Hydrostatic Force MCQ Quiz - Objective Question with Answer for Hydrostatic Force - Download Free PDF
Last updated on May 12, 2025
Latest Hydrostatic Force MCQ Objective Questions
Hydrostatic Force Question 1:
Which of the following statements is correct with respect to the difference between hydrostatics and hydrodynamics?
Answer (Detailed Solution Below)
Hydrostatic Force Question 1 Detailed Solution
Explanation:
Hydrostatics:
Hydrostatics deals with fluids at rest. This branch studies the conditions of fluids that are not in motion, particularly focusing on the forces exerted by fluids at rest. The main concepts include pressure distribution, buoyancy, and the effects of gravity on fluids.
-
Key Concept: In a fluid at rest, the pressure at any given point is due to the weight of the fluid above that point. The pressure increases with depth because the weight of the fluid above adds to the pressure experienced at lower points.
-
Applications:
-
Calculation of pressure in a fluid at a certain depth.
-
Understanding buoyancy (the force that allows objects to float).
-
Determining the forces acting on submerged bodies.
-
Hydrodynamics:
Hydrodynamics, on the other hand, is concerned with fluids in motion. It studies the behavior of fluids when they are moving, whether it’s the flow of water through pipes, the airflow over an aircraft wing, or ocean currents. This branch examines how forces, velocity, pressure, and energy interact in a moving fluid.
-
Key Concept: In hydrodynamics, the flow of the fluid is analyzed. It may be steady or unsteady, laminar (smooth and orderly) or turbulent (chaotic and irregular), and it often involves the study of velocity distributions, forces acting on boundaries (like pipe walls), and how pressure varies with speed.
-
Applications:
-
Analyzing fluid flow in pipes, ducts, and open channels.
-
Study of aerodynamics and hydrodynamics of vehicles (airplanes, ships).
-
Designing hydraulic systems and turbines.
-
Additional Information Key Differences between Hydrostatics and Hydrodynamics:
-
Fluid Motion:
-
Hydrostatics: Studies fluids at rest (no motion).
-
Hydrodynamics: Studies fluids in motion (flowing fluids).
-
-
Pressure Considerations:
-
Hydrostatics: Focuses on the pressure due to the weight of the fluid. Pressure at any point depends on the depth of the fluid.
-
Hydrodynamics: Focuses on how pressure changes with the velocity of the fluid and how forces act due to the fluid's motion.
-
-
Flow Conditions:
-
Hydrostatics: Since the fluid is not moving, concepts like flow type (turbulent or laminar) don't apply.
-
Hydrodynamics: It considers both laminar flow (where the fluid flows in smooth layers) and turbulent flow (where the fluid undergoes chaotic fluctuations).
-
Hydrostatic Force Question 2:
At what point within a submerged body in a fluid at rest will the pressure be greatest?
Answer (Detailed Solution Below)
Hydrostatic Force Question 2 Detailed Solution
In fluid mechanics, the pressure within a submerged body in a fluid at rest increases with depth due to the weight of the fluid above it. This is described by the hydrostatic pressure formula:
Hydrostatic pressure formula:
P = P0 + ρgh
Where:
-
P = Pressure at a certain depth
-
P0 = Atmospheric pressure at the surface
-
ρ = Density of the fluid
-
g = Acceleration due to gravity
-
h = Depth below the surface
From the formula, it is clear that pressure increases with depth (h). Therefore, the pressure will be greatest at the bottommost point of the submerged body.
Hydrostatic Force Question 3:
The location of the centre of pressure over a surface immersed in a liquid is
Answer (Detailed Solution Below)
Hydrostatic Force Question 3 Detailed Solution
Concept:
- According to Pascal’s Law, the pressure or intensity of pressure at a point in a static fluid is equal in all directions.
- For a plane surface of arbitrary shape immersed in a liquid in such a way that the plane of the surface makes an angle θ with the free surface of the liquid:
A = Total area of an inclined surface, h̅ = Depth of centre of gravity of inclined area from a free surface, h* = Distance of centre of pressure from the free surface of a liquid
\({h^*} = \frac{{{I_G}{{\sin }^2}\theta }}{{A̅ h}} + \bar h\)
For vertical plane surface: θ = 90°
\({h^*} = \frac{{{I_G}}}{{A \bar h}} + \bar h\)
The following facts can be concluded from the above equations:
- Center of pressure lies below the centroid because for any plane surface, the factor \(\frac{I_G}{A \bar h}\) is always positive
- Deeper the surface is lowered into the liquid (i.e. greater is the value of h̅), closer comes the centre of pressure to the centroid of the area
- Depth of the center of pressure is independent of the specific weight of the liquid and is consequently same for all liquids
Total Pressure:
\(F_h=\omega ∫ydA \cos \theta\)
\(F_y=\omega ∫ydA \sin \theta\)
The total pressure on a surface immersed in a liquid depends on its inclination to the liquid surface.
Hydrostatic Force Question 4:
A 5 m × 5 m closed tank of 10 m height contains water and oil and is connected to an overhead water reservoir as shown in the figure. Use γw = 10 kN/m3 and specific gravity of oil = 0.8.
The total force (in kN) due to pressure on the side PQR of the tank is (rounded off to the nearest integer).
Answer (Detailed Solution Below) 5580
Hydrostatic Force Question 4 Detailed Solution
Explanation:
Considering PQR
PR + 4 × (0.8γw) = 10γw
⇒ PR = (10 3.2)γw = 6.8γw
Now,
Considering force prism QRST
\(\rm F_{1}=\frac{(6.8+10)}{2} \gamma_{w} \times 4 \times 5=168 \gamma_{w}=1680 \mathrm{kN}\)
Considering force prism PQTU
\(\rm F_{2}=\frac{(10+16)}{2} \gamma_{w} \times 6 \times 5=3900 \mathrm{kN}\)
So, F = F1 + F2 = 1680 + 3900 = 5580 kN
Hydrostatic Force Question 5:
Determine the vertical component of hydrostatic force per unit width (in N/m) acting on the curved (quarter circle) gate as shown in the figure. The specific weight of water is given as 9810 N/m3.
Answer (Detailed Solution Below)
Hydrostatic Force Question 5 Detailed Solution
Vertical Hydrostatic Force: \( F_v = \gamma \cdot V \), Volume of Water above the Gate: \( V = \frac{\pi r^2}{4} \cdot 1 \, \text{(per unit width)} \), \( r = 1 \, \text{m} \), Substitute: \( F_v = 9810 \cdot \frac{\pi \cdot 1^2}{4} = 7704.75 \, \text{N/m} \)
Top Hydrostatic Force MCQ Objective Questions
A vertical triangular plane area, submerged in water, with one side in the free surface, vertex downward and latitude ‘h’ was the pressure centre below the free surface by
Answer (Detailed Solution Below)
Hydrostatic Force Question 6 Detailed Solution
Download Solution PDFCentre of pressure
\(\begin{array}{l} {{\rm{h}}^{\rm{*}}} = {\rm{\bar X}} + \frac{{{{\rm{I}}_{\rm{G}}}}}{{{\rm{A\bar X}}}} = \frac{{\rm{h}}}{3} + \frac{{\frac{{{\rm{b}}{{\rm{h}}^3}}}{{36}}}}{{\frac{{{\rm{bh}}}}{2}.\frac{{\rm{h}}}{3}}}\\ = \frac{{\rm{h}}}{3} + \frac{{\rm{h}}}{6} = \frac{{\left( {2 + 1} \right){\rm{h}}}}{6} = \frac{{\rm{h}}}{2} \end{array}\)
Important point:
Geometry |
Centre of pressure |
|
\(\frac{2h}{3}\) |
|
\(\frac{h}{2}\) |
|
\(\frac{{3h}}{4}\) |
|
\(\frac{{5h}}{8}\) |
|
\(\frac{{3\pi D}}{{32}}\) |
|
\(\frac{{3\pi D}}{{32}}\) |
|
\(\frac{{h\left( {a + 3b} \right)}}{{2\left( {a + 2b} \right)}}\) |
As the depth of immersion of a vertical plane surface increases, the location of centre of pressure
Answer (Detailed Solution Below)
Hydrostatic Force Question 7 Detailed Solution
Download Solution PDFConcept:
We know that the centre of pressure is the point at which resultant pressure force due to fluid acts and it is given by-
\({\rm{h}} = {\rm{\bar x}} + \frac{{{{\rm{I}}_{\rm{G}}}}}{{{\rm{A\bar x}}}}{\left( {\sin {\rm{θ }}} \right)^2}\)
Where, h represents the centre of pressure, IG = second moment of area, A = area, and x̅ = centre of gravity.
Note:
In case of vertical surface, θ = 90° and Sin θ = 1
\(\frac{{{{\rm{I}}_{\rm{G}}}}}{{{\rm{A\bar x}}}}{\left( {\sin {\rm{θ }}} \right)^2}\) = \(\frac{{{{\rm{I}}_{\rm{G}}}}}{{{\rm{A\bar x}}}}\)
\({\rm{h}} = {\rm{\bar x}} + \frac{{{{\rm{I}}_{\rm{G}}}}}{{{\rm{A\bar x}}}}\)
Now, as the depth of immersion increases, x̅ increases which results in an increase in h but the rate of increment would be decreased because the factor \(\frac{{{{\rm{I}}_{\rm{G}}}}}{{{\rm{A\bar x}}}}\) has x̅ in the denominator.
Hence h will come closer to x̅.
Important Points
For θ = 90°, which means vertical, the centre of pressure is farthest to the centre of gravity.
For θ = 0°, which means horizontal, the centre of pressure is coinciding with the centre of gravity.
What happens when a fluid is at rest?
Answer (Detailed Solution Below)
Hydrostatic Force Question 8 Detailed Solution
Download Solution PDFExplanation:
In a real fluid, the following forces are present
Gravity force (Fg) due to gravity
Pressure force (Fp) due to the pressure of the fluid
Viscous force (Fν) due to viscosity
Tension force (Fs) due to surface tension
Turbulent force (Ft) due to turbulence.
and (Fc) due to compressibility.
∴ Fnet = Fg + Fp + Fν + Fs + Ft + Fc
- If Fnet = Fg + Fp + Fν + Ft this is known as Reynold’s equations of motion.
- If Fnet = Fg + Fp + Fν this is known as the Navier-Stokes equation of motion.
- If Fnet = Fg + Fp this is known as Euler’s equation of motion.
∵ the fluid is in rest i.e. the velocity gradient is zero, therefore no shear stress/force will be acting and no external force is acting on it, the only force present is due to gravity (Fg) and pressure of the fluid (Fp) i.e. normal force only.
A 20 m high dam is filled with water up to the top. The force acting on the vertical dam wall (20 m high × 25 m wide) is given as (consider the density of water = 1000 kg/m3; g = acceleration due to gravity):
Answer (Detailed Solution Below)
Hydrostatic Force Question 9 Detailed Solution
Download Solution PDFConcept:
Total pressure is defined as the force exerted by a static fluid on a surface either on curved when the fluid comes in contact with the surface.
Centre of pressure is defined as the point of application of the total pressure on the surface.
The point of action of total hydrostatic force on the submerged surface is called the Centre of Pressure.
G = Centre of gravity of plane surface
P = Centre of pressure
F = ρgAx̅
Calculations:
Force on the dam wall (F) = Pressure at the center (P) × Projected area of the dam (A) = ρgAx̅
The pressure at the center (P) = ρ × g × x̅
Given:
Height of the dam (H) = 20 m
Height of the centroid from the water surface (x̅) = H/2 = 10 m
Density of water (ρ) = 1000 kg/m3
Width of dam (w) = 25 m
Area of dam (A) = 20 × 25 = 500 m2
F = P × A = (ρ × g × h̅) × 500 = (5 × 106 × g) = 5g MNDetermine the total water force (in kN) acting on the bottom of a tank which is completely filled with water. Height and width of the tank is 3 m and 2 m, respectively. Take length as 2 m
Answer (Detailed Solution Below)
Hydrostatic Force Question 10 Detailed Solution
Download Solution PDFConcept:
Water pressure at bottom of the tank (P) =ρgh
Where ρ = Density of water,
g = Acceleration due to gravity,
h = Height of the water in the tank.
Total water force at bottom of the tank (F) = Water pressures at bottom of the tank (P) × Area of the tank bottom (A)
Total water force at bottom of the tank (F) = ρghA
Calculation:
Given data:
Height of tank (h) = 3 m
Width of the tank (b) = 2 m and length of tank = 2 m
Total water force at bottom of the tank (F) =?
g = 9.81 m/s
ρ = 1000 kg/m3
Area of tank bottom (A) = 2 × 2 = 4 m2
Total water force at bottom of the tank (F) = 1000 × 9.81 × 3 × 4 = 117720 N
Total water force at bottom of the tank (F) = 117.720 kN
The water level in a dam is 10 m. The total force acting on vertical wall per metre length is:
Answer (Detailed Solution Below)
Hydrostatic Force Question 11 Detailed Solution
Download Solution PDFConcept:
Whenever a static mass of fluid comes into contact with a surface the fluid exerts force upon that surface. The magnitude of this force is known as the hydrostatic force or total pressure force.
The magnitude of hydrostatic force is given by F = ρgh̅A
Where ρ = Density of the fluid, h̅ = Depth of center of gravity of the surface from free liquid surface, A = Area of the surface
Calculation:
Given, h = 10 m
h̅ = 10/2 = 5 m
Area per metre length = 10 × 1 = 10 m2
Force acting on vertical wall = ρ × g × h̅ × A = 1000 × 9.81 × 5 × 10 = 490500 N = 490.5 kNDetermine the total pressure on a circular plate of diameter 2 m, which is placed vertically in water in such a way that the Centre of the plate is 2.5 m below the free surface of the water.
Answer (Detailed Solution Below)
Hydrostatic Force Question 12 Detailed Solution
Download Solution PDFConcept:
When any plate is immersed vertically in water having any arbitrary shape,
total horizontal force is given as
FH = ρgAx̅
Where,
ρ = density of the fluid (kg/m3),
g = acceleration due to gravity (m/s2)
x̅ = depth of the center of gravity of the submerged body from the free surface (m)
A = total area of the surface (m2)
Calculation:
Given: Diameter of circular plate d = 2 m, x̅ = 2.5 m, ρ = 1000 kg/m3
\(\begin{aligned} & A=\frac{\pi}{4} \times d^2=\frac{\pi}{4} \times 2^2=3.141 \\ & F_H=\rho g A \bar{x}=1000 \times 9.81 \times 3.141 \times 2.5 \\ & F_H=77047.55 \mathrm{~N} \approx 77048 \mathrm{~N} \end{aligned}\)
Additional Information
Note: When nothing is given about g, take g = 9.81 m/s2.
Don’t get confused between the centre of gravity and the centre of pressure while calculating FH.
While calculating FH centre of gravity is considered but the horizontal force acts at the centre of pressure.
Position of centre of pressure,
\(\bar{h}=\frac{I_{x x} \sin ^2 \theta}{A \cdot \bar{x}}+\bar{x}\)
Determine the factor when the hydrostatic force on one of the vertical sides of the beaker decreases when the height of the liquid column is reduced to one-fourth.
Answer (Detailed Solution Below)
Hydrostatic Force Question 13 Detailed Solution
Download Solution PDFConcept:
Hydrostatic force per unit width on the vertical side of beaker is given as
F = \(\frac 1 2\)ρgh2
Calculation:
Given:
h' = h/4
The hydrostatic force will be
F' = \(\frac 1 2\)ρg(h/4)2
F' = \(\frac{1}{16}\) ×\(\frac 1 2\)ρgh2
F' = \(\frac{1}{16}\)F
What is the depth of the centre of pressure for the rectangular lamina which is vertically inside the water of height h?
Answer (Detailed Solution Below)
Hydrostatic Force Question 14 Detailed Solution
Download Solution PDFConcept:
The Centre of Pressure is the average location of all of the pressure acting upon a body moving through a fluid.
The depth of the centre of pressure
\({h_{cp}} = \bar h + \frac{{{I_G}}}{{A\bar h}}\)
Calculation:
\({h_{cp}} = \frac{h}{2} + \frac{{b{h^3}/12}}{{\left( {b \times h} \right) \times \frac{h}{2}}}\)
\({h_{cp}} = \frac{{2h}}{3}\)When the depth of immersion of a plane surface is increased the centre of pressure will
Answer (Detailed Solution Below)
Hydrostatic Force Question 15 Detailed Solution
Download Solution PDFConcept:
We know that the centre of pressure is the point at which resultant pressure force due to fluid acts and it is given by-
\({\rm{h}} = {\rm{\bar x}} + \frac{{{{\rm{I}}_{\rm{G}}}}}{{{\rm{A\bar x}}}}{\left( {\sin {\rm{\theta }}} \right)^2}\)
Where, h represents the centre of pressure, IG = second moment of area, A = area, and x̅ = centre of gravity.
Now, as the depth of immersion increases, x̅ increases which results into increase in h but the rate of increment would be decreased because the factor \(\frac{{{{\rm{I}}_{\rm{G}}}}}{{{\rm{A\bar x}}}}{\left( {\sin {\rm{\theta }}} \right)^2}\) has x̅ in the denominator.
Hence h will come closer to x̅.
Important Points
For θ = 90°, means vertical, the centre of pressure is farthest to the centre of gravity.
For θ = 0°, means horizontal, the centre of pressure is coinciding with the centre of gravity.