Mechanical Properties of Fluids Formula

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Mechanical properties of fluids help in the study of the suitability of fluids for different applications. Fluids serve as a conduit for all biological processes in living things, including plants.

  • It is crucial to comprehend the behavior and characteristics of fluids.
  • The science under which mechanical properties of fluids work is called Hydrostatics.
  • A fluid is a substance that yields to the slightest pressure.

Key Terms: Pressure, Velocity, Liquids, Volume, Mass, Density, Molecules, viscosity, Force, Hydraulic brakes, incompressible fluid


What is Fluid Mechanics?

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The study of liquids, gasses, and particularly the forces they produce is known as fluid mechanics. Fluid mechanics is of importance to numerous scientific fields. 

  • It is the study of fluid systems, such as liquid or gas, under static and dynamic loads.
  • A subfield of continuous mechanics called fluid mechanics models the kinematics and mechanical behavior of materials as a continuous mass rather than as discrete particles.
  • Engineers are interested in fluid mechanics because fluids produce forces that can be used in practical applications.
  • Some well-known applications include jet propulsion, aerofoil design, wind turbines, and hydraulic brakes, but there are others that receive less attention, such as the design of mechanical heart valves.

Fluid Mechanics

Fluid Mechanics


What is Fluid Dynamics?

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Fluid dynamics can be expressed as "the branch of applied science that is concerned with the flow of liquids and gasses." One of the two subfields of fluid mechanics, the study of fluids and how forces influence them, is fluid dynamics.

Applications of Fluid Dynamics

Fluid Dynamics is studied by researchers from many different domains and offers approaches for investigating:

  • Evolution Of Stars
  • Ocean Currents
  • Weather Patterns
  • Plate Tectonics
  • Blood Circulation

Uses of Fluid Dynamics: Rocket engines, wind turbines, oil pipelines, and air conditioning units are a few significant technological uses of fluid dynamics. 

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What is Fluid Statics?

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The area of fluid mechanics known as fluid statics is responsible for studying incompressible fluids at rest. In contrast to fluid dynamics, which is the study of fluids in motion, it includes the study of the circumstances under which fluids are at rest in stable equilibrium.

  • Fluid statics is primarily concerned with comprehending these fluid equilibrium circumstances since the fact that these fluids are not in motion indicates that they have attained a stable equilibrium state. 
  • It is frequently referred to as hydrostatics when focusing on incompressible fluids (like liquids) as opposed to compressible fluids (like the majority of gasses).

Mechanical Properties of Fluids Formulas

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The formula of mechanical properties of fluids are:

The density of a sample:

ρ = m/V

Where, 

  • ρ: density of the fluid
  • m: Mass
  • V: Volume

The pressure exerted by fluid is given by:

p = F/A

Where,

  • p: Pressure
  • F: Force applied
  • A: The area affected

The pressure at a depth of h in a fluid of constant density:

p = p0 + ρgh

Where,

  • p: the pressure at height h
  • p0: the pressure at zero height
  • g: acceleration due to gravity
  • ρ: fluid density

Volume flow rate:

Q = dV/dt

Where,

  • Q: flow rate
  • dV: change in volume
  • dt: time period

Viscosity:

η = FL/vA

Where,

  • η: fluid viscosity
  • F: Force
  • L: distance between the plates
  • V: constant velocity
  • A: area of the plate

Solved Examples

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Ques. The force of 1200 N is applied on a circular disc of an area of 12 cm2. What would be the pressure on the surface of the disc?

Ans. Given

  • Force F = 1200 N 
  • Area A = 12 cm2 = 12 x 10-4

Pressure (P) = Force (F)/area (A)

P= 1200/12 x 10-4

Pressure, P = 106 N/m2 

Ques. A force of 500 N acts on an area of 1.0 m2. Calculate the pressure.

Ans. As per the given question,

  • Force (F) = 500N 
  • Area = (A) = 1.0m2

We have,

P = F/A 

⇒ P = 500/1

P = 500 N/m2 

P = 500 Pa


Previous Year Questions


Things to Remember

  • Fluid densities are independent of changes in pressure and remain constant. This property is known as incompressibility.
  • Tangential forces are not generated between two fluid surfaces when they are in contact. This property is known as non-viscosity.
  • The force per unit length in the plane of the liquid surface at right angles to either side of an imaginary line drawn on that surface is referred to as surface tension.
  • Cohesive force is the term used to describe the attraction between molecules of the same substance.
  • Adhesive force is the term used to describe the attraction between molecules of different substances.

Sample Questions

Ques. What is the Archimedes Principle? [3 marks]

Ans. The Archimedes principle states that whenever an object is submerged, whether fully or partially, it will be propelled upward by a buoyant force (FB) whose size is equal to the weight of the fluid displaced by the body.

The formula is - Fb = ρ × g × V 

  • Fb is the buoyant force
  • f is the density of the fluid
  • V is the submerged volume
  • g is the acceleration due to gravity

Ques. The density of steel is 9000 kg/m3 and the density of water is 1000 kg/m3. If a cube of steel that is 0.100 m on each side is placed in a tank of water and weighed while underwater, what is the apparent weight of the cube? [5 marks]

Ans. Here, the cube of steel has 0.100 m on each side 

The volume of the cube is 0.00100 m3

The density of steel is 9000 kg/m3

The mass of the cube is = volume x density = 9.00 kg. 

The weight of the cube when not submerged in water = (9.00 kg)(9.80 m/s2) = 88.2 N 

Thus,

The mass of water displaced by the cube - the volume of displaced water x density of water = 0.00100 m3 x 1000 kg/m3 = 1.00 kg 

The weight of the water displaced by the cube = mass of water x g= 9.80 N 

The buoyant force on the steel cube = 9.80 N 

Apparent weight of cube underwater = 88.2 N 9.80 N = 78.4 N

Ques. A hollow metal cube of 1.00 m on each side has a mass of 600 kg. How deep will this cube sink when placed in a vat of water? [3 marks]

Ans. Since the weight of the cube is 5880 N, it will need to displace 5880 N of water in order to float. 

Say the cube is deep by d meter. 

Volume of submerged portion of cube =(1.00 m)(1.00 m)(d m) = dm3

Mass of water displaced = Volume of the submerged portion of cube x density of water = d x1000 kg = 1000 d.kg 

Weight of water displaced =1000d x 9.8 = 9800d N 

So, 9800 d = 5880

d=0.600 m 

The cube will sink such that 0.6 m is underwater.

Ques. The force of 1000 N is applied on a circular disc of an area of 10 cm2. What would be the pressure on the surface of the disc? [3 marks]

Ans. It is Given that: 

Force F = 1000 N 

Area A = 10 cm2 = 10 x 104

= 10-3  m2

Pressure = Force/area 

= 1000/10-3 

Pressure 1.0 × 106 N/m2 

Ques. A force of 600 N acts on an area of 1.0 m2. Calculate the pressure. [2 marks]

Ans. As per the given question,

Force (F) = 600N 

Area = (A) = 1.0m2

P = F/A P = 600N/1.0m2

P = 600 N/m2 

P = 600 Pa

Ques. A 5 mm2 copper wire has a current of 5 mA of current flowing through it. Determine the current density. [3 marks]

Ans. As per the given question, it can be said that:

Total Current I is 5 mA

Total Area A is 5 mm2

The Current density J = I / A

= 5×10−3 / 5×10−3

= 1 A/m2

Ques. Determine the current density of if 50 Amperes of current flows through the battery in an area of 10 m2. [2 marks]

Ans. It is Given that:

  • Current I = 50 A,
  • Area A = 10 m2

The current density is given by J = 50 / 10

J = 5 A/m2.

Ques. A fluid with absolute viscosity of 0.98 Ns/m2 and kinematic viscosity of 3 m2/s. Determine the density of the fluid. [3 marks]

Ans. According to the given question:

Absolute viscosity μ = 0.98 Ns/m2

Kinematic viscosity ν = 3 m2/s

ν = μ/ρ

The density is given by,ρ = ν/μ

ρ = 3/(0.98)

ρ = 3.0612 kg/m3

Therefore, the density of a fluid is 3.0612 kg/m3.

Ques. Compute the flow rate of fluid if it is moving with the velocity of 20 m/s through a tube of diameter 0.03 m. [3 marks]

Ans. As per the given question:

The velocity of fluid flow v =20m/s

Diameter of pipe d = 0.03m

Area of the cross-section of the pipe = A = 4d2

A={(3.14)/4}(0.03)(0.03)

A =(0.785)(0.0009)

A = 0.000706 m2

Flow rate is given by Q = vA=(20)(0.000706)

Q = 0.014139 m3/s

Ques. Why do fluids exert pressure? [2 marks]

Ans. Because the fluid's particles can flow in all directions, the fluid applies pressure in all of them. These molecules are constantly colliding because of their mobility. In the aftermath of the collision, pressure is applied everywhere.

Ques. A water glass sitting on a table weighs 4 N. The bottom of the water glass has a surface area of 0.003 m2. Calculate the pressure the water glass exerts on the table. [3 marks]

Ans. As per the given question:

Force or thrust here = F= 4 N

Area A = 0.003 m2

Pressure exerted = P = F/A

= 4/0.003 Pa

= 1333.33 Pa

Ques. If you are holding a 2-kg book in the palm of your hand and if the area of contact between your hand and the book is 0.003 m2 then find out the pressure that is exerted on your hand by the book. [3 marks]

Ans. According to the given question:

Force = F = 2 x 9.8 N = 19.6 N 

Area = 0.003 m2 

Pressure = F/A 

= 19.6/0.003 Pa 

= 6533.33 Pa

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CBSE CLASS XII Related Questions

  • 1.
    Write the expression for the magnetic field due to a current element in vector form. Consider a 1 cm segment of a wire, centered at the origin, carrying a current of 10 A in positive x-direction. Calculate the magnetic field \( \mathbf{B} \) at a point \( (1 \, \text{m}, 1 \, \text{m}, 0) \).


      • 2.
        Photoemission of electrons occurs from a metal (\( \phi_0 = 1.96 \, \text{eV} \)) when light of frequency \( 6.4 \times 10^{14} \, \text{Hz} \) is incident on it. Calculate: Energy of a photon in the incident light, The maximum kinetic energy of the emitted electrons, and The stopping potential.


          • 3.
            Suppose a pure Si crystal has \( 5 \times 10^{28} \) atoms per \( \text{m}^3 \). It is doped with \( 5 \times 10^{22} \) atoms per \( \text{m}^3 \) of Arsenic. Calculate majority and minority carrier concentration in the doped silicon. (Given: \( n_i = 1.5 \times 10^{16} \, \text{m}^{-3} \))


              • 4.
                Two small identical metallic balls having charges \( q \) and \( -2q \) are kept far at a separation \( r \). They are brought in contact and then separated at distance \( \frac{r}{2} \). Compared to the initial force \( F \), they will now:

                  • attract with a force \( \frac{F}{2} \)
                  • repel with a force \( \frac{F}{2} \)
                  • repel with a force \( F \)
                  • attract with a force \( F \)

                • 5.
                  Two parallel plate capacitors X and Y are connected in series to a 6 V battery. They have the same plate area and same plate separation but capacitor X has air between its plates, whereas capacitor Y contains a material of dielectric constant 4. Calculate the capacitances of X and Y, if the equivalent capacitance of the combination of X and Y is \( 4 \, \mu\text{F} \). Calculate the potential difference across the plates of X and Y.


                    • 6.
                      The figure shows three point charges kept at the vertices of triangle ABC. The net electric field, due to this system of charges, at the midpoint M of base BC will be:

                        • \( \frac{q}{4 \pi \epsilon_0 l^2} \) pointing along MA
                        • \( \frac{q}{\pi \epsilon_0 l^2} \) pointing along AM
                        • \( \frac{q}{2 \pi \epsilon_0 l^2} \) pointing along AM
                        • Zero
                      CBSE CLASS XII Previous Year Papers

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