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**Mechanical Properties of Fluids**

**Mechanical Properties of Fluids****Chapter: 9**

Part – II |

**EXERCISE**

**1. Explain why:**

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

Ans: The height of the bloods column is quite large at feet than that at the brain. That is why, the blood exerts more pressure at the feet than at the brain.

**(b) Atmospheric pressure at a 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: Density of air decreases very rapidly with increase in height and reduces to nearly half its value at the sea level at a height of about 6 km. After 6 km, the density of air decreases but rather very slowly. This is the reason that at a height of about 6 km, atmospheric pressure reduces to nearly half its valve at the sea level.

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

Ans: When external force is applied on any fluid, the particles of the fluid are transmitted in all possible directions. According to Pascal’s law pressure is transmitted equally in all directions, that is the direction is not associated with the pressure. Hence it is not a scalar quantity.

**2. Explain why:**

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

Ans: When a liquid comes in contact with solid, three surfaces are formed namely liquid-air, solid-air and solid-liquid. The surface tension corresponding to these three interfaces i.e., S_{LA}, S_{SA} and S_{SL} are related of contact θ as,

Cos θ = S_{SA} – S_{SA} / S_{LA}.

In case of mercury glass, S_{SA}, <S_{SL}, therefore, cos θ is the negative or θ > 0, obtuse. In case of water glass S_{SA} > S_{SL}, therefore cos θ is positive < 90° I.e., 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: In the case of mercury the mercury glass, angle of contact is obtuse. To obtain this obtuse value of angle of contact, mercury tends to form a drop. In the case of water taken in a glass tube, adhesive forces are stronger than cohesive forces. Therefore, the water molecule wets the glass.

**(c) Surface tension of a liquid is independent of the area of the surface.**

Ans: The surface tension of a liquid is the force acting per unit length on a line drawn tangentially to the liquid surface which is at rest. This force is independent of the area of the liquid surface. Hence, surface tension is also independent of the area of the liquid surface.

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

Ans: As we know that, the clothes have narrow spaces in the form of capillaries. The rise of liquid in a capillary tube is directly proportional to cos θ. If θ is small, cos θ will be large. Due to which capillary rise will be more and so the detergent will penetrate more in clothes.

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

Ans: In the absence of external forces, the surface of the liquid drop tends to acquire the minimum surface area due to surface tension. The surface area of a sphere is the minimum for a given volume. Hence, under no external forces, liquid drops always take a spherical shape.

**3. 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.

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

Ans: Viscosity of gases increases with temperature, whereas viscosity of liquids decreases with 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: For solids with elastic modulus of rigidity, the shearing force is proportional to shear strain, while for fluids it is proportional to rate of shear strain.

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

Ans: Bernoulli’s principle.

**(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.

**4. Explain why:**

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

Ans: When we blow over the paper, the velocity of air below increases. As a result From the Bernoulli’ theorem, the pressure over it decreases whereas the atmospheric pressure below remains the same. Hence the paper stays horizontal.

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

Ans: When we try to close a tap of water with our fingers, fast jets of water gush through the openings between our fingers. The area of outlet of water eject is reduced, so velocity of water increases according to equation of continuity, av = constant. Hence, area and velocity are inversely proportional to each other.

**(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: Here, the size of the needle controls velocity of flow and thumb pressure controls pressure.

P + pgh + ½ pv2 = a constant, shows that p

Occurs with powers one and v occurs with power two, hence the velocity has more influence. The influence of the needle size on flow rate is thus much greater, and the velocity change due to constriction also highlights why needle size is a critical factor in controlling flow rate. That is why needle has a better control over flow.

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

Ans: **A fluid flowing out from a small hole has a large velocity according to the equation of continuity:**

Area × velocity = C

The backward thrust on the vessel is a result of the reaction to the fluid being expelled, which is an external force effect due to momentum conservation.

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

Ans: A spinning cricket ball would have followed a parabolic trajectory if there was no air. But because of air, the Magnus effect takes place. Thus, a pressure difference is created due to which the trajectory of the cricket ball does not remain parabolic. The parabolic trajectory remains operative only when gravitational force acts on the ball.

**5. 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: **The formula for pressure:**

P = F/A

**Calculate the force (F):**

F = m **×** g

For a mass m = 50 kg

g = 9.8 m/s^{2}

F = 50 kg **×** 9.8 m/s^{2 }= 490 N

Calculate the area (A)

Diameter = 1.0cm = 0.01m

**The radius r of the heel is:**

r = Diameter / 2

= 0.01m / 2

= 0.005 m

Calculation of A

A = 𝜋r^{2} = 𝜋(0.005 m)^{2}

= 𝜋 **×** 2.5 **×** 10^{-5} m^{2}

= 7.85 **×** 10^{-5} m^{2}

Pressure (p)

P = F/A = 490 N/ 7.85 **×** 10^{-5} M^{2}

= 6.24 **×** 10^{6} Pa.

**6. Toricelli’s barometer used mercury. Pascal duplicated it using French wine of density 984 kg m**^{–3}**. Determine the height of the wine column for normal atmospheric pressure.**

Ans: The formula

= P = p/gh

**Given that:**

P is the atmospheric pressure, we know that the normal atmospheric pressure the accepted value is, (101,325 pa)

*P *= 984 kg / m^{-3 }

g = 9.8 m/s^{2}

**Calculation for h:**

h = P/*p*g

h = 101,325 Pa / 984 kg /m^{-3} **×** 9.8 m/s^{2}

= 101325/9,633.2

= 10.52 m.

**7. A vertical off-shore structure is built to withstand a maximum stress of 109 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: **The formula:**

P = ρ **×** g **×** h

**where:**

P = pressure

ρ = density of seawater (approximately 1030 kg/m³)

g = 9.8 m/s²

h = depth of the ocean (3 km = 3000 m)

P = 1030 kg/m³ **×** 9.8 m/s² **×** 3000 m

(Since 1 Pa = 1 N/m²)

3.02 **×** 10^{7} Pa

= 3.02 **×** 10^{7} N/m².

This value of pressure is less than the stress of 10^{9} Pa which the structure can withstand. The structure is suitable for putting up on top of an oil well in the ocean.

**8. 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 425 cm**^{2}**. What maximum pressure would the smaller piston have to bear?**

Ans: **The formula:**

Pressure = Force / Area

Force exerted by car’s weight:

Force = Mass **×** Gravity

= 3000 kg **×** 9.8 m/s^{2}

= 29400 N

**Calculation of pressure (P):**

Pressure = Force / Area

(Area = 425 cm² = 0.0425 m²)

= 29400 N / 0.0425 m^{2 }

= 29400 N / 0.0425 m^{2}

= 692,059 Pa.

**9. A U-tube contains water and methylated spirit separated by mercury. The mercury columns in the two arms are 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: **Given that:**

Height of water (h_{w}) = 10.0cm = 0.1 m

Height of spirit column (h_{s}) = 12.cm = 0.125m

Density of water (p_{w}) = 1000 kg/m^{3}

Density of mercury (p_{m}) = 13,600 kg/m^{2}

= P_{w}gh_{w} = p_{s}gh_{s} + p_{m}gh_{m}

Here g, is the common, thus

= P_{w}h_{w} = p_{s}h_{s} + p_{m}h_{m}

Here, hm is the same in both arms thus,

= P_{w}h_{w} = p_{s}h_{s}

**The density of the spirit:**

= P_{s} = p_{w}h_{w} / h_{s}

= P_{s }= 1000kg/m^{3} **×** 0.1 m / 0.125 m

= P_{s} = 100 / 0.125

P_{s} = 800 kg/m^{3}

Calculation the specific gravity

= P_{s} / P_{w}

= 800 kg /m^{3} / 1000 kg / m^{3}

= 0.8.

**10. In the previous problem, 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:**

**Initial situation:**

Water level = 15.0 cm

Spirit level = 15.0 cm

Mercury level difference = 0

Density of water = 1 g/cm³

Density of spirit = 0.8 g/cm³

Density of mercury = 13.6 g/cm³

Pressure due to water column = height **×** density **×** gravity

= 15.0 cm **×** 1 g/cm³ **×** g

= 15.0 g/cm²

Pressure due to spirit column = height **×** density **×** gravity

= 15.0 cm **×** 0.8 g/cm³ **×** g

= 12.0 g/cm²

**Calculate the pressure difference:**

Pressure difference = 15.0cm – 12.0 g /cm3 = 3.0 g/cm2

**Calculate the height of mercury column:**

Pressure difference = density of mercury **×** height of mercury

= 3. 0 g/cm2 = 13.6 cm3 **×** height of mercury

= 3.0 g/cm2 / 13.6 g/cm3

= 0.220 cm.

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

Ans: Bernoulli’s equation can describe the flow of water through a rapid in terms of the relationship between pressure, velocity, and elevation. Since the flow of water in a river in a rapid way can not be treated as streamlined motion, Bernoulli’s equation cannot be accurately applied.

**12. Does it matter if one uses gauge instead of absolute pressures in applying Bernoulli’s equation? Explain.**

Ans: No, it does not matter whether you use gauge pressure or absolute pressure when applying Bernoulli’s equation, as long as you are consistent throughout the equation.

**13. Glycerine flows steadily through a horizontal tube of length 1.5 m and radius 1.0 cm. If the amount of glycerine collected per second at one end is 4.0 × 10**^{–3}** kg s**^{–1}** , what is the pressure difference between the two ends of the tube? (Density of glycerine = 1.3 × 10**^{3}** kg m**^{–3}** and viscosity of glycerine = 0.83 Pa s). [You may also like to check if the assumption of laminar flow in the tube is correct].**

Ans: **Given:**

L = 1.5m

r = 1.0cm = 0.01m

m = 4.0 **×** 10^{-3} kg s^{-1}

ρ = 1.3 **×** 10^{3} kg m^{-3}

η = 0.83 Pa s

**Calculate the volumetric flow rate:**

Q = m/p

= 4.0 **×** 10^{-3} / 1.3 **×** 10^{3}

= 3.08 **×** 10^{-6} m^{3}s^{-1}

**Use Poiseuille’s Law:**

Q = 𝜋r^{4}ΔP / 8ηL

**Rearranging to solve for the pressure difference ΔP:**

= ΔP = 8ηLQ / 𝜋r^{4}

ΔP = 8 **×** 0.83 Pa s **×** 1.5 m **×** 3.08 **×** 10^{-6} m^{3}s^{-1} / 𝜋 **×** (0.01m)^{4}

ΔP = 30.564 **×** 10^{-6} Pa m3 s^{-1} / 3.1416 **×** 10^{-8} m^{4}

ΔP = 9.73 **×** 10^{2} Pa

= 973 Pa.

**Check if the assumption of laminar flow is correct:**

**The average velocity v can be found using:**

v = Q/ 𝜋r^{2}

= 3.08 **×** 10^{-6} m^{3} s^{-1} / 𝜋 **×** (0.01m)^{2}

= 9.8 **×** 10^{-3} ms^{-1}

Re = 1.3 **×** 10^{3} **×** 9.8 **×** 10^{-3} **×** 2 **×** 0.01 / 0.83

= 0.31.

**14. In a test experiment on a model aeroplane in a wind tunnel, the flow speeds on the upper and lower surfaces of the wing are 70 m s**^{–1}** and 63 m **^{s-1}** 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: **Given:**

v_{1} = 70 ms^{-1}

v_{2} = 63 ms^{-1}

p = 1.3 kgm^{-3}

A = 2.5 m^{2}

Apply Bernoulli’s Equation

Rearrange to find the pressure difference:

ΔP = ½ x 1.3 **×** (70^{2} – 63^{2})

70^{2} = 4900

63^{2} = 3969

70^{2} – 63^{2} = 4900 – 3969 = 931

ΔP = 0.65 **×** 931

ΔP = 605.15 Pa

**Calculate the Lift Force:**

L = ΔP **×** A

L = 605.15 **×** 2.5

L = 1512.875 N.

**15. Figures 9.20(a) and (b) refer to the steady flow of a (non-viscous) liquid. Which of the two figures is incorrect ? Why?**

Ans: **As shown in the above image:**

l = 40 cm

= 40 cm **×** 10^{-2} m

Force = 4.5 **×** 10^{-2} N

Surface tension,

T = F/2*l*

= 4.5 **×** 10^{-2} / 2 **×** 40 **×** 10^{-2}

= 5.625 **×** 10^{-2} Nm^{-1}

In case of figures length is again, 40 **×** 10^{-2} m

Small weight that the film can support is

F = T **×** (2L)

= 5.625 **×** 10^{-2} **×** 2 **×** 40 **×** 10^{-2}

= 4.5 **×** 10^{-2} N.

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

Ans: Given that:

Tube cross-sectional area: 8.0 cm^{2} = 8.0 **×** 10^{-4} m^{2}

Flow speed inside the tube: 1.5 m/ min = 0.025 m/s

Number of holes: 40

Diameter of each holes: 1.0 mm = 1.0 **×** 10^{-3} m

**Calculate the total flow rate:**

Q = cross-sectional area x flow speed = 8.0 **×** 10^{-4} **×** 0.025 = 2.0 **×** 10^{-5} m^{3}/s

**Area of one hole:**

Area of one hole = 𝜋 (Diameter / 2)^{2}

= 𝜋 (1.0 **×** 10-3 / 2 )^{2} = 0.785 **×** 10^{-6} m^{2}

Total area of all holes = 40 **×** 0.785 **×** 10^{-6} = 31 .4 **×** 10 ^{-6} m^{2}

**Find the speed of ejection:**

= Total flow rate / Total area of all holes

= 2.0 **×** 10^{-5} / 31.4 **×** 10^{-6}

= 0.64 m/s.

**17. 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 × 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: **Given:**

W = 1.5 **×** 10^{-2} N

L = 30 cm = 0.3 m.

**Surface Tension Formula:**

F = T **×** Total length of the film.

W = T **×** 2L

T = W/2L

T = 1.5 **×** 10^{-2} / 2 **×** 0.3

T = 1.5 **×** 10^{-2} / 0.6

T = 0.025 N/m.

**18. Figure 9.21 (a) shows a thin liquid film supporting a small weight = 4.5 × 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: **As shown in image:**

Length L = 40cm = 10 **×** 10^{-2}m

Also, Force F = 4.5 **×** 10^{-2} N

Surface tension, T = F/2L

= 4.5 **×** 10^{-2} / 2 **×** 40 x 10^{-2}

= 5.625 **×** 10^{-2} Nm^{-1}

The figure length is again 40 x 10^{-2}m.

**Small weight that the film can support is:**

F = T **×** (2L)

= 5.625 **×** 10^{-2} **×** 2 **×** 40 **×** 10^{-2}

= 4.5 **×** 10^{-2} N.

**19. What is the pressure inside the drop of mercury of radius 3.00 mm at room temperature? Surface tension of mercury at that temperature (20 °C) is 4.65 × 10**^{–1}** N m**^{–1}**. The atmospheric pressure is 1.01 × 10**^{5}** Pa. Also give the excess pressure inside the drop. **

Ans: **Given:**

r = 3.00 mm = 3.00 **×** 10^{-3} m

Surface tension T = 4.65 **×** 10^{-1} N/m

Patm = 1.01 **×** 10^{5} Pa

**Calculate the Excess Pressure:**

ΔP = 2T / r

ΔP = 2 **×** 4.65 **×** 10^{-1} / 3.00 **×** 10^{-3}

ΔP = 0.93 / 0.003

ΔP = 310 Pa.

**Calculate the Pressure Inside the Drop:**

Pinside = Patm + ΔP

Pinside = 1.01 **×** 10^{5} + 310

Pinside = 1.0131 **×** 10^{5} Pa.

**20. What is the excess pressure inside a bubble of soap solution of radius 5.00 mm, given that the surface tension of soap solution at the temperature (20 °C) is 2.50 × 10**^{–2}** N m**^{–1}**? If an air bubble of the same dimension were formed at 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 × 10**^{5}** Pa).**

Ans: Excess Pressure = 4 **×** Surface Tension / Radius

**Given:**

Surface Tension (T) = 2.50 **×** 10^{–2} N m^{–1}

Radius (r) = 5.00 mm = 5.00 × 10^{–3} m

Excess Pressure = 4 **×** (2.50 **×** 10^{–2}) / (5.00 **×** 10^{–3})

= 20.0 Pa

Hydrostatic Pressure = ρ **×** g **×** h

**Given:**

Relative Density (ρ) = 1.20

g = 9.80 m s^{–2}

h = 40.0 cm = 0.40 m

Hydrostatic Pressure = 1.20 **×** 1000 kg m^{–3} **×** 9.80 m s^{–2} **×** 0.40 m

= 4704 Pa

**Adding the atmospheric pressure:**

Total Pressure = Hydrostatic Pressure + Atmospheric Pressure

= 4704 Pa + 1.01 **×** 10^{5} Pa

= 1.055 **×** 10^{5} Pa.