Buoyancy and Floatation MCQ Quiz - Objective Question with Answer for Buoyancy and Floatation - Download Free PDF

Explanation:

Longitudinal Vehicle Dynamics

• Longitudinal vehicle dynamics refers to the study of forces and motions that occur along the direction of travel of a vehicle. This includes forces that affect the acceleration, deceleration, and overall stability of the vehicle as it moves forward or backward.
• In the context of longitudinal vehicle dynamics, the buoyancy force is considered an insignificant external longitudinal force acting on the vehicle. Buoyancy force primarily relates to the force exerted by a fluid (like water) on an object submerged in it, causing the object to float or rise. Since typical vehicles (cars) operate on solid ground rather than in a fluid medium, the buoyancy force has minimal to no effect on their longitudinal dynamics.
• For a vehicle, the primary forces affecting its longitudinal motion include the longitudinal tire forces, aerodynamic drag force, and rolling resistance due to tires. The buoyancy force, however, is not a significant factor in this scenario.

Additional Information Longitudinal tire force

• Longitudinal tire force is a significant force in vehicle dynamics. It is the force generated by the interaction between the tires and the road surface, allowing the vehicle to accelerate, decelerate, and maintain traction. This force is crucial for the vehicle's ability to move forward or backward and is directly related to the performance and safety of the vehicle.

Aerodynamic drag force

• Aerodynamic drag force is another significant factor in longitudinal vehicle dynamics. This force opposes the motion of the vehicle as it moves through the air. It depends on the vehicle's shape, speed, and surface roughness. Reducing aerodynamic drag is essential for improving fuel efficiency and achieving higher speeds.

Rolling resistance due to tires

• Rolling resistance due to tires is also a critical force in longitudinal vehicle dynamics. It is the resistance that occurs when the tires roll over the road surface. This resistance is influenced by the tire material, tire pressure, road surface, and vehicle load. Minimizing rolling resistance is important for enhancing fuel efficiency and vehicle performance.

Explanation:

Stability of Floating body:

Consider a floating body tilted by an angle Δθ as shown below. For the untilted body the point G is the centre of gravity of the body where the body weight W acts. Point B is the centre of buoyancy (the centroid of the displaced volume of fluid) where the upward buoyancy force FB acts. 

• When the body is tilted the centre of buoyancy moves to a new position B' because the shape of the displaced volume changes.
• A new point M may be defined called the Metacentre.
• This is the point where a vertical line drawn upwards from the new centre of buoyancy B' of the tilted body intersects the line of symmetry of the body. The buoyancy force FB now acts through B'. 
• From the centre diagram in the figure we can see that W and FB give a Restoring Moment that rotates the body back to its untitled position.
• From the righthand diagram in the figure we can see that W and FB give an Overturning Moment that rotates the body even further in the tilted direction. 
• Hence we can say; if the metacentre M lies above the centre of gravity G then the body is stable. In other words the Metacentric height MG is positive.
• If the metacentre M lies below the centre of gravity G then the body is unstable. In other words the metacentric height MG is negative.​

Important Points

• A floating body is Stable if when it is displaced it returns to equilibrium.
• A floating body is Unstable if when it is displaced it moves to a new equilibrium. 

Concept:

For an object floating at the interface of two fluids, the buoyant force acting on the object equals its weight. The buoyant force is contributed by both the gasoline and water layers.

The fraction of the wood block submerged in gasoline is denoted by \( x \) and the fraction submerged in water is \( (1 - x) \).

The equilibrium condition can be written as:

\( \text{Weight of the wood block} = \text{Buoyant force due to gasoline} + \text{Buoyant force due to water} \)

Calculation:

Given data:

• Specific gravity of wood: \( S_w = 0.96 \)
• Specific gravity of gasoline: \( S_g = 0.74 \)
• Specific gravity of water: \( S_w = 1 \)

Using the equilibrium equation:

\( 0.96 = 0.74x + 1(1 - x) \)

Solving for \( x \):

\( 0.96 = 0.74x + 1 - x \)

\( 0.96 - 1 = -0.26x \)

\( -0.04 = -0.26x \)

\( x = \frac{0.04}{0.26} = \frac{2}{13} \)

Fraction of the wood block below the interface (submerged in water):

\( 1 - x = 1 - \frac{2}{13} = \frac{11}{13} \)

Closest to this value among options is \(\frac{5}{6} \)

Explanation:

Time period of oscillation of a floating body is given by

\({\rm{T}} = 2{\rm{\pi }}\sqrt {\frac{{{{\rm{K}}^2}}}{{{\rm{g}}\left( {{\rm{GM}}} \right)}}} \)

Where,

K = radius of gyration, and GM = Meta centre height

For a given ‘K’, \({\rm{T}} \propto \frac{1}{{\sqrt {{\rm{GM}}} }}\)

⇒ Higher GM, lesser will be oscillation of body.

EXPLANATION

Archimedes principle: 

• It is used to explain the law of flotation or upward thrust experienced when immersed in a fluid. 
• The principle of Archimedes states “When a body is immersed in a liquid, an upward thrust, equal to the weight of the liquid displaced, acts on it.”

• The flotation and sinking of an object are dependent upon the relative density of each other.
  • If the density of the object is more than the density of the liquid, the object will sink.
  • On the other side, if the density of an object is less than the liquid, then it will float over it.
  • If the density of the object and liquid is equal to each other, they are in equilibrium and float and sink both at the same time i.e the whole of the object will be immersed with its top surface at liquid level.
  • A piece of the metal having a specific gravity of 13.6 is placed in mercury of specific gravity 13.6, then the whole of the metal piece will be immersed with its top surface just at mercury level.

Explanation:

In the stability of floating bodies, the stable equilibrium is attained if the metacentre (M) point lies above the centre of gravity (G).

The partially submerged body resembles the floating body, where the weight of the body is balanced by the buoyancy force acting in the upwards direction. The stability of the floating is governed by the metacentre of the floating body.

Metacenter

• It is the point about which a body starts oscillating when the body is tilted by a small angle.
• It is the point where the line of action of buoyancy will meet the normal axis of the body when the body is given a small angular displacement.

Stability of floating bodies:

1. Stable equilibrium: The metacentre is above the centre of gravity of the body, then the disturbing couple is balanced by restoring couple, the body will be in stable equilibrium.

2. Unstable equilibrium: The metacentre is below the centre of gravity of the body, then the disturbing couple is supported by restoring couple, the body will be in unstable equilibrium.

3. Neutral equilibrium: The metacentre and the centre of gravity coincides at the same point, then the body is in neutral equilibrium.

Stability of submerged bodies: 

In the case of the submerged body the centre of gravity and centre of buoyancy is fixed, therefore the stability or instability is decided by the relative positions of the centre of buoyancy and the centre of gravity.

1. Stable equilibrium: For stable equilibrium, the body the centre of buoyancy is above the centre of gravity. The disturbing couple is countered by the restoring couple.

2. Unstable equilibrium: For unstable equilibrium, the centre of buoyancy is below the centre of gravity of the body, the disturbing couple is supported by the restoring couple.

3. Neutral equilibrium: when the centre of gravity and the centre of buoyancy coincide then it is the state of neutral equilibrium.

Centre of pressure

\(\begin{array}{l} {{\rm{h}}^{\rm{*}}} = {\rm{\bar X}} + \frac{{{{\rm{I}}_{\rm{G}}}}}{{{\rm{A\bar X}}}} = \frac{{\rm{h}}}{3} + \frac{{\frac{{{\rm{b}}{{\rm{h}}^3}}}{{36}}}}{{\frac{{{\rm{bh}}}}{2}.\frac{{\rm{h}}}{3}}}\\ = \frac{{\rm{h}}}{3} + \frac{{\rm{h}}}{6} = \frac{{\left( {2 + 1} \right){\rm{h}}}}{6} = \frac{{\rm{h}}}{2} \end{array}\)

Important point:

Concept:

When a body is either wholly or partially immersed in a fluid, a lift is generated due to the net vertical component of hydrostatic pressure forces experienced by the body. This lift is called the buoyant force and the phenomenon is called buoyancy.

The Archimedes principle states that the buoyant force on a submerged body is equal to the weight of the liquid displaced by the body and acts vertically upward through the centroid of the displaced volume.

Thus, the net weight of the submerged body, (the net vertical downward force experienced by it) is reduced from its actual weight by an amount that equals the buoyant force.

FB = ρghA = ρgV

Weight of cube = buoyancy force

ρiceViceg = ρwVVDg

Calculation:

Given:

ρice = 920 kg/m3, ρw =1000 kg/m3

Let the x be the volume of iceberg floats above the water surface.

ρiceVg = ρwg(V - x)

920 × V × g = 1000 × g × (V - x)

0.92V = (V - x)

x = 8%

The correct answer is the Density of the fluid and the volume of the body immersed in it.

Key Points

• The buoyant force depends on the volume of the liquid displaced. The buoyant force depends on the mass of the object, the weight of the object, and its density.
• The buoyant force depends directly upon:
  • The volume of the fluid displaced.
  • The density of the fluid in which the body is immersed.
  • Acceleration due to gravity at the place.

Important Points

• The three types of buoyancy are positive buoyancy, negative buoyancy, and neutral buoyancy.
  • Positive buoyancy is when the immersed object is lighter than the fluid displaced and this is the reason why the object floats.
  • Negative buoyancy is when the immersed object is denser than the fluid displaced which results in the sinking of the object.
  • Neutral buoyancy takes place when the weight of an immersed object is equal to the fluid displaced. 

Concept:

Weight(W) = mg = Vρg

where V = volume of body, ρ = density of body material, g = acceleration due to gravity

Buoyant force = FB = Vd × ρl × g

where Vd = Volume of fluid displaced by the body, ρl = density of fluid

Weight feels in fluid = W - FB

specific gravity = \(\frac {\rho}{\rho _w}\)

Calculation:

Given:

W = 100 N, Weight in water = 75 N

W - FB = 75 N 

100 - FB = 75

FB = 25

V × ρw × g = 25

\(V =\frac {25}{ρ_wg}\)

W = 100

Vρg = 100

\(\frac {25}{ρ_wg}ρ g=100\)

ρ = 4ρw

specific gravity = \(\frac {\rho}{\rho_w}=4\) 

Concept:

When the density of the body is less than that of the water, it floats. According to Archimedes' principle the upward buoyant force that is exerted on a body immersed in a fluid, whether fully or partially submerged, is equal to the weight of the fluid that the body displaces.

For floating bodies, the weight of the body = buoyancy force

Weight of the body = Immersed volume x Unit weight of water.

Calculations:

Given, 

L = 2m, B = 1 m, h = 1 m, γ = 10 kN/m³, d = 0.5 m

Weight of body = Buoyancy Force

Buoyancy (B) = Immersed volume × Unit weight of water

W = L x B x d × γ 

W = 2 x 1 x 0.5 × 10 = 10 kN 

Concept:

Archimedes principle:

The weight of a partial or fully submerged body is equal to the buoyancy force acting on the body if the fluid is at rest.

Weight of the block (W) = Buoyancy force (FB)

The buoyancy force FB = Weight of the fluid displaced by the body.

FB = ρfgVfd, where Vfd is the displaced fluid volume.

Calculation:

Given: 

H = 44 cm, L = B = 11 cm, ρb = 850

ρbgV = ρfgVfd

∴ 850 × 9.81 × 11 × 11 × 44 = 1000 × 9.81 × 11 × 11 × h...(h = height of the cuboid immersed in water)

∴ h = 37.4 cm

Explanation:

Types of equilibrium

Stable Equilibrium:

• If the body returns to its original position by retaining the originally vertical axis as vertical.

Unstable Equilibrium:

• If the body does not return to its original position but moves further from it.

Neutral Equilibrium:

• If the body neither returns to its original position nor increases its displacement further, it will simply adapt its new position.

Stability of Floating Bodies in Fluid:

When the body undergoes an angular displacement about a horizontal axis, the shape of the immersed volume changes and so the centre of buoyancy moves relative to the body.

Metacentre:

Meta Centre is defined as the point about which a body starts oscillating when the body is tilted by a small angle.

The meta-centre may also be defined as the point at which the line of action of the force of buoyancy will meet the normal axis of the body when the body is given a small angular displacement.

• For the body shown in the figure, M is above G, and the couple acting on the body in its displaced position is a restoring couple which tends to turn the body to its original position.
• If M were below G, the couple would be an overturning couple and the original equilibrium would have been unstable.
• When M coincides with G, the body will assume its new position without any further movement and thus will be in neutral equilibrium.

Hence the condition of stable equilibrium for a floating body can be expressed in terms of metacentric height as follows:

• GM > 0 (M is above G) ⇒ Stable equilibrium
• GM = 0 (M coinciding with G) ⇒ Neutral equilibrium
• GM < 0 (M is below G) ⇒ Unstable equilibrium

Concepts:

As per Archimedes principle, when any object is submerged partially or fully in any fluid (here fluid is water), it acts upon a additional force up to the submerged portion is called Buoyant force and it is given as:

F_b = Unit weight of fluid (here fluid is water) × Submerged volume of object

Further, Submerged volume of object, V_sub can be expressed as:

V_sub = Mass of object/Density of object

Calculation:

Here we are neglecting the buoyant force due to air and only buoyant force due to water is considered.

Weight of stone in air, W = 250 N

 (W = mg); m is the mass and g is acceleration due to gravity

Weight of stone in water, W_w = 150 N

Total Force on stone, when it is submerged in water fully,

F = mg – γ_w × V_sub

Since stone is fully submerged so its submerged volume is the volume of stone i.e. V_sub = V

Now,

F = W_w = 150 N (Given)

Mg = W_w = 250 N (Given)

Putting these values, we get

150 = 250 – 10000 × V

∴ V = 0.01 m³

Concept:

Archimedes principle:

• It states that a body when wholly or partially immersed in liquid experiences an upward thrust that is equal to the liquid displaced acts on it.
• Archimedes Principle is also known as the physical law of buoyancy.
• When a solid is fully immersed in a liquid, it loses weight, which is equal to the weight of the liquid it displaces.

Explanation:

• Archimedes' principle is used to explain the law of floatation.
• According to the Archimedes principle, a body continues to float, if the weight of the body is equal to the weight of the fluid by it.
• If the density of the body is greater than that of the fluid, it will sink.
• If the density of the body is lesser than that of the fluid, it will float.

Thus, the correct statement is only 2.

Key Points

•  The hydrometer basically consists of a weighted, sealed, long-necked glass bulb immersed in the measured liquid, the flotation depth shows the liquid density, and the neck can be calibrated to read the actual gravity value.
• A hydrometer used for determining the density of liquids is based on Archimedes' principle.