Important Questions for CBSE Class 11 Physics Chapter 9 - Mechanical Properties of Fluids

1 Mark Questions

1. State the law of floatation. 

Ans: Law of floatation states that a body would float in a liquid, if weight of the liquid that is displaced by the immersed part of the body is at least equal to or greater than the weight of the body.

2. Is the blood pressure of humans greater at the feet than at the brain?

Ans: It is quite obvious that the height of the blood column in the human body is more at the feet than at the brain. Since, we know that, pressure is directly dependent on the height of the liquid column, so pressure would be more at feet than that at the brain.

3. Define surface tension. 

Ans: Surface tension is measured as the force that is acting on a unit length of a line that is imagined to be drawn tangentially anywhere on the free surface of the liquid that is at rest.

4. Does Archimedes principle hold in a vessel in a free fall?

Ans: Archimedes Principle will not hold in a vessel in free fall. This is because, in this case, acceleration due to gravity is found to be zero and hence buoyant force will not exist in this case.

5. Oil is sprinkled on sea waves to calm them. Why?

Ans: We know that the surface tension of sea-water without oil is greater than oily water. So, the water without oil pulls the oily water against the direction of the breeze and hence, the sea waves would calm down.

6. A drop of oil placed on the surface of water spreads out, but a drop of water placed on oil contracts. Why?

Ans: Since the cohesive forces between the oil molecules are known to be less than the adhesive force that exists between the oil molecules, the drop of oil spreads out and would thus reverse holds for drop of water.

7. Water rises in a capillary tube but mercury falls in the same tube. Why?

Ans: The capillary rise could be given by:

$h=\frac{2T\cos \theta }{r\rho g}$

Where, 

$h=$height of capillary

$T=$Surface tension

$\theta =$Angle of contact

$r=$Radius of capillary

$\rho =$Density of liquid

$g=$Acceleration due to gravity

Now, for mercury–glass surface, $\theta $ is obtuse, hence $\cos \theta $ would be negative so will be the value of h.

Hence, we could conclude that mercury will depress below the level of surrounding liquid.

8. The diameter of ball A is half that of ball B. What will be their ratio of their terminal velocities in water?

Ans: We know that the terminal velocity is directly proportional to the square of radius of the ball, 

$v\propto {{r}^{2}}$

Now, the ratio of the terminal velocities of the two balls would be, 

$\frac{{{v}_{A}}}{{{v}_{B}}}=\frac{{{r}_{A}}^{2}}{{{r}_{B}}^{2}}$

But we are given, 

${{D}_{A}}=\frac{1}{2}{{D}_{B}}$

$\Rightarrow {{r}_{A}}=\frac{1}{2}{{r}_{B}}$

The ratio now becomes, 

$\frac{{{v}_{A}}}{{{v}_{B}}}=\frac{{{r}_{A}}^{2}}{{{r}_{B}}^{2}}=\frac{{{\left( \frac{1}{2}{{r}_{B}} \right)}^{2}}}{{{r}_{B}}^{2}}$

$\therefore \frac{{{v}_{A}}}{{{v}_{B}}}=\frac{1}{4}$

Hence, the ratio of terminal velocities is found to be 1:4.

9. Find out the dimensions of co-efficient of viscosity. 

Ans: We have the expression for viscous force as, 

$f=\eta A\frac{dv}{dx}$

Where, 

$A=area$

$\frac{dv}{dx}=\text{velocity gradient}$

$\eta =coefficient\text{ of viscosity}$

$\Rightarrow \eta =\frac{f}{A\frac{dv}{dx}}$

Now, substituting the respective dimensions of physical quantities into the above equation, we get, 

$\left[ \eta  \right]=\frac{\left[ ML{{T}^{-2}} \right]}{{{\left[ L \right]}^{2}}{{\left[ T \right]}^{-1}}}$

$\Rightarrow \left[ \eta  \right]=\left[ M{{L}^{-1}}{{T}^{-1}} \right]$

Therefore, we found the dimension of the coefficient of viscosity to be$\left[ \eta  \right]=\left[ M{{L}^{-1}}{{T}^{-1}} \right]$.

10. Define viscosity. 

Ans: Viscosity could be defined as the property of a fluid by virtue of which an internal frictional force comes into play when the fluid is in motion and would oppose the relative motion of its different layers.

11.What is the significance of Reynolds’s Number?

Ans: We know that, Reynolds’s Number $\left( {{N}_{R}} \right)$could be given by, 

${{N}_{R}}=\frac{\rho D{{v}_{c}}}{\eta }$

Where, 

$\rho =$ Density of liquid

$D=$Diameter of tube

${{v}_{c}}=$Critical velocity

$\eta =$Co-efficient of viscosity

Now, if the value of ${{N}_{R}}$ lies between 0 to 2000, the flow of liquid would be streamlined and if its value lies above 3000, the flow of liquid would be turbulent.

12. Give two areas where Bernoulli’s theorem is applied.

Ans: The two well-known areas where Bernoulli’s theorem is applied are: in atomizer and in lifting of an aeroplane wing.

13.What is conserved in Bernoulli’s theorem?

Ans: As per Bernoulli’s theorem, for an incompressible non–viscous liquid (fluid) that is undergoing steady flow the total energy of liquid at all points would be constant.

14. If the rate of flow of liquid through a horizontal pipe of length l and radius R is Q. What is the rate of flow of liquid if length and radius of tube is doubled?

Ans: We have the Poiseuille’s formula for the rate of flow of liquid through a tube of radius R is and length l as:

$Q=\frac{\pi \rho {{R}^{4}}}{8\eta l}$…………………………….. (1)

If R and l are doubled then the rate of flow $Q'$ would be, 

$Q'=\frac{\pi \rho {{\left( 2R \right)}^{4}}}{8\eta \left( 2l \right)}=\frac{16\pi \rho {{R}^{4}}}{2\left( 8\eta l \right)}$

Comparing with (1),

$\therefore Q'=8Q$

Therefore, we found the rate of flow to become 8 times under the given conditions.

15. Water is coming out of a hole made in the wall of tank filled with fresh water. If the size of the hole is increased, will the velocity of efflux change?

Ans: Velocity of efflux could be given by the following formula,

$V=\sqrt{2gh}$ 

Thus, we find that the velocity of efflux is independent of the area of the hole. Hence, the velocity of efflux would remain the same.

16. The accumulation of snow on an aeroplane wing may reduce the lift. Explain. 

Ans: Due to the accumulation of snow on the wings of the aeroplane, the structure of wings would change and will no longer remain as that of aerofoil. This would further cause the net upward force (i.e., lift) to be decreased.

17. The antiseptics used for cuts and wounds in human flesh have low surface tension. Why?

Ans: When the surface tension of antiseptics is less, they would spread more on cuts and wounds. As a result, the cut or wound is healed quickly.

18. Why should detergents have small angles of contact?

Ans: Capillary rise could be given by, 

$h=\frac{2T\cos \theta }{r\rho g}$

i.e., h is directly proportional to the cosine of angle $\theta $(Angle of contact)

Now, when $\theta $is small then $\cos \theta $ is large so will be the capillary rise. This further implies that detergents should have a smaller angle of contact for the detergent to penetrate more in the cloth and clean better.

19. Can Bernoulli's equation be used to describe the flow of water through a rapid in a river? Explain.

Ans: No, Bernoulli's equation cannot be used for describing the flow of water through a rapid in a river. This is because, in the mentioned case, the flow of water is turbulent and Bernoulli’s principle can only be applied to a streamline

flow.

20. Does it matter if one uses gauge instead of absolute pressures in applying Bernoulli's equation? Explain.

Ans: No, it doesn’t matter if one uses gauge pressure instead of absolute pressure while applying Bernoulli's equation. But the two points where Bernoulli's equation is applied should have atmospheric pressures that are significantly different.

21. Two vessels have the same base area but different shapes. The first vessel takes twice the volume of water that the second vessel requires to fill up to a particular common height. Is the force exerted by the water on the base of the vessel the same in the two cases? If so, why do the vessels filled with water to that same height give different readings on a weighing scale?

Ans: Yes, the two vessels having the same base area would definitely have identical force and equal pressure acting on their common base area. Since the shapes of the two vessels are given to be different, the force exerted on the sides of the vessels will have non-zero vertical components. When these vertical components are added, the total force on one vessel would come out to be greater than that on the other vessel. Hence, when these vessels are filled with water up to the same height, they would give different readings on a weighing scale.

2 Marks Questions

1. State the angle of contact and on what values the angle of contact depends.

Ans: Angle of contact between a liquid and a solid could be defined as the angle enclosed between the tangents to the liquid surface and the solid surface inside the liquid, both the tangents being drawn at the point of contact of liquid with the solid. Angle of contact is known to depend upon: 

1. Upon nature of liquid and solid in contact

2. The Medium which exists above the free surface of liquid.
   
   

2. Hydrostatic pressure is a scalar quantity even though pressure is force divided by area, and force is a vector. Explain.

Ans: When force is applied on liquid, the pressure is transmitted equally in all directions inside the liquid. As there is no fixed direction for the pressure here, we could classify it as a scalar quantity.

3. Find the work done in blowing a soap bubble of surface tension 0.06 N/m from 2cm radius to 5cm radius. 

Ans: We are given, Surface tension, \[=s=0.06N/m\]

\[{{r}_{1}}=2cm=0.02m\]

\[{{r}_{2}}=5cm=0.05m\]

Recall that a bubble has two surfaces, then the initial surface area of the bubble,  ${{A}_{1}}=2x4\pi \text{ }{{\text{r}}_{1}}^{2}=2\times 4\pi {{\left( 0.02 \right)}^{2}}=\text{ }32\pi \times {{10}^{-4}}{{m}^{2}}$

Final surface area of the bubble, 

${{A}_{2}}=2\times 4\pi {{r}_{2}}^{2}=2\times 4\pi {{\left( 0.05 \right)}^{2}}=\text{ }200\pi \times {{10}^{-4}}{{m}^{2}}$

Now, increase in surface area can be given by, \[\Delta A={{A}_{2}}-{{A}_{1}}=200\pi \times {{10}^{-4}}-32\pi \times {{10}^{-4}}=\text{ }168\pi \times {{10}^{-4}}{{m}^{2}}\]

Now we have, \[work\text{ }done\text{ }=\text{ }surface\text{ }tension\times Increase\text{ }in\text{ }surface\text{ }area\]

\[\therefore W=0.06\times 168\pi \times {{10}^{-4}}=0.003168J\]

Therefore, we found that, \[Work\text{ }done=0.003168J\].

4. Why does not the pressure of atmosphere break windows?

Ans: We know that the pressure of the atmosphere does not break windows. This is because atmospheric pressure is exerted equally on both sides of a window as a result of which no net force is exerted on the window and hence, uniform pressure does not break the window.

5. If a big drop of radius R is formed by 1000 small droplets of water, then find the radius of small drop.

Ans: Let r be the radius of small drop and R the radius of big drop. 

Now, Let $\rho =$Density of water

We know that, \[mass\text{ }of\text{ }1000\text{ }small\text{ }droplets=m=1000\times volume\times Density\]

$m=1000\times \frac{4}{3}\pi {{r}^{3}}\rho $

Also, 

\[Mass\text{ }of\text{ }Big\text{ }drop=M=\frac{4}{3}\pi {{R}^{3}}\rho \]

Now, we have, \[Mass\text{ }of\text{ }1000\text{ }small\text{ }droplets\text{ }=\text{ }Mass\text{ }of\text{ }Big\text{ }drop\]

$\Rightarrow 1000\times \frac{4}{3}\pi {{r}^{3}}\rho =\frac{4}{3}\pi {{R}^{3}}\rho $

$\Rightarrow 1000{{r}^{3}}={{R}^{3}}$

$\Rightarrow {{r}^{3}}=\frac{{{R}^{3}}}{1000}$

On taking cube root on both sides, we get,

$\therefore r=\frac{R}{10}$

Hence the radius of the small drop is found to be $\frac{1}{10}$ times the radius of the big drop.

6. A boulder is thrown into a deep lake. As it sinks deeper and deeper into water, does the buoyant force change?

Ans: The buoyant force is known to not change as the boulder sinks. This is because the boulder would displace the same volume of water at any depth and since water is practically incompressible, its density would be the same at all depth. So, the weight of water displaced or the buoyant force is the same at all depths.

7. At what depth in an ocean will a tube of air have one–fourth volume it will have on reaching the surface? Given Atmospheric Pressure = 76 cm of Hg and density of Hg = 13.6g/cc.

Ans: Let V be the volume of bubble on reaching the surface and h the height at which volume becomes $\frac{V}{4}$. 

Now, we have the Initial volume $={{V}_{1}}=V$

Final volume \[={{V}_{2}}=\frac{V}{4}\]

Pressure on the bubble at the surface would be, \[{{P}_{1}}=76cm\text{ }of\text{ }Hg\]

Pressure on the bubble at a depth of h cm would be, 

${{P}_{2}}=\left( 76+\frac{h}{13.6} \right)cm\text{ of Hg}$

Now, according to Boyle’s Law,

${{P}_{1}}{{V}_{1}}={{P}_{2}}{{V}_{2}}$

$\Rightarrow 76V=\left( 76+\frac{h}{13.6} \right)\frac{V}{4}$

$\Rightarrow 304-76=\frac{h}{13.6}$

$\therefore h=228\times 13.6=3100.8cm$

Therefore, we found the depth in an ocean at which a tube of air has one–fourth volume it will have on reaching the surface to be \[h=3100.8cm\].

8. Why is it painful to walk barefooted on a road covered with pebbles having sharp edges?

Ans: We know that it is painful to walk bare–footed on a road covered with pebbles having sharp edges. This is because of their small area. We know that, a \[Pressure\text{ }=\text{ }\frac{Force}{Area}\]

Since the area is less, the pressure is more and hence it will be painful.

9. A liquid stands at the same level in the U–tube when at rest. If A is the area of cross-section of tube and g is the acceleration due to gravity, what will be the difference in height of the liquid in the two limbs when the system is given acceleration ‘a’?

Ans:  Let L be the Length of the horizontal portion of the tube.

Mass of liquid in the portion CD \[=Volume\times Density\]

And let $\rho =Density\text{ }of\text{ }water$

\[Volume=Area\times Length\]

$V=AL$

Now, Mass of liquid in portion CD $M=Al\rho $

Now, from Newton’s second law,

\[Force\text{ }=F=Ma=AL\rho a\]………………………….. (1)

Also due to difference in height of liquid, the downward force exerted on liquid in the horizontal portion CD would be, 

Since, \[Pressure\text{ }=\text{ }\frac{Force}{Area}\]

And, \[Pressure=h\rho g\]

Where, h = height; P = Density; g = acceleration due to gravity

Now, \[Force=Pressure\times Area\]

$F=h\rho gA$…………………………………… (2)

Equating equations (1) and (2), we get, 

$Al\rho a=h\rho gA$

$\Rightarrow h=\frac{aL}{g}$

Therefore, we found the difference in height of the liquid in the two limbs when the system is given acceleration a to be $h=\frac{aL}{g}$.

10. Two balloons that have the same weight and volume contain equal amounts of helium. One is rigid and the other is free to expand as outside pressure decreases. When released, which balloon will rise higher?

Ans: The balloon that is free to expand would displace more air as it rises more than the balloon that is rigid and restrained from expanding. Since the balloon is free to expand it will experience more buoyant force and hence would rise higher.

11. An object floats on water with 20% of its volume above the water time. What is the density of object? Given Density of water \[=\text{ }\mathbf{1000kg}/{{\mathbf{m}}^{3}}.\]

Ans: Let volume of the entire object be V. 

Volume of object under water will be, ${{V}_{W}}=V-\frac{20}{100}V$

${{V}_{W}}=0.8V$

Let ${{\rho }_{W}}=density\text{ of water}$ 

Now, buoyant force, \[{{F}_{B}}={{V}_{W}}\times {{\rho }_{W}}\times g\]…………………….. (1)

Where, 

$g=$ acceleration due to gravity

$\rho =$Density of object

\[Weight\text{ }of\text{ }object\text{ }=\text{ }Mass\text{ }\times \text{ }Acceleration\text{ }due\text{ }to\text{ }gravity\]

Where, Mass of object = Volume of object × Density

\[M=V\rho \]

Weight of object \[=\rho Vg\]…………………. (2)

Now, according to the principle of floatation,

Buoyant force = Weight of object

From equations (1) and (2), 

$0.8V\times {{\rho }_{W}}g=\rho Vg$

$\Rightarrow \rho =0.8{{\rho }_{W}}$

But, density of water \[={{\rho }_{W}}=1000\]

\[\rho =1000\times 0.8=800kg/{{m}^{3}}\]

Therefore, we found the Density of the object to be \[800\text{ }Kg/{{m}^{3}}\].

12. A cubical block of iron 5cm on each side is floating on mercury in a vessel: Given Density of mercury \[=\text{ }\mathbf{13}.\mathbf{6g}/\mathbf{c}{{\mathbf{m}}^{3}}\] and Density of iron \[=\mathbf{7}.\mathbf{2g}/\mathbf{c}{{\mathbf{m}}^{3}}\]. 

a) What is the height of the block above mercury level?

Ans: Let \[h\text{ }=\text{ height of cubical block above mercury level}\]

Volume of cubical Iron Block, 

\[V=l\times b\times h=5\times 5\times 5=125c{{m}^{3}}\]

\[Mass\text{ }of\text{ }cubical\text{ }Iron\text{ }Block\text{ }=\text{ }Volume\text{ }\times \text{ }Density\text{ }of\text{ }Iron\]

$V=125\times 7.2=900g$………………………….. (1)

\[Volume\text{ }of\text{ }Mercury\text{ }displaced\text{ }=\text{ }Length\text{ }\times \text{ }Breadth\text{ }\times \text{ }Decreased\text{ }height\]

\[Volume\text{ }of\text{ }Mercury\text{ }displaced=l\times b\times \left( 5h \right)=5\times 5\times \left( 5h \right)\]

\[Mass\text{ }of\text{ }Mercury\text{ }displaced\text{ }=\text{ }Volume\text{ }of\text{ }Mercury\text{ }\times \text{ }Density\text{ }of\text{ }Mercury\]

\[Mass\text{ }of\text{ }Mercury\text{ }displaced=5\times 5\times \left( 5\text{ }\text{ }h \right)\times 13.6\] ………………………. (2)

Now, Principle of flotation, we have,\[Weight\text{ }of\text{ }Iron\text{ }Block\text{ }=\text{ }Weight\text{ }of\text{ }Mercury\text{ }Displaced\]

\[Mass\text{ }of\text{ }Iron\text{ }Block\text{ }\times g\text{ }=\text{ }Mass\text{ }of\text{ }Mercury\text{ }Displaced\text{ }\times g\]

From equation (1) and (2), we have, 

$900g=5\times 5\times \left( 5-h \right)\times 13.6g$

$\Rightarrow 900=1700\left( 5-h \right)$

\[\therefore h=2.35cm\]

Therefore, we found the height of the block above mercury level to be 2.35cm.

b) Water is poured into vessel so that it just covers the iron block. What is the height of the water column?

Ans: Let $x=$height of block in water when water is poured, 

Then, depth of block in mercury \[=\left( 5x \right)cm\]

Mass of water displaced will be \[=5\times 5\times x\times 1=25x\text{ }gm\]

$\rho =Density\text{ }of\text{ }water\text{ }in\text{ }g/c{{m}^{3}}$

\[Mass\text{ }of\text{ }Mercury\text{ }displaced=5\times 5\times \left( 5x \right)\times 13.6=25\times 13.6\left( 5x \right)gm\]

Now, according to the principle of floatation, we have, 

Weight of Iron Block = Weight of water displaced + Weight of mercury displaced.

$\Rightarrow 900=25x+\left( 25\times 13.6\left( 5-x \right) \right)$

$\Rightarrow 9008500=-1700x+25x$

$\therefore x=2.54\text{ }cm$

Hence, we found the height of the water column when water is poured into the vessel so that it just covers the iron block to be 2.54cm.

13. What should be the pressure inside a small air bubble of \[\mathbf{0}.\mathbf{1mm}\]radius situated just below the water surface? Surface tension of water$=\mathbf{7}.\mathbf{2}\times \mathbf{1}{{0}^{-2}}\mathbf{N}/\mathbf{m}$and atmospheric pressure\[=\mathbf{1}.\mathbf{013}\times \mathbf{1}{{\mathbf{0}}^{5}}\mathbf{N}/{{\mathbf{m}}^{2}}\]. 

Ans: We are given:

Radius of air bubble, \[R=0.1mm=0.1\times {{10}^{-3}}m\] 

Surface tension of water, \[T=7.2\times {{10}^{-2}}N/m\]

The excess pressure inside an air bubble could be given by, 

${{P}_{2}}-{{P}_{1}}=\frac{2T}{R}$

\[{{P}_{2}}=\] Pressure inside air bubble

${{P}_{1}}=$ Atmospheric pressure

Substituting the given values, we get, 

${{P}_{2}}-{{P}_{1}}=\frac{2\left( 7.2\times {{10}^{-2}} \right)}{0.1\times {{10}^{-3}}}=1.44\times {{10}^{3}}N/{{m}^{2}}$

$\Rightarrow {{P}_{2}}={{P}_{1}}+1.44\times {{10}^{3}}=1.013\times {{10}^{5}}+1.44\times {{10}^{3}}$

$\therefore {{P}_{2}}=1.027\times {{10}^{5}}N/{{m}^{2}}$

Hence, the pressure inside a small air bubble of \[\mathbf{0}.\mathbf{1mm}\] radius situated just below the water surface is found to be $1.027\times {{10}^{5}}N/{{m}^{2}}$.

14. Why is a soap solution a better cleansing agent than ordinary water?

Ans: Capillary rise in a cloth that has narrow spaces in the form of fine capillaries can be given by:

$h=\frac{2T\cos \theta }{r\rho g}$

Where, 

\[h=\]height of capillary

\[T=\]Tension surface

$\theta =$Angle of contact

\[r=\]Radius

$\rho =$Density

$g=$Acceleration due to gravity.

Now, the addition of soap to water would reduce the angle of contact $\theta $, this would increase \[cos\theta \] and so will the value of h, that is, the soap water would rise more in narrow spaces in the cloth and clean fabrics better than water alone.

15. If the radius of a soap bubble is r and surface tension of the soap solution is T. Keeping the temperature constant, what is the extra energy needed to double the radius of soap bubble?

Ans: We are given the initial radius \[=r\]

Surface Tension \[=\text{ }T\]

Surface Area \[=4\pi {{r}^{2}}\]

Energy that is required to blow a soap bubble of radius r, 

${{E}_{1}}=Surface\text{ }Tension\times 2\times Surface\text{ }Area$ (2 because bubble has two surfaces)

\[\Rightarrow {{E}_{1}}=T\times 2\times \left( 4\pi {{r}^{2}} \right)=8\pi {{r}^{2}}T\]

Final Radius \[=2r\]

Surface Tension \[=T\]

Surface Area \[=4\text{ }\pi {{\left( 2r \right)}^{2}}=16\pi {{r}^{2}}\]

Energy that is required to blow a soap bubble of Radius 2r, 

${{E}_{2}}=surface\text{ }Tension\times 2\times Surface\text{ }Area$

\[\Rightarrow {{E}_{2}}=T\times 2\times \left( 416\text{ }\pi \text{ }{{r}^{2}} \right)=32\text{ }\pi \text{ }{{r}^{2}}T\]

Now, the extra energy that is required, 

\[{{E}_{2}}{{E}_{1}}=32\pi {{r}^{2}}T8\pi {{r}^{2}}T\]

\[\therefore {{E}_{2}}-{{E}_{1}}=24\pi {{r}^{2}}T\]

Hence, the extra energy needed to double the radius of soap bubble is found to be \[{{E}_{2}}-{{E}_{1}}=24\pi {{r}^{2}}T\].

16. Find the work done in breaking a water drop of radius 1 mm into 1000 drops. Given the surface tension of water is 72 × 10-3 N/m. 

Ans: We are given:

Initial Radius \[=R={{10}^{-3}}m\left( =1mm \right)\]

Final Radius \[=r\]

Here, 1 drop is breaking into 1000 small droplets, so,

\[Initial\text{ }volume\text{ }=\text{ }1000\times Final\text{ }Volume\]

$\frac{4}{3}\pi {{R}^{3}}=1000\times \frac{4}{3}\pi {{r}^{3}}$

$\Rightarrow {{r}^{3}}=\frac{{{R}^{3}}}{{{10}^{3}}}$

On, taking cube root on both sides, we get, 

$r=\frac{R}{10}$…………………………. (1)

Initial Surface Area \[=4\pi {{R}^{2}}=4\pi \times {{10}^{-6}}{{m}^{2}}\]……………………. (2)

Final Surface Area \[=1000\times \left( 4\pi {{r}^{2}} \right)=4\pi \times {{10}^{-5}}{{m}^{2}}\]………………………… (3)

\[Increase\text{ }in\text{ }Surface\text{ }Area\text{ }=\text{ }Final\text{ }surface\text{ }Area\text{ }\text{ }Initial\text{ }surface\text{ }Area\]

$\Delta A=4\pi \left( {{10}^{-5}}-{{10}^{-6}} \right)$

Now, \[Work\text{ }Done\text{ }=\text{ }Surface\text{ }Tension\times Increase\text{ }in\text{ }surface\text{ }Area\]

$\Rightarrow W=72\times {{10}^{-3}}\times 4\pi \left( {{10}^{-5}}-{{10}^{-6}} \right)$

Work Done, 

$\Rightarrow W=8.14\times {{10}^{-6}}J$

Therefore, the work done in breaking a water drop of radius 1 mm into 1000 drops is \[8.14\times {{10}^{-6}}J\].

17. What is the energy stored in a soap bubble of diameter 4cm (given the surface tension\[=\mathbf{0}.\mathbf{07}\text{ }\mathbf{N}/\mathbf{m}\])?

Ans: We are given:

Diameter of soap bubble \[=4cm=4\times {{10}^{-2}}m\]

Radius of soap bubble \[=2\times {{10}^{-2}}m\]

Increase in surface Area $=2\times 4\pi {{R}^{2}}$ (2, a bubble has 2 surfaces\[Increase\text{ }in\text{ }Surface\text{ }Area=2\times 4\text{ }\pi \times {{\left( 2\times {{10}^{-2}} \right)}^{2}}=8\times 4\pi \times {{10}^{-4}}{{m}^{2}}\]

Now, \[energy\text{ }stored\text{ }=\text{ }Surface\text{ }Tension\text{ }\times \text{ }Increase\text{ }in\text{ }Surface\text{ }Area\]

$E=T\times 8\times 4\pi \times {{10}^{-4}}$

$\therefore E=7\times {{10}^{-4}}J$

Hence, the energy stored in the given soap bubble is found to be $E=7\times {{10}^{-4}}J$.

18. What is the work done in splitting a drop of water of 1 mm radius into 64 droplets? Given the surface tension of water is \[\mathbf{72}\times \mathbf{1}{{\mathbf{0}}^{-3}}\text{ }\mathbf{N}/{{\mathbf{m}}^{2}}\]. 

Ans: Let R be the radius of bigger drop \[=1mm={{10}^{-3}}m\]

\[r=\]radius of smaller drop

Bigger volume \[=64\times smaller\text{ }Volume\]

$\frac{4}{3}\pi {{R}^{3}}=64\times \frac{4}{3}\pi {{r}^{3}}$

$\Rightarrow {{r}^{3}}=\frac{{{R}^{3}}}{64}$

Taking cube root on both sides, we get, 

$r=\frac{R}{4}=\frac{{{10}^{-3}}}{4}$……………….. (1)

Initial Surface Area \[=4\pi {{R}^{2}}=4\pi \times {{\left( {{10}^{-3}} \right)}^{2}}\]………………. (2)

Final Surface Area $=4\pi {{r}^{2}}\times 64=4\pi \times \frac{{{\left( {{10}^{-3}} \right)}^{2}}}{16}\times 64$…………………… (3)

\[Increase\text{ }in\text{ }Surface\text{ }=\text{ }Final\text{ }Surface\text{ }Area\text{ }\text{ }Initial\text{ }Surface\text{ }Area\]

$\Rightarrow \Delta A=64\times 4\pi \times {{\left( \frac{{{10}^{-3}}}{4} \right)}^{2}}-4\pi \times {{\left( {{10}^{-3}} \right)}^{2}}$

\[Work\text{ }Done\text{ }=\text{ }Surface\text{ }Tension\times Increase\text{ }in\text{ }Surface\text{ }Area\]

$W=72\times {{10}^{-3}}\times \left( 64\times 4\pi \times \frac{{{10}^{-6}}}{4}-4\pi \times {{10}^{-6}} \right)$

$\therefore Work\text{ }Done\text{ }=W=\text{ }2.7\text{ }\times \text{ }{{10}^{-6}}\text{ }J$

Therefore, the work done in splitting a drop of water of 1 mm radius into 64 droplets is found to be $2.7\text{ }\times \text{ }{{10}^{-6}}\text{ }J$.

19. What is terminal velocity? What is the terminal velocity of a body in a freely falling system?

Ans: Terminal velocity can be defined as the maximum constant velocity that is acquired by the body while falling freely in a viscous medium. In a freely falling system, the acceleration due to gravity, \[g=0\]. Therefore, the terminal velocity of the body will also be zero.

20. What is the cause of viscosity in a fluid? How does the flow of fluid depend on viscosity?

Ans: The cause of viscosity of fluid is known to be internal friction. The flow of fluid decreases as viscosity increases, because we know that viscosity is a frictional force and greater the friction, lesser will be the flow of liquid.

21. If eight rain drops each of radius 1mm are falling through air at a terminal velocity of \[\mathbf{5}\text{ }\mathbf{cm}/\mathbf{s}\]. If they coalesce to form a bigger drop, what is the terminal velocity of bigger drop?

Ans: Let the radius of smaller drop be r and the radius of bigger drop be R, then, 

Volume of smaller drop$=\frac{4}{3}\pi {{r}^{3}}$

Volume of bigger drop $=\frac{4}{3}\pi {{R}^{3}}$

Now, we are given, 

Volume of bigger drop = Volume of 8 smaller drops.

$\Rightarrow \frac{4}{3}\pi {{R}^{3}}=8\times \frac{4}{3}\pi {{r}^{3}}$

$\Rightarrow {{R}^{3}}=8{{r}^{3}}$

Taking cube–root, we get, 

$R=2r=2\times 1mm=0.2cm$

Now, we know that, Terminal velocity of each small drop, 

${{N}_{T}}=\frac{2}{9}\times \frac{{{r}^{2}}}{\eta }\left( \rho -\sigma  \right)g$…………….. (1)

Terminal velocity of bigger drop would be, 

${{V}_{T}}=\frac{2}{9}\times \frac{{{R}^{2}}}{\eta }\left( \rho -\sigma  \right)g$……………………….. (2)

$\eta =$Co-efficient of viscosity

$\rho =$Density of body

$\sigma =$Density of fluid

$g=$acceleration due to gravity

Dividing equation (2) by (1),

$\frac{{{V}_{T}}}{{{N}_{T}}}=\frac{{{R}^{2}}}{{{r}^{2}}}$

$\Rightarrow {{V}_{T}}={{N}_{T}}\times \frac{{{R}^{2}}}{{{r}^{2}}}$

Substituting the given values, 

${{V}_{T}}=5\times \frac{{{\left( 0.2 \right)}^{2}}}{{{\left( 0.1 \right)}^{2}}}=5\times \frac{0.04}{0.01}$

\[\Rightarrow {{V}_{T}}=20cm/s\], which is the terminal velocity of the bigger drop.

22. Why does the cloud seem to be floating in the sky?

Ans: The terminal velocity of a raindrop is known to be directly proportional to the square of radius of drop. While falling, large drops will have high terminal velocities while small drops will have small terminal velocities. So, the small drops would fall so slowly that clouds would seem floating.

23. A metal plate 5 cm × 5 cm rests on a layer of castor oil 1 mm thick whose co-efficient of viscosity is \[\mathbf{1}.\mathbf{55Ns}{{\mathbf{m}}^{-2}}\]. What is the horizontal force required to move the plate with a speed of \[\mathbf{2cm}/\mathbf{s}\]?

Ans: We are given the following:

Length of metal plate \[=\text{ }5\text{ }cm\]

Breadth of metal plate \[=\text{ }5\text{ }cm\]

\[A=\] \[Area\text{ }of\text{ }metal\text{ }plate\text{ }=\text{ }Length\times Breadth\]

\[A=5\times 5=25c{{m}^{2}}\]

Co – efficient of viscosity $=\eta =1.55Ns/{{m}^{2}}$

\[dx=\]Small thickness of layer \[=1mm={{10}^{-3}}m\]

Small velocity \[=d\text{ }v=2\times {{10}^{-2}}m/s\]

Now, velocity gradient $=\frac{dv}{dx}=\frac{2\times {{10}^{-2}}}{{{10}^{-3}}}=20m$

Now, we have, horizontal force, $F=\eta A\frac{dv}{dx}$

$\Rightarrow F=1.55\times 25\times {{10}^{-4}}\times \frac{2\times {{10}^{-2}}}{{{10}^{-3}}}$

\[\therefore F=0.0775\text{ }N\]

Hence, we find the horizontal force required to move the plate with a speed of \[\mathbf{2cm}/\mathbf{s}\]to be \[\therefore F=0.0775\text{ }N\]. 

24. A small ball of mass ‘m’ and density ‘d’ dropped in a viscous liquid of density ‘d’. After some time, the ball falls with a constant velocity. What is the viscous force on the ball?

Ans: We have, \[Volume\text{ }=\frac{Mass}{Density}\]

Mass of ball \[=m\]

Density of ball \[=d\]

Volume of ball $=V=\frac{m}{d}$

Density of viscous liquid $={{d}_{1}}$

Now, Mass of liquid displaced by the ball, ${{m}_{1}}=\frac{m}{d}\times {{d}_{1}}$………………. (1)

For the ball falls with a constant terminal velocity, we have:

\[Viscous\text{ }force\text{ }F\text{ }=\text{ }weight\text{ }of\text{ }ball\text{ }in\text{ }water\]…………………….. (2)

\[Weight\text{ }of\text{ }ball\text{ }in\text{ }water\text{ }=\text{ }Weight\text{ }of\text{ }ball\text{ }\text{ }Weight\text{ }of\text{ }liquid\text{ }displaced\text{ }by\text{ }the\text{ }ball\]$\Rightarrow W=mg-{{m}_{1}}g=mg-\frac{mg{{d}_{1}}}{d}$

$\Rightarrow W=mg\left( 1-\frac{{{d}_{1}}}{d} \right)$

Thus, from equation (2),

Viscous force, 

$F=mg\left( 1-\frac{{{d}_{1}}}{d} \right)$

Therefore, we found the viscous force on the ball to be $F=mg\left( 1-\frac{{{d}_{1}}}{d} \right)$.

25. Water flows faster than honey. Why?

Ans: We have the Poiseuille’s formula given by,

$V=\frac{\pi \rho {{r}^{4}}}{8\eta l}$

Where, 

\[V=\] Volume of liquid flowing per second

\[R=\] Radius of narrow tube

\[P=\] Pressure difference across 2 ends of the tube.

$\eta =$ coefficient of viscosity

\[L=\]height of tube.

Thus, we found that, $V\propto \frac{1}{\eta }$, so, we could make this conclusion that, $\eta $for water is less than that of honey. Therefore, V for water is greater and hence it would flow faster.

26. What is Stoke’s law and what are the factors on which viscous drag depends?

Ans: We know Stoke’s law is given by:

$F=6\pi \eta rv$ 

Clearly, we find that the viscous drag force F would depend on:

i) $\eta =$ coefficient of viscosity

ii) r = radius of spherical body

iii) V = Velocity of body

27. Water flows through a horizontal pipe of which the cross–section is not constant. The pressure is 1cm of mercury where the velocity is \[\mathbf{0}.\mathbf{35m}/\mathbf{s}\]. Find the pressure at a point where the velocity is 0.65m/s.

Ans: We are given:

At one point, \[{{P}_{1}}=1cm\text{ }of\text{ }Hg=0.01m\text{ }of\text{ }Hg=0.01\times \left( 13.6\times {{10}^{3}} \right)\times 9.8\text{ }Pa\]

Velocity, ${{v}_{1}}=0.35m/s$

At another point, \[{{P}_{2}}=?\]

\[{{V}_{2}}=0.65m/s\]

Density of water, $\rho ={{10}^{3}}kg/m3$

Now, according to Bernoulli’s theorem,

${{P}_{1}}+\frac{1}{2}\rho {{v}_{1}}^{2}={{P}_{2}}+\frac{1}{2}\rho {{v}_{2}}^{2}$

$\Rightarrow {{P}_{2}}={{P}_{1}}-\frac{1}{2}\rho \left( {{v}_{2}}^{2}-{{v}_{1}}^{2} \right)$

Substituting the given values, we get, 

$\Rightarrow {{P}_{2}}=0.01\times 13.6\times {{10}^{3}}\times 9.8-\frac{1}{2}\times {{10}^{3}}\left( {{\left( 0.65 \right)}^{2}}-{{\left( 0.35 \right)}^{2}} \right)$

$\Rightarrow {{P}_{2}}=1332.8-0.15\times {{10}^{3}}=1182.8Pa=\frac{1182.8}{9.8\times 13.6\times {{10}^{3}}}mm\text{ of Hg}$

\[\therefore {{\mathbf{P}}_{2}}=\mathbf{0}.\mathbf{00887}\text{ }\mathbf{m}m\text{ o}\mathbf{f}\text{ }\mathbf{Hg}\]

Therefore, the pressure at a point where the velocity is 0.65m/s is found to be \[{{\mathbf{P}}_{2}}=\mathbf{0}.\mathbf{00887}\text{ }\mathbf{m}m\text{ o}\mathbf{f}\text{ }\mathbf{Hg}\].

28. Two pipes P and Q having diameters \[\mathbf{2}\times \mathbf{1}{{\mathbf{0}}^{-2}}\mathbf{m}\]and \[\mathbf{4}\times \mathbf{1}{{\mathbf{0}}^{-2}}\mathbf{m}\]respectively are joined in Series with the main supply line of water. What is the velocity of water flowing in pipe P?

Ans: We are given the following:

Diameter of pipe P \[=2\times {{10}^{-2}}m\]

Diameter of Pipe Q \[=4\times {{10}^{-2}}m\]

Now, according to the equation of continuity, we have, 

${{a}_{P}}{{v}_{P}}={{a}_{Q}}{{v}_{Q}}$

Where, \[{{a}_{P}},{{a}_{Q}}=\] Cross– section areas of pipe P and Q

\[{{v}_{P}},{{v}_{Q}}=\]Velocities of liquid at pipe P and Q

Now, we have, ${{a}_{Q}}=\pi {{r}^{2}}=\pi {{\left( \frac{{{d}_{Q}}}{2} \right)}^{2}}=\frac{\pi }{4}{{d}_{Q}}^{2}$……………….. (1)

Similarly, ${{a}_{P}}=\frac{\pi }{4}{{d}_{P}}^{2}$…………………………. (2)

Now, equation of continuity becomes, 

$\left( \frac{\pi }{4}{{d}_{P}}^{2} \right){{v}_{P}}=\left( \frac{\pi }{4}{{d}_{Q}}^{2} \right){{v}_{Q}}$

$\Rightarrow \frac{{{v}_{P}}}{{{v}_{Q}}}={{\left( \frac{{{d}_{Q}}}{{{d}_{P}}} \right)}^{2}}={{\left( \frac{4\times {{10}^{-2}}}{2\times {{10}^{-2}}} \right)}^{2}}={{2}^{2}}$

$\therefore {{v}_{P}}=4{{v}_{Q}}$

Hence, we found the velocity of water in pipe P to be four times the velocity of water in pipe Q.

29. A horizontal pipe of diameter 20 cm has a constriction of diameter 4 cm. The velocity of water in the pipe is 2m/s and pressure is \[\mathbf{10}\text{ }\mathbf{N}/{{\mathbf{m}}^{2}}\]. Calculate the velocity and pressure at the constriction?

Ans: We have the equation of continuity,

${{a}_{1}}{{v}_{1}}={{a}_{2}}{{v}_{2}}$

$\Rightarrow {{v}_{2}}=\frac{{{a}_{1}}{{v}_{1}}}{{{a}_{2}}}$……………………………. (1)

Now, we have, \[{{v}_{1}}\text{=}2m/s\]

\[{{v}_{2}}=?\]

\[{{a}_{1}}\text{, }{{a}_{2}}=Cross\text{ }\text{ }Sectional\text{ }Areas\]

${{a}_{2}}=\pi {{\left( 0.02 \right)}^{2}}{{m}^{2}}=2\times {{10}^{-2}}{{m}^{2}}$

${{a}_{1}}=\pi {{\left( 0.1 \right)}^{2}}{{m}^{2}}$

Now, from equation (1),

${{v}_{2}}=\frac{\pi \times {{\left( 0.1 \right)}^{2}}\times 2}{\pi \times {{\left( 0.02 \right)}^{2}}}=50m/s$

From Bernoulli’s theorem, for the horizontal pipeline, we have,

${{P}_{1}}+\frac{1}{2}\rho {{v}_{1}}^{2}={{P}_{2}}+\frac{1}{2}\rho {{v}_{2}}^{2}$

$\Rightarrow {{P}_{2}}={{P}_{1}}-\frac{1}{2}\rho \left( {{v}_{2}}^{2}-{{v}_{1}}^{2} \right)$

$\Rightarrow {{P}_{2}}={{10}^{7}}-\frac{1}{2}\times {{10}^{3}}\left( {{\left( 50 \right)}^{2}}-{{\left( 2 \right)}^{2}} \right)$

$\therefore {{P}_{2}}={{10}^{7}}-12.48\times {{10}^{5}}=8.752\times {{10}^{6}}N/{{m}^{2}}$

Therefore, we found the pressure at the constriction to be $8.752\times {{10}^{6}}N/{{m}^{2}}$.

30. The reading of a pressure metre attached to a closed is \[\mathbf{2}.\mathbf{5}\text{ }\times \mathbf{1}{{\mathbf{0}}^{5}}\mathbf{N}/{{\mathbf{m}}^{2}}\]. On opening the valve of pipe, the reading of the pressure metre reduces to\[\mathbf{2}.\mathbf{0}\times \mathbf{1}{{\mathbf{0}}^{5}}\mathbf{N}/{{\mathbf{m}}^{2}}\]. Calculate the speed of water flowing through the pipe. 

Ans: We are given:

Pressure at end 1, \[{{P}_{1}}=2.5\times {{10}^{5}}N/{{m}^{2}}\]

Pressure at end 2, \[{{P}_{2}}=2.0\times {{10}^{5}}N/{{m}^{2}}\]

As initially pipe was closed, \[{{v}_{1}}=0~\]

\[{{v}_{2}}=?\]

Density of water \[=\rho =1000kg/{{m}^{3}}\]

According to Bernoulli’s theorem for a horizontal pipe,

${{P}_{1}}+\frac{1}{2}\rho {{v}_{1}}^{2}={{P}_{2}}+\frac{1}{2}\rho {{v}_{2}}^{2}$

$\Rightarrow {{P}_{1}}={{P}_{2}}+\frac{1}{2}\rho {{v}_{2}}^{2}$

$\Rightarrow {{v}_{2}}^{2}=\frac{2\left( {{P}_{1}}-{{P}_{2}} \right)}{\rho }=\frac{2\left( 2.5\times {{10}^{5}}-2\times {{10}^{5}} \right)}{1000}$

$\Rightarrow {{v}_{2}}^{2}=\frac{2\times {{10}^{5}}\times 0.5}{1000}$

$\therefore {{v}_{2}}=\sqrt{100}=10m/s$

Therefore, we found the speed of water flowing through the pipe to be 10m/s.

31. A large bottle is fitted with a siphon made of capillary glass tubing. Compare the Coefficient of viscosity of water and petrol if the time taken to empty the bottle in the two cases is in the ratio 2:5. Given specific gravity of petrol = 0.8

Ans: We have the expression for rate of flow as, 

${{Q}_{1}}=\frac{\pi {{\rho }_{1}}{{R}^{4}}}{8{{\eta }_{1}}l}$

Where, 

\[{{Q}_{1}}=\]rate of flow of liquid in case 1

\[{{P}_{1}}=\]Pressure

\[R=\]Radius

\[{{s}_{1}}=\]specific gravity in 1st Case

${{\eta }_{1}}=$ Co-efficient of viscosity

\[{{s}_{2}}=\]Density of water

Also, ${{Q}_{1}}=\frac{V}{{{t}_{1}}}$

$V=$volume of liquid

\[{{t}_{1}}=\]time Case 1

Similarly, ${{Q}_{2}}=\frac{\pi {{P}_{2}}^{2}{{R}^{4}}}{8{{\eta }_{2}}l}=\frac{V}{{{t}_{2}}}$

Now, $\frac{{{Q}_{1}}}{{{Q}_{2}}}=\frac{{{t}_{2}}}{{{t}_{1}}}=\frac{5}{2}$……………………………. (1)

Also, $\frac{{{Q}_{1}}}{{{Q}_{2}}}=\frac{{{P}_{1}}{{\eta }_{2}}}{{{P}_{2}}{{\eta }_{2}}}$…………………………… (2)

Where, ${{P}_{1}}={{s}_{1}}gh$

${{P}_{2}}={{s}_{2}}gh$

$\Rightarrow \frac{{{P}_{1}}}{{{P}_{2}}}=\frac{{{s}_{1}}}{{{s}_{2}}}=\frac{{{10}^{3}}}{0.8\times {{10}^{3}}}=1.25$…………………… (3) 

Equating equations (1) and (2), 

$\frac{5}{2}=\frac{{{P}_{1}}{{\eta }_{2}}}{{{P}_{2}}{{\eta }_{2}}}=1.25\times \frac{{{\eta }_{2}}}{{{\eta }_{1}}}$

$\therefore \frac{{{\eta }_{1}}}{{{\eta }_{2}}}=1.25\times \frac{2}{5}=\frac{1}{2}$

Therefore, the ratio of Coefficient of viscosity of water and petrol is found to be 1:2.

32. Under a pressure head, the rate of flow of liquid through a pipe is Q. If the length of pipe is doubled and diameter of pipe is halved, what is the new rate of flow?

Ans: From Poiseuille’s equation for flow of liquid through a tube of radius R and length l:

$Q=\frac{\pi \rho {{R}^{4}}}{8\eta l}$

Now when diameter is halved, $D'=\frac{D}{2}$

For radius, $\Rightarrow R'=\frac{R}{2}$…………………….. (1)

Length is doubled, $l'=2l$………………….. (2)

Rate of flow liquid becomes,

$Q'=\frac{\pi PR{{'}^{4}}}{8\eta l'}=\frac{\pi P{{\left( \frac{R}{2} \right)}^{4}}}{8\eta \left( 2l \right)}=\frac{1}{32}\left( \frac{\pi P{{R}^{4}}}{8\eta l} \right)$

$\therefore Q'=\frac{Q}{32}$

Therefore, the new rate of flow is found to be $\frac{1}{32}$times the initial rate.

33. A U-shaped wire is dipped in a soap solution, and removed. The thin soap film formed between the wire and the light slider supports a weight of $1.5\times {{10}^{-2}}N$ (which includes the small weight of the slider). The length of the slider is 30 cm. What is the surface tension of the film?

Ans: We are given the following:

The weight that the soap film supports, $W=1.5\times {{10}^{-2}}N$  

Length of the slider, \[l=30cm=0.3m\]

A soap film has two free surfaces, 

So, \[Total\text{ }length=2l=2\times 0.3=0.6m\]

Surface tension, $S=\frac{weight}{2l}$

$\Rightarrow S=\frac{1.5\times {{10}^{-2}}}{0.6}=2.5\times {{10}^{-2}}N/m$

Hence, the surface tension of the film is found to be $2.5\times {{10}^{-2}}N/m$.

34. During blood transfusion the needle is inserted in a vein where the gauge pressure is 2000Pa. At what height must the blood container be placed so that blood may just enter the vein? (Use the density of whole blood from Table 10.1).

Ans: We are given:

Gauge pressure, \[P=2000Pa\]

Density of whole blood, $\rho =1.06\times {{10}^{3}}kg{{m}^{-3}}$

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

Height of the blood container \[=h\]

Pressure of the blood container, \[P=h\rho g\]

$\Rightarrow h=\frac{P}{\rho g}=\frac{2000}{1.60\times {{10}^{3}}\times 9.8}$

$\therefore h=0.1925\text{ }m$

Hence, we found that the blood may enter the vein if the blood container is kept at a height greater than 0.1925 m, i.e., about 0.2 m.

35. Torricelli’s barometer used mercury. Pascal duplicated it using French wine of density $984kg{{m}^{-3}}$. Determine the height of the wine column for normal atmospheric pressure.

Ans: We are given:

Density of mercury, ${{\rho }_{1}}=3.6\times {{10}^{3}}kg/{{m}^{3}}$

Height of the mercury column, \[{{h}_{1}}=0.76\text{ }m\]

Density of French wine, ${{\rho }_{2}}=984kg/{{m}^{3}}$

Height of the French wine column $={{h}_{2}}$

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

The pressure in both the columns is equal, i.e.,

Pressure in the mercury column = Pressure in the French wine column

${{\rho }_{1}}{{h}_{1}}g={{\rho }_{2}}{{h}_{2}}g$

$\Rightarrow {{h}_{2}}=\frac{{{\rho }_{1}}{{h}_{1}}}{{{\rho }_{2}}}=\frac{13.6\times {{10}^{3}}\times 0.76}{984}$

$\therefore {{h}_{2}}=\text{ }10.5\text{ }m$

Thus, the height of the French wine column for normal atmospheric pressure is found to be 10.5 m.

36. A hydraulic automobile lift is designed to lift cars with a maximum mass of 3000 kg. The area of cross-section of the piston carrying the load is \[\mathbf{425}c{{m}^{2}}\]. What maximum pressure would the smaller piston have to bear?

Ans: We are given, 

The maximum mass of a car that can be lifted, \[m=\text{ }3000\text{ }kg\]

Area of cross-section of the load-carrying piston, \[A=425c{{m}^{2}}=425\times {{10}^{-4}}{{m}^{2}}\]

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

Now, the maximum pressure exerted on the load-carrying piston,

$P=\frac{F}{A}=\frac{29400}{425\times {{10}^{-4}}}$

$\therefore P=6.917\times {{10}^{5}}Pa$

Hence, we found that the pressure is transmitted equally in all directions in a liquid. Thus, the maximum pressure that the smaller piston would have to bear is $6.917\times {{10}^{5}}Pa$.

3 Marks Questions

1. Calculate the radius of new bubble formed when two bubbles of radius \[{{r}_{1}}\] and \[{{r}_{2}}\] coalesce.

Ans: Assume, two soap bubble of radii \[{{r}_{1}}\] and \[{{r}_{2}}\]; and volumes as \[{{v}_{1}}\] and \[{{v}_{2}}\] respectively. Since bubble is in the form of a sphere,

\[{{v}_{1}}=\frac{4}{3}\pi {{r}_{1}}^{3};{{v}_{2}}=\frac{4}{3}\pi {{r}_{2}}^{3}\]

\[\text{If, }S=\]surface tension of the soap solution

\[{{P}_{1}}\And {{P}_{2}}=\]excess pressure inside the two soap bubbles

\[{{P}_{1}}=\frac{4S}{{{r}_{1}}};{{P}_{2}}=\frac{4S}{{{r}_{2}}}\]

Let r be the radius of the new soap bubble formed when the two soap bubble coalesce under and excess of pressure inside this new soap bubble then

\[V=\frac{4}{3}\pi {{r}^{3}}\]

\[P=\frac{4S}{r}\]

As the new bubble is formed under isothermal condition, so Boyle’s law holds good and hence

\[{{P}_{1}}{{v}_{1}}+{{P}_{2}}{{v}_{2}}=PV\]

\[\Rightarrow \frac{4S}{{{r}_{1}}}\times \frac{4}{3}\pi {{r}_{1}}^{3}+\frac{4S}{{{r}_{2}}}\times \frac{4}{3}\pi {{r}_{2}}^{3}=\frac{4S}{r}\times \frac{4}{3}\pi {{r}^{3}}\]

\[\Rightarrow 16S\pi r_{1}^{2}+16S\pi r_{2}^{2}=16S\pi {{r}^{2}}\]

\[\therefore r=\sqrt{r_{1}^{2}+r_{2}^{2}}\]

Therefore, the radius of the new bubble can be written as \[\sqrt{r_{1}^{2}+r_{2}^{2}}\].

2. A liquid drop of diameter \[4mm\] breaks into \[1000\]droplets of equal size. Calculate the resultant change in the surface energy. Surface tension of the liquid is \[0.07N/m\].

Ans: Given that, the diameter of drop, \[=4mm\]

Radius of drop, \[=2mm=2\times {{10}^{-3}}m\]

\[S=Surface\text{ tension}=0.07N/m\]

Let r be the radius of each of the small droplets. Then,

volume of big drop\[=1000\times \]volume of the small droplets 

\[\Rightarrow \] \[\frac{4}{3}\pi {{R}^{3}}=1000\times \frac{4}{3}\pi {{r}^{3}}\]

or \[R=10r\]

\[\Rightarrow R=\frac{2\times {{10}^{-3}}}{10}=2\times {{10}^{-4}}m\]

Original surface area of the drop\[=4\pi {{R}^{2}}\]

Total surface area of drop\[=1000\times 4\pi {{r}^{2}}-4\pi {{R}^{2}}=4\pi \left[ 1000{{r}^{2}}-{{R}^{2}} \right]\]

\[\Rightarrow \] Total surface area\[=4\times \frac{22}{2}\left( \left[ 1000\times {{\left( 2\times {{10}^{-4}} \right)}^{2}} \right]-\left[ {{\left( 2\times {{10}^{-3}} \right)}^{2}} \right] \right)\]

\[\Rightarrow \] Total surface area\[=4\times \frac{22}{4}\left[ 1000\times 4\times {{10}^{-3}}-4\times {{10}^{-6}} \right]=8\times \frac{22}{7}\left[ {{10}^{-5}}-{{10}^{-6}} \right]\]

\[\therefore \] Total surface area \[=\frac{3168}{7}\times {{10}^{-6}}{{m}^{2}}\]\[\text{Increase in surface energy = Surface tension}\times \text{ Increase in surface area}\]

\[\Rightarrow \] Increase in surface\[\text{=0}\text{.07}\times \frac{3168}{7}\times {{10}^{-6}}\]

\[\therefore \] Increase in surface \[=3168\times {{10}^{-8}}J\]

Therefore, the increase in surface energy of the drop is \[3168\times {{10}^{-8}}J\].

3. Two capillary tubes of length \[15cm\] and \[5cm\] and radii \[0.06cm\] and \[0.02cm\] respectively are connected in series. If the pressure difference across the end faces is equal to the pressure of \[15cm\] high water column, then find the pressure difference across the:

 1) first tube

 2) Second tube.

Ans: Poiseuille’s formula for flow of liquid through a tube of radius ‘\[r\]’ gives,

\[V=\frac{\pi {{\Pr }^{4}}}{8\eta l}\]

\[V=\] Volume of liquid

\[r=\] Radius of tube

\[\frac{P}{l}=\] Pressure Gradient

\[\eta =\] Co – efficient of viscosity

When two tubes are connected in series, then the volume of liquid through both the tubes is equal.

Radius of first tube \[=0.06cm\]

Radius of second tube \[=0.02cm\]

Length of first tube \[=15cm\]

Length of second tube \[=5cm\]

Now, Volume of liquid through first tube \[{{V}_{1}}=\frac{\pi {{\operatorname{P}}_{1}}{{r}_{1}}^{4}}{8\eta {{l}_{1}}}\]

Volume of liquid through second tube, \[{{V}_{2}}=\frac{\pi {{\operatorname{P}}_{2}}{{r}_{2}}^{4}}{8\eta {{l}_{2}}}\]

Equating above equations for tubes connected in Series,

\[{{V}_{1}}={{V}_{2}}\]

\[\Rightarrow \frac{\pi {{\operatorname{P}}_{1}}{{r}_{1}}^{4}}{8\eta {{l}_{1}}}=\frac{\pi {{\operatorname{P}}_{2}}{{r}_{2}}^{4}}{8\eta {{l}_{2}}}\]

Now, Pressure in first tube \[={{P}_{1}}=\left( 15-h \right)\rho g\]

\[\rho =\] Density of liquid

Pressure in Second tube \[={{P}_{2}}=h\rho g\]

\[15cm=\] height of water column.

Now, \[\frac{\pi \left( 15-h \right)\rho g\times {{r}_{1}}^{4}}{8\eta {{l}_{1}}}=\frac{\pi h\rho g\times {{r}_{2}}^{4}}{8\eta {{l}_{2}}}\]

\[\frac{\left( 15-h \right){{r}_{1}}^{4}}{{{l}_{1}}}=\frac{h{{r}_{2}}^{4}}{{{l}_{2}}}\]

\[\Rightarrow \frac{\left( 15-h \right)\times {{\left( 0.06 \right)}^{4}}}{15}=\frac{h\times {{\left( 0.02 \right)}^{4}}}{5}\]

\[\Rightarrow \frac{\left( 15-h \right)}{15}\times {{\left( 6 \right)}^{4}}=\frac{h}{5}\times {{\left( 2 \right)}^{4}}\]

\[\Rightarrow \frac{\left( 15-h \right)}{15}\times 1296=\frac{h}{5}\times 16\]

\[\Rightarrow \left( 15-h \right)27=h\]

\[\Rightarrow h=\frac{405}{28}=14.464cm\]

\[\therefore \] Pressure difference across first tube \[=15-14.464=0.536cm\]of water column

\[\therefore \] Pressure difference across second tube \[=14.464cm\] of water column.

Therefore, the Pressure differences across the first tube and the second tube is \[0.536cm\] and \[14.464cm\] respectively.

4. A metallic sphere of radius \[1\times {{10}^{-3}}m\]and density \[1\times {{10}^{4}}kg/{{m}^{3}}\] enters a tank of water after a free fall through a high ‘\[h\]’ in earth’s gravitational field. If its velocity remains unchanged after entering water, determine the value of \[h\]. Given: Co-efficient of viscosity of water \[=1\times {{10}^{-3}}Ns/{{m}^{2}};g=10m/{{s}^{2}}\]; density of water \[=1\times {{10}^{3}}kg/{{m}^{3}}\].

Ans: Given that, the velocity acquired by the sphere in falling freely through a height h is \[V=\sqrt{2gh}\]

As per the conditions of the problem, this is the terminal velocity of sphere in water,

That is, Terminal Velocity of sphere in water is \[{{V}_{T}}=\sqrt{2gh}\] \[...........(1)\]

By Stoke’s Law, the terminal velocity \[{{V}_{T}}\] of sphere in water is given by:

\[{{V}_{T}}=\frac{2\times {{r}^{2}}\times \left( P-\sigma  \right)g}{9\eta }\]

Radius of sphere, \[r=1\times {{10}^{-3}}m\]

Density of sphere, \[P=1\times {{10}^{4}}kg/{{m}^{3}}\]

Density of liquid, \[\sigma =1\times {{10}^{3}}kg/{{m}^{3}}\]

Acceleration due to gravity, \[g=10m/{{s}^{2}}\]

Coefficient of viscosity, \[\eta =1\times {{10}^{-3}}Ns/{{m}^{2}}\]

\[{{V}_{T}}=\frac{2\times {{\left( 1\times {{10}^{-3}} \right)}^{2}}\times \left( 1\times {{10}^{4}}-1\times {{10}^{3}} \right)\times 10}{9\times 1\times {{10}^{-3}}}\]

\[{{V}_{T}}=20m/s\]

From equation \[(1)\], 

\[{{V}_{T}}=\sqrt{2gh}\]

\[\Rightarrow {{V}_{T}}^{2}=2gh\]

\[\Rightarrow h=\frac{{{V}_{T}}^{2}}{2g}\]

\[\therefore h=\frac{{{20}^{2}}}{2\times 10}=20m\]

Therefore, the high h is \[20m\].

5. Define terminal velocity and derive an expression for it.

Ans: The maximum constant velocity acquired by the body which is falling freely in a viscous medium is called terminal velocity.

When a small spherical body falls freely through a viscous medium, three forces act on it as:

1. Weight of body acting vertically downwards

2. Upthrust due to buoyancy\[=\]weight of liquid displaced

3. Viscous drag (\[{{F}_{V}}\]) acting in the direction opposite to the motion of the body.

Let, \[\rho =\]Density of material

\[r=\]Radius of spherical body

So, \[{{\rho }_{0}}=\]Density of Medium.

\[\therefore \]True weight of the body\[=w=volume\times density\times g\]

\[W=\frac{4}{3}\pi {{r}^{3}}\rho g\]

Upward thrust, \[{{F}_{T}}\]= Volume of Medium displaced

\[{{F}_{T}}=\frac{4}{3}\pi {{r}^{3}}{{\rho }_{0}}g\]

\[V=\]Terminal velocity of body

According to Stoke’s law

\[{{F}_{V}}=6\pi \eta rv\]

When the body attains terminal velocity, then

\[{{F}_{T}}+{{F}_{V}}=W\]

\[\Rightarrow \frac{4}{3}\pi {{r}^{3}}{{\rho }_{0}}g+6\pi \eta rv=\frac{4}{3}\pi {{r}^{3}}\rho g\]

\[V=\frac{2{{r}^{2}}\left( s-{{\rho }_{0}} \right)g}{9\eta }\]

1) \[V\] directly depends on the radius of the body and the difference of the pressure of material and medium.

2) \[V\] inversely depends of coefficient of viscosity.

Therefore, terminal velocity can be described by expression \[V=\frac{2{{r}^{2}}\left( s-{{\rho }_{0}} \right)g}{9\eta }\].

6. What is equation of continuity? Water flows through a horizontal pipe of radius, \[1cm\] at a speed of \[2m/s\]. What should be the diameter of the nozzle if water is to come out at a speed of \[10m/s\]?

Ans: Assume a non-viscous liquid in streamline flow through a tube A B of varying cross-section.

Let, \[{{a}_{1}},{{a}_{2}}=area\text{ of cross-section at A and B}\]

\[{{v}_{1}},{{v}_{2}}=\] Velocity of flow of liquid at A and B

\[{{\rho }_{1}},{{\rho }_{2}}=\] Density of liquid at A and B

Volume of liquid entering per second at A \[={{a}_{1}}{{v}_{1}}\]

Mass of liquid entering per second at A \[={{a}_{1}}{{v}_{1}}{{\rho }_{1}}\]

Mass of liquid entering per second at B \[={{a}_{2}}{{v}_{2}}{{\rho }_{2}}\]

If there is no loss of liquid in tube and flow is steady, then

Mass of liquid entering per second at A \[=\] Mass of liquid leaving per second at B

\[{{a}_{1}}{{v}_{1}}{{\rho }_{1}}={{a}_{2}}{{v}_{2}}{{\rho }_{2}}\]

If the liquid is incompressible,

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

\[{{a}_{1}}{{v}_{1}}\rho ={{a}_{2}}{{v}_{2}}\rho \]

\[\Rightarrow {{a}_{1}}{{v}_{1}}={{a}_{2}}{{v}_{2}}\]

or \[av=\] constant

That is, \[V\propto \frac{1}{a}\]

It means the larger the area of cross-section, the smaller will be the flow of liquid.

Here \[{{D}_{1}}=2{{r}_{1}}=2\times 1=2cm\]

\[{{D}_{2}}=?\]

\[{{v}_{1}}=2m/s\]

\[{{v}_{2}}=10m/s\]

\[{{D}_{1}}=\] Diameter

\[{{r}_{1}}=\] Radius

\[v=\] velocity

\[a=\pi {{r}^{2}},\] \[D=2r\]

\[a=\pi \frac{{{d}^{2}}}{4},\] \[\frac{D}{2}=r\]

From equation of continuity

\[{{a}_{1}}{{v}_{1}}={{a}_{2}}{{v}_{2}}\]

\[\Rightarrow {{\left( \pi \frac{{{D}_{1}}^{2}}{4} \right)}^{2}}\times {{v}_{1}}={{\left( \pi \frac{{{D}_{2}}^{2}}{4} \right)}^{2}}\times {{v}_{2}}\]

\[\Rightarrow {{D}_{2}}={{D}_{1}}{{\left( \frac{{{v}_{1}}}{{{v}_{2}}} \right)}^{\frac{1}{2}}}\]

\[\Rightarrow {{D}_{2}}=2{{\left( \frac{2}{10} \right)}^{\frac{1}{2}}}=2\times \frac{1}{\sqrt{5}}\]

\[\therefore {{D}_{2}}=0.894cm\]

Therefore, the diameter of the nozzle is \[0.894cm\].

7. What is Bernoulli’s theorem? Show that the sum of pressure, potential and kinetic energy in the streamline flow is constant. 

Ans: According to this theorem, for the streamline flow of an ideal liquid, the total energy that is the sum of pressure energy, potential energy and kinetic energy per unit mass remains constant at every cross-section throughout the flow.

Assume a tube A B of varying cross – section.

\[{{p}_{1}}=\] Pressure applied on liquid at A

\[{{p}_{2}}=\] Pressure applied on liquid at B

\[{{a}_{1}},{{a}_{2}}=\] Area of cross – section at A & B

\[{{h}_{1}},{{h}_{2}}=\] height of section A and B from the ground.

\[{{v}_{1}},{{v}_{2}}=\] Normal velocity of liquid at A and B

\[\rho =\] Density of ideal liquid

Let \[{{p}_{1}}>{{p}_{2}}\]

\[m=\] Mass of liquid crossing per second through any section of tube.

\[{{a}_{1}}{{v}_{1}}\rho ={{a}_{2}}{{v}_{2}}\rho =m\]

or \[{{a}_{1}}{{v}_{1}}={{a}_{2}}{{v}_{2}}=\frac{m}{\rho }=v\]

As \[{{a}_{1}}>{{a}_{2}}\], \[\therefore {{v}_{2}}>{{v}_{1}}\]

Force of on liquid at A \[={{p}_{1}}{{a}_{1}}\]

Force on liquid at B \[={{p}_{2}}{{a}_{2}}\]

Work done per second on liquid at A \[={{p}_{1}}{{a}_{1}}\times {{v}_{1}}={{p}_{1}}v\] 

Work done per second on liquid at B \[={{p}_{2}}v\]

Net work done per second by pressure energy in moving the liquid from A to B \[={{p}_{1}}v-{{p}_{2}}v\] ………………………. (1)

If ‘\[m\]’ mass of liquid flows in one second from A to B then Increases in potential energy per second from A to B \[=mg{{h}_{2}}-mg{{h}_{1}}\] …………….. (2)

Increase in kinetic energy per second of liquid,

from A to B \[=\frac{1}{2}m{{v}_{2}}^{2}-\frac{1}{2}m{{v}_{1}}^{2}\] …………………… (3)

From, work energy principle:\[\text{Work done by pressure energy}=\text{Increase in P}\text{.E}\text{./sec}+\text{Increase in K}\text{.E}\text{./sec}\] From equation \[\text{1,2 }\!\!\And\!\!\text{ 3}\]

\[{{\text{p}}_{1}}v-{{p}_{2}}v=\left( mg{{h}_{2}}-mg{{h}_{1}} \right)+\frac{1}{2}m{{v}_{2}}^{2}-\frac{1}{2}m{{v}_{1}}^{2}\]

\[\Rightarrow {{\text{p}}_{1}}v+mg{{h}_{1}}+\frac{1}{2}m{{v}_{1}}^{2}={{p}_{2}}v+\frac{1}{2}m{{v}_{2}}^{2}+mg{{h}_{2}}\]

Dividing throughout by \[m,\]

\[\frac{{{\text{p}}_{1}}v}{m}+g{{h}_{1}}+\frac{1}{2}{{v}_{1}}^{2}=\frac{{{p}_{2}}v}{m}+\frac{1}{2}{{v}_{2}}^{2}+g{{h}_{2}}\]

\[\Rightarrow \frac{{{\text{p}}_{1}}}{\rho }+g{{h}_{1}}+\frac{1}{2}{{v}_{1}}^{2}=\frac{{{p}_{2}}}{\rho }+\frac{1}{2}{{v}_{2}}^{2}+g{{h}_{2}}\]

Density, \[{{\rho }_{1}}=\frac{m}{v}\]

So, \[\frac{p}{\rho }+gh+\frac{1}{2}{{v}^{2}}=\]Constant …………………………….. (4)

Where,

\[\frac{p}{\rho }=\] Pressure energy per unit mass

\[gh=\] potential energy per unit mass

\[\frac{1}{2}{{v}^{2}}=\] kinetic energy per unit mass

Therefore, from equation (4), that is \[\frac{p}{\rho }+gh+\frac{1}{2}{{v}^{2}}=\]a constant, Bernoulli’s theorem is proved.

8. In a horizontal pipeline of uniform area of cross – section, the pressure falls by \[5N/{{m}^{2}}\] between two points separated by a distance of \[1km\]. What is the change in kinetic energy per Kg of oil flowing at these points? Given the Density of oil \[=800kg/{{m}^{3}}\].

Ans: According to Bernoulli’s theorem, total energy is conserved,

\[\frac{p}{\rho }+gh+\frac{1}{2}{{v}^{2}}=\text{Constant}\]

For a horizontal pipe, \[h=0\]

\[\frac{p}{\rho }+\frac{1}{2}{{v}^{2}}=\text{Constant}\]

At ends 1 and 2,

\[\frac{{{p}_{1}}}{\rho }+\frac{{{v}_{1}}^{2}}{2}=\frac{{{p}_{2}}}{\rho }+\frac{{{v}_{2}}^{2}}{2}\]

\[\frac{{{p}_{1}}-{{p}_{2}}}{\rho }=\frac{1}{2}\left( {{v}_{2}}^{2}-{{v}_{1}}^{2} \right)\]    …………………………… (1)

Change in K. E. \[=\frac{1}{2}m\left( {{v}_{2}}^{2}-{{v}_{1}}^{2} \right)\]

Change in K. E. per Kg\[=\frac{1}{2}\left( {{v}_{2}}^{2}-{{v}_{1}}^{2} \right)\]

$\Rightarrow $Change in K. E. per Kg \[=\frac{{{p}_{1}}-{{p}_{2}}}{\rho }\]

Given that, \[{{p}_{1}}-{{p}_{2}}=5N/{{m}^{2}}\]

\[\rho =800kg/{{m}^{3}}\]

\[\therefore \text{Change in K}\text{.E}\text{.}=\frac{5}{800}=6.25\times {{10}^{-3}}J/kg\]

Therefore, the change in kinetic energy is \[6.25\times {{10}^{-3}}J/kg\].

9.

a) Water flows steadily along a horizontal pipe at a rate of \[8\times {{10}^{-3}}{{m}^{3}}/s\]. If the area of cross – section of the pipe is \[40\times {{10}^{-4}}{{m}^{2}}\], Calculate the flow velocity of water.

Ans: We know that,

Velocity of water \[=\frac{\text{Rate of flow}}{Area\text{ of cross-section}}\]

Given, Rate of flow \[=8\times {{10}^{-3}}{{m}^{3}}/s\]

Area of cross-Section \[=40\times {{10}^{-4}}{{m}^{2}}\]

So, Velocity of water \[=\frac{8\times {{10}^{-3}}}{40\times {{10}^{-4}}}=2m/s\]

Therefore, the velocity of water is \[2m/s\].

b) Find the total pressure in the pipe if the static pressure in the horizontal pipe is \[3\times {{10}^{4}}Pa\]. Density of water is \[1000kg/{{m}^{3}}\].

Ans: We know that,

\[\text{Total pressure = Static pressure + }\frac{1}{2}\rho {{v}^{2}}\]

So, total pressure\[=3\times {{10}^{4}}+\frac{1}{2}\times 1000\times {{\left( 2 \right)}^{2}}\]          \[\left( \because v=2m/s \right)\]

\[\therefore \text{Total pressure}=3.2\times {{10}^{4}}Pa\]

Therefore, the total pressure in the pipe is \[3.2\times {{10}^{4}}Pa\].

c) What is the net flow velocity if the total pressure is \[3.6\times {{10}^{4}}Pa\]?

Ans: We know that,

\[\text{Total pressure = Static pressure + }\frac{1}{2}\rho {{v}^{2}}\]

\[\Rightarrow \frac{1}{2}\rho {{v}^{2}}=\text{Total pressure}-\text{Static pressure }\]

\[\Rightarrow \frac{1}{2}\times 1000\times {{v}^{2}}=3.6\times {{10}^{4}}-3\times {{10}^{4}}\]

\[\Rightarrow \frac{1}{2}\times {{v}^{2}}=0.6\times {{10}^{4}}\]

\[\therefore v=\sqrt{\frac{2\times 0.6\times {{10}^{4}}}{1000}}=3.5m/s\]

Therefore, the net flow velocity with total pressure of \[3.6\times {{10}^{4}}Pa\] is \[3.5m/s\].

10. A \[50\,kg\] girl wearing high heel shoes balances on a single heel. The heel is circular with a diameter \[1.0\,cm\]. What is the pressure exerted by the heel on the horizontal floor?

Ans: Given that, 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:

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

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

11. A vertical off-shore structure is built to withstand a maximum stress of \[{{10}^{9}}\,Pa\]. Is the structure suitable for putting up on top of an oil well in the ocean? Take the depth of the ocean to be roughly \[3\,km\], and ignore ocean currents.

Ans: Yes.

Given that, 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,

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

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

12. Figure 10.24 (a) shows a thin liquid film supporting a small weight\[4.5\times {{10}^{-2}}N\]. What is the weight supported by a film of the same liquid at the same temperature in Fig. (b) and (c)? Explain your answer physically.

Ans: For case (a): 

From given, the length of the liquid film is supported by the weight \[l\text{ }=\text{ }40\text{ }cm\text{ }=\text{ }0.4\text{ }m\].
The weight supported by the film, \[W=4.5\times \,{{10}^{-2}}N\] 

Every liquid film has two free surfaces. Hence the surface tension can be calculated as 

\[S=\frac{Force}{2l}=\frac{4.5\times \,{{10}^{-2}}N}{2\times \,0.4m}=5.625\times \,{{10}^{-2}}N/m\]

As the properties of the liquid are the same in all the cases, the surface tension in all three cases is equal. An,

Therefore, the weight supported is also the same in all three cases, that is, \[5.625\times \,{{10}^{-2}}N/m\].

13. What is the pressure inside the drop of mercury of radius \[3.0mm\] at room temperature? Surface tension of mercury at that temperature \[(20{}^\circ \,C)\] is \[4.65\times {{10}^{-1}}N{{m}^{-1}}\]. The atmospheric pressure \[1.01\times {{10}^{5}}Pa\]. Also give the excess pressure inside the drop.

Ans: Given that, the radius of the mercury drop, \[r=3\times {{10}^{-3}}m\] 

The surface tension of mercury, \[S=4.65\times \,{{10}^{-1}}N/m\] 

The atmospheric pressure, \[{{P}_{0}}=1.01\times \,{{10}^{5}}Pa\] 

Total pressure inside the mercury drop \[=\]Excess pressure inside mercury \[+\]Atmospheric pressure 

Total pressure \[=\frac{2S}{r}+{{P}_{0}}=\frac{2\times 4.65\times \,{{10}^{-1}}}{3\times {{10}^{-3}}}+1.01\times \,{{10}^{5}}=1.013\times \,{{10}^{5}}Pa\]

Therefore, the excess pressure is obtained as \[=\frac{2S}{r}=310Pa\].

14. In deriving Bernoulli's equation, we equated the work done on the fluid in the tube to its change in the potential and kinetic energy.

a) What is the largest average velocity of blood flow in an artery of diameter \[2.0\times {{10}^{-3}}m\] if the flow must remain laminar? 

Ans: Given that, 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 \[1.966m/s\].

b) Do the dissipative forces become more important as the fluid velocity increases? Discuss qualitatively.

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

4 Marks Questions

1. Explain why 

a) The blood pressure in humans is greater at the feet than at the brain.

Ans: The pressure of a liquid with density\[\rho \], with the liquid column of height \[h\] 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. Therefore, the blood pressure at the feet is more than what it is at the brain.

b) Atmospheric pressure at the height of about \[6\,km\] decreases to nearly half of its value at the sea level, though the height of the atmosphere is more than \[100\,km\]. 

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. Thereafter, 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.

c) Hydrostatic pressure is a scalar quantity even though pressure is force divided by area.

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.

So 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.

2. Fill in the blanks using the word(s) from the list appended with each statement: 

a) Surface tension of liquids generally ……with temperatures. (increases / decreases)

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.

b) Viscosity of gases…… with temperature, whereas viscosity of liquids …… with temperature. (increases / decreases) 

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.

Therefore, the viscosity decreases with an increase in temperature.

c) For solids with elastic modulus of rigidity, the shearing force is proportional to….., while for fluids it is proportional to ……(shear strain / rate of shear strain) 

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.

d) For a fluid in a steady flow, the increase in flow speed at a constriction follows……. (conservation of mass / Bernoulli's principle) 

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.

Therefore, 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.

3. A U-tube contains water and methylated spirit separated by mercury. The mercury columns in the two arms are in level with \[10.0\,cm\]of water in one arm and \[12.5\,cm\]of spirit in the other. What is the specific gravity of spirit?

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

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

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

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

4. In problem 10.9, if \[15.0\,cm\] of water and spirit each are further poured into the respective arms of the tube, what is the difference in the levels of mercury in the two arms? (Specific gravity of mercury \[=13.6\])

Ans: Given 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 \[0.221\text{ }cm\].

5. Figures 10.23 (a) and (b) refer to the steady flow of a (non-viscous) liquid. Which of the two figures is incorrect? Why?

Ans: Figure (a) is incorrect.

In the case given in figure (b),

Where,

\[{{A}_{1}}=\]Area of pipe1

\[{{A}_{2}}=\]Area of pipe 2

\[{{V}_{1}}=\]Speed of the fluid in pipe1

\[{{V}_{2}}=\]Speed of the fluid in pipe 2

From the law of continuity, we have:

\[{{A}_{1}}{{V}_{1}}={{A}_{2}}{{V}_{2}}\]

When the area of cross-section in the middle of the venturimeter is small, the speed of the flow of liquid through this part is more. According to Bernoulli's principle, if speed is more, then pressure is less. Pressure is directly proportional to height. So, the level of water in pipe 2 is less.

Therefore, figure (a) is not possible.

6.

a) What is the largest average velocity of blood flow in an artery of radius \[2.0\times {{10}^{-3}}m\]if the flow must remain laminar? 

Ans: Given 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:

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

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

b) What is the corresponding flow rate? (Take viscosity of blood to be \[2.084\times {{10}^{-3}}Pa\,s\])

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

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

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

7. Mercury has an angle of contact equal to \[\mathbf{140}{}^\circ \]with soda lime glass. A narrow tube of radius \[\mathbf{1}.\mathbf{00}\text{ }\mathbf{mm}\] made of this glass is dipped in a trough containing mercury. By what amount does the mercury dip down in the tube relative to the liquid surface outside? Surface tension of mercury at the temperature of the experiment is \[4.65\times {{10}^{-1}}N{{m}^{-1}}\]. Density of mercury is \[13.6\times {{10}^{3}}\,kg/{{m}^{3}}\].

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

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

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. 

Therefore, the mercury level drops by \[5.34mm\].

8. The cylindrical tube of a spray pump has a cross-section of \[8.0c{{m}^{2}}\] one end of which has \[40\] fine holes each of diameter \[1.0mm\]. If the liquid flow inside the tube is \[1.5m{{\min }^{-1}}\], what is the speed of ejection of the liquid through the holes?

Ans: Given that, area of cross-section of the spray pump, \[{{A}_{1}}=8c{{m}^{2}}=8\times {{10}^{-4}}{{m}^{2}}\]

Number of holes, \[n=40\]

Diameter of each hole, \[d=1mm=1\times {{10}^{-3}}m\]

Radius of each hole, \[r=d/2=0.5\times {{10}^{-3}}m\]

Area of cross-section of each hole, \[a=\pi {{r}^{2}}=\pi {{\left( 0.5\times {{10}^{-3}} \right)}^{2}}{{m}^{2}}\]

Total area of 40 holes, \[{{A}_{2}}=n\times a\]

\[\Rightarrow {{A}_{2}}=40\times \pi {{\left( 0.5\times {{10}^{-3}} \right)}^{2}}{{m}^{2}}\]

\[\Rightarrow {{A}_{2}}=31.41\times {{10}^{-6}}{{m}^{2}}\]

Speed of flow of liquid inside the tube, \[{{V}_{1}}=1.5m/\min =0.025m/s\]

Speed of ejection of liquid through the holes \[={{V}_{2}}\]

According to the law of continuity, we have:

\[{{A}_{1}}{{V}_{1}}={{A}_{2}}{{V}_{2}}\]

\[{{V}_{2}}=\frac{{{A}_{1}}{{V}_{1}}}{{{A}_{2}}}\]

\[\Rightarrow {{V}_{2}}=\frac{8\times {{10}^{-4}}\times 0.025}{31.41\times {{10}^{-6}}}\]

\[\therefore {{V}_{2}}=0.636m/s\]

Therefore, the speed of ejection of the liquid through the holes is \[0.636m/s\].

9. In Millikan's oil drop experiment, what is the terminal speed of an uncharged drop of radius \[2.0\times {{10}^{-5}}\,m\]and density\[1.2\times {{10}^{3}}\,kg/{{m}^{3}}\]. Take the viscosity of air at the temperature of the experiment to be \[1.8\times {{10}^{-5}}Pa\,s\]. How much is the viscous force on the drop at that speed? Neglect buoyancy of the drop due to air.

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:

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

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

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

5 Marks Questions

1. State the principle on which Hydraulic lift work and explain its working.

Ans: The hydraulic lift works on the principle of the Pascal’s law. According to this law, in the absence of gravity, the pressure is same at all points inside the liquid lying at the same horizontal plane.

Working of Hydraulic effect:

\[a=\] Area of cross \[-\] section of piston at C

\[A=\] Area of cross \[-\] section of piston at D.

Let a downward force f be applied on the piston C. Then the pressure exerted on the liquid, \[P=\frac{f}{a}\]

According to Pascal’s law, this pressure is transmitted equally to the piston of cylinder D.

\[\therefore \] Upward force acting on the piston of cylinder D will be,

\[F=PA\]

\[\Rightarrow F=\frac{f}{a}A\]

As \[A\gg a\],\[F\gg f\].

That is, small force applied on the smaller piston will appear as a very large force on the large piston. As a result of this heavy load placed on larger piston is easily lifted upwards.

2. Show that if two soap bubbles of radii a and b coalesce to form a single bubble of radius c. If the external pressure is P, show that the surface tension T of soap solution is \[T=\frac{P\left( {{c}^{3}}-{{a}^{3}}-{{b}^{3}} \right)}{4\left( {{a}^{2}}+{{b}^{2}}-{{c}^{2}} \right)}\].

Ans: We can say that, pressure inside the bubble of radius, \[a={{P}_{1}}=P+\frac{4T}{a}\]

Volume of bubble of radius a, \[{{V}_{1}}=\frac{4}{3}\pi {{a}^{3}}\]

Pressure inside the bubble of radius, \[b={{P}_{2}}=P+\frac{4T}{b}\]

Volume of the bubble of radius, \[{{V}_{2}}=\frac{4}{3}\pi {{b}^{3}}\]

Pressure inside the bubble of radius, \[C={{P}_{3}}=P+\frac{4T}{c}\]

Volume of bubble of radius C, \[{{V}_{3}}=\frac{4}{3}\pi {{c}^{3}}\]

Since, temperature remains the same during the change, from Boyle’s Law:

\[{{P}_{1}}{{V}_{1}}+{{P}_{2}}{{V}_{2}}={{P}_{3}}{{V}_{3}}\]

\[\left( P+\frac{4T}{a} \right)\times \frac{4}{3}\pi {{a}^{3}}+\left( P+\frac{4T}{b} \right)\times \frac{4}{3}\pi {{b}^{3}}=\left( P+\frac{4T}{c} \right)\times \frac{4}{3}\pi {{c}^{3}}\]

\[\frac{4}{3}P\pi {{a}^{3}}+\frac{16T\pi {{a}^{2}}}{3}+\frac{4P\pi {{b}^{3}}}{3}+\frac{16T\pi {{b}^{2}}}{3}=\frac{4P\pi {{c}^{3}}}{3}+\frac{16T\pi {{c}^{2}}}{3}\]

\[\frac{4}{3}P\pi {{a}^{3}}+\frac{16T\pi {{a}^{2}}}{3}+\frac{4P\pi {{b}^{3}}}{3}+\frac{16T\pi {{b}^{2}}}{3}-\frac{4P\pi {{c}^{3}}}{3}-\frac{16T\pi {{c}^{2}}}{3}=0\]

By taking \[\frac{4\pi }{3}\] common from above equation,

\[P{{a}^{3}}+4T{{a}^{2}}+4{{b}^{3}}+4T{{b}^{2}}-P{{c}^{3}}-4T{{c}^{2}}=0\]

\[\Rightarrow P\left( {{a}^{3}}+{{b}^{3}}-{{c}^{3}} \right)+4T\left( {{a}^{2}}+{{b}^{2}}-{{c}^{2}} \right)=0\]

\[\Rightarrow -P\left( {{a}^{3}}+{{b}^{3}}-{{c}^{3}} \right)=4T\left( {{a}^{2}}-{{b}^{2}}-{{c}^{2}} \right)\]

\[4T\left( {{a}^{2}}+{{b}^{2}}-{{c}^{2}} \right)=P\left( {{c}^{3}}-{{a}^{3}}-{{b}^{3}} \right)\]

\[\therefore T=\frac{P\left( {{c}^{3}}-{{a}^{3}}-{{b}^{3}} \right)}{4\left( {{a}^{2}}+{{b}^{2}}-{{c}^{2}} \right)}\]

Hence proved.

3. Explain why 

a) The angle of contact of mercury with glass is obtuse, while that of water with glass is acute

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.

b) Water on a clean glass surface tends to spread out, while mercury on the same surface tends to form drops. (Put differently, water wets glass while mercury does not.) 

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.

Therefore, the water molecules stay closer to glass molecules as well, keeping the glass surface wet.

c) The surface tension of a liquid is independent of the area of the surface.

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. 

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

d) Water with detergent dissolved in it should have small angles of contact. 

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. 

That is, when \[{{S}_{sl}}<{{S}_{la}}\] and the value of \[\cos \theta =\frac{{{S}_{sa}}-{{S}_{sl}}}{{{S}_{la}}}\] increases, it leads 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. 

Therefore, a droplet will tend to become spherical.

4. Explain why 

a) To keep a piece of paper horizontal, you should blow over, not under, it.

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. Because of this, one should blow on the upper side of the paper.

b) When we try to close a water tap with our fingers, fast jets of water gush through the openings between our fingers.

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 so the water gushes out with a lot of speed.

c) The size of the needle of a syringe controls flow rate better than the thumb pressure exerted by a doctor while administering an injection. 

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

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

d) A fluid flowing out of a small hole in a vessel results in a backward thrust on the vessel.

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.

So, 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. So, the ball deviates from the expected parabolic trajectory.

5. Glycerin flows steadily through a horizontal tube of length \[1.5\,m\] and radius \[1.0\,cm\]. If the amount of glycerin collected per second at one end is\[4.0\times {{10}^{-3}}kg/s\], what is the pressure difference between the two ends of the tube? (Density of glycerin \[1.3\times {{10}^{3}}\,kg/{{m}^{3}}\] and viscosity of glycerin\[=0.83\,Pa\,s\]). (You may also like to check if the assumption of laminar flow in the tube is correct.)

Ans: Given that,

Length of the horizontal tube,  \[l=1.5\text{ }m\]

The radius of the tube, \[r=1\,cm=\,0.01m\] 

Thus, the diameter of the tube, \[\therefore d=0.02m\] 

As given in the question, Glycerine is flowing at a rate of \[M=4.0\times {{10}^{-3}}kg\,{{s}^{-1}}\] 

The density of Glycerine, \[\rho =1.3\times {{10}^{3}}kg\,{{m}^{-3}}\]

Viscosity of Glycerine\[\eta =0.83\text{ }Pa\text{ }s\]

The volume of Glycerine flowing per sec: 

\[V=M/\rho =\frac{4.0\times {{10}^{-3}}}{1.3\times {{10}^{3}}}=3.08\times {{10}^{-6}}{{m}^{3}}{{s}^{-1}}\]

According to Poiseuille's Law, we have the expression for the rate of flow:

\[V=\frac{\pi P{{r}^{4}}}{8\eta l}\]

here, P is the pressure difference between the two ends of the tube

\[P=\frac{V8\eta l}{\pi {{r}^{4}}}=\text{ }9.8\text{ }\times \text{ }{{10}^{2}}Pa\] 

Reynolds' number is given by the following expression: 

\[\frac{4\rho V}{\pi d\eta }=\frac{4(1.3\times {{10}^{3}})(3.08\times {{10}^{-6}})}{\pi (0.02)0.83}=0.3\]

Reynolds' number is about \[0.3\]; Therefore, the flow is laminar.

6. In a test experiment on a model aero plane in a wind tunnel, the flow speeds on the upper and lower surfaces of the wing are \[70\,m/s\,\]and  \[63\,m/s\] respectively. What is the lift on the wing if its area is\[2.5{{m}^{2}}\]? Take the density of air to be\[1.3\,kg{{m}^{-3}}\].

Ans: The given speed of wind on the wing's upper surface, \[{{v}_{1}}=70m\text{ }{{s}^{-1}}\]

and the speed of wind relative to the wing's lower surface, \[{{v}_{2}}=63m\text{ }{{s}^{-1}}\].

The area of the wing, \[A=2.5{{m}^{2}}\]

The density of air, \[\rho =1.3kg/{{m}^{3}}\]

Now, according to Bernoulli's theorem, we have the relation: \[{{P}_{1}}+\frac{1}{2}\rho {{v}_{1}}^{2}={{P}_{2}}+\frac{1}{2}\rho {{v}_{2}}^{2}\]

In the current case, let \[{{P}_{2}}\] be the pressure on the lower side of the wing and \[{{P}_{1}}\] be the pressure on the upper side, and the v-s are the velocities mentioned initially. 

The required lift is the force that comes from the pressure difference in the lower and the upper surface. 

This can be obtained as:

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

\[Lift\,\,=\frac{1}{2}\rho ({{v}_{2}}^{2}-{{v}_{1}}^{2})A=\frac{1}{2}1.3({{70}^{2}}-{{63}^{2}})2.5=1.51\times {{10}^{3}}N\]

Therefore, \[1.51\times {{10}^{3}}N\]is the lift on the wing of the airplane.

7. What is the excess pressure inside a bubble of soap solution of radius\[5.0mm\], given that the surface tension of soap solution at the temperature \[(20{}^\circ \,C)\] is \[4.65\times {{10}^{-1}}N{{m}^{-1}}\]. If an air bubble of the same dimension were formed at the depth of \[40.0\,cm\]inside a container containing the soap solution (of relative density\[1.20\]), what would be the pressure inside the bubble? (1 atmospheric pressure is \[1.01\times {{10}^{5}}Pa\].)

Ans: Given that,

The excess pressure inside the soap bubble is \[20\text{ }Pa\]

The pressure inside the air bubble is \[=1.06\times \,{{10}^{5}}Pa\]

A soap bubble is of radius, \[r=5\times {{10}^{-3}}m\]

The surface tension of the soap solution, \[S=2.5\times \,{{10}^{-2}}N/m\]

Also, the relative density of the soap solution \[=\text{ }1.20~\]

Thus, the density of the soap solution, \[\rho =1.2\times {{\rho }_{air}}=1.2\times {{10}^{3}}kg/{{m}^{2}}\]

It is given that the air bubble is formed at a depth, \[h\text{ }=\text{ }40\text{ }cm\text{ }=\text{ }0.4\text{ }m\]

The radius of the air bubble, \[r=5\times {{10}^{-3}}m\] 

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

Therefore, it can be noted that the excess pressure inside the soap bubble is given by:

\[P=\frac{4S}{r}=20Pa\]

The excess pressure inside the air bubble can be written as: 

\[P=\frac{2S}{r}=10Pa\].

The total pressure inside the air bubble under consideration is the sum of the atmospheric pressure, the pressure due to the column of the liquid and the excess pressure:

\[{{P}_{total}}={{P}_{0}}+hg\rho +P=1.057\times \,{{10}^{5}}Pa\]

Therefore, the pressure inside the air bubble is \[1.057\times \,{{10}^{5}}Pa\].

8. A tank with a square base of area \[1.0{{m}^{2}}\]is divided by a vertical partition in the middle. The bottom of the partition has a small-hinged door of area \[20.0\,c{{m}^{2}}\] . The tank is filled with water in one compartment, and an acid (of relative density $1.7$) in the other, both to a height of \[4.0\,m\]. Compute the force necessary to keep the door closed. 

Ans: Given 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: \[\Delta P={{P}_{2}}-{{P}_{1}}=2.744\times 1\,{{0}^{4}}Pa\]

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

Therefore, the force necessary to keep the door closed is \[54.88N\].

9. A manometer reads the pressure of a gas in an enclosure as shown in figure When a pump removes some of the gas, the manometer reads as in figure (b) The liquid used in the manometers is mercury and the atmospheric pressure is 76 cm of mercury

a) Give the absolute and gauge pressure of the gas in the enclosure for cases (a) and (b), in units of cm of mercury. 

Ans: In the first case, that is, 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. 

Therefore, the gauge pressure is $20cm$of Hg. 

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

Therefore, absolute pressure is \[96\text{ }cm\text{ }Hg\].

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

Therefore, gauge pressure is \[=18\text{ }cm\] Hg. 

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

Therefore, absolute pressure is \[\text{58 }cm\text{ }Hg\].

b) How would the levels change in case (b) if 13.6 cm of water (immiscible with mercury) are poured into the right limb of the manometer? (Ignore the small change in the volume of the gas.)

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: 

\[77=58+h\] 

$\Rightarrow h=19cm$
Therefore, the mercury levels differ by \[19cm\].

10. A plane is in level flight at a constant speed and each of its two wings has an area of \[25{{m}^{2}}\]. If the speed of the air is \[180\,km/h\]over the lower wing and \[234\,km/h\]over the upper wing surface, determine the plane's mass. (Take air density to be\[\,1kg/{{m}^{3}}\].)

Ans: Given that, 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: 

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

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

11. Two narrow bores of diameters \[\mathbf{3}.\mathbf{0}\text{ }\mathbf{mm}\] and \[6.\mathbf{0}\text{ }\mathbf{mm}\] are joined together to form a U-tube open at both ends. If the U-tube contains water, what is the difference in its levels in the two limbs of the tube? Surface tension of water at the temperature of the experiment is \[7.31\times {{10}^{-2}}N{{m}^{-1}}\]. Let the angle of contact to be zero and density of water be \[1.0\times {{10}^{3}}\,kg/{{m}^{3}}\]. 

Ans: Given 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: 

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

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

12.

a) It is known that density \[\mathbf{\rho }\] of air decreases with height \[y\] as \[{{\mathbf{\rho }}_{0}}{{e}^{y/{{y}_{0}}}}\]. Where \[{{\mathbf{\rho }}_{0}}=1.25kg\,{{m}^{-3}}\]the density at sea is level, and \[{{y}_{0}}\] is a constant. This density variation is called the law of atmospheres. Obtain this law assuming that the temperature of atmosphere remains a constant (isothermal conditions). Also assume that the value of \[g\] remains constant. 

Ans: Given that, volume of the balloon is, \[V\text{ }=\text{ }1425\text{ }{{m}^{3}}\]

Mass of the payload, \[m=400kg\] 

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

\[{{y}_{0}}=8000m\]

The density of helium is\[{{\rho }_{He}}=0.18kg\,{{m}^{-3}}\] , and that of air at sea level is \[{{\rho }_{0}}=1.25kg\,{{m}^{-3}}\]

Let the density of the balloon be ρ and the height to which the balloon can rise be y. 

The air density (\[\rho \]) decreases with increasing height (\[y\]) as: 

\[\rho /{{\rho }_{0}}={{e}^{-y/{{y}_{0}}}}\]

This relation is also called the law of atmospherics.

It can be inferred from the above law-equation that the rate of decrease of density with height is directly proportional to \[\rho \],

\[-\frac{d\rho }{dy}\propto \rho \]

\[\Rightarrow -\frac{d\rho }{dy}=k\rho \] 

\[\Rightarrow -\frac{d\rho }{\rho }=kdy\]

Here, k is a proportionality constant. 

We set the limits such that the height changes from \[0\] to \[y\], and the density varies from \[{{\rho }_{0}}\]to \[\rho \].

\[\int\limits_{{{\rho }_{0}}}^{\rho }{\frac{d\rho }{\rho }=}\int\limits_{0}^{y}{kdy}\]

Next, integrating the sides between these limits, we obtain:

\[\left[ \ln \,\rho  \right]_{{{\rho }_{0}}}^{\rho }=-ky\]

\[\Rightarrow \rho /{{\rho }_{0}}={{e}^{-ky}}\]

Comparing the above equation with the law of atmospherics, we get

\[\Rightarrow k=\frac{1}{{{y}_{0}}}\]

Therefore, we have been able to derive the expression for the law of atmospherics.

b) A large He balloon of volume \[1425{{m}^{3}}\] is used to lift a payload of\[\mathbf{400}\text{ }\mathbf{kg}\]. Assume that the balloon maintains a constant radius as it rises. How high does it rise? 

[Take \[{{y}_{0}}=8000m\]and \[{{\mathbf{\rho }}_{He}}=0.18kg\,{{m}^{-3}}\] ]

Ans: We know that, the generic expression for mass density is:

\[\rho =\frac{M}{V}\]

In the current case, it becomes the total mass (including the payload and the helium) per unit volume of the balloon:

\[\rho =\frac{({{M}_{payload}}+{{M}_{He}})}{V}=\frac{m+V{{\rho }_{He}}}{V}\]

\[\Rightarrow \rho =0.46kg/{{m}^{3}}\]

Using the law of atmospherics, we can obtain the value of \[y\]:

\[\rho /{{\rho }_{0}}={{e}^{-y/{{y}_{0}}}}\]

\[\Rightarrow \ln \,\rho /{{\rho }_{0}}=-y/{{y}_{0}}\]

\[\therefore y=-8000\times ln(0.46/1.25)=8000m=8km\] 

Therefore, the balloon can rise to a height of \[8km\].

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