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A tank containing water upto a depth of 650 mm is stationary. Find the force exerted by fluid of specific gravity .55 on the side of tank,width of tank is 1.5m
- a)1709.9 N
- b)3419.4N
- c)6838.8 N
- d)1367.75 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 650 mm is stationary. Find the...
Explanation: p=ρ*g*h
=550*9.81*0.65
=3507 N/m2
F=wAĥ
= 0.5*0.65*3507*1.5
= 1709.7N.
=550*9.81*0.65
=3507 N/m2
F=wAĥ
= 0.5*0.65*3507*1.5
= 1709.7N.
Most Upvoted Answer
A tank containing water upto a depth of 650 mm is stationary. Find the...
To find the force exerted by the fluid on the side of the tank, we can use the concept of hydrostatic pressure.
Hydrostatic pressure is the pressure exerted by a fluid at rest due to the force of gravity acting on the fluid column above. It can be calculated using the formula:
Pressure = density × gravity × height
In this case, the height of the fluid column is given as 650 mm or 0.65 m. The specific gravity of the fluid is given as 0.55, which means it is 0.55 times as dense as water.
Let's calculate the force exerted by the fluid on the side of the tank using the given data:
1. Convert the height of the fluid column from millimeters to meters:
Height = 650 mm = 0.65 m
2. Calculate the density of the fluid:
Density = specific gravity × density of water
Density = 0.55 × 1000 kg/m³ (density of water)
Density = 550 kg/m³
3. Calculate the force exerted by the fluid:
Force = pressure × area
The area of the side of the tank can be calculated using the width of the tank, which is given as 1.5 m. Since the force is exerted uniformly on the entire side of the tank, we can take the area as the product of the width and the height of the tank.
Area = width × height
Area = 1.5 m × 0.65 m
Area = 0.975 m²
Now, we can calculate the force:
Force = pressure × area
Force = density × gravity × height × area
Force = 550 kg/m³ × 9.8 m/s² × 0.65 m × 0.975 m²
Force ≈ 1709.9 N
Therefore, the force exerted by the fluid of specific gravity 0.55 on the side of the tank is approximately 1709.9 N.
Hydrostatic pressure is the pressure exerted by a fluid at rest due to the force of gravity acting on the fluid column above. It can be calculated using the formula:
Pressure = density × gravity × height
In this case, the height of the fluid column is given as 650 mm or 0.65 m. The specific gravity of the fluid is given as 0.55, which means it is 0.55 times as dense as water.
Let's calculate the force exerted by the fluid on the side of the tank using the given data:
1. Convert the height of the fluid column from millimeters to meters:
Height = 650 mm = 0.65 m
2. Calculate the density of the fluid:
Density = specific gravity × density of water
Density = 0.55 × 1000 kg/m³ (density of water)
Density = 550 kg/m³
3. Calculate the force exerted by the fluid:
Force = pressure × area
The area of the side of the tank can be calculated using the width of the tank, which is given as 1.5 m. Since the force is exerted uniformly on the entire side of the tank, we can take the area as the product of the width and the height of the tank.
Area = width × height
Area = 1.5 m × 0.65 m
Area = 0.975 m²
Now, we can calculate the force:
Force = pressure × area
Force = density × gravity × height × area
Force = 550 kg/m³ × 9.8 m/s² × 0.65 m × 0.975 m²
Force ≈ 1709.9 N
Therefore, the force exerted by the fluid of specific gravity 0.55 on the side of the tank is approximately 1709.9 N.
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A tank containing water upto a depth of 650 mm is stationary. Find the force exerted by fluid of specific gravity .55 on the side of tank,width of tank is 1.5ma)1709.9 Nb)3419.4Nc)6838.8 Nd)1367.75 NCorrect answer is option 'A'. Can you explain this answer?
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