Mechanical Properties Of Fluids: Important Questions

A fluid, such as liquid or gas is any substance that can flow continuously and deform under applied stress.

• The study of the mechanical properties of fluids at rest is called Hydrostatics.
• Each fluid has its own composition and specific characteristics, but some characteristics are common to them all.
• Mechanical properties of fluid are determined to see whether a fluid is suitable or not for different purposes.

The mechanical properties of fluids can be classified as follows:

• Kinematic properties: These properties refer to the understanding of the motion of fluid.

For example: The velocity and acceleration of fluids are kinematic properties.

• Thermodynamic properties: These properties help in determining the thermodynamic state of the fluid.

For example: Temperature, density, pressure, and specific enthalpy.

• Physical properties: The physical properties tell us about the physical state of any fluid, like colour and odour.

---

Very Short Answer Questions [1 Mark Questions]

Ques. Can we estimate what is the exact fractional volume of an ice cube in a system that is freely falling under gravity?

Ans. No, we cannot estimate the exact fractional volume of an ice cube in a system that is freely falling under gravity because the value of g acting downwards, in this case, is zero and the pseudo force acting upwards is zero. So the net value of g is zero which clearly means that it is a gravityless surface.

Ques. What would be the effect on barometric height if a few drops of water were added to the barometer tube?

Ans. The barometric height will come down when a few drops are added to the barometer tube. When we add the water drops it exerts downward pressure on the mercury and hence the height of the mercury column decreases.

Ques. Why are two holes made to empty an oil tin? 

Ans. The two holes are made to empty the oil tin so that the air keeps on entering the tin through the other hole and the pressure built inside the tin becomes equal to atmospheric pressure while a single hole would impact the pressure inside the tin to be less than the atmospheric pressure and soon the oil will stop flowing out.

Ques. What do you mean by 1 torr of pressure? 

Ans. The pressure exerted by a 1 mm column of mercury is known as 1 torr of pressure.

1 torr = 10^-3 m of Hg
1 torr = ρgh = 13.6 × 10³ × 9.8 × 10^-3 N/m² (Pascal)
= 133.3 N/m²

Read More:

Short Answer Questions [2 Marks Questions]

Ques. Why is the blood pressure in humans greater at the feet than at the brain?

Ans. The height of the blood column in humans is concentrated more at the feet rather than at the brain which causes more force to be applied in order to push the blood back to the heart against the gravity due to the height column. Due to less distance, the pressure on the brain is less, and more force is required to carry the blood back to the heart.

Ques. Why does density change with the change in temperature?

Ans. The change of volume with temperature is the reason for the change in density with temperature. The density is expressed as mass divided by volume.

The volume increases when we heat up something because of the fact that the molecules that move fast are further apart. Since volume is inversely proportional to density, therefore increasing the volume decreases the density.

Ques. What are the systolic and diastolic blood pressure values of a healthy human being?

Ans. Normal systolic blood pressure is 120 mm of Hg which measures the pressure in your arteries when your heart beats.. Diastolic blood pressure indicates the pressure in the arteries when the heart rests between beats. A normal diastolic blood pressure number of a healthy human being is 80 mm of Hg.

Ques. Find the height of sap in trees, which mainly consists of water. In summer, the sap rises in a system of capillaries of radius r = 2.5 × 10⁻⁵ m. Given that the surface tension of sap is T = 7.28 × 10⁻² Nm⁻¹ and the angle of contact 0°. Is the surface tension alone responsible for the supply of water to the top of all trees?

Ans. The sap will rise to a height that can be calculated as

\(\mathrm{h}=\frac{2 \mathrm{~T} \cos 0^{\circ}}{\rho \mathrm{gr}}=\frac{2 \times 7.28 \times 10^{-2}}{10^{3} \times 9.8 \times 2.5 \times 10^{-5}}=0.6 \mathrm{~m}\)

The sap can rise to this maximum height due to surface tension and since many trees are taller in height than this, therefore capillary action alone cannot account for the rise of water in all trees.

Ques. State Pascal’s law.

Ans. Pascal’s law states that the pressure in a fluid at rest is the same at all points which are at the same height. A change in pressure applied to an enclosed fluid is transmitted undiminished to every point of the fluid and the walls of the containing vessel.

Ques. Define Viscosity.

Ans. The property of the fluids by virtue of which an opposing force comes into play between different parallel layers of a liquid whenever there is a relative motion between these layers of the liquid is called Viscosity.

Read More:

Long Answer Questions [3 Marks Questions]

Ques. Given that the surface tension and vapor pressure of water at 20° C are 7.28 × 10⁻² Nm⁻¹ and 2.33 × 10³ Pa respectively. Find out the radius of the smallest spherical water droplet which can form without evaporating at 25°C

Ans. Given

• Surface tension, S = 7.28 × 10⁻² Nm⁻¹
• Vapour pressure, P = 2.33 × 10³ Pa

If the water pressure is greater than the vapor pressure P, then the drop will evaporate

We have

\(P =\frac{2S}{R}\) 

\(\Rightarrow S = \frac {2S}{P}\)

Where R is the radius of the water drop

On substituting the values, we get

R = \(\frac{2S}{P}\) = \(\frac{2 \times 7.28 \times 10^{-2}}{2.33 \times 10^3}\) = 6.25 × 10⁻⁵ m

Ques. What amount of energy is released or absorbed when two mercury droplets of radii 0.1 cm and 0.2 cm collapse into one single drop? Given that the surface tension of mercury T = 435.5 × 10⁻³ Nm⁻¹.

Ans: Let R be the radius of a big single drop formed when two small drops of radii r₁ and r₂ collapse together.

Then the volume of the big drop = volume of two small drops

Therefore, \(\frac{4}{3}\) πR³ = \(\frac{4}{3}\) πr₁³+ \(\frac{4}{3}\) πr₂³

⇒ R³ = r₁³ + r₂³

Given

• r₁ = 0.1 cm
• r₂ = 0.2 cm
• Surface tension, T = 435.5 × 10⁻³ Nm⁻¹

On substituting the values, we get
⇒ R³ = (0.1)³ + (0.2)³ = 0.001 + 0.008 = 0.009 cm³
⇒ R = 0.21 cm

Change in surface area = 4πR² − (4πr₁² + 4π₂²)

The Energy released, E = Surface Tension × change in surface area

⇒ E = T × [4πR² − (4πr₁² + 4π₂²)]

⇒ E = 4πT [R² − (r₁² + r₂²)]

⇒ E = 4 × 3.14 × 435.5 × 10⁻³ [(0.21)² − {(0.1)² + (0.2)²}] = −32.27 × 10⁻³ Joule

Ques. Write three characteristics of fluids.

Ans. The characteristics of fluids are:

• They have a very small shearing stress i.e. a little application of force along the tangents to their surfaces, bringing them in motion along the force. So they start flowing along the direction of force applied.
• Gases are highly compressible but for liquids, the external pressure causes a very small change in their volumes.
• Liquids have free surfaces of their own while gases have not.

Read More:

Very Long Answer Questions [5 Marks Questions]

Ques. A drop of liquid breaks into smaller droplets resulting in the lowering of the temperature of the droplets. Assuming that a drop of liquid of radius R breaks into N small droplets each of radius r. What will be the decrease in the temperature?

Ans. We have

ΔE = σ(Final area − initial area of the surface)

Also, ΔE = msΔt

According to the law of conservation of mass,
Volume of the bigger droplet of radius R = Number of smaller droplet x Volume of one smaller droplet of radius r

Therefore, \(\frac{4}{3}\) πR³ = N x \(\frac{4}{3}\)πr³

⇒ R³ = Nr³

⇒  \(r= \frac{R}{N^{\frac{1}{3}}}\)

ΔE = σΔA = σ[Area of N drops of radius r − Area of the big drop]

msΔt = σ[N.4πr² − 4πR²]

⇒ VρsΔt = 4πσ[Nr² − R²]

N.(\(\frac{4}{3}\) πr³)ρsΔt = 4πσ[Nr² − R²]

Where

• m = mass of smaller drops
• ρ = density of liquid
• s = specific heat of liquid
• Δt = change in temperature

\(\Delta t = \frac {4 \pi \sigma \times 3}{N \times 4 \pi r^3 \rho s}[Nr^2-R^2]\)

\(\Delta t = \frac {3 \sigma}{N \rho s}[\frac {Nr^2}{r^3} - \frac {R^2}{r^2}]\)

Where \(r^3 = \frac {R^3}{N}\)

\(\Delta t = \frac {3 \sigma N}{N \rho s}[\frac {1}{r} - \frac {1}{R}]\)

\(\Delta t = \frac {3 \sigma}{\rho s}[\frac {1}{r} - \frac {1}{R}]\)

Therefore, Δt will be positive which indicates \(\frac{1}{R}\) < \(\frac{1}{r}\)

Hence, the increase in the temperature is required for the formation of smaller drops, and The energy required is taken from the surroundings whose temperature decreases.

Ques. As one ascends the atmosphere, the pressure decreases. Given that the density of air is ρ, then what is the pressure change dp over a differential height dh?

Ans. Let us consider a part of the atmosphere of thickness dh. As we know the pressure at a point in fluid is equal in all directions, so the pressure on the upper layer is p acting downward and pressure on the lower layer is (p + dp) that is acting upward. Therefore, the force due to pressure is balanced by the Buoyant force of the air

(p + dp) A − p.A = −Vρg

⇒ p.A + dpA − pA = −A(dh)ρg

 ⇒ dp.A = −ρpg(dh)A

⇒ dp/p = −ρg(dh)

The Negative sign indicates that the pressure decreases with the increase in height.

The underlying assumption that limits this model which works for relatively small distances is that the temperature T remains constant only near the surface of the earth and not at a greater height.

Read More:

Ques. The free surface of the oil in a tanker which is at rest is horizontal. If the free surface is tilted by an angle θ and the acceleration of the tanker is α ms⁻¹, what will be the slope of the free surface?

Ans. Let us consider the tanker is being pulled by force F which produces acceleration in the truck in the forward direction. Now let us consider only a small element of mass δm at P.

When a tanker is being pulled by forward acceleration then this element of mass also experiences the force in the forward direction. Due to the inertia of rest, the tanker tries to remain at rest. Due to the same acceleration that was generated, it moves in a backward direction as it is free and not rigidly connected to the tanker. Hence, all the forces acting on mass δ are F₁ = δma in a horizontally backward direction due to the tanker's acceleration a, F₂ = δmg vertically downward due to gravity.

We resolve the components of F₁ and F₂ along a perpendicular to the inclined surface of the oil. The Normal is balanced by the component δma sinθ of F₁ and when the surface is inclined at the maximum angle, then

δma cosθ = δmg sinθ

The required slope is \(\frac{\cos \theta} {\sin \theta} = \frac {g}{a}\)

Therefore, \(tan \theta = \frac{a}{g}\) is the required slope.

Ques. Six raindrops each of radius 1.5 mm, come down with a terminal velocity of 6 cm s^-1. They coalesce to form bigger drops. What is the terminal velocity of a bigger drop?

Ans. Given,

• Terminal velocity of each drop, v_l = 6 cm s^-1 = 6 x 10^-2 ms^-1
• Radius of each small drop, r = 1.5 mm = 1.5 x 10^-3 m

We have, terminal velocity given by

\(v_t=\frac{2}{9} \frac{r^2}{\nu}(\rho-\sigma )g\)

\(\Rightarrow 6 \times 10^{-2}=\frac{2}{9} \frac{(1.5 \times 10^{-3})^2}{\nu}(\rho-\sigma )g\)

\(\Rightarrow \frac {(\rho - \sigma)g}{\nu}=\frac {6 \times 10^{-2} \times 9}{2(1.5 \times 10^{-3})^{2}}\)

Now, volume of six drops = 6 x 4/3 πr³

Let the volume of bigger drops = 4/3 πR³

⇒ 6 x 4/3 πr³ = 4/3 πR³

⇒ R³ = 6r³

⇒ R = 6^1/3r

⇒ R = 1.818 x 1.5 x 10^-3 = 2.727 x 10^-3 m

Now, the terminal velocity of the bigger drop is given by

\(v'_t=\frac{2}{9} \frac{R^2}{\nu}(\rho-\sigma )g=\frac{2R^2}{9} \frac {(\rho - \sigma)g}{\nu}\)

\(\Rightarrow v'_t= \frac {2 \times(2.727 \times 10^{-3})^2}{9} \frac {6 \times 10^{-2} \times 9}{2(1.5 \times 10^{-3})^{2}}\)

\(\Rightarrow v'_t=19.84 \times 10^{-2}m/s =19.84\: cm\,s^{-1}\)

---

Also Read: