Chapter 9: Mechanical Properties of Fluids

Ans: Provided that,

Base area of the given tank,\[A=1.0{{m}^{2}}\]

Area of the hinged door, \[a=0.02{{m}^{2}}\]

The density of water, \[{{\rho }_{1}}=1\,{{0}^{3}}kg\,{{m}^{-3}}\]

The density of acid, \[{{\rho }_{2}}=1.7\times 1\,{{0}^{3}}kg\,{{m}^{-3}}\]

Height of the water column, \[{{h}_{1}}=4m\]

Height of the acid column, \[{{h}_{2}}=4m\]

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

Now, the pressure due to water is given as: \[{{P}_{1}}={{\rho }_{1}}{{h}_{1}}g=4\times 1\,{{0}^{3}}\times 9.8=3.92\times {{10}^{4}}Pa\]

The pressure due to acid is given as: \[{{P}_{2}}={{\rho }_{2}}{{h}_{2}}g=4\times 1.7\times 1\,{{0}^{3}}\times 9.8=6.662\times {{10}^{4}}Pa\]

 The pressure difference between the two columns: \[\Delta P={{P}_{2}}-{{P}_{1}}=2.744\times 1\,{{0}^{4}}Pa\]

Thus, the force exerted on the door \[F=P\times a=54.88N\]

It can be concluded that the force necessary to keep the door closed is \[54.88N\].

Ans: In the first case i.e., fig (a)

Atmospheric pressure, \[{{P}_{0}}=76cm\,Hg\]

 The pressure (or height) difference between the levels of mercury in the two limbs denotes the gauge pressure. 

Hence, the gauge pressure is 20 cm of Hg. 

\[{{P}_{absolute}}\text{ }=\text{ }{{P}_{0}}\text{ }+\text{ }{{P}_{Gauge}}\text{ }=\text{ }76\text{ }+\text{ }20\text{ }=\text{ }96\text{ }cm\text{ }Hg\]

 In the second figure (figure (b)), the difference between the levels of mercury in the two limbs \[=\text{ }18\text{ }cm\] Hence, gauge pressure is \[=\text{ }18\text{ }cm\] Hg. 

Again, the \[{{P}_{absolute}}\text{ }=\text{ }{{P}_{0}}\text{ }+\text{ }{{P}_{Gauge}}\text{ }=\text{ }76\text{ + (-18)  = 58 cm Hg}\]

Ans: Now, \[13.6\,cm\] water is filled into the right limb of figure (b). 

The relative density of mercury \[13.6\,cm\]

In other words, a column of \[13.6\,cm\] water is equivalent to \[1\,cm\] mercury. 

Consider \[h\] being the difference in the mercury levels in the two limbs. 

The pressure experienced in the right limb is given as: 

\[{{P}_{right}}\text{ }=\text{ }{{P}_{0}}\text{ }+\text{ }1cm\,Hg\text{ }=\text{ }76\text{ + 1  = 77 cm Hg}\]

The level of the mercury column will go up in the left limb. 

Hence, pressure in the left limb, \[{{P}_{L}}=58+h\] 

Equating the pressures in both the limbs and simplifying for h, we get: 

\[77=58+h\] 

$\Rightarrow h=19cm$
It can be hence concluded that the mercury levels differ by \[19cm\]. 

Ans: Yes, 

Pressure in a liquid is given as:

\[P=\rho hg\].

The heights up to which the vessels are filled are equal, the densities are equal, and g is the same for both cases. Hence the pressures are also the same

Now, let the force that is experienced at the base of an area \[A\] be \[F=P\times A\]

Therefore, the forces in both cases are equal.

Ans: The given gauge pressure, \[P\text{ }=\text{ }2000\text{ }Pa\] 

The density of whole blood,  \[\rho =1.06\times 1\,{{0}^{3}}kg\,{{m}^{-3}}\] 

Let \[h\] be the height of the blood container

The pressure inside the blood container is \[P=\rho hg\]

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

\[\Rightarrow h=\frac{P}{g\rho }=\frac{2000}{9.8\times 1.06\times 1\,{{0}^{3}}}=0.1925m\]

This means that the blood would enter the vein only if the container is kept at a height of \[0.1925m\]higher than the height at which the needle is inserted.

Ans: The diameter of the given artery, \[d=2\times {{10}^{-3}}m\]

The viscosity of blood, \[\eta =2.084\times {{10}^{-3}}Pa\,\,s\]

The density of blood, \[\rho =1.06\times {{10}^{3}}kg/{{m}^{2}}\]

 Reynolds' number for laminar flow, \[{{N}_{R}}\text{ }=\text{ }2000\]

The maximum average velocity of blood flowing through the d-diameter orifice is as follows: 

\[V=\frac{{{N}_{R}}\eta }{\rho d}=\frac{2000\times 2.084\times {{10}^{-3}}}{(1.06\times {{10}^{3}})(2\times {{10}^{-3}})}=1.966m/s\]

Therefore, the largest average velocity of blood is \[1.966m/s\].

Ans: The dissipative forces become more prominent with an increase in the fluid velocity.  As the velocity increases, the turbulence also increases. The turbulent nature of the flow causes dissipative loss in a fluid.

Ans: Provided that,

Radius of the artery, \[r=2\times {{10}^{-3}}m\]. 

Diameter of the artery, \[d=2(2\times {{10}^{-3}}m)\]

The viscosity of blood, \[\eta =2.084\times {{10}^{-3}}Pa\,\,s\]

The density of blood, \[\rho =1.06\times {{10}^{3}}kg/{{m}^{2}}\]

The Reynolds number of Laminar flow, \[{{N}_{R}}\text{ }=\text{ }2000\]

The largest value of average velocity can be given as:

\[V=\frac{{{N}_{R}}\eta }{\rho d}=\frac{2000\times 2.084\times {{10}^{-3}}}{(1.06\times {{10}^{3}})(4\times {{10}^{-3}})}=0.983m/s\]

Hence the largest average velocity of blood flow in an artery of the radius\[2\times {{10}^{-3}}m~\] is \[0.983m/s\].

Ans: The flow rate is written as the time derivative of the volume that flowed past a cross-section: 

\[R=\pi {{r}^{2}}V{}_{Avg}=3.14\,{{(2\times {{10}^{-3}})}^{2}}\,(9.83)=1.235\times {{10}^{-5}}{{m}^{3}}/s\]

Hence the corresponding flow rate is \[1.235\times {{10}^{-5}}{{m}^{3}}/s\].

Ans: As provided in the question, the area of the wings is \[A=2\times 25=50{{m}^{^{2}}}\]

Speed of air over the lower wing, \[{{v}_{1}}=180km/h=50m/s\]

Speed of air over the upper wing, \[{{v}_{2}}=234km/h=65m/s\] 

The density of air, \[\rho =1\,kg/{{m}^{3}}\]

The pressure of air over the lower wing\[={{P}_{1}}\] 

The pressure of air over the upper wing \[={{P}_{2}}\]

The upward force on the plane can be obtained using Bernoulli's equation as: \[{{P}_{1}}+\frac{1}{2}\rho {{v}_{1}}^{2}={{P}_{2}}+\frac{1}{2}\rho {{v}_{2}}^{2}\]

Thus, the upward force is obtained as:

\[Lift\,=({{P}_{1}}-{{P}_{2}})A=\left( \frac{1}{2}\rho {{v}_{2}}^{2}-\frac{1}{2}\rho {{v}_{1}}^{2} \right)A=43125N\]

The mass of the plane can be obtained as: 

\[m=F/g=4400.5kg\]

Hence, the mass of the plane is nearly \[4400.5kg\].

Ans: The given terminal speed \[=5.8cm/s\,\], the viscous force \[\,\,=3.9\times {{10}^{-10}}N\]

The radius of the given uncharged drop, \[r=2.0\times {{10}^{-5}}m\]

The density of the uncharged drop, \[\rho =1.2\times 1\,{{0}^{3}}kg\,{{m}^{-3}}\]

It is known that the viscosity of air, \[\eta =1.8\times {{10}^{-5}}\,Pa\,s\]

In this case, the density of air (\[{{\rho }_{0}}\] ) can be taken as zero to neglect buoyancy of air. Acceleration due to gravity is \[g=9.8m/{{s}^{2}}\] 

Terminal velocity (v) is given as: 

\[v=\frac{2{{r}^{2}}(\rho -{{\rho }_{0}})g}{9\eta }=\frac{2{{(2.0\times {{10}^{-5}})}^{2}}(1.2\times 1\,{{0}^{3}}-0)9.8}{91.8\times {{10}^{-5}}}=5.8\times {{10}^{-2}}m/s\]

Hence, the terminal velocity of the drop is \[5.8cm/s\]

The viscous force experienced by the drop is given by:

\[F=6\pi \eta rv\]

\[\Rightarrow F=3.9\times {{10}^{-10}}N\]

Hence the viscous force is calculated to be \[3.9\times {{10}^{-10}}N\].

Ans: As provided in the question, the angle of contact between mercury and soda-lime glass, \[\theta =140{}^\circ \]

The radius of the narrow tube \[r=1.0\times {{10}^{-3}}m\]

The given surface tension of mercury, \[S=0465N/m\] 

The density of mercury, \[\rho =13.6\times 1\,{{0}^{3}}kg\,{{m}^{-3}}\]

Dip in the height of the mercury \[=h\]

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

The surface tension can be written in terms of the angle of contact and the dip in height:

\[S=\frac{h\rho gr}{2\cos \theta }\]

\[\Rightarrow h=\frac{2\cos \theta }{S\rho gr}=-0.00534m=-5.34mm\]

Notice the negative sign; it shows the dropping level of mercury. 

Hence, the mercury level drops by 5.34 mm.

Ans: Provided that,

Diameter of the first bore, \[{{d}_{1}}=3.0\times {{10}^{-3}}m\]

Hence the radius of the first bore, \[{{r}_{1}}={{d}_{1}}/2=1.5\times {{10}^{-3}}m\] 

Diameter of the second bore, \[{{d}_{2}}=6.0\times {{10}^{-3}}m\]

Hence, the radius of the second bore,  \[{{r}_{2}}={{d}_{2}}/2=3.0\times {{10}^{-3}}m\]

The surface tension of water, \[S=7.3\times {{10}^{-2}}N/m\]

The angle of contact between the water surface and the bore surface, \[\theta =0{}^\circ \]

The density of water, \[\rho =1.0\times 1\,{{0}^{3}}kg\,{{m}^{-3}}\]

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

Let\[{{h}_{1}}\]and \[{{h}_{2}}\] be the heights to which water rises in the first and second tubes, respectively. These are given as: 

\[{{h}_{1}}=\frac{2S\cos \theta }{\rho g{{r}_{1}}}\]and \[{{h}_{2}}=\frac{2S\cos \theta }{\rho g{{r}_{2}}}\]

The difference in the water levels in the two limbs of the tube can be calculated as: 

\[\Delta h=\frac{2S\cos \theta }{\rho g{{r}_{1}}}-\frac{2S\cos \theta }{\rho g{{r}_{2}}}=\frac{2\times 7.3\times {{10}^{-2}}\times \cos 0}{1.0\times 1\,{{0}^{3}}\times 9.8}=4.97mm\]

Hence, the difference between the levels is obtained as \[=4.97mm\].