This chapter delves into the fascinating world of fluids, exploring how they behave under various conditions. We’ll uncover the concepts of pressure, buoyancy, and fluid flow.
Pressure
Pressure: The force exerted per unit area.
Atmospheric Pressure: Air column’s weight.
Gauge Pressure: The pressure relative to atmospheric pressure.
Pascal’s Law: When pressure is applied to a confined fluid, the pressure is transmitted uniformly throughout the fluid.
Archimedes’ Principle: When you put something in a fluid, it gets a lift equal to the weight of the fluid it pushes aside.
Floatation: Whether something floats or sinks depends on how heavy it is compared to the fluid.
Fluid Flow
Streamline Flow: Smooth, orderly fluid flow.
Turbulent Flow: Chaotic, irregular fluid flow.
Equation of Continuity: For incompressible fluids, the product of cross-sectional area and velocity remains constant.
Bernoulli’s Principle: As fluid velocity increases, its pressure decreases.
Viscosity
Viscosity: A fluid’s resistance to flow.
Stokes’ Law: Describes the viscous force acting on a spherical object moving through a fluid.
By understanding these principles, we gain insights into the mechanics of fluids, from everyday phenomena to complex engineering applications.
1. Explain why
(a) The blood pressure in humans is greater 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
(c) Hydrostatic pressure is a scalar quantity even though pressure is force divided by area.
Ans : (a) Blood Pressure and Height:
Blood pressure is the force exerted by the blood against the walls of blood vessels. The heart pumps blood throughout the body, and gravity acts on this blood. As a result, the pressure in the blood vessels increases with depth. Since the feet are at a lower level compared to the brain, the blood pressure is higher at the feet.
(b) Atmospheric Pressure and Height:
As we move higher in the atmosphere, the amount of air above us decreases, leading to a decrease in pressure. The majority of the Earth’s atmosphere is concentrated in the lower layers, so a significant portion of the pressure drop occurs within the first few kilometers.
(c) Hydrostatic Pressure as a Scalar Quantity:
While pressure is defined as force per unit area, it is a scalar quantity because it does not have a specific direction. Hydrostatic pressure acts equally in all directions at a given point within a fluid. This is why it’s considered a scalar quantity, even though it’s derived from force and area, which are vector quantities.
2. Explain why
(a) The angle of contact of mercury with glass is obtuse, while that of water with glass 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.)
(c) Surface tension of a liquid is independent of the area of the surface
(d) Water with detergent disolved in it should have small angles of contact.
(e) A drop of liquid under no external forces is always spherical in shape
Ans : (a) Angle of Contact:
The angle of contact between a liquid and a solid surface depends on the balance between cohesive and adhesive forces.
Mercury and Glass: Mercury’s strong cohesive forces dominate over its adhesive forces with glass, resulting in a high contact angle and a non-wetting behavior.
Water and Glass: Water’s strong adhesive forces with glass overcome its cohesive forces, leading to a low contact angle and wetting behavior.
(b) Wetting and Non-Wetting:
Water and Glass: The strong adhesive forces between water and glass promote spreading, resulting in wetting.
Mercury and Glass: The weak adhesive forces between mercury and glass cause it to minimize contact, forming droplets and exhibiting non-wetting behavior.
(c) Surface Tension and Area:
Surface tension is an intrinsic property of a liquid, independent of its surface area. It arises from the cohesive forces between liquid molecules, which tend to minimize the surface area.
(d) Detergent and Angle of Contact:
Detergents reduce the surface tension of water by disrupting its surface structure. This weakens the cohesive forces, allowing water to spread more easily and reduce the contact angle.
(e) Spherical Shape of a Liquid Drop:
A liquid drop, in the absence of external forces, tends to minimize its surface energy. A spherical shape achieves this minimum surface area for a given volume.
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)
(b) Viscosity of gases … with temperature, whereas viscosity of liquids … with temperature (increases / decreases)
(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)
(d) For a fluid in a steady flow, the increase in flow speed at a constriction follows (conservation of mass / 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 :
(a)Ans: Decreases
(b) Ans: Increases ,Decreases
(c) Ans: Shear strain , Rate of shear strain
(d) Ans: Conservation of mass ,Bernoulli’s principle
(e)Ans: Greater.
4. Explain why
(a) To keep a piece of paper horizontal, you should blow over, not under, it
(b) When we try to close a water tap with our fingers, fast jets of water gush through the openings between our fingers
(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
(d) A fluid flowing out of a small hole in a vessel results in a backward thrust on the vessel
(e) A spinning cricket ball in air does not follow a parabolic trajectory
Ans : (a) Blowing over the paper increases the air velocity above it. According to Bernoulli’s principle, this leads to a decrease in air pressure above the paper. As the air pressure below the paper remains atmospheric, the paper is lifted upwards.
(b) When we partially close a water tap, the reduced cross-sectional area forces the water to flow faster. This is due to the principle of continuity, which states that the product of the cross-sectional area and the flow velocity remains constant.
(c) In Bernoulli’s equation (P + 1/2 ρv² = constant), the velocity term (v²) has a greater impact on the overall pressure than the pressure term (P). Therefore, a small change in the needle diameter, which significantly affects the flow velocity, has a more pronounced effect on the flow rate than a change in thumb pressure.
(d) The backward thrust on the vessel is a consequence of Newton’s third law of motion. As the fluid exits the vessel with momentum, an equal and opposite momentum is imparted to the vessel, pushing it backward.
(e) A spinning cricket ball experiences the Magnus effect, which causes it to deviate from its parabolic trajectory. This effect arises from the pressure difference between the sides of the spinning ball, resulting in a curved flight path.
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 : *Understanding the Problem:**
We have a girl standing on a single high-heel shoe.
Determine pressure exerted by the heel on the floor.
Solution:
1. Calculate the Area of the Heel:
The area of the circular heel is given by:
A = πr²
Where:
* r is the radius of the heel (0.5 cm = 0.005 m)
A = π × (0.005 m)²
≈ 7.85 × 10⁻⁵ m²
2. Calculate the Force Exerted by the Heel:
The force exerted by the heel is equal to the weight of the girl:
F = mg = 50 kg × 9.8 m/s² = 490 N
3. Calculate the Pressure:
Pressure (P) =force per unit area:
P = F / A
Substituting the values:
P = 490 N / 7.85 × 10⁻⁵ m²
≈ 6.24 × 10⁶ Pa
Therefore, the pressure exerted by the heel on the floor is approximately 6.24 × 10⁶ 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 : Understanding the Problem:
We’re asked to find the height of a wine column that would balance atmospheric pressure, similar to Torricelli’s mercury barometer.
Solution:
Given,
The pressure exerted by a fluid column
Pressure = Density × Gravity × Height
For mercury and wine, we can equate the pressures:
Density_mercury × g × Height_mercury = Density_wine × g × Height_wine
Since ‘g’ is the same for both, we can cancel it out:
Density_mercury × Height_mercury = Density_wine × Height_wine
We know the density of mercury (ρ_mercury) is 13.6 × 10³ kg/m³, and the density of wine (ρ_wine) is 984 kg/m³.
We also know the height of the mercury column in a standard barometer is 0.76 m.
Substituting the values:
(13.6 × 10³ kg/m³) × 0.76 m = (984 kg/m³) × Height_wine
Solving for Height_wine:
Height_wine = (13.6 × 10³ kg/m³ × 0.76 m) / 984 kg/m³
≈ 10.5 m
hence, the height of the wine column for normal atmospheric pressure would be approximately 10.5 meters.
7 .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 : Understanding the Problem:We need to determine if a structure can withstand the pressure at a depth of 3 km in the ocean.
Solution:
1. Calculate the Pressure at the Given Depth:
given,
Pressure at a depth h in a fluid
Pressure = Density × Gravity × Height
Density of seawater (ρ) ≈ 1030 kg/m³
* Gravity (g) = 9.81 m/s²
* Height (h) = 3 km = 3000 m
Pressure = 1030 kg/m³ × 9.81 m/s² × 3000 m
≈ 3.03 × 10⁷ Pa
2. Compare the Pressure with the Maximum Stress:
The pressure at the given depth (3.03 × 10⁷ Pa) is significantly lower than the maximum stress the structure can withstand (10⁹ Pa).
Conclusion
Yes, the structure is suitable for putting up on top of the oil well. It can easily withstand the hydrostatic pressure at that depth.
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 cm2 . What maximum pressure would the smaller piston have to bear ?
Ans : *Understanding the Problem:**
We have a hydraulic lift with two pistons: a larger one carrying the load and a smaller one applying the force. We need to find the maximum pressure on the smaller piston to lift the car.
1. Calculate the Weight of the Car:
Weight of the car, W = mass × acceleration due to gravity = 3000 kg × 9.8 m/s² = 29,400 N
2. Determine Force on the Larger Piston:
The force on the larger piston (F₁) is equal to the weight of the car:
F₁ = 29,400 N
3. Apply Pascal’s Law:
Pascal’s law states that the pressure applied to an enclosed fluid is transmitted uniformly throughout the fluid. Therefore, the pressure on the smaller
piston (P₂) is equal to the pressure on the larger piston (P₁):
P₁ = P₂
4. Determine Pressure on the Smaller Piston:
Pressure is defined as force per unit area:
P₁ = F₁ / A₁
Where:
* F₁ is the force on the larger piston (29,400 N)
* A₁ is the area of the larger piston (425 cm² = 0.0425 m²)
Substituting the values:
P₁ = 29,400 N / 0.0425 m²
≈ 6.9 × 10⁵ Pa
Therefore, the maximum pressure the smaller piston would have to bear is approximately 6.9 × 10⁵ Pa.
9.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 : *Understanding the Problem:**
We have a U-tube with water and methylated spirit separated by mercury. The difference in the heights of the liquid columns is due to the difference in their densities. We need to find the specific gravity of the spirit.
1. Pressure Balance:At the interface between the mercury and the water/spirit, the pressure must be the same. This is because the mercury column is continuous and the pressure at a given depth in a fluid is the same at all points at that depth.
2. Pressure Equation:
Given,
( ρ )Pressure at a depth
P = ρgh
Where:
* P = pressure
* ρ = density of the fluid
* g = acceleration due to gravity
* h = depth
3. Equating Pressures:
For the water and spirit columns:
ρ_water * g * h_water = ρ_spirit * g * h_spirit
Since g is the same on both sides, we can cancel it out:
ρ_water * h_water = ρ_spirit * h_spirit
4. Calculating Specific Gravity:Specific gravity is the ratio of a substance’s density to the density of a reference substance, typically water.
Specific gravity of spirit = ρ_spirit / ρ_water
Substituting the values:
Specific gravity of spirit = (h_water / h_spirit) = 10 cm / 12.5 cm
= 0.8
Therefore, the specific gravity of the spirit is 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 : **Understanding the Problem:**
We have a U-tube with water and spirit separated by mercury. Initially, the difference in the heights of the water and spirit columns was 2.5 cm. After adding 15 cm of each liquid, we need to find the new difference in the mercury levels.
*Solution:
1. Final Heights:* Height of the water column: h₁ = 10 cm + 15 cm = 25 cm
* Height of the spirit column: h₂ = 12.5 cm + 15 cm = 27.5 cm
2. Density Ratio:
* Density of water: ρ₁ = 1 g/cm³
* Density of spirit: ρ₂ = 0.8 g/cm³
* Density of mercury: ρ_mercury = 13.6 g/cm³
3. Pressure Balance:
At the interface between the mercury and the water/spirit, the pressure must be equal. This can be expressed as:
ρ₁gh₁ = ρ₂gh₂
4. Difference in Mercury Levels:
The difference in mercury levels, h_mercury, can be calculated using the pressure difference between the two arms of the U-tube:
ρ_mercury * g * h_mercury = ρ₁gh₁ – ρ₂gh₂
Substituting the values and simplifying:
13.6 * h_mercury = 1 * 25 – 0.8 * 27.5
Solving for h_mercury, we get:
h_mercury ≈ 0.221 cm
Therefore, the difference in the levels of mercury in the two arms is approximately 0.221 cm.
11. 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 directly used to describe the flow of water through a rapid in a river.
Bernoulli’s equation is applicable to ideal fluid flow, which assumes the fluid to be incompressible, non-viscous, and flowing steadily. However, in a rapid, the flow is turbulent and viscous forces play a significant role.
In turbulent flow, energy is dissipated due to friction and eddies, which violates one of the assumptions of Bernoulli’s equation. Therefore, it cannot accurately predict the pressure and velocity changes in such a complex flow scenario.
To analyze the flow through a rapid, more complex fluid dynamics models that consider turbulence and energy dissipation would be necessary.
12. 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 or absolute pressure in Bernoulli’s equation.
Bernoulli’s equation relates the pressure, velocity, and height of a fluid in steady, incompressible flow. The key point is that it’s used to analyze pressure differences, not absolute pressures. Whether we use gauge or absolute pressure, as long as we’re consistent throughout the equation, the calculated pressure differences will be accurate.
Gauge pressure, which measures pressure relative to atmospheric pressure, is often more practical to use in real-world applications. By using gauge pressure in Bernoulli’s equation, we can still obtain meaningful results for fluid flow analysis.
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 × 103 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:
Length of capillary (l) = 1.5 m
Radius of capillary (r) = 1 x 10^-2 m
Volume flow rate (V) = 4 x 10^-3 m^3/s
Density of fluid = 1.3 x 10^3 kg/m^3
Viscosity of fluid (η) = 0.83 Pa s
Calculation:
Calculate the pressure difference (p):
Using Poiseuille’s law for laminar flow:
V = (πpr^4) / (8ηl)
Rearranging for pressure:
p = (8Vηl) / (πr^4)
Substituting the given values:
p = (8 * (4/1.3) * 10^-6 * 0.83 * 1.5 * (7/22) * (1/10^-8)) Pa
Solving this equation gives:
p = 9.75 x 10^2 Pa
Check Reynolds number (Re):
Reynolds number helps determine if the flow is laminar or turbulent.
Re = (ρVd) / η
Here, d is the diameter of the capillary (2r).
Calculating Re:
Re = (1.3 * 10^3 * (4/1.3) * 10^-6 * 2 * 10^-2) / 0.83
This gives:
Re = 0.3
Conclusion:
Since the Reynolds number is less than 1, the flow is laminar.
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–1and 63 m s-1 respectively. What is the lift on the wing if its area is 2.5 m2 ? Take the density of air to be 1.3 kg m–3 .
Ans : Let v1 and v2 be the speeds on the upper and lower surfaces of the wing of an aeroplane, respectively. Let P1 and P2 be the pressures on the upper and lower surfaces of the wing, respectively.
We are given:
v1 = 70 m/s
v2 = 63 m/s
ρ = 1.3 kg/m³
According to Bernoulli’s theorem:
P1/ρ + (v1^2)/2 + gh = P2/ρ + (v2^2)/2 + gh
Assuming the height difference (h) is negligible:
P1/ρ – P2/ρ = (v2^2 – v1^2)/2
Simplifying:
P1 – P2 = (ρ/2) * (v2^2 – v1^2) = (1.3/2) * (63^2 – 70^2) Pa = 605.15 Pa
This pressure difference provides the lift to the aeroplane. The lift force on the aeroplane is calculated as:
Lift = Pressure difference × Area of wings
Lift = 605.15 Pa × 2.5 m² = 1512.875 N
≈ 1.51 × 10³ 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 : Incorrect Figure is (a). According to Bernoulli’s principle, as the fluid flows through a narrower section (the kink), its velocity increases and its pressure decreases. This lower pressure would result in a lower liquid level in the vertical tube connected to the narrower section. Therefore, the liquid level should be lower in the tube connected to the kink, not higher.
16. The cylindrical tube of a spray pump has a cross-section of 8.0 cm2 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 : *Understanding the Problem:**
We have a cylindrical tube with a certain cross-sectional area and a number of small holes at one end. We’re given the flow rate of liquid through the tube and need to find the speed of ejection through the holes.
Solution:
1. Calculate the Total Area of the Holes:* Area of one hole = πr² = π(0.0005 m)² ≈ 7.85 × 10⁻⁷ m²
* Total area of 40 holes = 40 × 7.85 × 10⁻⁷ m² ≈ 3.14 × 10⁻⁵ m²
2. Apply the Equation of Continuity:
The equation of continuity states that the product of the cross-sectional area (A) and the flow velocity (v) remains constant for an incompressible fluid:
A₁v₁ = A₂v₂
Where:
* A₁ is the cross-sectional area of the tube (8.0 cm² = 8 × 10⁻⁴ m²)
* v₁ is the flow velocity inside the tube (1.5 m/min = 0.025 m/s)
* A₂ is the total area of the holes (3.14 × 10⁻⁵ m²)
* v₂ is the ejection velocity through the holes
Substituting the values:
(8 × 10⁻⁴ m²) × (0.025 m/s) = (3.14 × 10⁻⁵ m²) × v₂
Solving for v₂:
v₂ ≈ 0.64 m/s
Hence, the speed of ejection of the liquid through the holes is approximately 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 : The force due to surface tension balances the weight of the slider and the additional weight.
So, the force due to surface tension,
F = 1.5 × 10⁻² N.
The total length of the liquid film, l = 2 × 30 cm = 0.6 m, as there are two surfaces (top and bottom).
Surface tension, T = F/l = (1.5 × 10⁻² N) / 0.6 m
= 2.5 × 10⁻² 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 : The weight supported by the films in figures (b) and (c) will also be 4.5 x 10^-2 N.
The weight supported by the film depends on the total length of the film’s boundary. In all three figures, the length of the film’s boundary is the same (40 cm + 40 cm = 80 cm).
The surface tension force acting along this boundary supports the weight. Since the boundary length and the surface tension of the soap film are the same in all three cases, the weight supported will also be the same.
Therefore, regardless of the shape of the film, the weight supported remains constant.
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 × 105 Pa. Also give the excess pressure inside the drop.
Ans : Understanding the Problem:**
We have a mercury droplet with a specific radius and surface tension. We need to calculate the pressure inside the droplet, both the absolute pressure and the excess pressure (pressure above atmospheric pressure).
Solution:
1. Excess Pressure:
The excess pressure inside a liquid drop is given by the formula:
Excess pressure = 2T/R
Where:
* T is the surface tension of the liquid (4.65 × 10⁻¹ N/m)
* R is the radius of the droplet (3.00 × 10⁻³ m)
Substituting the values:
Excess pressure = 2 × (4.65 × 10⁻¹ N/m) / (3.00 × 10⁻³ m) ≈ 310 Pa
2. Absolute Pressure:The absolute pressure inside the droplet is the sum of the atmospheric pressure and the excess pressure:
Absolute pressure = Atmospheric pressure + Excess pressure
Substituting the values:
Absolute pressure = 1.01 × 10⁵ Pa + 310 Pa
≈ 1.0131 × 10⁵ Pa
Therefore, the excess pressure inside the mercury droplet is approximately 310 Pa, and the absolute pressure is approximately 1.0131 × 10⁵ 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 × 105 Pa)
Ans : Understanding the Problem:
We have two scenarios: a soap bubble and an air bubble inside a soap solution. We need to calculate the excess pressure inside each bubble.
**Solution:**
1. Excess Pressure Inside the Soap Bubble:
Given,
excess pressure inside a soap bubble by formula
Excess pressure = 4T/R
Where:
* T = surface tension of the soap solution (2.50 × 10⁻² N/m)
* R= radius of the bubble (5.00 × 10⁻³ m)
Substituting the values:
Excess pressure = 4 × (2.50 × 10⁻² N/m) / (5.00 × 10⁻³ m) = 20 Pa
2. Excess Pressure Inside the Air Bubble:
The excess pressure inside the air bubble will be the sum of the pressure due to the surface tension and the hydrostatic pressure due to the depth of the bubble in the soap solution.
Excess pressure = 4T/R + ρgh
Where:
* ρ = density of the soap solution (1.20 × 10³ kg/m³)
*g denote acceleration due to gravity (approximately 9.81 m/s²)
* h = depth of the bubble (0.4 m)
Substituting the values:
Excess pressure = 20 Pa + (1.20 × 10³ kg/m³) × (9.81 m/s²) × (0.4 m) ≈ 4705 Pa
3. Absolute Pressure Inside the Air Bubble:The absolute pressure inside the air bubble is the sum of the atmospheric pressure and the excess pressure:
Absolute pressure = Atmospheric pressure + Excess pressure
= 1.01 × 10⁵ Pa + 4705 Pa
≈ 1.06 × 10⁵ Pa
Therefore, the excess pressure inside the soap bubble is 20 Pa, and the absolute pressure inside the air bubble is approximately 1.06 × 10⁵ Pa.
