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All questions of Hydrostatic Forces on Surfaces for Civil Engineering (CE) Exam
A floating body is in stable equilibrium when
- a)the centre of gravity is above the centre of buoyancy
- b)the metacentre is below the centre of gravity but above centre of buoyancy
- c)when the metacentre is below the centre of buoyancy
- d)when
is positive
Correct answer is option 'D'. Can you explain this answer?
A floating body is in stable equilibrium when
a)
the centre of gravity is above the centre of buoyancy
b)
the metacentre is below the centre of gravity but above centre of buoyancy
c)
when the metacentre is below the centre of buoyancy
d)
when is positive
is positive
|
Lavanya Menon answered |
Floating body will be under stable equilibrium when metacentre is positive
A vertical rectangular plane surface is submerged in water such that its top and bottom surfaces are 1.5 m and 6.0 m respectively below the free surface. The position of centre of pressure below the free surface will be at a distance of- a)3.75 m
- b)4,0 m
- c)4.2 m
- d)4.5 m
Correct answer is option 'C'. Can you explain this answer?
A vertical rectangular plane surface is submerged in water such that its top and bottom surfaces are 1.5 m and 6.0 m respectively below the free surface. The position of centre of pressure below the free surface will be at a distance of
a)
3.75 m
b)
4,0 m
c)
4.2 m
d)
4.5 m
|
Sanya Agarwal answered |
An isosceles triangular plate of base 3 m and altitude 3 m is immersed vertically in an oil of specific gravity 0.8. The base of the plate coincides with the free surface of oil. The centre of pressure will lie at a distance of (from free surface)- a)2.5 m
- b)2 m
- c)1.5 m
- d)1 m
Correct answer is option 'C'. Can you explain this answer?
An isosceles triangular plate of base 3 m and altitude 3 m is immersed vertically in an oil of specific gravity 0.8. The base of the plate coincides with the free surface of oil. The centre of pressure will lie at a distance of (from free surface)
a)
2.5 m
b)
2 m
c)
1.5 m
d)
1 m
|
Priyanka Shah answered |
Depth of pressure (h*) from free oil surface is given by,
A cubic tank is completely filled with water. What will be the ratio of the hydrostatic force exerted on the base and on any one of the vertical sides?- a)1:1
- b)2:1
- c)1:2
- d)3:2
Correct answer is option 'B'. Can you explain this answer?
A cubic tank is completely filled with water. What will be the ratio of the hydrostatic force exerted on the base and on any one of the vertical sides?
a)
1:1
b)
2:1
c)
1:2
d)
3:2
|
Arjun Unni answered |
Hydrostatic Force in a Cubic Tank
1. Hydrostatic Force on the Base:
- The hydrostatic force exerted on the base of the cubic tank is directly proportional to the depth of the water.
- Since the tank is completely filled with water, the depth is the same as the height of the tank.
- The formula for calculating hydrostatic force on the base is F_base = ρ*g*A*h, where ρ is the density of water, g is the acceleration due to gravity, A is the area of the base, and h is the height of the water column.
- As the height is the same as the side length of the cubic tank, the area of the base is (side length)^2.
2. Hydrostatic Force on a Vertical Side:
- The hydrostatic force exerted on a vertical side of the tank is also directly proportional to the depth of the water.
- Since the tank is completely filled, the depth of the water is the same across all sides.
- The formula for calculating hydrostatic force on a vertical side is F_side = ρ*g*A*h, where A is the area of the side (side length * height) and h is the height of the water column.
- The area of a vertical side is the product of the side length and the height of the tank.
Therefore, the ratio of the hydrostatic force exerted on the base to that on a vertical side is:
F_base : F_side = (ρ*g*(side length)^2*h) : (ρ*g*side length*height) = h : side length = 1 : 1
Thus, the correct ratio of the hydrostatic force exerted on the base to that on any one of the vertical sides of the cubic tank is 1:1.
1. Hydrostatic Force on the Base:
- The hydrostatic force exerted on the base of the cubic tank is directly proportional to the depth of the water.
- Since the tank is completely filled with water, the depth is the same as the height of the tank.
- The formula for calculating hydrostatic force on the base is F_base = ρ*g*A*h, where ρ is the density of water, g is the acceleration due to gravity, A is the area of the base, and h is the height of the water column.
- As the height is the same as the side length of the cubic tank, the area of the base is (side length)^2.
2. Hydrostatic Force on a Vertical Side:
- The hydrostatic force exerted on a vertical side of the tank is also directly proportional to the depth of the water.
- Since the tank is completely filled, the depth of the water is the same across all sides.
- The formula for calculating hydrostatic force on a vertical side is F_side = ρ*g*A*h, where A is the area of the side (side length * height) and h is the height of the water column.
- The area of a vertical side is the product of the side length and the height of the tank.
Therefore, the ratio of the hydrostatic force exerted on the base to that on a vertical side is:
F_base : F_side = (ρ*g*(side length)^2*h) : (ρ*g*side length*height) = h : side length = 1 : 1
Thus, the correct ratio of the hydrostatic force exerted on the base to that on any one of the vertical sides of the cubic tank is 1:1.
A house-top water tank is made of flat plates and is full to the brim. Its height is twice that of any side. The ratio of total thrust force on the bottom of the tank to that on any side will be:- a)4
- b)2
- c)1
- d)0.5
Correct answer is option 'C'. Can you explain this answer?
A house-top water tank is made of flat plates and is full to the brim. Its height is twice that of any side. The ratio of total thrust force on the bottom of the tank to that on any side will be:
a)
4
b)
2
c)
1
d)
0.5
|
|
Sanvi Kapoor answered |
Concept:
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 = ρ g h̅̅ A N
Where ρ = Density of the fluid, h̅ = Depth of center of gravity of the surface from free liquid surface, A = Area of the surface
Calcualtion:
Given tank has the height twice as the width of the tank.
Pressure force acting on the vertical surface:
The center of gravity of the vertical surface from the free surface will be
h¯= 2x/2 = xm
Area of the vertical surface = 2x × x = 2x2
Now the Pressure force ρ × g × x × 2x2 = 2ρgx3
Now, the force on the bottom surface:
The center of gravity of the bottom surface from the free liquid surface = hieht of the tank = 2x
Area of the bottom surface = x × x = x2
Now the pressure force = ρ × g × 2x × x2 = 2ρgx3
Therefore the ratio of the forces will be one as the forces on the bottom side and the vertical sides are equal.
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 = ρ g h̅̅ A N
Where ρ = Density of the fluid, h̅ = Depth of center of gravity of the surface from free liquid surface, A = Area of the surface
Calcualtion:
Given tank has the height twice as the width of the tank.
Pressure force acting on the vertical surface:
The center of gravity of the vertical surface from the free surface will be
h¯= 2x/2 = xm
Area of the vertical surface = 2x × x = 2x2
Now the Pressure force ρ × g × x × 2x2 = 2ρgx3
Now, the force on the bottom surface:
The center of gravity of the bottom surface from the free liquid surface = hieht of the tank = 2x
Area of the bottom surface = x × x = x2
Now the pressure force = ρ × g × 2x × x2 = 2ρgx3
Therefore the ratio of the forces will be one as the forces on the bottom side and the vertical sides are equal.
A square lamina (each side equal to 2m) is submerged vertically in water such that the upper edge of the lamina is at a depth of 0.5 m from the free surface. What will be the total water pressure (in kN) on the lamina?
- a)19.62
- b)39.24
- c)58.86
- d)78.48
Correct answer is option 'C'. Can you explain this answer?
A square lamina (each side equal to 2m) is submerged vertically in water such that the upper edge of the lamina is at a depth of 0.5 m from the free surface. What will be the total water pressure (in kN) on the lamina?
a)
19.62
b)
39.24
c)
58.86
d)
78.48
|
|
Lavanya Menon answered |
Total liquid pressure on the lamina = F = 
,
where γ = specific weight of the liquid,
= depth of centroid of the lamina from the free surface,
A= area of the centroid.
Now, γ = 9:81 x 103 N / m3;
= 0.5 + 1 ⁄ 2 x 2m = 1.5 m,
A = 2 * 2 m2 = 4 m2.
Hence, F = 58.86 kN.
,where γ = specific weight of the liquid,
= depth of centroid of the lamina from the free surface,A= area of the centroid.
Now, γ = 9:81 x 103 N / m3;
= 0.5 + 1 ⁄ 2 x 2m = 1.5 m,
A = 2 * 2 m2 = 4 m2.
Hence, F = 58.86 kN.
One end of a two dimensional water tank has the shape of a quadrant of a circle of radius 2m. when the tank is full, the vertical component of the force per unit length on the curved surface will be - a)19.62 π kN
- b)39.24 π kN
- c)9.81 π kN
- d)29.43 π kN
Correct answer is option 'C'. Can you explain this answer?
One end of a two dimensional water tank has the shape of a quadrant of a circle of radius 2m. when the tank is full, the vertical component of the force per unit length on the curved surface will be
a)
19.62 π kN
b)
39.24 π kN
c)
9.81 π kN
d)
29.43 π kN
|
|
Ishani Basu answered |
We can use the formula for the hydrostatic force on a curved surface:
F = γhA
where F is the hydrostatic force, γ is the specific weight of water (9810 N/m3), h is the depth of the water, and A is the area of the curved surface.
Since the tank is full, the depth of the water is 2 m (equal to the radius of the curved surface). To find the area of the curved surface, we can use the formula for the area of a quadrant of a circle:
A = (1/4)πr^2
where r is the radius of the circle. Substituting r = 2 m, we get:
A = (1/4)π(2^2) = π m^2/4
Now we can plug in the values:
F = (9810 N/m3)(2 m)(π m^2/4) = 9810π/2 N
Simplifying:
F ≈ 15,318.5 N
So the vertical component of the force per unit length on the curved surface is:
F/2πr = 15,318.5 N/(2π(2 m)) ≈ 1,941.5 N/m ≈ 19.62 kN/m (rounded to two decimal places)
Therefore, the answer is (a) 19.62.
F = γhA
where F is the hydrostatic force, γ is the specific weight of water (9810 N/m3), h is the depth of the water, and A is the area of the curved surface.
Since the tank is full, the depth of the water is 2 m (equal to the radius of the curved surface). To find the area of the curved surface, we can use the formula for the area of a quadrant of a circle:
A = (1/4)πr^2
where r is the radius of the circle. Substituting r = 2 m, we get:
A = (1/4)π(2^2) = π m^2/4
Now we can plug in the values:
F = (9810 N/m3)(2 m)(π m^2/4) = 9810π/2 N
Simplifying:
F ≈ 15,318.5 N
So the vertical component of the force per unit length on the curved surface is:
F/2πr = 15,318.5 N/(2π(2 m)) ≈ 1,941.5 N/m ≈ 19.62 kN/m (rounded to two decimal places)
Therefore, the answer is (a) 19.62.
A body is floating as shown in the given figure. The centre of buoyancy, centre of gravity and metacentre are labelled respectively as B, G and M. The body is- a)vertically stable
- b)vertically unstable
- c)rotationally stable
- d)rotationally unstable
Correct answer is option 'D'. Can you explain this answer?
A body is floating as shown in the given figure. The centre of buoyancy, centre of gravity and metacentre are labelled respectively as B, G and M. The body is
a)
vertically stable
b)
vertically unstable
c)
rotationally stable
d)
rotationally unstable
|
|
Swati Gupta answered |
Since metacentre is below centre of gravity the body is in unstable equilibrium in rotation.
Consider a frictionless, massless and leak-proof plug blocking a rectangular hole of dimensions 2R × L at the bottom of an open tank as shown in the figure. The head of the plug has the shape of a semi-cylinder of radius R. The tank is filled with a liquid of density ρ up to the tip of the plug. The gravitational acceleration is g. Neglect the effect of the atmospheric pressure.
The force F required to hold the plug in its position is- a)
- b)
- c)πR2ρgL
- d)
Correct answer is option 'A'. Can you explain this answer?
Consider a frictionless, massless and leak-proof plug blocking a rectangular hole of dimensions 2R × L at the bottom of an open tank as shown in the figure. The head of the plug has the shape of a semi-cylinder of radius R. The tank is filled with a liquid of density ρ up to the tip of the plug. The gravitational acceleration is g. Neglect the effect of the atmospheric pressure.
The force F required to hold the plug in its position is
The force F required to hold the plug in its position is
a)
b)
c)
πR2ρgL
d)
|
Pioneer Academy answered |
Concept:
The force applied from the bottom should be such that it must balance the force applied due to water above the cap.
Force F here is vertical hydrostatic force
F = ρgV
Where V = volume of fluid above the curved surface
So, F = ρgV
V = [Volume of a cuboid – Volume of hemisphere]
The force applied from the bottom should be such that it must balance the force applied due to water above the cap.
Force F here is vertical hydrostatic force
F = ρgV
Where V = volume of fluid above the curved surface
So, F = ρgV
V = [Volume of a cuboid – Volume of hemisphere]
One end of a two-dimensional water tank has the shape of a quadrant of a circle of radius 2 m. When the tank is full, the vertical component of the force per unit length on the curved surface will be- a)9.81π kN
- b)19.62 π kN
- c)29.43π kN
- d)39.24π kN
Correct answer is option 'A'. Can you explain this answer?
One end of a two-dimensional water tank has the shape of a quadrant of a circle of radius 2 m. When the tank is full, the vertical component of the force per unit length on the curved surface will be
a)
9.81π kN
b)
19.62 π kN
c)
29.43π kN
d)
39.24π kN
|
Aditya Jain answered |
B)Zero
c)6.28
d)3.14
The correct answer is d) 3.14.
Explanation:
The vertical component of the force per unit length on the curved surface is given by:
F = ρgH
where ρ is the density of water, g is the acceleration due to gravity, and H is the height of the water column above the curved surface.
In this case, the curved surface is the quadrant of a circle of radius 2 m. The height of the water column above the curved surface will be equal to the height of the quadrant, which is also 2 m.
Therefore, H = 2 m.
Substituting this value into the equation for F, we get:
F = ρg(2 m) = 2ρg
Since ρ and g are constants, the vertical component of the force per unit length on the curved surface is also constant and equal to 2ρg.
Now, we need to determine the value of 2ρg.
The density of water is approximately 1000 kg/m³, and the acceleration due to gravity is approximately 9.81 m/s².
Substituting these values into the equation for F, we get:
F = 2ρg = 2(1000 kg/m³)(9.81 m/s²) = 19620 N/m
This is the total force per unit length on the curved surface. However, we are only interested in the vertical component of this force.
Since the curved surface is a quadrant of a circle, we can use the formula for the area of a quadrant to find the length of the curved surface:
A = (1/4)πr²
where r is the radius of the circle.
Substituting r = 2 m, we get:
A = (1/4)π(2 m)² = π m²
Therefore, the vertical component of the force per unit length on the curved surface is:
Fv = F/A = (19620 N/m)/(π m²) ≈ 3.14 N/m
So the correct option is d) 3.14.
c)6.28
d)3.14
The correct answer is d) 3.14.
Explanation:
The vertical component of the force per unit length on the curved surface is given by:
F = ρgH
where ρ is the density of water, g is the acceleration due to gravity, and H is the height of the water column above the curved surface.
In this case, the curved surface is the quadrant of a circle of radius 2 m. The height of the water column above the curved surface will be equal to the height of the quadrant, which is also 2 m.
Therefore, H = 2 m.
Substituting this value into the equation for F, we get:
F = ρg(2 m) = 2ρg
Since ρ and g are constants, the vertical component of the force per unit length on the curved surface is also constant and equal to 2ρg.
Now, we need to determine the value of 2ρg.
The density of water is approximately 1000 kg/m³, and the acceleration due to gravity is approximately 9.81 m/s².
Substituting these values into the equation for F, we get:
F = 2ρg = 2(1000 kg/m³)(9.81 m/s²) = 19620 N/m
This is the total force per unit length on the curved surface. However, we are only interested in the vertical component of this force.
Since the curved surface is a quadrant of a circle, we can use the formula for the area of a quadrant to find the length of the curved surface:
A = (1/4)πr²
where r is the radius of the circle.
Substituting r = 2 m, we get:
A = (1/4)π(2 m)² = π m²
Therefore, the vertical component of the force per unit length on the curved surface is:
Fv = F/A = (19620 N/m)/(π m²) ≈ 3.14 N/m
So the correct option is d) 3.14.
The centre of gravity of the volume of liquid displaced is called- a)centre of pressure
- b)centre of buoyancy
- c)metacentre
- d)None of these
Correct answer is option 'B'. Can you explain this answer?
The centre of gravity of the volume of liquid displaced is called
a)
centre of pressure
b)
centre of buoyancy
c)
metacentre
d)
None of these
|
Akshat Datta answered |
The Centre of Buoyancy
The centre of buoyancy is the point at which the buoyant force acting on an object is considered to be concentrated. It is the centre of gravity of the volume of liquid displaced by the object. This concept is important in the field of fluid mechanics, particularly in the study of buoyancy and stability of floating and submerged objects.
Explanation
When an object is submerged in a fluid (liquid or gas), it experiences an upward force called the buoyant force. This force is equal to the weight of the fluid displaced by the object. According to Archimedes' principle, the buoyant force acts through the centre of gravity of the displaced fluid.
The centre of gravity of an object is the point where its weight can be considered to act. Similarly, the centre of buoyancy is the point where the buoyant force can be considered to act. In simple terms, it is the average position of all the buoyant forces acting on different parts of the submerged object.
Significance
The centre of buoyancy plays a crucial role in determining the stability of floating and submerged objects. If the centre of gravity of an object is below the centre of buoyancy, the object will be stable and tend to return to its original position when displaced. On the other hand, if the centre of gravity is above the centre of buoyancy, the object will be unstable and tend to capsize or topple over.
For example, in a ship, the centre of gravity is usually located above the centre of buoyancy. This ensures the stability of the ship and prevents it from capsizing. The design of ships takes into account the position of the centre of buoyancy relative to the centre of gravity to ensure safe and stable operation.
Conclusion
In conclusion, the centre of buoyancy is the point at which the buoyant force is considered to act on a submerged object. It is the centre of gravity of the volume of fluid displaced by the object. Understanding the concept of the centre of buoyancy is essential for analyzing the stability and behavior of floating and submerged objects.
The centre of buoyancy is the point at which the buoyant force acting on an object is considered to be concentrated. It is the centre of gravity of the volume of liquid displaced by the object. This concept is important in the field of fluid mechanics, particularly in the study of buoyancy and stability of floating and submerged objects.
Explanation
When an object is submerged in a fluid (liquid or gas), it experiences an upward force called the buoyant force. This force is equal to the weight of the fluid displaced by the object. According to Archimedes' principle, the buoyant force acts through the centre of gravity of the displaced fluid.
The centre of gravity of an object is the point where its weight can be considered to act. Similarly, the centre of buoyancy is the point where the buoyant force can be considered to act. In simple terms, it is the average position of all the buoyant forces acting on different parts of the submerged object.
Significance
The centre of buoyancy plays a crucial role in determining the stability of floating and submerged objects. If the centre of gravity of an object is below the centre of buoyancy, the object will be stable and tend to return to its original position when displaced. On the other hand, if the centre of gravity is above the centre of buoyancy, the object will be unstable and tend to capsize or topple over.
For example, in a ship, the centre of gravity is usually located above the centre of buoyancy. This ensures the stability of the ship and prevents it from capsizing. The design of ships takes into account the position of the centre of buoyancy relative to the centre of gravity to ensure safe and stable operation.
Conclusion
In conclusion, the centre of buoyancy is the point at which the buoyant force is considered to act on a submerged object. It is the centre of gravity of the volume of fluid displaced by the object. Understanding the concept of the centre of buoyancy is essential for analyzing the stability and behavior of floating and submerged objects.
The water level in a dam is 10 m. The total force acting on vertical wall per metre length is:- a)49.05 kN
- b)490.5 kN
- c)981 kN
- d)490 kN
Correct answer is option 'B'. Can you explain this answer?
The water level in a dam is 10 m. The total force acting on vertical wall per metre length is:
a)
49.05 kN
b)
490.5 kN
c)
981 kN
d)
490 kN
|
Nitya Sharma answered |
Given:
Water level in the dam = 10 m
To find:
Total force acting on vertical wall per metre length
Solution:
The total force acting on the vertical wall of the dam is the product of the pressure exerted by the water and the area of the wall.
1. Pressure exerted by the water:
The pressure exerted by a fluid depends on its density and the depth of the fluid column. The pressure at a certain depth in a fluid can be calculated using the equation:
P = ρgh
Where:
P = pressure (N/m² or Pa)
ρ = density of the fluid (kg/m³)
g = acceleration due to gravity (9.81 m/s²)
h = depth of the fluid column (m)
In this case, the fluid is water and its density is approximately 1000 kg/m³. The depth of the fluid column is given as 10 m. Substituting these values into the equation, we can calculate the pressure exerted by the water:
P = (1000 kg/m³) x (9.81 m/s²) x (10 m) = 98,100 N/m² or 98.1 kPa
2. Area of the wall:
The area of the vertical wall is the height of the water level multiplied by the length of the wall. Since we are considering the force per meter length, we can assume the length of the wall to be 1 m. Therefore, the area of the wall is:
A = 10 m x 1 m = 10 m²
3. Total force:
The total force acting on the vertical wall per meter length is the product of the pressure and the area of the wall:
Total force = Pressure x Area = 98,100 N/m² x 10 m² = 981,000 N or 981 kN
Therefore, the correct option is (c) 981 kN.
Water level in the dam = 10 m
To find:
Total force acting on vertical wall per metre length
Solution:
The total force acting on the vertical wall of the dam is the product of the pressure exerted by the water and the area of the wall.
1. Pressure exerted by the water:
The pressure exerted by a fluid depends on its density and the depth of the fluid column. The pressure at a certain depth in a fluid can be calculated using the equation:
P = ρgh
Where:
P = pressure (N/m² or Pa)
ρ = density of the fluid (kg/m³)
g = acceleration due to gravity (9.81 m/s²)
h = depth of the fluid column (m)
In this case, the fluid is water and its density is approximately 1000 kg/m³. The depth of the fluid column is given as 10 m. Substituting these values into the equation, we can calculate the pressure exerted by the water:
P = (1000 kg/m³) x (9.81 m/s²) x (10 m) = 98,100 N/m² or 98.1 kPa
2. Area of the wall:
The area of the vertical wall is the height of the water level multiplied by the length of the wall. Since we are considering the force per meter length, we can assume the length of the wall to be 1 m. Therefore, the area of the wall is:
A = 10 m x 1 m = 10 m²
3. Total force:
The total force acting on the vertical wall per meter length is the product of the pressure and the area of the wall:
Total force = Pressure x Area = 98,100 N/m² x 10 m² = 981,000 N or 981 kN
Therefore, the correct option is (c) 981 kN.
A rectangular plane surface of width 2 m and height 3 m is placed vertically in water. What will be the location of center of pressure of the surface when its upper edge is horizontal and lies 2.5 m below the free surface of water?- a)4.0755 m
- b)4.0125 m
- c)4.2525 m
- d)4.1875 m
Correct answer is option 'D'. Can you explain this answer?
A rectangular plane surface of width 2 m and height 3 m is placed vertically in water. What will be the location of center of pressure of the surface when its upper edge is horizontal and lies 2.5 m below the free surface of water?
a)
4.0755 m
b)
4.0125 m
c)
4.2525 m
d)
4.1875 m
|
|
Yash Joshi answered |
Given data:
Width of surface (b) = 2 m
Height of surface (h) = 3 m
Distance of upper edge from free surface of water (h1) = 2.5 m
To calculate the location of center of pressure, we need to find the area of the surface (A) and the height of the center of pressure (h2) from the bottom edge of the surface.
Calculation:
The area of the rectangular surface can be calculated as:
A = b × h
A = 2 × 3
A = 6 m²
The height of the center of pressure can be calculated using the following formula:
h2 = (Iy / Ax) + h1
Where,
Iy = Moment of inertia of the area about the horizontal axis passing through the centroid of the area
Ax = Area of the water plane
The moment of inertia of the rectangular surface about the horizontal axis passing through the centroid of the surface can be calculated as:
Iy = (b × h³) / 12
Iy = (2 × 3³) / 12
Iy = 2.25 m⁴
The area of the water plane can be calculated as:
Ax = b × x
Where x is the distance of the centroid of the water plane from the bottom edge of the surface.
The centroid of the water plane will be at a distance of (2.5 + (3/2)) = 3.5 m from the bottom edge of the surface.
Ax = 2 × 3.5
Ax = 7 m²
Now, we can calculate the height of the center of pressure as:
h2 = (Iy / Ax) + h1
h2 = (2.25 / 7) + 2.5
h2 = 0.3214 + 2.5
h2 = 2.8214 m
Therefore, the location of the center of pressure is 2.8214 m from the bottom edge of the surface. The correct answer is option D.
Width of surface (b) = 2 m
Height of surface (h) = 3 m
Distance of upper edge from free surface of water (h1) = 2.5 m
To calculate the location of center of pressure, we need to find the area of the surface (A) and the height of the center of pressure (h2) from the bottom edge of the surface.
Calculation:
The area of the rectangular surface can be calculated as:
A = b × h
A = 2 × 3
A = 6 m²
The height of the center of pressure can be calculated using the following formula:
h2 = (Iy / Ax) + h1
Where,
Iy = Moment of inertia of the area about the horizontal axis passing through the centroid of the area
Ax = Area of the water plane
The moment of inertia of the rectangular surface about the horizontal axis passing through the centroid of the surface can be calculated as:
Iy = (b × h³) / 12
Iy = (2 × 3³) / 12
Iy = 2.25 m⁴
The area of the water plane can be calculated as:
Ax = b × x
Where x is the distance of the centroid of the water plane from the bottom edge of the surface.
The centroid of the water plane will be at a distance of (2.5 + (3/2)) = 3.5 m from the bottom edge of the surface.
Ax = 2 × 3.5
Ax = 7 m²
Now, we can calculate the height of the center of pressure as:
h2 = (Iy / Ax) + h1
h2 = (2.25 / 7) + 2.5
h2 = 0.3214 + 2.5
h2 = 2.8214 m
Therefore, the location of the center of pressure is 2.8214 m from the bottom edge of the surface. The correct answer is option D.
In calculating the drag force using CD the area used is- a)always the frontal area
- b)the planform area when the body is flat like an airfoil
- c)the planform area when the body is bluff like a sphere
- d)always the planform area
Correct answer is option 'B'. Can you explain this answer?
In calculating the drag force using CD the area used is
a)
always the frontal area
b)
the planform area when the body is flat like an airfoil
c)
the planform area when the body is bluff like a sphere
d)
always the planform area
|
|
Aaditya Jain answered |
Planform area is the area of body as seen from above. This area is used for thin, flat surfaces where frictional forces are predominant as for example: flat plate, flow, airplane wings and hydrofoils.
The boundary or profile drag is essentially made up of- a)shear drag due to pure skin friction
- b)pressure drag and shear drag
- c)shear drag and wake drag
- d)pressure drag due to unbalanced pressure distribution
Correct answer is option 'B'. Can you explain this answer?
The boundary or profile drag is essentially made up of
a)
shear drag due to pure skin friction
b)
pressure drag and shear drag
c)
shear drag and wake drag
d)
pressure drag due to unbalanced pressure distribution
|
|
Juhi Choudhary answered |
The total drag (sometimes called as profile drag) of a body is made up of two parts: Frictional drag and pressure drag (or from drag), depending on whether the shear stresses or pressure differences cause it.
A vertical Gate 6 m × 6 m holds water on one side with free surface at its top. The moment about the bottom edge of the gate of the water force will be (γw is the specific weight of water)
- a)18γ
- b)216y
- c)36y
- d)72y
Correct answer is option 'B'. Can you explain this answer?
A vertical Gate 6 m × 6 m holds water on one side with free surface at its top. The moment about the bottom edge of the gate of the water force will be (γw is the specific weight of water)
a)
18γ
b)
216y
c)
36y
d)
72y
|
|
Sanvi Kapoor answered |
Concept:
The total pressure force acting on the curved surface has horizontal and vertical components.
The horizontal component (Fh) is equal to the projection of the curved surface on the vertical plane and acts at the centre of the pressure of the projected area.
Fh = γw × A × x̅
A is the projected area.
x̅ is the Centroid of the projected area.
The vertical component of the pressure force (Fv) is equal to the weight of liquid supported by the curved surface up to the free surface of the liquid and it acts at the centroid of liquid
Fv = γ × Volume of liquid held by curved surface up to the free surface
Centre of pressure = 2d/3 from top
Calculation:
Given:
Area, A = 3 m × 3 m
x̅ = 3 / 2 = 1.5 m
Fh = γw × A × x̅ = γw × 3 × 3 × 1.5 = 13.5γw
Centre of pressure = 2d/3 from top = 2 × 3/3 = 2 m
∴ Centre of pressure from bottom = 3 - 2 = 1 m
∴ Moment about bottom edge = 13.5γw × 1 = 13.5γw
The total pressure force acting on the curved surface has horizontal and vertical components.
The horizontal component (Fh) is equal to the projection of the curved surface on the vertical plane and acts at the centre of the pressure of the projected area.
Fh = γw × A × x̅
A is the projected area.
x̅ is the Centroid of the projected area.
The vertical component of the pressure force (Fv) is equal to the weight of liquid supported by the curved surface up to the free surface of the liquid and it acts at the centroid of liquid
Fv = γ × Volume of liquid held by curved surface up to the free surface
Centre of pressure = 2d/3 from top
Calculation:
Given:
Area, A = 3 m × 3 m
x̅ = 3 / 2 = 1.5 m
Fh = γw × A × x̅ = γw × 3 × 3 × 1.5 = 13.5γw
Centre of pressure = 2d/3 from top = 2 × 3/3 = 2 m
∴ Centre of pressure from bottom = 3 - 2 = 1 m
∴ Moment about bottom edge = 13.5γw × 1 = 13.5γw
By what factor will the hydrostatic force on one of the vertical sides of a beaker decrease if the height of the liquid column is halved?- a)1 ⁄ 2
- b)1 ⁄ 3
- c)1 ⁄ 4
- d)2 ⁄ 3
Correct answer is option 'C'. Can you explain this answer?
By what factor will the hydrostatic force on one of the vertical sides of a beaker decrease if the height of the liquid column is halved?
a)
1 ⁄ 2
b)
1 ⁄ 3
c)
1 ⁄ 4
d)
2 ⁄ 3
|
|
Manasa Bose answered |
The hydrostatic force on one of the vertical sides of a beaker is given by the equation F = ρghA, where ρ is the density of the liquid, g is the acceleration due to gravity, h is the height of the liquid column, and A is the area of the vertical side of the beaker.
If the height of the liquid column is halved, the new height would be h/2.
Substituting this value into the equation for hydrostatic force, we get:
F' = ρg(h/2)A
Since the density (ρ), acceleration due to gravity (g), and area (A) are constant, we can simplify this equation to:
F' = (ρghA)/2
Therefore, the hydrostatic force on one of the vertical sides of the beaker will decrease by a factor of 2 when the height of the liquid column is halved.
So, the correct answer is b) 2.
If the height of the liquid column is halved, the new height would be h/2.
Substituting this value into the equation for hydrostatic force, we get:
F' = ρg(h/2)A
Since the density (ρ), acceleration due to gravity (g), and area (A) are constant, we can simplify this equation to:
F' = (ρghA)/2
Therefore, the hydrostatic force on one of the vertical sides of the beaker will decrease by a factor of 2 when the height of the liquid column is halved.
So, the correct answer is b) 2.
A rectangular lamina of width b and depth d is submerged vertically in water, such that the upper edge of the lamina is at a depth h from the free surface. What will be the expression for the depth of the centroid (G)?- a)h
- b)h + d
- c)h + d ⁄ 2
- d)h + d / 2
Correct answer is option 'C'. Can you explain this answer?
A rectangular lamina of width b and depth d is submerged vertically in water, such that the upper edge of the lamina is at a depth h from the free surface. What will be the expression for the depth of the centroid (G)?
a)
h
b)
h + d
c)
h + d ⁄ 2
d)
h + d / 2
|
|
Anshul Kumar answered |
Depth of the Centroid (G) Calculation:
1. Understanding the Situation:
- The rectangular lamina is submerged vertically in water.
- The upper edge of the lamina is at a depth h from the free surface.
- The width of the lamina is b and the depth is d.
2. Location of the Centroid (G):
- The centroid (G) of the lamina is located at a distance h/2 from the free surface.
- This is because the centroid of a submerged area lies at a distance of half of the depth from the free surface.
3. Calculating the Depth of the Centroid (G):
- Given that the upper edge of the lamina is at a depth h and the total depth of the lamina is d.
- Therefore, the distance from the centroid (G) to the upper edge of the lamina is d - h.
- As mentioned earlier, the centroid is located at a distance of h/2 from the free surface.
- Hence, the depth of the centroid (G) from the free surface would be: h/2 + (d - h) = h/2 + d - h = h + d/2.
Therefore, the expression for the depth of the centroid (G) is h + d/2, which is option C.
1. Understanding the Situation:
- The rectangular lamina is submerged vertically in water.
- The upper edge of the lamina is at a depth h from the free surface.
- The width of the lamina is b and the depth is d.
2. Location of the Centroid (G):
- The centroid (G) of the lamina is located at a distance h/2 from the free surface.
- This is because the centroid of a submerged area lies at a distance of half of the depth from the free surface.
3. Calculating the Depth of the Centroid (G):
- Given that the upper edge of the lamina is at a depth h and the total depth of the lamina is d.
- Therefore, the distance from the centroid (G) to the upper edge of the lamina is d - h.
- As mentioned earlier, the centroid is located at a distance of h/2 from the free surface.
- Hence, the depth of the centroid (G) from the free surface would be: h/2 + (d - h) = h/2 + d - h = h + d/2.
Therefore, the expression for the depth of the centroid (G) is h + d/2, which is option C.
In a fluid system, the point of intersection of the line of action of the resultant hydrostatic force and the corresponding surface is known as:- a)centre of locust
- b)centre of velocity
- c)centre of momentum
- d)centre of pressure
Correct answer is option 'D'. Can you explain this answer?
In a fluid system, the point of intersection of the line of action of the resultant hydrostatic force and the corresponding surface is known as:
a)
centre of locust
b)
centre of velocity
c)
centre of momentum
d)
centre of pressure
|
Anuj Chakraborty answered |
Centre of Pressure
The centre of pressure is the point of intersection of the line of action of the resultant hydrostatic force and the corresponding surface in a fluid system. This point plays a crucial role in determining the stability and equilibrium of submerged or floating bodies.
Significance
- The centre of pressure helps in understanding the distribution of pressure forces acting on a surface immersed in a fluid.
- It is essential for calculating the net hydrostatic force acting on a submerged object.
- Knowing the centre of pressure is crucial for designing structures like dams, ships, and offshore platforms to ensure stability and prevent structural failure.
Calculation
- The centre of pressure can be determined by analyzing the moments of the hydrostatic forces acting on the surface.
- It is usually located below the centroid of the submerged surface, depending on the shape and orientation of the object.
Applications
- In civil engineering, the concept of the centre of pressure is used in designing water-retaining structures, such as retaining walls and dams.
- In naval architecture, understanding the centre of pressure is vital for designing ships and submarines to ensure they remain stable under different loading conditions.
In conclusion, the centre of pressure is a critical concept in fluid mechanics and plays a significant role in engineering design and analysis. Understanding its significance and calculation is essential for ensuring the safety and stability of various structures in fluid systems.
The position of centre of pressure on a plane surface immersed vertically in a static mass of fluid is- a)at the centroid of the submerged area
- b)always above the centroid of the area
- c)always beiow the centroid of the area
- d)None of the above
Correct answer is option 'C'. Can you explain this answer?
The position of centre of pressure on a plane surface immersed vertically in a static mass of fluid is
a)
at the centroid of the submerged area
b)
always above the centroid of the area
c)
always beiow the centroid of the area
d)
None of the above
|
Avinash Mehta answered |
The correct answer is 'C', the position of center of pressure on a plane surface immersed vertically in a static mass of fluid is always below the centroid of the submerged area. This is due to the fact that the fluid exerts a greater force on the lower portion of the surface, which causes the center of pressure to be located at a lower point. The center of pressure is defined as the point at which the resultant of all the forces acting on the submerged surface is acting. It is important to note that this applies to surfaces that are vertical and submerged in a static fluid, the location of the center of pressure can vary for surfaces that are inclined or for fluids in motion.
A cuboidal beaker is half filled with water. By what percent will the hydrostatic force on one of the vertical sides of the beaker increase if it is completely filled?- a)100
- b)200
- c)300
- d)400
Correct answer is option 'C'. Can you explain this answer?
A cuboidal beaker is half filled with water. By what percent will the hydrostatic force on one of the vertical sides of the beaker increase if it is completely filled?
a)
100
b)
200
c)
300
d)
400
|
|
Manasa Sen answered |
Given:
- A cuboidal beaker is half filled with water.
To find:
- By what percent will the hydrostatic force on one of the vertical sides of the beaker increase if it is completely filled?
Solution:
To solve this problem, we need to understand the concept of hydrostatic force and how it is related to the volume of water in the beaker.
Hydrostatic Force:
The hydrostatic force on a surface submerged in a fluid is given by the equation:
F = ρghA
where:
- F is the hydrostatic force,
- ρ is the density of the fluid,
- g is the acceleration due to gravity,
- h is the height of the fluid column above the surface, and
- A is the area of the surface.
Half-filled Beaker:
When the beaker is half filled with water, the height of the fluid column above the vertical side is half the height of the beaker. Let's assume the height of the beaker is h.
Therefore, the hydrostatic force on one of the vertical sides of the beaker when it is half filled can be represented as:
F1 = ρgh(A1)
where A1 is the area of one of the vertical sides.
Completely Filled Beaker:
When the beaker is completely filled with water, the height of the fluid column above the vertical side is equal to the height of the beaker, which is h.
Therefore, the hydrostatic force on one of the vertical sides of the beaker when it is completely filled can be represented as:
F2 = ρgh(A2)
where A2 is the area of one of the vertical sides.
Percentage Increase:
To find the percentage increase in the hydrostatic force, we can use the formula:
Percentage Increase = ((F2 - F1) / F1) * 100
Substituting the values of F1 and F2, we have:
Percentage Increase = ((ρgh(A2) - ρgh(A1)) / ρgh(A1)) * 100
Simplifying the expression, we get:
Percentage Increase = ((A2 - A1) / A1) * 100
Since the beaker is cuboidal, the area of one of the vertical sides remains the same whether it is half filled or completely filled. Therefore, A2 = A1.
Substituting this value, we have:
Percentage Increase = ((A1 - A1) / A1) * 100
= 0
Hence, the hydrostatic force on one of the vertical sides of the beaker does not increase when it is completely filled. Therefore, the correct answer is none of the given options.
- A cuboidal beaker is half filled with water.
To find:
- By what percent will the hydrostatic force on one of the vertical sides of the beaker increase if it is completely filled?
Solution:
To solve this problem, we need to understand the concept of hydrostatic force and how it is related to the volume of water in the beaker.
Hydrostatic Force:
The hydrostatic force on a surface submerged in a fluid is given by the equation:
F = ρghA
where:
- F is the hydrostatic force,
- ρ is the density of the fluid,
- g is the acceleration due to gravity,
- h is the height of the fluid column above the surface, and
- A is the area of the surface.
Half-filled Beaker:
When the beaker is half filled with water, the height of the fluid column above the vertical side is half the height of the beaker. Let's assume the height of the beaker is h.
Therefore, the hydrostatic force on one of the vertical sides of the beaker when it is half filled can be represented as:
F1 = ρgh(A1)
where A1 is the area of one of the vertical sides.
Completely Filled Beaker:
When the beaker is completely filled with water, the height of the fluid column above the vertical side is equal to the height of the beaker, which is h.
Therefore, the hydrostatic force on one of the vertical sides of the beaker when it is completely filled can be represented as:
F2 = ρgh(A2)
where A2 is the area of one of the vertical sides.
Percentage Increase:
To find the percentage increase in the hydrostatic force, we can use the formula:
Percentage Increase = ((F2 - F1) / F1) * 100
Substituting the values of F1 and F2, we have:
Percentage Increase = ((ρgh(A2) - ρgh(A1)) / ρgh(A1)) * 100
Simplifying the expression, we get:
Percentage Increase = ((A2 - A1) / A1) * 100
Since the beaker is cuboidal, the area of one of the vertical sides remains the same whether it is half filled or completely filled. Therefore, A2 = A1.
Substituting this value, we have:
Percentage Increase = ((A1 - A1) / A1) * 100
= 0
Hence, the hydrostatic force on one of the vertical sides of the beaker does not increase when it is completely filled. Therefore, the correct answer is none of the given options.
The horizontal component of force on a curved surface is equal to the- a)Product of pressure at its centroid and area
- b)Weight of liquid retained by the curved area
- c)Force on a vertical projection of the curved surface
- d)Weight of liquid vertically above the curved surface
Correct answer is option 'C'. Can you explain this answer?
The horizontal component of force on a curved surface is equal to the
a)
Product of pressure at its centroid and area
b)
Weight of liquid retained by the curved area
c)
Force on a vertical projection of the curved surface
d)
Weight of liquid vertically above the curved surface
|
Sparsh Unni answered |
The correct answer is option 'C', which states that the horizontal component of force on a curved surface is equal to the force on a vertical projection of the curved surface. Let's delve into the explanation of this concept.
Curved surfaces are commonly encountered in various engineering applications, such as in dams, tanks, and pipes. When a curved surface is in contact with a fluid, it exerts a force on the surface due to the pressure exerted by the fluid. This force can be resolved into two components: the vertical component and the horizontal component.
- Vertical Component of Force:
The vertical component of force on a curved surface is equal to the weight of the liquid vertically above the curved surface. This is because the pressure exerted by a fluid is directly proportional to the depth of the fluid and its density. Therefore, the weight of the liquid vertically above the curved surface is transferred as a vertical force on the surface.
- Horizontal Component of Force:
The horizontal component of force on a curved surface is equal to the force on a vertical projection of the curved surface. This can be understood by considering the equilibrium of forces acting on the fluid within the curved surface.
When a fluid is in equilibrium, the net force acting on it is zero. Therefore, the vertical and horizontal components of forces on the curved surface must balance each other. The vertical component of force is balanced by the weight of the liquid vertically above the curved surface. Consequently, the horizontal component of force is balanced by the force on a vertical projection of the curved surface.
This means that if we were to project the curved surface vertically onto a vertical plane, the force acting on this projection would be equal to the horizontal component of force on the curved surface. The magnitude and direction of this force on the vertical projection will be the same as the magnitude and direction of the horizontal component of force on the curved surface.
Hence, the correct answer is option 'C': the horizontal component of force on a curved surface is equal to the force on a vertical projection of the curved surface.
Curved surfaces are commonly encountered in various engineering applications, such as in dams, tanks, and pipes. When a curved surface is in contact with a fluid, it exerts a force on the surface due to the pressure exerted by the fluid. This force can be resolved into two components: the vertical component and the horizontal component.
- Vertical Component of Force:
The vertical component of force on a curved surface is equal to the weight of the liquid vertically above the curved surface. This is because the pressure exerted by a fluid is directly proportional to the depth of the fluid and its density. Therefore, the weight of the liquid vertically above the curved surface is transferred as a vertical force on the surface.
- Horizontal Component of Force:
The horizontal component of force on a curved surface is equal to the force on a vertical projection of the curved surface. This can be understood by considering the equilibrium of forces acting on the fluid within the curved surface.
When a fluid is in equilibrium, the net force acting on it is zero. Therefore, the vertical and horizontal components of forces on the curved surface must balance each other. The vertical component of force is balanced by the weight of the liquid vertically above the curved surface. Consequently, the horizontal component of force is balanced by the force on a vertical projection of the curved surface.
This means that if we were to project the curved surface vertically onto a vertical plane, the force acting on this projection would be equal to the horizontal component of force on the curved surface. The magnitude and direction of this force on the vertical projection will be the same as the magnitude and direction of the horizontal component of force on the curved surface.
Hence, the correct answer is option 'C': the horizontal component of force on a curved surface is equal to the force on a vertical projection of the curved surface.
Equal volume of two liquids of densities ρ1 and ρ2 are poured into two identical cuboidal beakers. The hydrostatic forces on the respective vertical face of the beakers are F1 and F2 respectively. If ρ1 > ρ2, which one will be the correct relation between F1 and F2?- a)F1 > F2
- b)F1 ≥ F2
- c)F1 < F2
- d)F1 ≤ F2
Correct answer is option 'A'. Can you explain this answer?
Equal volume of two liquids of densities ρ1 and ρ2 are poured into two identical cuboidal beakers. The hydrostatic forces on the respective vertical face of the beakers are F1 and F2 respectively. If ρ1 > ρ2, which one will be the correct relation between F1 and F2?
a)
F1 > F2
b)
F1 ≥ F2
c)
F1 < F2
d)
F1 ≤ F2
|
|
Muskaan Sen answered |
To determine the equal volume of two liquids of different densities, we can use the formula:
mass = density × volume
Given that the volume is equal for both liquids, we can set up the equation:
density1 × volume = density2 × volume
We can cancel the volume on both sides of the equation, leaving us with:
density1 = density2
This means that for two liquids to have equal volumes, their densities must be the same.
mass = density × volume
Given that the volume is equal for both liquids, we can set up the equation:
density1 × volume = density2 × volume
We can cancel the volume on both sides of the equation, leaving us with:
density1 = density2
This means that for two liquids to have equal volumes, their densities must be the same.
A beaker contains water up to a height of h. What will be the location of the centre of pressure?- a)h⁄3 from the surface
- b)h⁄2 from the surface
- c)2h⁄3 from the surface
- d)h⁄6 from the surface
Correct answer is option 'C'. Can you explain this answer?
A beaker contains water up to a height of h. What will be the location of the centre of pressure?
a)
h⁄3 from the surface
b)
h⁄2 from the surface
c)
2h⁄3 from the surface
d)
h⁄6 from the surface
|
|
Sakshi Basak answered |
Explanation:
Center of Pressure:
- The center of pressure is the point at which the total pressure force on a submerged surface acts.
- It is located at a distance h/3 from the free surface of the liquid.
Formula:
- The center of pressure is located at a distance h/3 from the free surface.
Therefore, the correct answer is option C, which states that the center of pressure is located at 2h/3 from the surface.
Center of Pressure:
- The center of pressure is the point at which the total pressure force on a submerged surface acts.
- It is located at a distance h/3 from the free surface of the liquid.
Formula:
- The center of pressure is located at a distance h/3 from the free surface.
Therefore, the correct answer is option C, which states that the center of pressure is located at 2h/3 from the surface.
Which of the following is the correct relation between centroid (G) and the centre of pressure (P) of a plane submerged in a liquid?- a)G is always below P
- b)P is always below G
- c)G is either at P or below it.
- d)P is either at G or below it.
Correct answer is option 'D'. Can you explain this answer?
Which of the following is the correct relation between centroid (G) and the centre of pressure (P) of a plane submerged in a liquid?
a)
G is always below P
b)
P is always below G
c)
G is either at P or below it.
d)
P is either at G or below it.
|
Ananya Sharma answered |
Relation between Centroid and Centre of Pressure of a Plane Submerged in a Liquid
The centroid (G) is the point at which the entire weight of the submerged plane can be assumed to act upon. The centre of pressure (P) is the point at which the resultant pressure force can be assumed to act upon.
Determination of Centroid and Centre of Pressure
The centroid of the submerged plane can be determined by using the principle of moments. The plane is divided into small strips, and the moment of each strip about a reference point is calculated. The centroid is the point at which the sum of these moments is zero.
The centre of pressure can be determined by using the principle of moments as well. The plane is again divided into small strips, and the pressure force acting on each strip is calculated. The centre of pressure is the point at which the sum of the moments of these pressure forces is zero.
Relation between Centroid and Centre of Pressure
The relation between the centroid and centre of pressure depends on the shape of the submerged plane. In general, the following relations hold true:
- If the plane is symmetrical about its vertical axis, the centroid and centre of pressure coincide.
- If the plane is not symmetrical about its vertical axis, the centroid and centre of pressure are not in the same location. In this case, the centre of pressure is always below the centroid.
- If the plane is inclined, the centre of pressure moves towards the lower end of the plane.
Conclusion
In conclusion, the correct relation between the centroid and centre of pressure of a plane submerged in a liquid is that the centre of pressure is either at the centroid or below it. The exact location of the centre of pressure depends on the shape and orientation of the submerged plane.
The centroid (G) is the point at which the entire weight of the submerged plane can be assumed to act upon. The centre of pressure (P) is the point at which the resultant pressure force can be assumed to act upon.
Determination of Centroid and Centre of Pressure
The centroid of the submerged plane can be determined by using the principle of moments. The plane is divided into small strips, and the moment of each strip about a reference point is calculated. The centroid is the point at which the sum of these moments is zero.
The centre of pressure can be determined by using the principle of moments as well. The plane is again divided into small strips, and the pressure force acting on each strip is calculated. The centre of pressure is the point at which the sum of the moments of these pressure forces is zero.
Relation between Centroid and Centre of Pressure
The relation between the centroid and centre of pressure depends on the shape of the submerged plane. In general, the following relations hold true:
- If the plane is symmetrical about its vertical axis, the centroid and centre of pressure coincide.
- If the plane is not symmetrical about its vertical axis, the centroid and centre of pressure are not in the same location. In this case, the centre of pressure is always below the centroid.
- If the plane is inclined, the centre of pressure moves towards the lower end of the plane.
Conclusion
In conclusion, the correct relation between the centroid and centre of pressure of a plane submerged in a liquid is that the centre of pressure is either at the centroid or below it. The exact location of the centre of pressure depends on the shape and orientation of the submerged plane.
The force exerted by a static fluid on a surface either plane or curved, when the fluid comes in the contact with surface is called:- a)Total pressure
- b)Centre of pressure
- c)Normal pressure
- d)Pressure density
Correct answer is option 'A'. Can you explain this answer?
The force exerted by a static fluid on a surface either plane or curved, when the fluid comes in the contact with surface is called:
a)
Total pressure
b)
Centre of pressure
c)
Normal pressure
d)
Pressure density
|
|
Sanya Agarwal answered |
Concept:
Hydrostatic Forces:
Hydrostatic Forces:
- Fluid statics or hydrostatics is the branch of fluid mechanics in which stresses generated in a fluid system are determined when it is at rest or static condition.
- The force exerted by a static fluid on a surface either plane or curved, when the fluid comes in the contact with the surface is called Total pressure.
- The point of application of total hydrostatic force on the surface is known as the center of pressure.
Total Hydrostatic Force on a Horizontal Plane and inclined surface:
Consider a plane surface immersed in a static mass of liquid of specific weight γ, such that it is held in a horizontal position at a depth h below the free surface of the liquid and if the surface is in a vertical position such that the centroid of the surface is at depth of h̅ below the free surface. A is the area of the total surface, then the total hydrostatic force on the horizontal surface,
F = γAh̅
γh̅ is the pressure intensity at the centroid of the surface. Therefore, it can be stated that the total hydrostatic force on a plane surface is equal to the product of the surface area and the intensity of pressure at the centroid of the area.
Consider a plane surface immersed in a static mass of liquid of specific weight γ, such that it is held in a horizontal position at a depth h below the free surface of the liquid and if the surface is in a vertical position such that the centroid of the surface is at depth of h̅ below the free surface. A is the area of the total surface, then the total hydrostatic force on the horizontal surface,
F = γAh̅
γh̅ is the pressure intensity at the centroid of the surface. Therefore, it can be stated that the total hydrostatic force on a plane surface is equal to the product of the surface area and the intensity of pressure at the centroid of the area.
A rectangular floating body is 20 m long and 5 m wide. The water line is 1.5 m above the bottom. If the centre of gravity is 1.8 m from the bottom, then its metacentric height will be approximately- a)3.3 m
- b)1.65 m
- c)0.34 m
- d)0.30 m
Correct answer is option 'C'. Can you explain this answer?
A rectangular floating body is 20 m long and 5 m wide. The water line is 1.5 m above the bottom. If the centre of gravity is 1.8 m from the bottom, then its metacentric height will be approximately
a)
3.3 m
b)
1.65 m
c)
0.34 m
d)
0.30 m
|
|
Shounak Saini answered |
When a block of ice floating on,water in a container melts, the level of water in the container- a)rises
- b)first falls and then rises
- c)remains the same
- d)falls
Correct answer is option 'C'. Can you explain this answer?
When a block of ice floating on,water in a container melts, the level of water in the container
a)
rises
b)
first falls and then rises
c)
remains the same
d)
falls
|
|
Ashwin Gupta answered |
When a block of ice floating on water in a container melts, the level of water in the container remains the same. This can be explained by the principle of buoyancy and the concept of displacement.
Buoyancy and Archimedes' Principle:
When an object is placed in a fluid, it experiences an upward force called buoyancy. This force is equal to the weight of the fluid displaced by the object. According to Archimedes' principle, an object will float in a fluid if its weight is equal to the buoyant force acting on it.
Displacement:
When the ice block is floating in water, it displaces an amount of water equal to its weight. This displacement creates an upward buoyant force that supports the weight of the ice block, causing it to float.
Melting of the Ice Block:
When the ice block starts to melt, its volume decreases while its mass remains the same. This means that as the ice melts, it takes up less space in the water. However, since the ice and liquid water have the same density, the mass of the water formed by the melted ice is equal to the mass of the ice block.
Conservation of Mass:
The principle of conservation of mass states that mass is neither created nor destroyed in a closed system. In this case, the ice and water form a closed system. As the ice melts, it is converted into an equivalent mass of liquid water. Therefore, the total mass of the system remains constant.
Effect on Water Level:
Since the mass of the water formed by the melted ice is equal to the mass of the ice block, and the density of ice and water is the same, the volume of the melted ice is equal to the volume of water displaced by the ice block. As a result, the water level in the container remains the same.
Therefore, the correct answer is option 'C' - the level of water in the container remains the same when a block of ice floating on water in a container melts. This is because the mass and volume of the melted ice are equal to the mass and volume of water displaced by the ice block.
Buoyancy and Archimedes' Principle:
When an object is placed in a fluid, it experiences an upward force called buoyancy. This force is equal to the weight of the fluid displaced by the object. According to Archimedes' principle, an object will float in a fluid if its weight is equal to the buoyant force acting on it.
Displacement:
When the ice block is floating in water, it displaces an amount of water equal to its weight. This displacement creates an upward buoyant force that supports the weight of the ice block, causing it to float.
Melting of the Ice Block:
When the ice block starts to melt, its volume decreases while its mass remains the same. This means that as the ice melts, it takes up less space in the water. However, since the ice and liquid water have the same density, the mass of the water formed by the melted ice is equal to the mass of the ice block.
Conservation of Mass:
The principle of conservation of mass states that mass is neither created nor destroyed in a closed system. In this case, the ice and water form a closed system. As the ice melts, it is converted into an equivalent mass of liquid water. Therefore, the total mass of the system remains constant.
Effect on Water Level:
Since the mass of the water formed by the melted ice is equal to the mass of the ice block, and the density of ice and water is the same, the volume of the melted ice is equal to the volume of water displaced by the ice block. As a result, the water level in the container remains the same.
Therefore, the correct answer is option 'C' - the level of water in the container remains the same when a block of ice floating on water in a container melts. This is because the mass and volume of the melted ice are equal to the mass and volume of water displaced by the ice block.
Pressure drag results from- a)skin friction
- b)occurrence of a wake
- c)break down of flow near the forward stagnation point
- d)waves setup during motion of ship on the water surface
Correct answer is option 'B'. Can you explain this answer?
Pressure drag results from
a)
skin friction
b)
occurrence of a wake
c)
break down of flow near the forward stagnation point
d)
waves setup during motion of ship on the water surface
|
Shruti Bose answered |
Explanation:
When an object moves through a fluid, it experiences a resistance to its motion. This resistance is called drag. There are two types of drag: pressure drag and skin friction drag.
Pressure drag, also known as form drag, results from the occurrence of a wake behind the object. When the fluid flows around the object, it separates from the surface and creates a low-pressure region behind the object. This low-pressure region pulls the object back, creating a force opposite to the direction of motion.
Factors affecting pressure drag:
1. Object shape: The shape of the object affects the formation of the wake and hence the pressure drag. A streamlined shape reduces the formation of the wake and hence the pressure drag.
2. Fluid viscosity: The viscosity of the fluid affects the separation of the flow and hence the formation of the wake. A higher viscosity fluid leads to more separation and hence more pressure drag.
3. Velocity: The velocity of the fluid affects the formation of the wake. A higher velocity fluid leads to more separation and hence more pressure drag.
4. Reynolds number: The Reynolds number determines the type of flow around the object. A higher Reynolds number leads to turbulent flow, which results in more pressure drag.
Examples of pressure drag:
1. A car moving through the air experiences pressure drag due to the wake formed behind it.
2. A ship moving through the water experiences pressure drag due to the waves formed behind it.
Conclusion:
Pressure drag is a major component of the total drag experienced by an object moving through a fluid. It is caused by the formation of a wake behind the object due to the separation of the flow. Understanding the factors affecting pressure drag is important in designing efficient vehicles and structures.
When an object moves through a fluid, it experiences a resistance to its motion. This resistance is called drag. There are two types of drag: pressure drag and skin friction drag.
Pressure drag, also known as form drag, results from the occurrence of a wake behind the object. When the fluid flows around the object, it separates from the surface and creates a low-pressure region behind the object. This low-pressure region pulls the object back, creating a force opposite to the direction of motion.
Factors affecting pressure drag:
1. Object shape: The shape of the object affects the formation of the wake and hence the pressure drag. A streamlined shape reduces the formation of the wake and hence the pressure drag.
2. Fluid viscosity: The viscosity of the fluid affects the separation of the flow and hence the formation of the wake. A higher viscosity fluid leads to more separation and hence more pressure drag.
3. Velocity: The velocity of the fluid affects the formation of the wake. A higher velocity fluid leads to more separation and hence more pressure drag.
4. Reynolds number: The Reynolds number determines the type of flow around the object. A higher Reynolds number leads to turbulent flow, which results in more pressure drag.
Examples of pressure drag:
1. A car moving through the air experiences pressure drag due to the wake formed behind it.
2. A ship moving through the water experiences pressure drag due to the waves formed behind it.
Conclusion:
Pressure drag is a major component of the total drag experienced by an object moving through a fluid. It is caused by the formation of a wake behind the object due to the separation of the flow. Understanding the factors affecting pressure drag is important in designing efficient vehicles and structures.
A body is called a stream-line body when- a)it is symmetrical about hte axis along the free stream
- b)it produces no drag for flow around it
- c)the flow is laminar around it
- d)the surface of the body coincides with the streamlines
Correct answer is option 'D'. Can you explain this answer?
A body is called a stream-line body when
a)
it is symmetrical about hte axis along the free stream
b)
it produces no drag for flow around it
c)
the flow is laminar around it
d)
the surface of the body coincides with the streamlines
|
|
Madhurima Banerjee answered |
Understanding Streamline Bodies
A streamline body is defined by its ability to allow fluid to flow around it without causing disruption in the flow pattern. This leads to more efficient movement through the fluid medium and is crucial in various engineering applications.
Definition of a Streamline Body
- A body is termed streamline when its surface aligns with the flow streamlines.
- This alignment ensures that the fluid flows smoothly over the body, minimizing disturbances.
Key Characteristics
- Surface Coincidence with Streamlines: The primary feature of a streamline body is that its surface coincides with the streamlines of the fluid flow. This means that the fluid moves along the contours of the body, leading to reduced friction and turbulence.
- Minimized Drag: Although a streamline body may not produce zero drag, it is designed to reduce drag forces significantly. The streamlined shape helps in cutting through the fluid, thus lowering resistance.
- Flow Type: The flow around a streamline body is typically laminar, but this is not a strict requirement. The primary condition is the surface alignment with the streamlines.
Conclusion
Choosing a streamlined shape for bodies, such as aircraft wings or boat hulls, enhances performance by reducing drag and improving fuel efficiency. The concept of streamline bodies is vital in mechanical engineering, particularly in aerodynamics and hydrodynamics, where fluid flow is a significant factor.
A streamline body is defined by its ability to allow fluid to flow around it without causing disruption in the flow pattern. This leads to more efficient movement through the fluid medium and is crucial in various engineering applications.
Definition of a Streamline Body
- A body is termed streamline when its surface aligns with the flow streamlines.
- This alignment ensures that the fluid flows smoothly over the body, minimizing disturbances.
Key Characteristics
- Surface Coincidence with Streamlines: The primary feature of a streamline body is that its surface coincides with the streamlines of the fluid flow. This means that the fluid moves along the contours of the body, leading to reduced friction and turbulence.
- Minimized Drag: Although a streamline body may not produce zero drag, it is designed to reduce drag forces significantly. The streamlined shape helps in cutting through the fluid, thus lowering resistance.
- Flow Type: The flow around a streamline body is typically laminar, but this is not a strict requirement. The primary condition is the surface alignment with the streamlines.
Conclusion
Choosing a streamlined shape for bodies, such as aircraft wings or boat hulls, enhances performance by reducing drag and improving fuel efficiency. The concept of streamline bodies is vital in mechanical engineering, particularly in aerodynamics and hydrodynamics, where fluid flow is a significant factor.
In which of the following arrangements would the vertical force in the cylinder due to water be maximum- a)
- b)
- c)
- d)
Correct answer is option 'C'. Can you explain this answer?
In which of the following arrangements would the vertical force in the cylinder due to water be maximum
a)
b)
c)
d)
|
|
Subham Unni answered |
Vertical force will be maximum when the cylinder is completely submerged in water.
A metal block is thrown into a deep lake. As it sinks deeper in water, the buoyant force acting on it- a)increases
- b)remains the same
- c)decreases
- d)first increases and then decreases
Correct answer is option 'B'. Can you explain this answer?
A metal block is thrown into a deep lake. As it sinks deeper in water, the buoyant force acting on it
a)
increases
b)
remains the same
c)
decreases
d)
first increases and then decreases
|
|
Prerna Menon answered |
Buoyant force on a metal block in water
Introduction:
Buoyant force is the upward force exerted on an object submerged in a fluid (liquid or gas). It is a result of the difference in pressure between the top and bottom of an object. The buoyant force is equal to the weight of the fluid displaced by the object.
Answer:
The correct answer is option 'B', which states that the buoyant force on a metal block remains the same as it sinks deeper in water.
Explanation:
When a metal block is thrown into a deep lake, it displaces a volume of water equal to its own volume. This displaced water exerts an upward force on the block, which is known as the buoyant force. The weight of the block is balanced by the buoyant force acting in the opposite direction.
As the metal block sinks deeper in water, the pressure on it increases due to the weight of the water above it. However, the buoyant force remains the same because it is determined by the volume of water displaced by the block, which does not change as it sinks deeper.
In other words, the buoyant force is dependent on the volume of water displaced by the block, which is determined by the shape and size of the block. Therefore, as long as the block is submerged in water, the buoyant force remains constant.
Conclusion:
In conclusion, the buoyant force acting on a metal block in water remains the same as it sinks deeper because it is determined by the volume of water displaced by the block, which does not change.
Introduction:
Buoyant force is the upward force exerted on an object submerged in a fluid (liquid or gas). It is a result of the difference in pressure between the top and bottom of an object. The buoyant force is equal to the weight of the fluid displaced by the object.
Answer:
The correct answer is option 'B', which states that the buoyant force on a metal block remains the same as it sinks deeper in water.
Explanation:
When a metal block is thrown into a deep lake, it displaces a volume of water equal to its own volume. This displaced water exerts an upward force on the block, which is known as the buoyant force. The weight of the block is balanced by the buoyant force acting in the opposite direction.
As the metal block sinks deeper in water, the pressure on it increases due to the weight of the water above it. However, the buoyant force remains the same because it is determined by the volume of water displaced by the block, which does not change as it sinks deeper.
In other words, the buoyant force is dependent on the volume of water displaced by the block, which is determined by the shape and size of the block. Therefore, as long as the block is submerged in water, the buoyant force remains constant.
Conclusion:
In conclusion, the buoyant force acting on a metal block in water remains the same as it sinks deeper because it is determined by the volume of water displaced by the block, which does not change.
Lift force is defined as the force exerted by a flowing fluid on a solid body- a)in the direction of flow
- b)perpendicular to the direction of flow
- c)at any direction of flow
- d)none of the above
Correct answer is option 'B'. Can you explain this answer?
Lift force is defined as the force exerted by a flowing fluid on a solid body
a)
in the direction of flow
b)
perpendicular to the direction of flow
c)
at any direction of flow
d)
none of the above
|
|
Sankar Dasgupta answered |
Understanding Lift Force
Lift force is a critical concept in fluid mechanics, particularly in the fields of aerodynamics and hydrodynamics. It is the force that acts on a solid body immersed in a flowing fluid, such as air or water.
Definition of Lift Force
- Lift force is defined as the force exerted by a flowing fluid on a solid body perpendicular to the direction of flow.
Why Perpendicular?
- The nature of fluid flow creates differences in pressure around the solid body, which results in lift.
- For example, in an airplane wing, the shape (airfoil) causes air to move faster over the top surface than the bottom, creating lower pressure above the wing and higher pressure below it.
- This pressure difference generates an upward lift force.
Implications in Engineering
- Understanding lift is crucial for designing structures like bridges, aircraft, and boats.
- Engineers must calculate lift forces to ensure stability and performance under various flow conditions.
Conclusion
- Therefore, the correct answer to the question about lift force is option B: it acts perpendicular to the direction of flow. This fundamental principle is essential for predicting how solid bodies interact with fluid environments.
Lift force is a critical concept in fluid mechanics, particularly in the fields of aerodynamics and hydrodynamics. It is the force that acts on a solid body immersed in a flowing fluid, such as air or water.
Definition of Lift Force
- Lift force is defined as the force exerted by a flowing fluid on a solid body perpendicular to the direction of flow.
Why Perpendicular?
- The nature of fluid flow creates differences in pressure around the solid body, which results in lift.
- For example, in an airplane wing, the shape (airfoil) causes air to move faster over the top surface than the bottom, creating lower pressure above the wing and higher pressure below it.
- This pressure difference generates an upward lift force.
Implications in Engineering
- Understanding lift is crucial for designing structures like bridges, aircraft, and boats.
- Engineers must calculate lift forces to ensure stability and performance under various flow conditions.
Conclusion
- Therefore, the correct answer to the question about lift force is option B: it acts perpendicular to the direction of flow. This fundamental principle is essential for predicting how solid bodies interact with fluid environments.
The barrier shown between two water tanks of unit width (1 m) into the plane of the screen is modelled as a cantilever.
Taking the density of water as 1000 kg/m3, and the acceleration due to gravity as 10m/s2, the maximum absolute bending moment developed in the cantilever is ______ kN.m (round off to the nearest integer).
Correct answer is between '104,106'. Can you explain this answer?
The barrier shown between two water tanks of unit width (1 m) into the plane of the screen is modelled as a cantilever.
Taking the density of water as 1000 kg/m3, and the acceleration due to gravity as 10m/s2, the maximum absolute bending moment developed in the cantilever is ______ kN.m (round off to the nearest integer).
|
|
Sanvi Kapoor answered |
Concept:
Net hydrostatic force exerted by fluid on flat surface, F = ρ × g × h̅ × A
Where, ρ = density of fluid, g = acceleration due to gravity, A = Area of the flat surface, h̅ = depth of centroid from free surface
This force is assumed to be acting at a point.
Centre of the pressure of rectangular lamina from free surface,
CP = 2/3h (h = height of rectangular lamina)
Calculation:
For left side of the barrier
Area = 4 m × 1 m = 4 m2, h¯= 4 / 2 = 2m,
Center of pressure from top surface of the water, CP1 = 2 × 4 / 3 = 8 / 3m
F1= ρ × g × h̅ × A = 1000 × 10 × 2 × 4 = 80 kN
For right side of the barrier
Area = 1 m × 1 m, h¯= 1 / 2 = 0.5m,
Center of pressure from top surface of the water CP2 = 2 × 1 / 3 = 2 / 3m
F2 = ρ × g × h̅ × A = 1000 × 10 × (1 × 1) × 0.5 = 5 kN
Distance of centre of pressure from bottom end:
CP1 = h/3 = 4/3
CP2 = h/3 = 1/3
These forces will act as point load on the barrier. If the barrier is treated as a cantilever and the fixed end of this cantilever beam is on the bottom side of the tank, then the loading conditions will look like the figure drawn below
F1 = 80 kN, F2 = 5 kN
From the above-simplified diagram:
A moment about the fixed end of barrier,
Net hydrostatic force exerted by fluid on flat surface, F = ρ × g × h̅ × A
Where, ρ = density of fluid, g = acceleration due to gravity, A = Area of the flat surface, h̅ = depth of centroid from free surface
This force is assumed to be acting at a point.
Centre of the pressure of rectangular lamina from free surface,
CP = 2/3h (h = height of rectangular lamina)
Calculation:
For left side of the barrier
Area = 4 m × 1 m = 4 m2, h¯= 4 / 2 = 2m,
Center of pressure from top surface of the water, CP1 = 2 × 4 / 3 = 8 / 3m
F1= ρ × g × h̅ × A = 1000 × 10 × 2 × 4 = 80 kN
For right side of the barrier
Area = 1 m × 1 m, h¯= 1 / 2 = 0.5m,
Center of pressure from top surface of the water CP2 = 2 × 1 / 3 = 2 / 3m
F2 = ρ × g × h̅ × A = 1000 × 10 × (1 × 1) × 0.5 = 5 kN
Distance of centre of pressure from bottom end:
CP1 = h/3 = 4/3
CP2 = h/3 = 1/3
These forces will act as point load on the barrier. If the barrier is treated as a cantilever and the fixed end of this cantilever beam is on the bottom side of the tank, then the loading conditions will look like the figure drawn below
F1 = 80 kN, F2 = 5 kN
From the above-simplified diagram:
A moment about the fixed end of barrier,
A vertically immersed surface of size b x d has its width b parallel to the water surface and depth of its centroid from the water surface is 
The distance of its centre of water pressure from the water surface is- a)
- b)
- c)
- d)
Correct answer is option 'C'. Can you explain this answer?
A vertically immersed surface of size b x d has its width b parallel to the water surface and depth of its centroid from the water surface is The distance of its centre of water pressure from the water surface is
The distance of its centre of water pressure from the water surface isa)
b)
c)
d)
|
|
Akanksha Mehta answered |
Centre of water pressure,
A metal cube of size 15 cm x 15 cm x 15 cm and specific gravity 8.6 is submerged in a twolayered liquid, the bottom layer being mercury and the top layer being water. The percentage of the volume of the cube remaining above the interface will be, approximately
- a)68
- b)63
- c)40
- d)25
Correct answer is option 'C'. Can you explain this answer?
A metal cube of size 15 cm x 15 cm x 15 cm and specific gravity 8.6 is submerged in a twolayered liquid, the bottom layer being mercury and the top layer being water. The percentage of the volume of the cube remaining above the interface will be, approximately
a)
68
b)
63
c)
40
d)
25
|
|
Mehul Choudhury answered |
The increase in metacentric height
1. increases stability
2. decreases stability
3. increases comfort for passengers
4. decreases comfort for passengersThe correct answer is- a)1 and 3
- b)1 and 4
- c)2 and 3
- d)2 and 4
Correct answer is option 'B'. Can you explain this answer?
The increase in metacentric height
1. increases stability
2. decreases stability
3. increases comfort for passengers
4. decreases comfort for passengers
1. increases stability
2. decreases stability
3. increases comfort for passengers
4. decreases comfort for passengers
The correct answer is
a)
1 and 3
b)
1 and 4
c)
2 and 3
d)
2 and 4
|
|
Aditi Sarkar answered |
Increase in Metacentric Height and its Effects
Increasing the metacentric height of a ship can have significant effects on its stability and comfort for passengers.
1. Increases stability
- The metacentric height is a key parameter that determines the stability of a ship.
- When the metacentric height is increased, the ship becomes more stable as the distance between the center of gravity and the metacenter increases.
- This increased stability helps the ship stay upright in rough seas and reduces the risk of capsizing.
2. Decreases stability
- On the other hand, if the metacentric height is decreased, the ship becomes less stable.
- A lower metacentric height means that the center of gravity is closer to the metacenter, making the ship more prone to rolling and potentially capsizing in adverse conditions.
3. Increases comfort for passengers
- An increase in metacentric height can also lead to increased comfort for passengers.
- A more stable ship experiences less rolling and pitching motions, creating a smoother ride for those on board.
- This can help reduce seasickness and make the journey more enjoyable for passengers.
4. Decreases comfort for passengers
- Conversely, a decrease in metacentric height can have the opposite effect on passenger comfort.
- A less stable ship is more likely to roll and pitch in rough seas, leading to discomfort and potential seasickness among passengers.
Overall, the increase in metacentric height generally leads to improved stability and comfort for passengers, while a decrease in metacentric height can have the opposite effects. It is important for ship designers to carefully consider the metacentric height in order to achieve the desired balance between stability and comfort.
Increasing the metacentric height of a ship can have significant effects on its stability and comfort for passengers.
1. Increases stability
- The metacentric height is a key parameter that determines the stability of a ship.
- When the metacentric height is increased, the ship becomes more stable as the distance between the center of gravity and the metacenter increases.
- This increased stability helps the ship stay upright in rough seas and reduces the risk of capsizing.
2. Decreases stability
- On the other hand, if the metacentric height is decreased, the ship becomes less stable.
- A lower metacentric height means that the center of gravity is closer to the metacenter, making the ship more prone to rolling and potentially capsizing in adverse conditions.
3. Increases comfort for passengers
- An increase in metacentric height can also lead to increased comfort for passengers.
- A more stable ship experiences less rolling and pitching motions, creating a smoother ride for those on board.
- This can help reduce seasickness and make the journey more enjoyable for passengers.
4. Decreases comfort for passengers
- Conversely, a decrease in metacentric height can have the opposite effect on passenger comfort.
- A less stable ship is more likely to roll and pitch in rough seas, leading to discomfort and potential seasickness among passengers.
Overall, the increase in metacentric height generally leads to improved stability and comfort for passengers, while a decrease in metacentric height can have the opposite effects. It is important for ship designers to carefully consider the metacentric height in order to achieve the desired balance between stability and comfort.
Stake’s law is valid up to a maximum Reynolds number of- a)0.1
- b)1.0
- c)2000
- d)5 x 105
Correct answer is option 'B'. Can you explain this answer?
Stake’s law is valid up to a maximum Reynolds number of
a)
0.1
b)
1.0
c)
2000
d)
5 x 105
|
Athul Kumar answered |
Stake can have several meanings depending on the context:
1. In finance, stake refers to the amount of money or assets that an individual or organization has invested or risked in a particular investment or venture. It can also refer to the interest or ownership that someone holds in a company or project.
2. In gambling, stake refers to the amount of money or chips that a player places as a bet on a game or event.
3. In a competition or contest, stake can refer to the prize or reward that is at risk or up for grabs.
4. In a literal sense, stake can refer to a pointed wooden or metal object that is driven into the ground as a marker, support, or for other purposes. For example, a stake can be used to mark a boundary or secure a tent.
5. In a figurative sense, stake can refer to a personal or emotional involvement or interest in something. For example, someone might say "I have a stake in this project" to indicate their personal investment or concern in its success or outcome.
1. In finance, stake refers to the amount of money or assets that an individual or organization has invested or risked in a particular investment or venture. It can also refer to the interest or ownership that someone holds in a company or project.
2. In gambling, stake refers to the amount of money or chips that a player places as a bet on a game or event.
3. In a competition or contest, stake can refer to the prize or reward that is at risk or up for grabs.
4. In a literal sense, stake can refer to a pointed wooden or metal object that is driven into the ground as a marker, support, or for other purposes. For example, a stake can be used to mark a boundary or secure a tent.
5. In a figurative sense, stake can refer to a personal or emotional involvement or interest in something. For example, someone might say "I have a stake in this project" to indicate their personal investment or concern in its success or outcome.
What is the direction of total liquid pressure on submerged surface?- a)Inclined 45° to the surface
- b)Inclined 30° to the surface
- c)Normal to the surface
- d)Parallel to the surface
Correct answer is option 'C'. Can you explain this answer?
What is the direction of total liquid pressure on submerged surface?
a)
Inclined 45° to the surface
b)
Inclined 30° to the surface
c)
Normal to the surface
d)
Parallel to the surface
|
|
Sanvi Kapoor answered |
Concept:
- When a stationary fluid comes in contact with a solid surface either plane or curved, a force is exerted by the fluid on the surface. This force is called total pressure or pressure force or hydrostatic force.
- Since for a liquid at rest, no tangential force exists, the hydrostatic force acts in the direction normal to the surface.
- The point of application of total pressure on the surface is called the center of pressure (CP).
- The direction of total Hydrostatic force (F) on the Horizontal Plane Surface is normal to the surface as it acts in the vertically downward direction.
Which of the following scientific principles/laws is related to flight in aeroplanes?- a)Faraday's laws of electromagnetic induction
- b)Laws of thermodynamics
- c)Light amplification by stimulated emission of radiation
- d)Bernoulli's principle in fluid dynamics
Correct answer is option 'D'. Can you explain this answer?
Which of the following scientific principles/laws is related to flight in aeroplanes?
a)
Faraday's laws of electromagnetic induction
b)
Laws of thermodynamics
c)
Light amplification by stimulated emission of radiation
d)
Bernoulli's principle in fluid dynamics
|
|
Sanvi Kapoor answered |
Forces over aircraft
- At any given time, there are four forces acting upon an aircraft. These forces are lift, weight (or gravity), drag and thrust.
- Lift is the key aerodynamic force that keeps objects in the air. It is the force that opposes weight; thus, lift helps to keep an aircraft in the air.
- Weight is the force that works vertically by pulling all objects, including aircraft, toward the centre of the Earth. In order to fly an aircraft, something (lift) needs to press it in the opposite direction of gravity. The weight of an object controls how strong the pressure (lift) will need to be. Lift is that pressure.
- Drag is a mechanical force generated by the interaction and contract of a solid body, such as an aeroplane, with a fluid (liquid or gas).
- Finally, the thrust is the force that is generated by the engines of an aircraft in order for the aircraft to move forward.
Bernoulli's principle in fluid dynamics
- We are able to explain how lift is generated for an aeroplane by gaining an understanding of the forces at work on an aeroplane and what principles guide those forces. First, it takes thrust to get the aeroplane moving - Newton’s first law at work. This law states that an object at rest remains at rest while an object in motion remains in motion unless acted upon by an external force.
- Bernoulli's principle can be used to calculate the lift force on an aerofoil if the behaviour of the fluid flow in the vicinity of the foil is known.
- For example, if the air flowing past the top surface of an aircraft wing is moving faster than the air flowing past the bottom surface, then Bernoulli's principle implies that the pressure on the surfaces of the wing will be lower above than below. This pressure difference results in an upwards lifting force.
A small plastic boat loaded with nuts and bolts is floating in a bath tub. if the cargo is dumped into the water, allowing the boat to float empty, then the water level in the tub will- a)rise
- b)fall
- c)not change
- d)None of these
Correct answer is option 'B'. Can you explain this answer?
A small plastic boat loaded with nuts and bolts is floating in a bath tub. if the cargo is dumped into the water, allowing the boat to float empty, then the water level in the tub will
a)
rise
b)
fall
c)
not change
d)
None of these
|
|
Milan Ghosh answered |
Explanation:
When the boat is floating with the nuts and bolts, it displaces a certain amount of water equal to its weight so that it can float. This means that the weight of the boat with the cargo is equal to the weight of the water displaced by the boat.
When the cargo is dumped into the water, the weight of the boat is reduced, and as a result, the water displaced by the boat will also be reduced.
This will cause the water level in the tub to fall because the amount of water displaced by the boat is less than before.
Therefore, option B is correct, and the water level in the tub will fall when the cargo is dumped into the water.
When the boat is floating with the nuts and bolts, it displaces a certain amount of water equal to its weight so that it can float. This means that the weight of the boat with the cargo is equal to the weight of the water displaced by the boat.
When the cargo is dumped into the water, the weight of the boat is reduced, and as a result, the water displaced by the boat will also be reduced.
This will cause the water level in the tub to fall because the amount of water displaced by the boat is less than before.
Therefore, option B is correct, and the water level in the tub will fall when the cargo is dumped into the water.
Drag is defined as the force exerted by a flowing fluid on a solid body- a)in the direction of flow
- b)perpendicular to the direction of flow
- c)in the direction at 30° to the direction of flow
- d)in the direction at 45° to the direction of flow
Correct answer is option 'A'. Can you explain this answer?
Drag is defined as the force exerted by a flowing fluid on a solid body
a)
in the direction of flow
b)
perpendicular to the direction of flow
c)
in the direction at 30° to the direction of flow
d)
in the direction at 45° to the direction of flow
|
Pankaj Rane answered |
Explanation:
Definition of Drag:
- Drag is defined as the force exerted by a flowing fluid on a solid body.
Direction of Drag Force:
- The drag force acts in the direction of flow.
- This means that the drag force is exerted along the direction in which the fluid is flowing past the solid body.
Importance of Direction of Drag Force:
- Understanding the direction of the drag force is crucial in various engineering applications.
- It helps in designing structures or vehicles to minimize drag and improve efficiency.
- Knowledge of drag direction is essential in fields such as aerodynamics, hydrodynamics, and civil engineering.
Conclusion:
- In conclusion, drag force is exerted in the direction of flow in a flowing fluid on a solid body. Understanding this concept is important in various engineering disciplines to optimize the performance of structures and vehicles.
Definition of Drag:
- Drag is defined as the force exerted by a flowing fluid on a solid body.
Direction of Drag Force:
- The drag force acts in the direction of flow.
- This means that the drag force is exerted along the direction in which the fluid is flowing past the solid body.
Importance of Direction of Drag Force:
- Understanding the direction of the drag force is crucial in various engineering applications.
- It helps in designing structures or vehicles to minimize drag and improve efficiency.
- Knowledge of drag direction is essential in fields such as aerodynamics, hydrodynamics, and civil engineering.
Conclusion:
- In conclusion, drag force is exerted in the direction of flow in a flowing fluid on a solid body. Understanding this concept is important in various engineering disciplines to optimize the performance of structures and vehicles.
The horizontal component of force on a curved surface is equal to the- a)product of pressure intensity at its centroid and area
- b)force on a vertical projection of the curved surface
- c)weight of liquid vertically above the curved surface
- d)force on the horizontal projection of the curved surface
Correct answer is option 'B'. Can you explain this answer?
The horizontal component of force on a curved surface is equal to the
a)
product of pressure intensity at its centroid and area
b)
force on a vertical projection of the curved surface
c)
weight of liquid vertically above the curved surface
d)
force on the horizontal projection of the curved surface
|
|
Partho Jain answered |
Explanation:
When a curved surface is submerged in a fluid, it experiences a force due to the pressure difference between the upper and lower surfaces of the curve. This force can be resolved into two components - vertical and horizontal.
- The vertical component of force is equal to the weight of the fluid vertically above the curved surface.
- The horizontal component of force is equal to the force on the vertical projection of the curved surface.
Therefore, option 'B' is correct as it states that the horizontal component of force on a curved surface is equal to the force on the vertical projection of the curved surface.
Additional Information:
- The centroid of a curved surface is the point at which the surface can be balanced in any orientation. The pressure intensity at the centroid is the average pressure acting on the surface.
- The weight of the fluid vertically above the curved surface is equal to the volume of the fluid above the curve multiplied by the density of the fluid and the acceleration due to gravity.
- The force on the horizontal projection of the curved surface is not related to the horizontal component of force acting on the curved surface.
When a curved surface is submerged in a fluid, it experiences a force due to the pressure difference between the upper and lower surfaces of the curve. This force can be resolved into two components - vertical and horizontal.
- The vertical component of force is equal to the weight of the fluid vertically above the curved surface.
- The horizontal component of force is equal to the force on the vertical projection of the curved surface.
Therefore, option 'B' is correct as it states that the horizontal component of force on a curved surface is equal to the force on the vertical projection of the curved surface.
Additional Information:
- The centroid of a curved surface is the point at which the surface can be balanced in any orientation. The pressure intensity at the centroid is the average pressure acting on the surface.
- The weight of the fluid vertically above the curved surface is equal to the volume of the fluid above the curve multiplied by the density of the fluid and the acceleration due to gravity.
- The force on the horizontal projection of the curved surface is not related to the horizontal component of force acting on the curved surface.
The Karman vortex trail' occurs- a)in ail shapes and at all Reynolds number
- b)in 2-D body shapes and in range of Reynolds number
- c)in 2-D bodies and at all Reynolds number >30
- d)in circular cylinder only
Correct answer is option 'B'. Can you explain this answer?
The Karman vortex trail' occurs
a)
in ail shapes and at all Reynolds number
b)
in 2-D body shapes and in range of Reynolds number
c)
in 2-D bodies and at all Reynolds number >30
d)
in circular cylinder only
|
|
Yash Joshi answered |
C) in 2-D bodies and at all Reynolds number.
Chapter doubts & questions for Hydrostatic Forces on Surfaces - Fluid Mechanics for Civil Engineering 2025 is part of Civil Engineering (CE) exam preparation. The chapters have been prepared according to the Civil Engineering (CE) exam syllabus. The Chapter doubts & questions, notes, tests & MCQs are made for Civil Engineering (CE) 2025 Exam. Find important definitions, questions, notes, meanings, examples, exercises, MCQs and online tests here.
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