Learning Objectives:

• Define the terms in relation to fluid mechanics such as density, specific gravity, viscosity and vapor pressure.

Solved Example: 60-1-02

Practical fluids:

A. Are viscous

B. Possess surface tension

C. Are compressible

D. Possess all the above properties

Practically, fluids have properties such as viscosity, compressibility as well as surface tension.

Correct Answer: D

Solved Example: 60-1-03

Density of water is maximum at:

A. 0$^\circ$C

B. 0$^\circ$K

C. 4$^\circ$C

D. 100$^\circ$C

Water exists in liquid format between 0-100 C. Generally density of liquid increases as temperature is reduced. However, water has a small exception. Water displays abnormal behavior between 0-4 Degree Celsius when the density actually decreases as the temperate decreases.

Correct Answer: C

Solved Example: 60-1-10

Property of a fluid by which molecules of different kinds of fluids are attracted to each other is called:

A. Adhesion

B. Cohesion

C. Viscosity

D. Compressibility

Adhesion is attraction between unlike molecules whereas cohesion is attraction between like molecules.

Correct Answer: A

Solved Example: 60-1-11

Specific weight of water in S.I. units is equal to:

A. $1000\ N/m^3 $

B. $10000\ N/m^3$

C. $9.81 \times 10^3\ N/m^3$

D. $9.81 \times 10^6\ N/m^3$

Specific weight is given by: \[\gamma = \rho g\] For water, \[\gamma = 1000 \times 9.81 = 9.81 \times 10^3\ N/m^3\]

Correct Answer: C

Solved Example: 60-1-12

Which of the following is dimensionless?

A. Specific weight

B. Specific volume

C. Specific speed

D. Specific gravity

Specific gravity is given by: \[SG = \dfrac{\rho_{\mathrm{substance}}}{\rho_{\mathrm{reference}}}\] Since, it is a ratio of two similar parameters (density), it is a dimentionless number.

Correct Answer: D

Solved Example: 60-1-21

If 850 kg liquid occupies volume of one cubic meter, means 0.85 represents its:

A. Specific weight

B. Specific mass

C. Specific gravity

D. Specific density

Density, \[\rho = \dfrac{m}{V} = \dfrac{850}{1} = 850\ kg/m^3\] Specific Gravity. \[\mathrm{sp.\ gr.} = \dfrac{850}{1000} = 0.85\]

Correct Answer: C

Solved Example: 60-1-23

The process of diffusion of one liquid into the other through a semi-permeable membrane is called:

A. Viscosity

B. Osmosis

C. Surface tension

D. Cohesion

Correct Answer: B

Solved Example: 60-1-26

A piece of metal of specific gravity 7 floats in mercury of specific gravity 13.6. What fraction of its volume is under mercury?

A. 0.5

B. 0.4

C. 0.515

D. 0.6

\[\mathrm{Volume\ fraction\ under\ Mercury} = \dfrac{7}{13.6} = 0.515\]

Correct Answer: C

Solved Example: 60-1-27

If 5.6m$^3$ of oil weighs 46800 N, What is the specific gravity of the oil?

A. 852

B. 0.0852

C. 85.2

D. 0.852

\[\mathrm{Mass} = \dfrac{W}{g} = \dfrac{46800}{9.8} =4770.61\ kg\] \[\mathrm{Density} = \dfrac{m}{V} = \dfrac{4770.61}{5.6} =851.90\ kg/m^3\] \[\mathrm{Specific\ gravity} = \dfrac{\rho_{\ \mathrm{fluid}}}{\rho_{\ \mathrm{water}}} = \dfrac{851.9}{1000} = 0.852\]

Correct Answer: D

Solved Example: 60-1-29

Which of the following forces does NOT act on fluid which is at rest?

A. Viscous

B. Hydrostatic

C. Gravity

D. Surface tension

Correct Answer: A

Learning Objectives:

• Explain and use the following terms: shear stress, velocity gradient, viscosity, Newtonian liquid.
• Define the coefficient of viscosity in terms of this relationship (Newton’s Law of viscosity).

Solved Example: 60-1-01

Fluid is a substance that:

A. Continually deforms (flows) under an applied shear stress.

B. Always expands until it fills any container

C. Has the same shear stress at a point regardless of its motion

D. Remain at rest under action of any shear force

Fluids include liquids and gases. While gases always expand till they fill any container, liquids do not.
One of the defining characteristics of fluids is they cannot resist shear stresses (like what solids can), and on application of shear stresses, they start deforming or flowing.

Correct Answer: A

Solved Example: 60-1-04

Poise is the unit of:

A. Surface tension

B. Capillarity

C. Viscosity

D. Buoyancy

In CGS (Centimeter, Gram, Second) system, the unit of viscosity is Poise, whereas the unit of kinematic viscosity is Stokes.

Correct Answer: C

Solved Example: 60-1-05

Choose the correct relationship:

A. Specific gravity = gravity $\times$ density

B. Dynamic viscosity = kinematic viscosity $\times$ density

C. Kinematic viscosity = dynamic viscosity $\times$ density

D. Hydrostatic force = surface tension $\times$ gravity

\[\mathrm{Kinematic\ viscosity} = \dfrac{\mathrm{Dynamic\ Viscosity}}{\mathrm{density}}\] \[\mathrm{Dynamic\ viscosity} = \mathrm{Kinematic\ Viscosity} \times \mathrm{density}\]

Correct Answer: B

Solved Example: 60-1-06

Choose the wrong statement:

A. Viscosity of a fluid is that property which determines the amount of its resistance to a shearing force

B. Viscosity of liquids decreases with increase in temperature

C. Viscosity of liquids is appreciably affected by change in pressure

D. Viscosity is expressed as Poise, Stoke, or Saybolt seconds.

Viscosity of a liquid decreases with temperature, mostly because due to increased energy, the molecules freely move within the liquid and the resistance to motion is now less.

However, viscosity of fluid is generally independent of pressure. Liquids are not that compressible as of gases

Correct Answer: C

Solved Example: 60-1-07

Kinematic viscosity is equal to:

A. $\dfrac{\mathrm{Dynamic\ viscosity}}{\mathrm{density}}$

B. Dynamic viscosity $\times$ density

C. $\dfrac{\mathrm{Density}}{\mathrm{dynamic\ viscosity}}$

D. $\dfrac{1}{{\mathrm{Dynamic\ viscosity} \times \mathrm{density}}}$

Kinematic viscosity is the ratio of dynamic viscosity and density. \[\nu = \dfrac{\mu}{\rho}\] where,
$\nu$ = Kinematic viscosity
$\mu$ = Dynamic viscosity
$\rho$ = Density

Correct Answer: A

Solved Example: 60-1-08

Kinematic viscosity of water in comparison to mercury is:

A. Higher

B. Lower

C. Same

D. Higher/lower depending on temperature

Dynamic viscosity of water at 27$^\circ C$ = 0.89 centipoise whereas that of mercury at same temperature = 1.53 centipoise, so dynamic viscosity of mercury is higher than water.

However, kinematic viscosity is ratio of dynamic viscosity to density and mercury is 13.6 denser than water.

Once you divide by higher density of mercury, the kinematic viscosity of mercury is much lesser than water. (water = $1 \times 10^{-6} m^2/s$ compared to mercury = $0.115 \times 10^{-6} m^2/s$)

Correct Answer: A

Solved Example: 60-1-14

A fluid in equilibrium can't sustain:

A. Tensile stress

B. Compressive stress

C. Shear stress

D. Bending stress

A solid can resist axial forces or shear forces, by offering resistance within the body. On the other hand, fluids cannot resist shear forces and start to deform. Ideal fluids offer abosolutely no resistace to shear forces, whereas practical fluids, due to their voscosity, offer some resistance.

Correct Answer: C

Solved Example: 60-1-16

The property of fluid by virtue of which it offers resistance to shear is called:

A. Surface tension

B. Adhesion

C. Cohesion

D. Viscosity

Ideal fluid offer no resistance to shear force and they have zero viscosity.

Real fluid offer some resistance to shear force. Viscosity is a measure of this resistance. Due to viscosity, the relative motion between the layers of fluid is opposed to some extent.

Correct Answer: D

Solved Example: 60-1-17

Kinematic viscosity is dependent upon:

A. Pressure

B. Level

C. Flow

D. Density

Kinematic viscosity is given by the formula, \[\nu = \dfrac{\mu}{\rho}\]
where,
$\nu$ = Kinematic viscosity
$\mu$ = Dynamic viscosity
$\rho$ = Density

Correct Answer: D

Solved Example: 60-1-19

Choose the wrong statement:

A. Fluids are capable of flowing

B. Fluids conform to the shape of the containing vessels

C. Fluids have weaker intermolecular forces compared to solids.

D. When in equilibrium, fluids can sustain shear forces

Let us see each statement individually.

• Fluids are capable of flowing. TRUE, unlike solids which occupy fixed space, fluids which comprise of liquids and gases can flow.
• Fluids conform to the shape of the containing vessels. TRUE, Fluids do not have fixed shape of thier own and take the shape of the container.
• Fluids have weaker intermolecular forces compared to solids. TRUE, Since the intermolecular forces in fluids are weaker than solids, the molecules are relatively free to occupy different locations and hence fluids can flow.
• When in equilibrium, fluids can sustain shear forces. FALSE, Fluids cannot sustain shear forces, and when a shear force is applied their layers start having relative velocity with respect to each other.

Correct Answer: D

Solved Example: 60-1-22

The property by virtue of which a liquid opposes relative motion between its different layers is called:

A. Surface tension

B. Co-efficient of viscosity

C. Viscosity

D. Osmosis

Correct Answer: C

Solved Example: 60-1-25

The kinematic viscosity is the:

A. Ratio of absolute viscosity to the density of the liquid

B. Product of absolute viscosity and density of the liquid

C. Ratio of density of the liquid to the absolute viscosity

D. Product of absolute viscosity and mass of the liquid

The kinematic viscosity of a fluid is the ratio of absolute (dynamic) viscosity to its density. \[\nu = \dfrac{\mu}{\rho}\]

Correct Answer: A

Solved Example: 60-2-01

Which of the following is an example of Newtonian fluid?

A. Blood

B. Soap solution

C. Water

D. Ketchup

Newtonian fluids are defined as fluids for which the shear stress is linearly proportional to the shear strain rate.

Correct Answer: C

Solved Example: 60-2-02

Kinematic viscosity:

A. Does not depend upon dynamic viscosity

B. Does not depend upon flow rate

C. Does not depend upon density

D. Depends upon absolute pressure

Kinematic viscosity, \[\nu = \dfrac{\mu}{\rho}\] So, kinematic viscosity depends upon dynamic viscosity and density. Hence, options A and C are incorrect. Kinematic viscosity does not depend upon absolute pressure. Hence, option D is also incorrect. Kinematic viscosity does not depend upon flow rate. Hence, option B is correct.

Correct Answer: B

Solved Example: 60-2-03

Find the kinematic viscosity of oil having density 980 kg/m$^3$ when at at a certain point in the oil the shear stress is 0.25 N/m$^2$ and velocity gradient is 0.3 sec$^{-1}$.

A. 7.773 $\times 10^{-2}$ $\dfrac{m^2}{s}$

B. 0.193 $\times 10^{-3}$ $\dfrac{m^2}{s}$

C. 8.503 $\times 10^{-4}$ $\dfrac{m^2}{s}$

D. 8.503 $\times 10^{-2}$ $\dfrac{m^2}{s}$

\begin{align*} \tau &= \mu \dfrac{du}{dy}\\ 0.25 &= \mu \times 0.3\\ \mu &= \dfrac{0.25}{0.3}\\ &= 0.83\ \mathrm{Pa.s} \end{align*} \[\nu = \dfrac{\mu}{\rho} = \dfrac{0.83}{980} = 8.503 \times 10^{-4} \dfrac{m^2}{s}\]

Correct Answer: C

Solved Example: 60-2-04

Which is the fluid whose viscosity does NOT change with the rate of deformation?

A. Ideal fluid

B. Real fluid

C. Newtonian fluid

D. Non-Newtonian fluid

If you plot shear stress vs. rate of deformation, the slope of the curve is viscosity.

For a Newtonian fluid, the graph is a straight line, which means the slope (viscosity) always remains constant. This is not true for non-Newtonian fluids. Since they are represented by curves, their viscosity keeps on changing.

Correct Answer: C

Solved Example: 60-2-05

A Newtonian fluid fills the clearance between a shaft and a sleeve. When a force of 0.9 kN is applied to the shaft parallel to the sleeve, the shaft attains a speed of 1.25 cm/s. What will be the speed of the shaft if a force of 3 kN is applied?

A. 4.16 cm/s

B. 5.19 cm/s

C. 5.26 cm/s

D. 6.32 cm/s

\[\tau = \mu \dfrac{du}{dy}\] \[\tau = \dfrac{F}{A}\] \[\dfrac{F_1}{A} = \mu \dfrac{du_1}{dy}\] \[\dfrac{F_2}{A} = \mu \dfrac{du_2}{dy}\] Dividing, \[\dfrac{F_1}{F_2} = \dfrac{du_1}{du_2}\] \[\dfrac{0.9}{3} = \dfrac{1.25}{du_2}\] \[du_2 = 4.16 \ \mathrm{m/s}\]

Correct Answer: A

Solved Example: 60-2-06

Consider two flat parallel plates placed in horizontal condition 1.3 cm apart and the space between them is filled with the oil of viscosity 15.0 poise. The upper plate is moved with a velocity of 3.25 m/s, then the shear stress in the oil is:

A. Between 100 - 150 N /m$^2$

B. Between 151 - 200 N/m$^2$

C. Between 201 - 350 N/m$^2$

D. Between 351 - 450 N/m$^2$

\begin{align*} \tau &= \mu \dfrac{du}{dy}\\ &= 15 \times 10^{-1} \left(\dfrac{3.25}{1.3 \times 10^{-2}}\right)\\ &= 375\ \mathrm{N/m}^2 \end{align*}

Correct Answer: D

Solved Example: 60-2-07

For a Newtonian fluid:

A. Shear stress is proportional to shear strain

B. Rate of shear stress is proportional to shear strain

C. Shear stress is proportional to rate of shear strain

D. Rate of shear stress is proportional to rate of shear strain

A newtonian fluid follows Newton's Law of Viscosity which says the shear stress is proportional to the rate of shear strain. \[\tau \propto \dfrac{du}{dy}\]

Correct Answer: C

Solved Example: 60-2-08

According to the power-law model, $\tau = K {\left({\dfrac{{du}}{{dy}}} \right)^{n}}$, what is the flow behavior index 'n' for pseudoplastic fluids?

A. n = 0

B. n = 1

C. n < 1

D. n > 1

\[\tau = K\left(\dfrac{\partial u}{\partial y}\right)^n\] For Pseudoplastic, n < 1
For Newtonian fluid n = 1
For Dilatant n > 1

Correct Answer: C

Solved Example: 60-2-09

A cubic wooden block of edge 100 mm and weight 1 kN is sliding down on an inclined plane of inclination 30° with the horizontal. A Newtonian fluid with the viscosity 0.2 Ns/m$^2$ is layered on the inclined plane. If the thickness of the layer is 0.02 mm, then the terminal velocity of the block in m/s.

A. 0.25

B. 2.5

C. 5

D. 0.5

Correct Answer: C

Solved Example: 60-2-10

Newton's law of viscosity depends upon:

A. Shear stress, pressure and velocity

B. Shear stress and strain in the fluid

C. Shear stress and velocity gradient

D. Viscosity and shear stress

Correct Answer: C

Solved Example: 60-2-11

A fluid, which is incompressible and having no viscosity is:

A. Ideal Plastic Fluid

B. Non-Newtonian Fluid

C. Real Fluid

D. Ideal Fluid

Correct Answer: D

Solved Example: 60-2-13

Consider two parallel plates separated by a distance of 1 cm filled with a Newtonian fluid of viscosity 10$^{-3}$ Pa.s. The top plate is moving with a velocity of 1 m/s whereas the bottom plate is stationary. The shear stress (in Pa, rounded off to one decimal place) on the top plate is:

A. $0.05\ \mathrm{Pa}$

B. $0.1\ \mathrm{Pa}$

C. $0.2\ \mathrm{Pa}$

D. $0.25\ \mathrm{Pa}$

\begin{align*} \tau &= \mu \dfrac{du}{dy}\\ &= 10^{-3} \times \dfrac{(1 - 0)}{1 \times 10^{-2}}\\ &= 0.1\ \mathrm{Pa} \end{align*}

Correct Answer: B

Learning Objectives:

• Define the terms surface tension and capillarity.

Solved Example: 60-1-09

Surface tension has the units of:

A. N/m

B. N/$m^2$

C. N

D. N.m

Surface tension is force per unit length, hence its units will be N/m.

Correct Answer: A

Solved Example: 60-1-13

The tendency of a liquid surface to contract is due to the following property:

A. Cohesion

B. Adhesion

C. Viscosity

D. Surface tension

If a molecule is within liquid, it is attracted by equal forces from all direction.
However, if you comare that, with a molecule on the surface of liquid, it is attracted unequally, as the forces of attraction are different from within the liquid compared to outside the surface.
Due to this, the molecules on the surface of liquid are in constant state of pull, or tension, which is referred as, surface tension.

Correct Answer: D

Solved Example: 60-1-15

Surface tension:

A. Acts in the plane of the interface normal to any line in the surface

B. Is also known as capillarity

C. Is a function of the curvature of the interface

D. Decreases with fall in temperature

At the surface of liquid, a liquid molecule is attracted by two types of forces:

• Adhesion: Force of attraction between different molecules
• Cohesion: Force of attraction between similar molecules

Correct Answer: A

Solved Example: 60-1-18

Surface energy per unit area of a surface is numerically equal to:

A. Atmospheric pressure

B. Surface tension

C. Force of adhesion

D. Viscosity

\[\sigma = \dfrac{F}{L}\]

Correct Answer: B

Solved Example: 60-1-24

Free surface of a liquid tends to contract to the smallest possible area due to force of:

A. Surface tension

B. Viscosity

C. Cohesion

D. Adhesion

Let's see each option individually.

• Surface tension: State of liquid surfacec to remain in the constant state of tension by occupying smaller surface area.
• Viscosity: A measure of resistance between layers of fluid.
• Adhesion: Force of attraction between dissimilar molecules.
• Cohesion: Force of attraction between similar or like molecules.

Correct Answer: A

Solved Example: 60-3-01

Property of a fluid by which its own molecules are attracted is called:

A. Adhesion

B. Cohesion

C. Cohesion

D. Surface tension

Similar or own molecules = Cohesion
Dissimilar molecules (water and glass) = Adhesion

Correct Answer: B

Solved Example: 60-3-02

The pressure of air in a soap bubble of 0.7 cm diameter is 8 mm of water above the pressure outside. The surface tension of the soap solution is:

A. 0.10 N/m

B. 0.07 N/m

C. 0.14 N/m

D. 0.15 N/m

Surface tension $\sigma$ is the force per unit contact length, \[\sigma = \dfrac{F}{L}\] For soap bubble, \[F = P \times A = h \rho g \times \dfrac{\pi}{4} d^2\] Also, since we have two surfaces, \[L = 2 \times \pi \times d\] Substituting, \begin{align*} \sigma &= \dfrac{h \rho g \times \left(\dfrac{\pi}{4} d^2\right)}{2 \times \pi \times d}\\ &= \dfrac{h \rho g d}{8}\\ &= \dfrac{8 \times 10^{-3} \times 1000 \times 9.81 \times 0.7 \times 10^{-2}}{8}\\ &=0.07\ N/m \end{align*}

Correct Answer: B

Solved Example: 60-3-03

The rise of a Liquid in a capillary tube is due to:

A. Viscosity

B. Osmosis

C. Diffusion

D. Surface Tension

Correct Answer: D

Solved Example: 60-3-03

A capillary tube of radius r can support a liquid of weight 6.28 $\times$ 10$^{-4}$ N. if the surface tension of the liquid is 5 $\times$ 10$^{-2}$ N/m. the radius of capillary must be:

A. 2.5 $\times$ 10$^{-3}$ m

B. 2.0 $\times$ 10$^{-4}$ m

C. 1.5 $\times$ 10$^{-3}$ m

D. 2.0 $\times$ 10$^{-3}$ m

\[\sigma = \dfrac{F}{L}\] \begin{align*} L &= \dfrac{F}{\sigma}\\ L &= \dfrac{6.28 \times 10^{-4}}{5 \times 10^{-2}}\\ L &= 0.01256\\ 2 \pi r &= 0.01256\\ r &= \dfrac{0.01256}{2 \pi}\\ r &= 0.002\ \mathrm{m} \end{align*}

Correct Answer: D