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.

Concept:

The pitch diameter of a Pelton wheel is calculated by equating the tangential velocity derived from geometry and speed ratio. The tangential velocity is:

and also

Equating both:

Final Answer:

Explanation:

Circulation

• Circulation is a concept in fluid dynamics that measures the total "rotational effect" of a fluid flow around a closed contour or loop. It is mathematically defined as the line integral of the velocity vector along a closed curve. The dimension of circulation can be derived from its physical definition and mathematical representation.

Circulation is defined as the line integral of velocity over a closed loop:

\( \Gamma = \oint \vec{V} \cdot d\vec{l} \)

Where:

• \( \vec{V} \) = velocity (dimension = \( \frac{L}{T} \))
• \( d\vec{l} \) = length element (dimension = \( L \))

Therefore,

\( \text{Dimension of } \Gamma = \frac{L}{T} \times L = \frac{L^2}{T} \)

Concept:

For laminar flow:

Friction factor (f) \(= \frac{{64}}{{{{\rm{R}}_{\rm{e}}}}}\)

For turbulent flow:

Case I: Smooth Pipe

Friction factor (f) \( = \frac{{0.316}}{{{{\left( {{{\rm{R}}_{\rm{e}}}} \right)}^{1/4}}}}\)

Case II: Rough Pipe

\(\frac{1}{{\sqrt {\rm{f}} }} = 2{\log _{10}}\left( {\frac{{\rm{r}}}{{\rm{k}}}} \right) + 1.74\)

Case III: In between smooth & rough (Transition)

Friction (f) = \(\rm{\phi (R_e, \frac{k}{D})}\)

∴ For laminar flow friction factor depends only upon the Reynolds number of the flow and is independent of the contact surface.

As can observe, the friction factor is inversely proportional to Reynold’s number. Also, the velocity distribution of laminar flow through the pipe is parabolic.

∴ Statement 3 and 4 are correct.

Mistake Points

• For the Laminar flow, the friction factor depends only on the Reynolds number and is independent of the internal condition of the pipe. 
• For turbulent flow, when the Reynolds number Re > 4000, the friction factor f depends not only on Re but also on the internal roughness of the pipe.

Explanation:

Theoretical Power Required to Drive a Single-Acting Reciprocating Pump:

• The theoretical power required to drive a single-acting reciprocating pump is derived based on the fundamental principles of fluid mechanics and pump operation. The power required is a function of the work done by the pump in lifting the liquid through the given head, considering the weight of the liquid, the stroke volume, and the number of strokes per minute.

The theoretical power for a single-acting reciprocating pump is calculated by multiplying the volume of fluid delivered per second by the specific weight and total head (suction + delivery).

\( P = \frac{w \times A \times L \times N}{60} (H_s + H_d) \)

Where:

• \( w \): Specific weight of the fluid
• \( A \): Cross-sectional area of the piston
• \( L \): Length of stroke
• \( N \): Number of delivery strokes per minute
• \( H_s \): Suction head
• \( H_d \): Delivery head

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:

Positive displacement pump:

• Positive displacement pumps are those pumps in which the liquid is sucked and then it is pushed or displaced to the thrust exerted on it by a moving member, which results in lifting the liquid to the required height.
• Reciprocating pump, Vane pump, Lobe pump are the examples of positive displacement pump whereas the centrifugal pump is the non-positive displacement pump.​