Nd)4024.5 NCorrect answer is option 'A'. Can you explain this answer?
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