Question 1

Q1. 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

Accepted Answer

Here are the explanations for the given phenomena:

**(a) Blood pressure at the feet vs. the brain:**
The pressure in a fluid column is given by the formula for gauge pressure, P = hρg, where 'h' is the height of the fluid column, 'ρ' is the density of the fluid, and 'g' is the acceleration due to gravity. The heart acts as a pump, maintaining blood pressure throughout the body.

- The feet are at a greater vertical distance (height 'h') below the heart compared to the brain.
- This means the height of the blood column from the heart to the feet is significantly larger than the height of the blood column from the heart to the brain.
- Consequently, the hydrostatic pressure (hρg) due to the weight of the blood column is much greater at the feet. This adds to the pressure pumped by the heart, resulting in a higher overall blood pressure at the feet than at the brain.

**(b) Decrease in atmospheric pressure with height:**
The atmosphere is a fluid (air), but its density is not uniform. The density of air is highest at the sea level and decreases as we go up in altitude. This is because the lower layers of air are compressed by the weight of the layers above them.

- The formula P = hρg assumes a constant density 'ρ', which is not true for the atmosphere.
- Since the density of air decreases significantly with height, the pressure also decreases exponentially, not linearly.
- Most of the mass of the atmosphere is concentrated in the lower altitudes. Therefore, a significant drop in pressure (to nearly half) occurs within the first 6 km, even though the atmosphere extends much further.

**(c) Hydrostatic pressure as a scalar quantity:**
Pressure is defined as force per unit area (P = F/A). While force is a vector and area can be treated as a vector (with its direction normal to the surface), pressure itself is a scalar quantity.

- The reason is that at any point inside a fluid at rest, the pressure is exerted equally in all directions. It does not have a specific direction associated with it.
- The force exerted by the fluid on any surface in contact with it is always perpendicular (normal) to that surface, regardless of the orientation of the surface.
- Since the force's direction depends on the orientation of the area it acts upon, and the magnitude of pressure at a point is the same in all directions, pressure is defined as a scalar quantity.

Question 2

Q2. 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 mercur

Accepted Answer

Here are the explanations based on the properties of fluids:

**(a) Angle of contact for mercury-glass and water-glass:**
The angle of contact depends on the relative strength of cohesive forces (attraction between molecules of the same liquid) and adhesive forces (attraction between molecules of the liquid and the solid surface).
- **Water-Glass:** The adhesive force between water molecules and glass molecules is stronger than the cohesive force between water molecules. This causes the water molecules near the glass surface to be pulled towards the glass, resulting in a concave meniscus and an acute angle of contact (< 90°).
- **Mercury-Glass:** The cohesive force between mercury molecules is much stronger than the adhesive force between mercury and glass molecules. This causes the mercury molecules to pull each other inwards, away from the glass, resulting in a convex meniscus and an obtuse angle of contact (> 90°).

**(b) Wetting and non-wetting of surfaces:**
This is a direct consequence of the interplay between cohesive and adhesive forces, as explained in part (a).
- **Water on Glass:** Since the adhesive force is stronger, water is attracted to the glass surface more than to itself. This causes the water to spread out and 'wet' the glass.
- **Mercury on Glass:** Since the cohesive force is stronger, mercury molecules are attracted to each other more strongly than to the glass. This causes the mercury to pull itself together into droplets to minimize contact with the glass surface, thus it does not 'wet' the glass.

**(c) Surface tension and surface area:**
Surface tension is an intrinsic property of a liquid at a given temperature. It is defined as the force per unit length acting perpendicular to an imaginary line drawn on the liquid surface (S = F/L) or as the surface energy per unit area (S = E/A). While it is defined in terms of area (in the energy definition), it does not depend on the total area of the liquid surface. It is a characteristic of the intermolecular forces within the liquid. Changing the surface area requires work to be done against these forces, but the value of surface tension (energy per unit area) remains constant for a given liquid and temperature.

**(d) Detergents and angle of contact:**
Detergents are surfactants (surface-active agents). When dissolved in water, they reduce the surface tension of the water. According to the Young-Laplace equation, a lower surface tension of the liquid-air interface leads to a smaller angle of contact. A small angle of contact means the water can spread more easily and penetrate the pores of fabrics, which enhances its cleaning action. This property is called wetting.

**(e) Spherical shape of a liquid drop:**
For a given volume, a sphere is the geometric shape that has the minimum possible surface area. A liquid surface possesses potential energy called surface energy, which is directly proportional to its surface area (E = S × A). Due to surface tension, a liquid always tends to minimize its surface energy. In the absence of external forces like gravity (which would distort the shape), the drop assumes a spherical shape to achieve the minimum surface area and thus the minimum surface energy, leading to a state of maximum stability.

Question 3

Q3. 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) Visc

Accepted Answer

(a) Surface tension of liquids generally **decreases** with temperatures.
(b) 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 **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 from **conservation of mass** (Equation of Continuity), while the decrease of pressure there follows from **Bernoulli’s principle**.
(e) For the model of a plane in a wind tunnel, turbulence occurs at a **greater** speed for turbulence for an actual plane.

Question 4

Q4. 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 th

Accepted Answer

Here are the explanations for the given observations based on principles of fluid dynamics:

**(a) Blowing over a piece of paper:**
This is a classic demonstration of Bernoulli's principle, which states that for a fluid in motion, an increase in speed occurs simultaneously with a decrease in pressure. When you blow over the top surface of the paper, you create a region of fast-moving air. This high-speed air has lower pressure compared to the relatively still air below the paper. The higher pressure of the still air underneath exerts an upward force on the paper, lifting it and keeping it horizontal.

**(b) Fast jets of water from a tap:**
This phenomenon is explained by the principle of conservation of mass, expressed as the Equation of Continuity (A₁v₁ = A₂v₂), where A is the cross-sectional area and v is the fluid velocity. When you partially close the tap opening with your fingers, you significantly decrease the effective cross-sectional area (A₂) through which the water can flow. To maintain a constant volume flow rate (Av), the velocity of the water (v₂) must increase proportionally. This results in fast-moving jets of water gushing through the small openings.

**(c) Syringe needle size and flow rate:**
The flow of a fluid through a narrow tube (like a syringe needle) is governed by Poiseuille's Law. The volume flow rate (Q) is given by Q = (πΔP r⁴) / (8ηL), where ΔP is the pressure difference, r is the radius of the tube, η is the viscosity, and L is the length of the tube. From this formula, we can see that the flow rate is proportional to the fourth power of the radius (Q ∝ r⁴) but only directly proportional to the pressure difference (Q ∝ ΔP). Therefore, a small change in the radius of the needle has a much more significant impact on the flow rate than a similar proportional change in the thumb pressure. This is why the needle size is a more effective controller of the flow rate.

**(d) Backward thrust on a vessel:**
This is an application of Newton's Third Law of Motion and the principle of conservation of linear momentum. As the fluid is pushed out of the hole, it gains momentum in the forward direction. To conserve the total momentum of the system (vessel + fluid), the vessel must gain an equal and opposite momentum. This means the exiting fluid exerts a force (thrust) on the vessel in the backward direction, causing it to recoil.

**(e) A spinning cricket ball's trajectory (Magnus Effect):**
A non-spinning ball follows a parabolic path due to gravity. However, a spinning ball deviates from this path due to the Magnus effect. As the ball spins, it drags a thin layer of air around with it. On one side of the ball, the direction of this dragged air is the same as the direction of the airflow past the ball. This results in a higher relative speed and, by Bernoulli's principle, lower pressure. On the opposite side, the dragged air opposes the airflow, resulting in a lower relative speed and higher pressure. This pressure difference between the two sides creates a net sideways force (a dynamic lift) on the ball, causing it to curve or swerve away from the standard parabolic trajectory.

Question 5

Q5. 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?

Accepted Answer

The pressure exerted by the heel on the horizontal floor is calculated by dividing the force exerted by the girl by the area of the heel in contact with the floor.

**Given:**
- Mass of the girl (m) = 50 kg
- Diameter of the circular heel (d) = 1.0 cm

**Step-by-step solution:**

1.  **Calculate the force exerted:** The force is the weight of the girl.
    F = m × g
    Assuming g = 9.8 m/s²
    F = 50 kg × 9.8 m/s² = 490 N

2.  **Calculate the radius of the heel:**
    Radius (r) = Diameter (d) / 2 = 1.0 cm / 2 = 0.5 cm

3.  **Convert the radius to SI units (meters):**
    r = 0.5 cm = 0.5 × 10⁻² m = 0.005 m

4.  **Calculate the area of the circular heel:**
    Area (A) = π × r²
    A = 3.14159 × (0.005 m)²
    A = 3.14159 × 2.5 × 10⁻⁵ m²
    A ≈ 7.854 × 10⁻⁵ m²

5.  **Calculate the pressure:**
    Pressure (P) = Force (F) / Area (A)
    P = 490 N / (7.854 × 10⁻⁵ m²)
    P ≈ 6.24 × 10⁶ N/m² or 6.24 × 10⁶ Pa

**Result:**
The pressure exerted by the heel on the horizontal floor is approximately 6.24 × 10⁶ Pa.

Question 6

Q6. 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 o

Accepted Answer

The specific gravity of spirit is 0.8.

Question 7

Q7. 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 o

Accepted Answer

The specific gravity of the spirit is 0.8.

Question 8

Q8. In a hydraulic lift, the two pistons are of area 60 cm² and 1500 cm². What force is required on the smaller piston to lift a car of mass 2000 kg on the larger piston? (g = 9.8 N/kg

Accepted Answer

The force required on the smaller piston is 784 N.

Question 9

Q9. A U-tube of uniform bore contains some mercury. If water is poured into one arm and spirit into the other, the levels of the liquids are found to be 10 cm and 12.5 cm respectively,

Accepted Answer

The specific gravity of the spirit is 0.8.

Question 10

Q10. In a U-tube, water is poured into one arm and oil into the other. The oil column is 15 cm high and the water column is 12 cm high. What is the specific gravity of the oil?

Accepted Answer

The specific gravity of the oil is 0.8.

Question 11

Q11. 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 o

Accepted Answer

The specific gravity of the spirit is 0.8.

Question 12

Q12. 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 o

Accepted Answer

The specific gravity of methylated spirit is 0.8.

Question 13

Q13. In a hydraulic lift, the two pistons are of area 60 cm² and 1500 cm². What is the force exerted on the larger piston when a force of 50 N is applied to the smaller piston?

Accepted Answer

The force exerted on the larger piston is 1250 N.

Question 14

Q14. A U-tube of uniform bore (cross-sectional area A) contains a liquid of density ρ. Initially, the liquid in both arms stands at a height h. Now the right arm is evacuated, and the l

Accepted Answer

The difference in the levels of the liquid in the two arms will be P_atm / (ρg), where P_atm is the atmospheric pressure, ρ is the density of the liquid, and g is the acceleration due to gravity.

Question 15

Q15. Two narrow bores of diameters 3.0 mm and 6.0 mm are joined together to form a U-tube open at both ends. If the U-tube contains water, what is the difference in the levels of water

Accepted Answer

The difference in the levels of water in the two bores is approximately 4.97 mm.

Question 16

Q16. 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 o

Accepted Answer

The specific gravity of the spirit is 0.8.

Question 17

Q17. 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 o

Accepted Answer

The specific gravity of the spirit is 0.8.

Question 18

Q18. 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 o

Accepted Answer

The specific gravity of spirit is 0.8.

Question 19

Q19. Can Bernoulli’s equation be used to describe the flow of water through a rapid in a river? Explain.

Accepted Answer

No, Bernoulli's equation cannot be accurately used to describe the flow of water through a rapid in a river.

Question 20

Q20. Does it matter if one uses gauge pressure or absolute pressure in applying Bernoulli’s equation? Explain.

Accepted Answer

No, it does not matter whether one uses gauge pressure or absolute pressure in applying Bernoulli's equation, as long as the choice is consistent throughout the equation.

Question 21

Q21. A U-tube of uniform bore contains some mercury. Two liquids are poured into the two arms. The first liquid of density ρ₁ is poured into one arm and the second liquid of density ρ₂

Accepted Answer

The relation between the densities of the liquids (ρ₁, ρ₂) and the heights of the mercury levels from the bottom (h₁, h₂) is given by: ρ₁ (H - h₁) + ρ_m h₁ = ρ₂ (H - h₂) + ρ_m h₂ where H is the total height of the liquid columns from the bottom (assuming they reach the top of the U-tube or a common reference point for the open ends) and ρ_m is the density of mercury. A more practical relation considering the height of the liquid columns above the mercury interface is: ρ₁ (h_liquid1) = ρ₂ (h_liquid2). If we consider the total height of the liquid columns from the bottom of the U-tube to the free surface of the liquids, and the mercury levels are h₁ and h₂ from the bottom, then the heights of the liquid columns above the mercury interface are (H_total - h₁) and (H_total - h₂) respectively, where H_total is the height of the free surface of the liquids from the bottom. Assuming the free surfaces of both liquids are at the same height from the bottom, let this height be H_top. Then the height of liquid 1 above mercury is (H_top - h₁) and height of liquid 2 above mercury is (H_top - h₂). The pressure balance at the mercury-liquid interface gives: P_atm + ρ₁g(H_top - h₁) = P_atm + ρ₂g(H_top - h₂). This simplifies to ρ₁(H_top - h₁) = ρ₂(H_top - h₂).

Question 22

Q22. Can Bernoulli's equation be used to describe the flow of water through a rapid in a river? Explain.

Accepted Answer

No, Bernoulli's equation cannot be used to accurately describe the flow of water through a rapid in a river. This is because the flow in a rapid violates several key assumptions required for Bernoulli's equation to be valid.

Question 23

Q23. Does it matter if one uses gauge pressure or absolute pressure in applying Bernoulli's equation? Explain.

Accepted Answer

No, it does not matter whether one uses gauge pressure or absolute pressure in applying Bernoulli's equation, as long as consistency is maintained throughout the calculation. The reason is that Bernoulli's equation deals with pressure differences, and the atmospheric pressure (which differentiates absolute from gauge pressure) cancels out.

Question 24

Q24. 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 o

Accepted Answer

The relative density of spirit is 0.8.

Question 25

Q25. In problem 10.24, 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? (Rel

Accepted Answer

The difference in the levels of mercury in the two arms is approximately 0.22 cm.

Question 26

Q26. Can Bernoulli's equation be used to describe the flow of water through a rapid in a river? Explain.

Accepted Answer

No, Bernoulli's equation cannot be used to describe the flow of water through a rapid in a river.

Question 27

Q27. Does it matter if one uses gauge pressure or absolute pressure in applying Bernoulli's equation? Explain.

Accepted Answer

No, it does not matter whether one uses gauge pressure or absolute pressure in applying Bernoulli's equation, as long as consistency is maintained throughout the calculation.

Chapter 10: Mechanical Properties of Fluids