Chapter 9: Mechanical Properties of Fluids

Ans: The pressure of a liquid with density\[\rho \], with the liquid column of \[h\] is given by the relation:
\[P=\text{ }\rho hg\]
here \[g\] is the acceleration due to the gravity 

It can be inferred that pressure is directly proportional to height. In the case of humans, the circulatory system can be considered as the liquid (blood)-column. 

The height of the column is the least at the head level and the maximum at the feet; hence, the pressure (blood pressure) in human vessels depends on this height. Thus, the blood pressure at the feet is more than what it is at the brain.

Ans: The density of a fluid depends on how much it has been pressed by the amount of fluid over it. This means that the density of air is the maximum near the sea level. 

The density of air increases with a decrease in the height from the sea surface. At the height of about 6 km, the total mass squeezes the layer of air over here to nearly half of the pressure value at sea level. 

In the case of just small changes in altitude, the atmospheric pressure is proportional to density. However, when larger height scales are considered, the density itself depends on the height, and the pressure isn't linearly dependent anymore. 

Moreover, the pressure increases faster than the linear dependence expected in the case of small changes in altitudes.

Ans: Mathematically, pressure is the perpendicular force per unit area. 

So, only the force component that is perpendicular to the surface or along the area vector is used. The direction of this component and the direction of the area vector are the same.

Hence effectively, there is no direction involved anymore. 

In other words, when force is applied to a liquid, the pressure in the liquid is transferred in all directions. Thus, the hydrostatic pressure does not have a fixed direction, and it is a scalar physical quantity.

Ans: The angle between the surface inside the liquid and the tangent to the liquid surface at the point of contact is the angle of contact\[(\theta )\], as shown in the following figure

Let \[{{S}_{la}}\] be the interfacial tension at the liquid-air interface, \[{{S}_{sl}}\] be at the solid-liquid interface, and \[{{S}_{sa}}\] be at the solid-air interface. At all the points of contact, the surface forces between the three media must be in equilibrium, i.e.,

\[\cos \theta =\frac{{{S}_{sa}}-{{S}_{sl}}}{{{S}_{la}}}\]

In the case of a mercury drop on glass, \[{{S}_{sl}}>{{S}_{sa}}\], the angle of contact \[(\theta )\] is obtuse. 

In the case of water on the glass,\[{{S}_{sl}}<{{S}_{la}}\], the angle of contact\[(\theta )\] is acute.

Ans: The Mercury molecules have a strong force of attraction between themselves and a weak force of attraction toward solids. Hence, the angle of contact tends to become obtuse, and the molecules come closer and tend to form a drop.
While in the case of water, the ratio of attraction between water molecules and the attraction between water molecules and the glass is lower. Hence the water molecules stay closer to glass molecules as well, keeping the glass surface wet. 

Ans: Surface tension can be defined as the force acting per unit length at the interface between the liquid and any other material. 

There is no dependency of this force on the area of the surface. 

Hence, surface tension is independent of the area of the surface of the liquid.

Ans: Detergent water has small angles of contact \[(\theta )\] because the detergent molecules are sticky; in other words, they are significantly attracted to the water molecules and the solid which is in contact with the detergent water. 

Therefore, the molecules of water can get closer to the solid. 

In other words, \[{{S}_{sl}}<{{S}_{la}}\] and the value of \[\cos \theta =\frac{{{S}_{sa}}-{{S}_{sl}}}{{{S}_{la}}}\] increases, leading to a decrease in the value of \[\theta \]making it acute or small.

(e) A drop of liquid under no external forces is always spherical in shape

Ans: Surface tension pulls the surface together as much as possible. 

This means that in the absence of any other external force/pressure. The surface tension will try to shrink the area as much as possible. 

Beyond a certain extent, the volume cannot be shrunk, and the shape with minimum surface area for the unchanged volume is a sphere. 

Hence a droplet will tend to become spherical.

Ans: decreases. 

The surface tension of a liquid surface is inversely proportional to the motion of the liquid molecules. An increase in the temperature leads to an increase in the random motion of the molecules, and hence the surface tension drops.

Ans: increases; decreases

All fluids experience resistance to their motion when they are made to flow. This resistance is known as viscosity. 

The motion of the gas molecules becomes very random, and giving them a direction gets difficult. 

The viscosity of gases increases with the increasing random motion of the molecules. The molecules of the liquid are more restricted than and not as free as those of gas. Increasing the temperature of liquids increases the motion of the molecules, making them more mobile and improves their ability to flow.

Hence the viscosity decreases with an increase in temperature.

Ans: Shear strain; Rate of shear strain 

For elastic solids, shearing force is proportional to the shear strain.
However, the shearing force is proportional to the rate of shear strain in the case of fluids.

Ans: Conservation of mass/Bernoulli’s principle

Bernoulli's principle works in the conservation of mass and energy, which includes the effects of the continuity equation.  Hence, Conservation of mass and Bernoulli's Equation is the correct options

(e) For the model of a plane in a wind tunnel, turbulence occurs at a …... speed for turbulence for an actual plane (greater / smaller)

Ans: Greater 

As provided in the question, there is a model of a plane in a wind tunnel. Following Bernoulli's Principle and referring to the relevant Reynolds number it can be seen that the turbulence occurs at a greater speed.

Ans: As per Bernoulli's principle, the fluid pressure reduces if the velocity is increased. 

The paper falls under gravity, and to make it stay up and horizontal, we need to reduce the pressure on the top of it. For this reason. one should blow on the upper side of the paper.

Ans: According to the continuity equation, for a flowing fluid, the rate of volume of fluid crossing any cross-section is constant. This means that the velocity of the flowing fluid is high if the cross-section of the passage is reduced.  

Covering the tap opening reduces the cross-section, and hence the water gushes out with a lot of speed.

Ans: The liquid velocity at the needle is much higher than that inside the syringe. And higher velocity means much lower pressure. 

Furthermore, the changes in the pressures at the different cross-sections are proportional to the pressures in the regions.
A human thumb cannot constantly provide constant pressure.  So the significant change in the pressures is carried to the needle end as well. The pressure over here changes proportionally. 

But as mentioned before, the changes in the pressure at the needle end are proportional to the pressure at the needle end, which is very small. 

Hence the absolute change in the pressure at the needle end is also minimal. Thus it can be said that the consistency of the pressure is controlled by the cross-section of the needle.

Ans: As the fluid flows out of a small orifice, its velocity is significantly high according to the continuity equation. 

This outgoing liquid generates momentum.

Going by the law of conservation of momentum, a counter momentum is generated on the vessel, making it experience a backward thrust.

(e) A spinning cricket ball in air does not follow a parabolic trajectory

Ans: A thrown spinning cricket ball has two motions – rotational and linear.  The falling ball experiences viscosity in the upwards direction. 

A rotating ball experiences viscosity due to the relative motion of the surface of the ball and the air in the vicinity. So at any instance, the face that is moving downwards faces an upwards viscosity force while the face going upwards faces a downwards viscosity force. 

These viscous forces add up in such a way that the velocity of the air on one side of the ball is much higher than on their side, reducing the pressure on the side which has a higher velocity. 

This pressure imbalance causes a force directed towards the low-pressure region. Hence the ball deviates from the expected parabolic trajectory.

Ans: As provided, the mass of the girl, \[m\text{ }=\text{ }50\text{ }kg\], the diameter of the heel \[d\text{ }=\text{ }1\text{ }cm\text{ }=\text{ }0.01\text{ }m\] , the radius of the heel, \[r=\frac{d}{2}=0.005m\]

Therefore, the area of the heel, \[A=\pi {{r}^{2}}\]

\[\Rightarrow A=\pi \times {{0.005}^{2}}\]

\[\Rightarrow A=7.85\times {{10}^{-5}}{{m}^{2}}\]

The force exerted by the heel on the floor is:

\[~F\text{ }=\text{ }mg\text{ }=\text{ }50\text{ }\times \text{ }9.8\text{ }=490\text{ }N\]

The pressure exerted on the floor by the heel:

\[P=\frac{F}{A}=\frac{490}{7.85\times {{10}^{-5}}}=6.24\times {{10}^{6}}N{{m}^{-2}}\]

Hence clearly, the pressure exerted by the heel on the horizontal floor is \[6.24\times {{10}^{6}}N{{m}^{-2}}\].

Ans: Provided that,

Density of mercury, \[{{\rho }_{1}}=13.6\times {{10}^{3}}kg/{{m}^{3}}\]

The height of the mercury column, 

\[{{h}_{1}}=0.76m\]

Provided density of French wine, 

\[{{\rho }_{2}}=0.984\times {{10}^{3}}kg/{{m}^{3}}\]

Let the height of the column of wine \[={{h}_{2}}\]

The acceleration due to gravity\[=g=9.8m{{s}^{-2}}\] 

The pressure in both the columns (given by\[\rho hg\]) is equal at equal altitudes of the same liquid, which means:

\[{{\rho }_{1}}{{h}_{1}}{{g}_{1}}={{\rho }_{2}}{{h}_{2}}{{g}_{2}}\]

\[\Rightarrow {{h}_{2}}=\frac{{{\rho }_{1}}{{h}_{1}}g}{{{\rho }_{2}}g}=10.5m\]

 Thus, the height of the column of wine at normal atmospheric pressure is \[10.5m\].

Ans: Yes 

As provided, the maximum allowable stress for the structure,  \[P={{10}^{9}}Pa\]

Depth of the ocean, \[d=3\times {{10}^{3}}m\]

The density of water, \[\rho ={{10}^{3}}kg/{{m}^{3}}\]

The acceleration due to gravity, \[g=9.8m/{{s}^{2}}\]

 The pressure exerted because of water at depth,

\[P=\rho dg=3\times {{10}^{3}}\times {{10}^{3}}\times 9.8=2.94\times {{10}^{7}}Pa\]

Hence it has been calculated that the maximum allowable stress for the structure is significantly greater than the pressure of the water and hence can be used safely.

Ans: As provided, the mass limit on the mass of a car that can be lifted, \[m\text{ }=\text{ }3000\text{ }kg\] 

The area of cross-section of the load-carrying piston, \[A=0.0425{{m}^{2}}\] 

The maximum force exerted by the load, \[F\text{ }=\text{ }mg\text{ }=\text{ }3000\text{ }\times \text{ }9.8\text{ }=\text{ }29400\text{ }N\] 

The maximum pressure experienced by the piston that is carrying the load is, \[P=\frac{F}{A}=\frac{29400}{425\times {{10}^{-4}}}\text{= }6.917\text{ }\times {{10}^{-4}}\text{ }Pa\] 

Since the pressure is transferred/transmitted equally in all directions in a liquid, the maximum pressure on the smaller piston would be\[\text{ }6.917\text{ }\times {{10}^{-4}}\text{ }Pa\].

Ans: The provided system of water, mercury, and methylated spirit can be shown as follows:

The height of the spirit column, \[{{h}_{1}}=12.5cm=0.125m\]

The height of the water column is: \[{{h}_{2}}=10cm=0.1m\] 

\[{{P}_{0}}=\] Atmospheric pressure, \[{{\rho }_{1}}=\] Density of spirit, \[{{\rho }_{2}}=\]Density of water 

Pressure at point A, \[{{P}_{0}}+{{\rho }_{1}}{{h}_{1}}g\]

Pressure at point B, \[{{P}_{0}}+{{\rho }_{2}}{{h}_{2}}g\] 

The values of pressures at points A and B are equal

\[{{P}_{0}}+{{\rho }_{1}}{{h}_{1}}g={{P}_{0}}+{{\rho }_{2}}{{h}_{2}}g\]

\[\Rightarrow \frac{{{\rho }_{1}}}{{{\rho }_{2}}}=\frac{{{h}_{2}}}{{{h}_{1}}}=0.8\]

Hence, clearly, the specific gravity of spirit is \[0.8\].

Ans: Provided that,

Height of the water column, \[{{h}_{1}}=10+15=25cm\] 

Height of the spirit column,  \[{{h}_{2}}=12.5+15=27.5cm\]

The density of water, \[{{\rho }_{1}}=1\,g\,c{{m}^{-3}}\] ; that of spirit, \[{{\rho }_{2}}=0.8g\,c{{m}^{-3}}\] and that of mercury is \[13.6\,g\,c{{m}^{-3}}\] 

Now, let \[h\] be the difference between the levels of mercury in the two arms. The pressure exerted by the height \[h\] of the mercury column: \[=\rho hg=h\times 13.6\] 

The difference between the pressures exerted by water and spirit: \[{{\rho }_{1}}{{h}_{1}}g-{{\rho }_{2}}{{h}_{2}}g=g\left( 25\times 1\text{ }\text{ }27.5\times 0.8 \right)=3g\]

Equating the pressures, we get: 

\[13.6\text{ }hg\text{ }=\text{ }3g\text{ }\]

\[\Rightarrow h\text{ }=\text{ }0.220588\text{ }\approx \text{ }0.221\text{ }cm\]

The difference between the levels of mercury in the two arms is \[0.221\text{ }cm\].