Exercises
Q9.1. 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) Pressure in a fluid at rest increases with depth: P = P₀ + ρgh. The feet are much lower than the brain, so the column of blood above the feet is taller. This greater height of blood column means blood pressure at the feet is considerably higher than at the brain.
(b) The atmosphere is not uniform — its density decreases rapidly with altitude. Most of the atmosphere's mass is concentrated in the lower layers. Atmospheric pressure at any height equals the weight of air per unit area above that height. Since air is much denser near the surface, the weight of air above 6 km is about half the total atmospheric weight, so pressure has halved. The very thin upper atmosphere (6–100 km) contributes only half the total air mass.
(c) Pressure at a point in a fluid acts equally in all directions — it has no specific direction of its own. The force associated with pressure always acts normal to whatever surface it acts on, and the direction depends on the orientation of that surface. Since pressure does not have a unique associated direction, it is a scalar quantity.
Q9.2. 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.
(c) Surface tension of a liquid is independent of the area of the surface
(d) Water with detergent dissolved in it should have small angles of contact.
(e) A drop of liquid under no external forces is always spherical in shape
Ans.
(a) For mercury on glass, cohesive forces between mercury molecules are much stronger than adhesive forces between mercury and glass → mercury surface curves upward near glass → obtuse contact angle. For water, adhesive forces (water-glass) are greater than cohesive forces (water-water) → water clings to glass → concave meniscus → acute contact angle.
(b) Since water wets glass (adhesion > cohesion), it spreads to maximize water-glass contact, minimizing surface energy. Mercury does not wet glass (cohesion > adhesion); it forms drops to minimize mercury-glass contact area and surface energy.
(c) Surface tension is the force per unit length along the surface (or energy per unit area). It arises from the nature of intermolecular forces at the surface, not from how large the surface is. Therefore surface tension is independent of surface area.
(d) Detergent reduces the surface tension of water significantly. With lower surface tension, the liquid can spread more easily → the contact angle decreases. Hence water with detergent has a small contact angle and wets surfaces better.
(e) For a given volume, a sphere has the minimum possible surface area. A liquid drop minimizes its surface energy (= surface tension × area) by taking the shape with minimum area. Under no external forces, the equilibrium shape with minimum surface energy is a sphere.
Q9.3. Fill in the blanks
(a) Surface tension of liquids generally ... with temperatures
(b) Viscosity of gases ... with temperature, whereas viscosity of liquids ... with temperature
(c) For solids: shearing force proportional to ...; for fluids: proportional to ...
(d) Increase in flow speed at constriction follows ...
(e) Model plane turbulence occurs at a ... speed than actual plane
Ans.
(a) decreases — Higher temperature weakens surface cohesion.
(b) Gases: increases; Liquids: decreases — In gases, higher T → more random motion → more momentum transfer → higher η. In liquids, higher T → molecules overcome cohesive bonds more easily → lower η.
(c) Solids: shear strain; Fluids: rate of shear strain — Solids resist deformation; fluids resist the rate of deformation.
(d) conservation of mass (equation of continuity: A₁v₁ = A₂v₂).
(e) greater — Reynolds number Re = ρvL/η. Smaller L (model) requires greater v to reach the same critical Re for turbulence.
Q9.4. Explain why
(a) Blow over paper to keep horizontal
(b) Fast jets through finger-blocked tap
(c) Needle size controls injection flow rate better
(d) Hole in vessel causes backward thrust
(e) Spinning cricket ball doesn't follow parabola
Ans.
(a) Blowing over paper increases air speed above it → Bernoulli's principle: higher speed = lower pressure above. Greater atmospheric pressure below provides net upward force keeping paper horizontal.
(b) Fingers reduce the cross-sectional area. By continuity (A₁v₁ = A₂v₂), reduced area forces water to flow at much higher velocity through the tiny gaps between fingers.
(c) Poiseuille's equation: Q = πPr⁴/(8ηl). Flow rate Q ∝ r⁴ — extremely sensitive to radius. A tiny change in needle radius dramatically changes Q. Pressure P affects Q only linearly and is harder to control precisely. So needle size gives far better flow control.
(d) By Newton's third law, the fluid ejected in one direction carries momentum → vessel receives equal and opposite momentum (backward thrust/recoil). This is the rocket principle.
(e) The spinning ball creates a Magnus effect — spin drags air, creating higher speed on one side (lower pressure) and lower speed on other side (higher pressure). This transverse pressure difference exerts a sideways force, deflecting the ball from a parabolic 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?
Answer
Given: m = 50 kg, d = 1.0 cm
⇒ r = 5 × 10⁻³ m
F = mg = 50 × 9.8 = 490 N
A = πr² = π × (5 × 10⁻³)² = 7.854 × 10⁻⁵ m²
P = F/A = 490 / (7.854 × 10⁻⁵) = 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.
Answer
P₀ = ρgh
⇒ h = P₀/(ρg) = (1.013 × 10⁵) / (984 × 9.8) = (1.013 × 10⁵) / 9643.2 = 10.50 m
(Much taller than a mercury barometer at 0.76 m, since wine is far less dense.)
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.
Answer
Pressure at 3 km depth:
P = ρgh = 1000 × 9.8 × 3000 = 2.94 × 10⁷ Pa
Since 2.94 × 10⁷ Pa << 10⁹ Pa (maximum stress), yes, the structure is suitable for the ocean 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?
Answer
F = 3000 × 9.8 = 2.94 × 10⁴ N, A = 425 × 10⁻⁴ = 4.25 × 10⁻² m²
By Pascal's law, pressure throughout fluid is uniform:
P = F/A = (2.94 × 10⁴) / (4.25 × 10⁻²) = 6.92 × 10⁵ Pa
9. A U-tube contains water and methylated spirit separated by mercury. Mercury columns in both 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?
Answer
At equal mercury levels, pressures are equal:
ρ_water × h_water = ρ_spirit × h_spirit
1000 × 10.0 = ρ_spirit × 12.5
ρ_spirit = 800 kg/m³
Specific gravity = 800/1000 = 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)
Answer
New columns: water = 25.0 cm, spirit = 27.5 cm
Let mercury fall by x cm in water arm and rise x cm in spirit arm.
Equating pressures at lower mercury level:
1 × (25 + x) = 0.8 × (27.5 − x) + 13.6 × (2x)
25 + x = 22 − 0.8x + 27.2x
25 + x = 22 + 26.4x
3 = 25.4x → x = 0.118 cm
Difference in mercury levels = 2x = 0.236 cm ≈ 0.24 cm
11. Can Bernoulli's equation be used to describe the flow of water through a rapid in a river? Explain.
Answer
No. Bernoulli's equation requires steady, non-viscous, incompressible flow along a streamline. Flow through rapids is highly turbulent and unsteady — velocity fluctuates randomly — and viscous energy dissipation is significant. These conditions violate Bernoulli's assumptions.
12. Does it matter if one uses gauge instead of absolute pressures in applying Bernoulli's equation? Explain.
Answer
No, it does not matter. Bernoulli's equation involves pressure differences. When gauge pressures (P_absolute − P_atm) are used, P_atm cancels from both sides, leaving the same equation. The result is identical whether absolute or gauge pressures are used.
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 is 4.0 × 10⁻³ kg s⁻¹, what is the pressure difference between the two ends? (ρ = 1.3 × 10³ kg m⁻³, η = 0.83 Pa s)
Answer
Q = 4.0 × 10⁻³ / (1.3 × 10³) = 3.077 × 10⁻⁶ m³/s
Poiseuille's equation: P = 8ηlQ/(πr⁴)
= (8 × 0.83 × 1.5 × 3.077 × 10⁻⁶) / (π × (0.01)⁴)
= (3.065 × 10⁻⁵) / (3.142 × 10⁻⁸) = 9.76 × 10² Pa ≈ 976 Pa
Check: v = Q/A = 9.79 × 10⁻³ m/s; Re = ρvd/η = 0.31 << 2000
⇒ laminar flow confirmed ✓
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 m²? (ρ_air = 1.3 kg m⁻³)
Answer
P_lower − P_upper = ½ρ(v_upper² − v_lower²)
= ½ × 1.3 × (70² − 63²) = ½ × 1.3 × 931 = 605.2 Pa
Lift = ΔP × A = 605.2 × 2.5 = 1513 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?
Answer
Figure (a) is incorrect.
By the equation of continuity (A₁v₁ = A₂v₂), the velocity must increase in the narrower section. Streamlines must crowd closer together in the constriction. Figure (a) shows equally spaced streamlines throughout — implying constant velocity — which violates the continuity equation. Figure (b) correctly shows closer streamlines in the narrower section, indicating increased velocity.
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?
Answer
A₁ = 8.0 × 10⁻⁴ m², v₁ = 1.5/60 = 0.025 m/s
Each hole area = π × (5 × 10⁻⁴)² = 7.854 × 10⁻⁷ m²
Total holes area A₂ = 40 × 7.854 × 10⁻⁷ = 3.14 × 10⁻⁵ m²
By continuity: v₂ = A₁v₁/A₂ = (8.0 × 10⁻⁴ × 0.025) / (3.14 × 10⁻⁵) = 0.637 m/s ≈ 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. The length of the slider is 30 cm. What is the surface tension of the film?
Answer
Soap film has two surfaces, so: F = 2TL
T = W/(2L) = (1.5 × 10⁻²) / (2 × 0.30) = 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.
Answer
The weight supported is 4.5 × 10⁻² N in all three figures (a), (b), and (c).
Reason: The force due to surface tension = 2T × L, where L is the length of the contact line. The shape of the film doesn't matter; only the perimeter length matters. Since the same length of wire is in contact with the film in all three cases, and the surface tension T is the same (same liquid, same temperature), the weight supported is identical.
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.
Answer
r = 3.00 × 10⁻³ m, T = 0.465 N/m
Excess pressure (single surface — drop): ΔP = 2T/r = (2 × 0.465) / (3.00 × 10⁻³) = 310 Pa
Total pressure inside = P₀ + ΔP = 1.01 × 10⁵ + 310 = 1.013 × 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 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 atm = 1.01 × 10⁵ Pa).
Answer
r = 5.00 × 10⁻³ m, T = 2.50 × 10⁻² N/m
Soap bubble (two surfaces): Excess pressure = 4T/r = (4 × 2.50 × 10⁻²) / (5.00 × 10⁻³) = 20 Pa
Air bubble at 40 cm depth
ρ = 1.20 × 10³ = 1200 kg/m³
Pressure at depth: P_d = 1.01 × 10⁵ + 1200 × 9.8 × 0.40 = 1.01 × 10⁵ + 4704 = 1.057 × 10⁵ Pa
Air bubble excess (single surface): ΔP = 2T/r = 10 Pa
Pressure inside = 1.057 × 10⁵ + 10 = 1.058 × 10⁵ Pa
Important Extra Questions
State and derive Pascal's law. Give one practical application.
Answer
Pascal's Law: A change in pressure applied to an enclosed fluid is transmitted undiminished to every portion of the fluid and to the walls of the containing vessel.
Basis: In a static fluid, consider a small element in equilibrium. Net force = 0 in all directions. For any horizontal direction, pressure must be equal at all points at the same height. The pressure only varies vertically: P = P₀ + ρgh. An externally applied pressure increase ΔP adds equally to pressure at every point at every depth throughout the fluid.
Application — Hydraulic Press
Two pistons of areas A₁ (small) and A₂ (large) connected by fluid.
Force F₁ on small piston → pressure P = F₁/A₁ → large piston force F₂ = P × A₂ = F₁ × (A₂/A₁)
Mechanical advantage = A₂/A₁. Small force → large force. Used in: hydraulic brakes, car jacks, industrial presses.
State Bernoulli's theorem and write its equation. State the conditions for validity.
Answer
Bernoulli's Theorem: For steady, non-viscous, incompressible fluid flow along a streamline, the sum of pressure energy per unit volume, kinetic energy per unit volume, and potential energy per unit volume is constant.
P + ½ρv² + ρgh = constant
Between points 1 and 2: P₁ + ½ρv₁² + ρgh₁ = P₂ + ½ρv₂² + ρgh₂
Conditions
(i) Steady flow — velocity at each point constant in time
(ii) Non-viscous (inviscid) — no internal friction
(iii) Incompressible — constant density
(iv) Irrotational flow
(v) Flow along a streamline
Applications: Aerofoil lift, Venturimeter, Atomizer, Magnus effect.
A tank is filled with water to height H. A hole is made at height h from the bottom. Find the velocity of efflux (Torricelli's theorem).
Answer
Depth of hole below water surface = (H − h).
Apply Bernoulli's between water surface (1) and hole (2):
P₀ + ½ρv₁² + ρgH = P₀ + ½ρv₂² + ρgh
Since tank is large, v₁ ≈ 0. Both at atmospheric pressure P₀:
ρgH = ½ρv₂² + ρgh
½v₂² = g(H − h)
v₂ = √(2g(H − h)) — Torricelli's theorem
This equals the velocity a body would acquire by falling freely from height (H − h).
Derive terminal velocity using Stokes' law. A steel ball of radius 2 mm falls through glycerine (η = 1.5 Pa·s, ρ_f = 1260 kg/m³). Density of steel = 8000 kg/m³. Find terminal velocity.
Answer
Stokes' Law: Viscous drag force on a sphere of radius r moving at velocity v: F = 6πηrv
Terminal velocity v_t: When net force = 0:
Weight = Buoyancy + Viscous drag
(4/3)πr³ρ_s g = (4/3)πr³ρ_f g + 6πηr v_t
v_t = 2r²(ρ_s − ρ_f)g / (9η)
Given: r = 2 × 10⁻³ m, η = 1.5 Pa·s, ρ_s = 8000, ρ_f = 1260 kg/m³
v_t = 2 × (2 × 10⁻³)² × (8000 − 1260) × 9.8 / (9 × 1.5)
= 2 × 4 × 10⁻⁶ × 6740 × 9.8 / 13.5
= (527,264 × 10⁻⁶) / 13.5 = 0.039 m/s = 3.9 cm/s
What is surface tension? Derive excess pressure inside a soap bubble.
Answer
Surface Tension: Property of a liquid surface that makes it act like a stretched elastic membrane, tending to minimize its area. Defined as force per unit length along the surface:
T = F/L (SI unit: N/m = J/m²)
Excess pressure inside a soap bubble
A soap bubble of radius r has two surfaces. Let inside pressure = P_in, outside = P₀.
Virtual expansion by dr:
Work by excess pressure = (P_in − P₀) × 4πr² × dr
Increase in surface area (two surfaces) = 2 × 8πr dr = 16πr dr
Work against surface tension = T × 16πr dr
Equating:
(P_in − P₀) × 4πr² × dr = 16πrT dr
P_in − P₀ = 4T/r
Excess pressure inside soap bubble = 4T/r
(For a liquid drop with single surface: excess pressure = 2T/r)