Fluid Mechanics MCQ Quiz - Objective Question with Answer for Fluid Mechanics - Download Free PDF

The correct answer is Mumbai.

Key Points

• Mumbai defeated Madhya Pradesh by five wickets in the Syed Mushtaq Ali Trophy final held in Bengaluru.
• The team successfully chased a target of 175, finishing at 180 for five in just 17.5 overs.
• This win marked Mumbai's second Syed Mushtaq Ali Trophy title, with their first victory coming in 2022.
• Key contributions came from Suryakumar Yadav (48), Ajinkya Rahane (37), and a crucial partnership between Suryansh Shedge (36*) and Atharva Ankolekar (16*).

Additional Information

• The Syed Mushtaq Ali Trophy is India's premier domestic T20 cricket tournament, named after the legendary Indian cricketer.
• Suryakumar Yadav and Ajinkya Rahane played crucial roles in stabilizing Mumbai’s innings after early setbacks.
• Rajat Patidar was the standout performer for Madhya Pradesh, scoring an unbeaten 81 runs, his fifth fifty of the tournament.
• Madhya Pradesh is yet to win its first Syed Mushtaq Ali Trophy title.
• The match showcased exceptional performances in a high-stakes environment, with Mumbai ultimately prevailing due to their collective batting strength.

Explanation:

Vapour Pressure:

• Vapour pressure is a fundamental concept in the study of liquids and their phase transitions. It is defined as the pressure exerted by the vapour in equilibrium with its liquid at a given temperature. This means that at a specific temperature, a certain number of liquid molecules will have enough kinetic energy to escape the liquid phase and enter the vapour phase, creating a dynamic equilibrium where the rate of evaporation equals the rate of condensation.
• When a liquid is placed in a closed container, molecules continuously move between the liquid and vapour phases. Initially, the rate of evaporation exceeds the rate of condensation because there are few molecules in the vapour phase. As more molecules enter the vapour phase, the rate of condensation increases. Eventually, a state of equilibrium is reached where the number of molecules evaporating equals the number of molecules condensing. The pressure exerted by the vapour at this equilibrium state is called the vapour pressure.

Factors Affecting Vapour Pressure:

• Temperature: Vapour pressure increases with temperature because higher temperatures provide more energy for molecules to escape the liquid phase.
• Nature of the Liquid: Different liquids have different vapour pressures at the same temperature. For example, volatile liquids like ether have higher vapour pressures compared to less volatile liquids like water.

Explanation:

Priming of a Pump

• Priming of a pump refers to the process of removing air from the pump casing and suction line to ensure that the pump operates efficiently. This process is crucial for the proper functioning of the pump, especially in cases where the pump is used to lift fluids from a lower level to a higher level.
• Pumps are designed to move fluids by creating a pressure differential. However, when air is trapped in the pump casing or suction line, it can disrupt the pressure differential and prevent the pump from functioning correctly. Priming involves filling the pump casing and suction line with the fluid to be pumped, ensuring that there is no air trapped in the system. This allows the pump to create the necessary pressure differential and move the fluid efficiently.

Methods of Priming:

• Manual Priming: This method involves manually filling the pump casing and suction line with the fluid to be pumped. This can be done using a priming pump, a priming chamber, or by pouring the fluid directly into the pump.
• Automatic Priming: Some pumps are equipped with automatic priming systems that use a small auxiliary pump or a priming chamber to remove air from the pump casing and suction line. These systems are designed to maintain the prime automatically, ensuring continuous operation.

Importance of Priming:

• Prevents Cavitation: Cavitation occurs when air bubbles form in the pump and collapse, causing damage to the pump components. Priming ensures that there is no air in the system, preventing cavitation and extending the life of the pump.
• Ensures Efficient Operation: Air in the pump casing or suction line can disrupt the pressure differential and reduce the pump's efficiency. Priming ensures that the pump operates at its optimal efficiency, providing consistent performance.
• Prevents Damage: Running a pump without priming can cause it to run dry, leading to overheating and damage to the pump components. Priming ensures that the pump is always filled with fluid, preventing damage and ensuring reliable operation.

Explanation:

Single Volute Pump Casing:

• A single volute pump casing is a type of pump casing design in which the volute is a single spiral-shaped chamber that surrounds the impeller. The primary function of the volute is to convert the kinetic energy imparted to the fluid by the impeller into pressure energy.
• In a centrifugal pump, the fluid enters the pump impeller along or near to the rotating axis and is accelerated by the impeller, flowing radially outward into a diffuser or volute chamber, from where it exits into the downstream piping system. The single volute design ensures that the fluid is smoothly transitioned from the high-velocity region near the impeller to the lower-velocity region in the volute, helping in maintaining the flow characteristics.

Advantages:

• Uniform Flow Distribution: The single volute design helps maintain a uniform flow distribution around the impeller, reducing the chances of flow separation and energy losses. This uniformity in flow minimizes the radial forces acting on the impeller, leading to smoother operation and reduced wear and tear on the pump components.
• Efficiency: By maintaining a uniform flow distribution, the single volute design helps in achieving higher pump efficiencies. The smooth transition of fluid reduces turbulence and hydraulic losses, ensuring that a significant portion of the energy imparted by the impeller is converted into pressure energy.
• Cost-Effectiveness: Single volute pump casings are generally simpler and less expensive to manufacture compared to more complex designs like double volute casings. This cost-effectiveness makes them a preferred choice for many industrial applications where cost considerations are critical.

Disadvantages:

• Limited Operating Range: Single volute pumps may have a restricted operating range compared to double volute designs. At off-design conditions, single volute pumps can experience higher radial forces, leading to increased vibration and wear.
• Potential for Cavitation: In certain conditions, single volute designs can be more prone to cavitation, especially at low flow rates. Cavitation can cause significant damage to the pump impeller and casing, leading to reduced pump life and efficiency.

Explanation:

Double Volute Casings

• A double volute casing is a type of pump casing where the volute (the spiral-shaped casing that collects fluid discharged from the impeller) is divided into two separate channels or volutes. These channels are designed to balance the hydraulic forces on the impeller, thereby reducing radial loads and prolonging the life of the pump components.
• In a double volute casing, the fluid is discharged from the impeller into two separate volute channels. These channels help to balance the pressure around the impeller, reducing radial thrust and minimizing the mechanical stress on the pump bearings and shaft. This design is especially beneficial in high-capacity and high-pressure applications.

Advantages:

• Reduced radial thrust on the impeller, leading to longer bearing and shaft life.
• Improved hydraulic balance, which enhances the overall reliability and performance of the pump.
• Better handling of high-pressure and high-flow applications.

Disadvantages:

• Increased complexity in design and manufacturing, which can lead to higher production costs.
• More challenging alignment and assembly processes due to the additional components and tighter tolerances required.

Applications: Double volute casings are commonly used in industrial and municipal applications where high flow rates and pressures are required, such as in water treatment plants, chemical processing, and power generation.

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.

The correct answer is Pascal's law.

• The Pascal's law states that in a fluid which is at rest in a container, the pressure applied to one part of the fluid is uniformly transmitted to all the parts of the fluid.

Key Points

• A hydraulic lift employs this principle to lift heavy objects.
• When pressure is applied to a fluid through one piston, it results in an equivalent pressure on another piston in the system which is then able to lift objects.
• With the increase in the area of the second piston, the force exerted by it also increases thus enabling lifting of heavier objects.

Additional Information

• Hooke's law states that force needed to extend or compress a spring by some distance is directly proportional to that distance.
• Newton's first law of motion - A body at rest remains at rest, or if in motion, remains in motion at constant velocity unless acted on by a net external force. 
• Archimedes' principle states that 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.

CONCEPT:

Bernoulli's principle: For a streamlined flow of an ideal liquid in a varying cross-section tube the total energy per unit volume remains constant throughout the fluid.

• This means that in steady flow the sum of all forms of mechanical energy in a fluid along a streamline is the same at all points on that streamline.

From Bernoulli's principle

\(\frac{{{{\rm{P}}_1}}}{{\rm{\rho }}} + {\rm{g}}{{\rm{h}}_1} + \frac{1}{2}{\rm{v}}_1^2 = \frac{{{{\rm{P}}_2}}}{{\rm{\rho }}} + {\rm{g}}{{\rm{h}}_2} + \frac{1}{2}{\rm{v}}_2^2\)

\(\frac{{\rm{P}}}{{\rm{\rho }}} + {\rm{gh}} + \frac{1}{2}{{\rm{v}}^2} = {\bf{constant}}.\)

EXPLANATION:

• From above it is clear that Bernoulli's equation states that the summation of pressure head, kinetic head, and datum/potential head is constant for steady, incompressible, rotational, and non-viscous flow.
• In other words, an increase in the speed of the fluid occurs simultaneously with a decrease in pressure or a decrease in the fluid's potential energy i.e. the total energy of a flowing system remains constant until an external force is applied.
• So Bernoulli’s equation refers to the conservation of energy.
• All of the above are the measuring devices like Venturimeter, Orifice meter, and Pitot tube meter works on the Bernoulli’s theorem. Therefore option 4 is correct.

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:

For flow along a stream line acceleration is given as

If V = f(s, t)

Then, \(dV = \frac{{\partial V}}{{\partial s}}ds + \frac{{\partial V}}{{\partial t}}dt\)

\(a = \frac{{dV}}{{dt}} = \;\frac{{\partial V}}{{\partial s}} \times \frac{{ds}}{{dt}} + \frac{{\partial V}}{{\partial t}}\) 

For steady flow \(\frac{{\partial V}}{{\partial t}} = 0\)

Then \(a = \frac{{\partial V}}{{\partial s}} \times \frac{{ds}}{{dt}}\) 

Since V = f(s) only for steady flow therefore \(\frac{{\partial v}}{{\partial s}} = \frac{{dv}}{{ds}}\)

Therefore \(a = V \times \frac{{dV}}{{ds}}\)

Calculation:

Given, V_A = 2 m/s, V_B = 5 m/s, and distance s = 1 m

\(\frac{{dV}}{{ds}} = \frac{{\left( {5 - 2} \right)}}{1} = 3\)

So acceleration of fluid at B is

\({a_B} = {V_B} \times \frac{{dV}}{{ds}} = 5 \times 3 = 15\)

Concept:

Most efficient channel: A channel is said to be efficient if it carries the maximum discharge for the given cross-section which is achieved when the wetted perimeter is kept a minimum.

Rectangular Section:

Area of the flow, A = b × d

Wetted Perimeter, P = b + 2 × d  

For the most efficient Rectangular channel, the two important conditions are

1. b = 2 × d
2.  \(R = \frac{A}{P} = \frac{{b\times d}}{{b + 2 \times d}} = \frac{{2{d^2}}}{{4d}} = \frac{d}{2}= \frac{b}{4}\)

Calculation

Given: b = 2 m

 \(R = \frac{2}{4}\)

⇒ R = 0.5

Where R = hydraulic radius, A = Area of flow, P = wetted perimeter, y = depth of flow, T = Top width

Concept:

Specific gravity

• Specific gravity is also termed as relative density.
• The relative density/specific gravity of a substance is defined as the ratio of the density, mass or weight of the substance to the density, mass or weight of water at 4° C

\(Specific~gravity,S = \frac{\rho }{{{\rho _{water}}}}\)

Calculation:

Given:

Volume, V = 1 liter = 10^-3 m³,   mass, m = 7.5 kg

\(\rho = \frac{m}{V} = \frac{{7.5}}{{{{10}^{ - 3}}}} = 7500~\frac{{kg}}{{{m^3}}}\)

\(Specific~gravity,S = \frac{\rho }{{{\rho _{water}}}} = \frac{{7500}}{{1000}} = 7.5\)

Explanation:

The total energy of a flowing fluid can be represented in terms of head, which is given by

\(\frac{p}{ρ{g}}\;+\;\frac{V^2}{2g}\;+\;z\; \)

The sum of the pressure head and hydrostatic pressure head is called the piezometric head. It is given by 

Piezometric head = \(\mathbf{\frac{P}{ρ{g}}+z}\)

where \(P\over\gamma \)= pressure energy per unit weight or pressure head

\(V^2\over{2g}\)= kinetic energy per unit weight or kinetic energy head

z = potential energy per unit weight or elevation head

The pressure at any point in a static fluid is obtained by Hydro-static law which is given by -

\(\frac{dP}{dz}=-ρ{g}\)

∴ P = -ρgz

∴ P = ρgh

where P = pressure above atmospheric pressure and h = height of the point from the free surface.

At point A, pressure head = \(P_A\over\gamma\) = hA  and datum head = zA

At point B, pressure head = \(P_B\over\gamma\) = hB  and datum head = zB

Piezometric head at point A = \(\frac{P}{ρ{g}}+z\) = hA + zA = H

Piezometric head at point B = \(\frac{P}{ρ{g}}+z\) = hB + z₀ = H

∴ piezometric head remains constant at all points in the liquid.

Explanation:

Pitot Tube is a device used for calculating the velocity of flow at any point in a pipe or a channel. 

The pitot tube is used to measure velocity at a point.

In the question velocity at the stagnation point is given which is zero. So here stagnation pressure will be the correct answer. Because this stagnation pressure head is used to calculate the velocity at a point.

V = \(\sqrt{2gh}\)

It is based on the principle that if the velocity of flow at a point becomes zero, the pressure there is increased due to the conversion of the kinetic energy into pressure energy. 

Working:

• The liquid flows up the tube and when equilibrium is attained, the liquid reaches a height above the free surface of the water stream
• Since the static pressure, under this situation, is equal to the hydrostatic pressure due to its depth below the free surface, the difference in level between the liquid in the glass tube and the free surface becomes the measure of dynamic pressure \(p_0-p=\frac{\rho V^2}{2} = h\rho g\)  where p₀, p and V are the stagnation pressure, static pressure and velocity respectively at point A
• Such a tube is known as a Pitot tube and provides one of the most accurate means of measuring the fluid velocity
• For an open stream of liquid with a free surface, this single tube is sufficient to determine the velocity, but for a fluid flowing through a closed duct, the Pitot tube measures only the stagnation pressure and so the static pressure must be measured separately.

Mistake PointsIn the option velocity at the stagnation point is mentioned, at the stagnation point velocity is already zero there is no need to measure velocity at the stagnation point. the pitot tube is used to measure velocity at any point by measuring the stagnation pressure. Hence the best possible option out of the provided options is option B.

Explanation:

A manometer is a device that measures pressure by balancing a column of liquid against a column of gas or another liquid. The liquid used in a manometer should have the following characteristics:

• The liquid should stick on the walls: The liquid used in a manometer should stick to the walls of the tube to prevent it from flowing back and forth due to vibration or turbulence.
• Low surface tension: The liquid used in a manometer should have low surface tension to ensure that the meniscus does not significantly affect the pressure reading.
• It should be immiscible: The liquid used in a manometer should be immiscible with the gas or liquid being measured, to prevent the two fluids from mixing and affecting the accuracy of the pressure measurement.
•  High viscosity is NOT a desirable characteristic for a manometer liquid. A highly viscous liquid will not respond quickly to changes in pressure, leading to slow and inaccurate readings. Thus, manometer liquids are typically chosen to have relatively low viscosity.