Chapter 9: Mechanical Properties of Fluids Application Questions | physics Class 11 CBSE

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Chapter 9: Mechanical Properties of Fluids Application Questions | physics Class 11 CBSE | ByteNotes.in

Application Question

Medium difficulty • Concept in a practical situation

Medium

Question 1

Applied Concept

A surgeon uses a syringe with a thin needle to inject medicine. With reference to fluid dynamics, explain why the needle controls the flow rate better than the thumb pressure.

The equation of continuity Av = constant means that at the narrow needle (very small A), the fluid speed v is very large even for a small volume flow rate; the needle cross-section dominates flow rate.
Even a slight change in needle diameter causes a large change in fluid velocity and hence flow rate (since A is proportional to r²); this makes the needle a precise flow-rate controller.
Thumb pressure changes the total pressure but affects flow rate less sensitively, since the syringe barrel is much wider; the pressure difference required to drive flow through the narrow needle (by Poiseuille's law) is much larger than through the barrel.
In clinical practice, selecting a needle of appropriate gauge (diameter) gives the doctor precise control over the rate of injection, which is critical for administering drugs safely.

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Application Question

Hard difficulty • Concept in a practical situation

Hard

Question 2

Applied Concept

A submarine is designed to operate at a depth of 1000 m in the ocean. With reference to pressure at depth, calculate the gauge pressure at this depth and explain the design requirement for the hull. (Density of sea water = 1.03 × 10³ kg m⁻³, g = 10 m s⁻²)

Gauge pressure at depth h = 1000 m: Pg = ρgh = 1.03 × 10³ × 10 × 1000 = 103 × 10⁵ Pa ≈ 103 atm.
This means the pressure acting inward on the submarine hull is about 103 times atmospheric pressure; the absolute pressure is approximately 104 atm.
The hull must be designed to withstand this enormous compressive force; even a small window of 20 cm × 20 cm (area 0.04 m²) experiences a net inward force of 103 × 10⁵ × 0.04 ≈ 4.12 × 10⁵ N.
Engineers must use very thick, high-strength steel for the hull and design all penetrations (windows, hatch covers) to resist the net pressure difference between the ocean outside and the atmospheric pressure maintained inside the submarine.

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Application Question

Medium difficulty • Concept in a practical situation

Medium

Question 3

Applied Concept

A water tap is being partially closed with a finger. Fast jets of water gush through the narrow gaps. Explain this observation using Bernoulli's principle and the equation of continuity.

When fingers partially close the tap, the cross-sectional area A available for water flow decreases significantly at the finger gaps.
By the equation of continuity (Av = constant), the reduced area forces a large increase in water velocity through the narrow gaps.
By Bernoulli's principle (P + ½ρv² = constant at the same height), the higher velocity at the gaps corresponds to lower pressure there.
The pressure difference between the water behind the fingers (higher P) and at the gap (lower P) further accelerates the water, producing fast, thin jets — the same principle used in spray nozzles and atomisers.

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Application Question

Hard difficulty • Concept in a practical situation

Hard

Question 4

Applied Concept

A fully loaded Boeing aircraft has a mass of 3.3 × 10⁵ kg and a total wing area of 500 m². Estimate the pressure difference needed between the lower and upper surfaces of the wings to keep the aircraft in level flight. (g = 9.8 m s⁻²)

For the aircraft to fly in level flight, the upward aerodynamic force due to pressure difference must balance the weight of the aircraft.
Weight of aircraft: W = mg = 3.3 × 10⁵ × 9.8 = 3.234 × 10⁶ N.
Required pressure difference: ΔP = W/A = 3.234 × 10⁶ / 500 ≈ 6.5 × 10³ N m⁻² (Pa).
This pressure difference arises because air flows faster over the curved upper surface of the aerofoil than under the flatter lower surface; the speed above the wing needs to be only about 8% higher than below to generate this lift, as calculated in the chapter.

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Application Question

Hard difficulty • Concept in a practical situation

Hard

Question 5

Applied Concept

In a car lift used in a service garage, compressed air exerts force on a small piston of radius 5 cm. This pressure is transmitted to a second piston of radius 15 cm to lift a car of mass 1350 kg. Calculate the force required on the smaller piston and the air pressure needed. (g = 9.8 m s⁻²)

Weight of car: F₂ = mg = 1350 × 9.8 = 13230 N ≈ 1.323 × 10⁴ N.
By Pascal's Law: F₁/A₁ = F₂/A₂, so F₁ = F₂ × (A₁/A₂) = F₂ × (r₁/r₂)² = 13230 × (5/15)² = 13230 × (1/9) = 1470 N ≈ 1.5 × 10³ N.
Air pressure required: P = F₁/A₁ = 1470 / (π × (5 × 10⁻²)²) = 1470 / (7.854 × 10⁻³) ≈ 1.87 × 10⁵ Pa.
This is nearly double atmospheric pressure; the mechanical advantage of the device is A₂/A₁ = 9, meaning the force is multiplied by a factor of 9, allowing a small compressor to lift a heavy vehicle.

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Application Question

Medium difficulty • Concept in a practical situation

Medium

Question 6

Applied Concept

A swimmer dives to a depth of 10 m in a lake. Calculate the pressure acting on the swimmer and explain why submarines must be specially designed to operate at 1 km depth. (ρ_water = 1000 kg m⁻³, g = 10 m s⁻²)

Pressure at 10 m depth: P = Pa + ρgh = 1.01 × 10⁵ + 1000 × 10 × 10 = 1.01 × 10⁵ + 1 × 10⁵ = 2.01 × 10⁵ Pa ≈ 2 atm.
This represents a 100% increase in pressure from the surface level, which a human body can tolerate with some discomfort but no structural damage.
At 1 km depth: pressure increase = ρgh = 1000 × 10 × 1000 = 10⁷ Pa = 100 atm — 100 times more than at the surface.
At such enormous pressures, ordinary vessel hulls would be crushed; submarines must use extremely thick high-strength steel hulls and specially designed sealed hatches, windows, and plumbing to survive the crushing external pressure while maintaining 1 atm internally for crew comfort.

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Application Question

Medium difficulty • Concept in a practical situation

Medium

Question 7

Applied Concept

A girl wearing high-heel shoes balances on a single circular heel of diameter 1.0 cm. If the girl's mass is 50 kg, calculate the pressure on the floor and compare it with the pressure of an elephant standing on all four feet, assuming each foot has an area of 400 cm². (Mass of elephant = 2000 kg, g = 10 m s⁻²)

Pressure from the girl's heel: F = mg = 50 × 10 = 500 N; A = π(0.5 × 10⁻²)² = π × 0.25 × 10⁻⁴ ≈ 7.85 × 10⁻⁵ m²; P_girl = 500/(7.85 × 10⁻⁵) ≈ 6.37 × 10⁶ Pa.
Total foot area of elephant: A = 4 × 400 cm² = 4 × 400 × 10⁻⁴ m² = 0.16 m²; F = 2000 × 10 = 20000 N; P_elephant = 20000/0.16 = 1.25 × 10⁵ Pa.
The girl's heel exerts about 50 times more pressure than the elephant, demonstrating that a large force spread over a wide area is less damaging than a small force concentrated on a tiny area.
This is the physics behind the danger of high-heel shoes on soft floors and the safety of wide-footed animals — pressure, not force alone, determines the impact on a surface.

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Application Question

Hard difficulty • Concept in a practical situation

Hard

Question 8

Applied Concept

A capillary tube of radius 0.5 mm is dipped in water. Using the formula for capillary rise, find the height of water that rises in the tube. (S = 0.073 N m⁻¹, ρ = 1000 kg m⁻³, θ ≈ 0°, g = 9.8 m s⁻²)

Capillary rise formula: h = 2S cos θ / (ρga); for water on glass, θ ≈ 0°, so cos θ = 1.
Substituting: h = 2 × 0.073 × 1 / (1000 × 9.8 × 0.5 × 10⁻³) = 0.146 / 4.9 ≈ 0.0298 m = 2.98 cm.
This means water rises nearly 3 cm in a capillary of 0.5 mm radius; in narrower tubes (e.g., 0.05 mm radius), the rise would be 10 times larger at about 30 cm.
This capillary action is crucial in biological systems — plants use it to draw water from roots up to leaves through xylem vessels of very fine diameter, working against gravity.

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Application Question

Hard difficulty • Concept in a practical situation

Hard

Question 9

Applied Concept

A copper ball of radius 2 mm falls through oil at 20°C with a terminal velocity of 6.5 cm s⁻¹. Use Stokes' Law to find the viscosity of the oil. (ρ_copper = 8.9 × 10³ kg m⁻³, ρ_oil = 1.5 × 10³ kg m⁻³, g = 9.8 m s⁻²)

Using terminal velocity formula: vt = 2a²(ρ − σ)g / (9η), where a = 2 × 10⁻³ m, ρ = 8.9 × 10³, σ = 1.5 × 10³, vt = 6.5 × 10⁻² m s⁻¹.
Rearranging: η = 2a²(ρ − σ)g / (9vt) = 2 × (2 × 10⁻³)² × (8.9 − 1.5) × 10³ × 9.8 / (9 × 6.5 × 10⁻²).
Numerator: 2 × 4 × 10⁻⁶ × 7.4 × 10³ × 9.8 = 2 × 4 × 7.4 × 9.8 × 10⁻³ = 579.68 × 10⁻³.
Denominator: 9 × 6.5 × 10⁻² = 0.585; η = 0.5797/0.585 ≈ 9.9 × 10⁻¹ Pa·s — a reasonable viscosity for oil at 20°C, confirming that the terminal velocity method can accurately determine fluid viscosity.

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Application Question

Hard difficulty • Concept in a practical situation

Hard

Question 10

Applied Concept

A U-tube contains water in one arm and methylated spirit in the other, with the mercury columns at the same level. The water column is 10.0 cm and the spirit column is 12.5 cm. Find the specific gravity of the spirit.

For the mercury levels to be equal in both arms, the pressure at the base of each mercury column must be equal.
Pressure from water column: P_water = ρ_water × g × h_water = 1000 × g × 0.10.
Pressure from spirit column: P_spirit = ρ_spirit × g × h_spirit = ρ_spirit × g × 0.125.
Setting equal: 1000 × 0.10 = ρ_spirit × 0.125, giving ρ_spirit = 100/0.125 = 800 kg m⁻³; specific gravity of spirit = 800/1000 = 0.8.

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Chapter sections

Chapter 9: Mechanical Properties of Fluids | Application Questions

A surgeon uses a syringe with a thin needle to inject medicine. With reference to fluid dynamics, explain why the needle controls the flow rate better than the thumb pressure.

A submarine is designed to operate at a depth of 1000 m in the ocean. With reference to pressure at depth, calculate the gauge pressure at this depth and explain the design requirement for the hull. (Density of sea water = 1.03 × 10³ kg m⁻³, g = 10 m s⁻²)

A water tap is being partially closed with a finger. Fast jets of water gush through the narrow gaps. Explain this observation using Bernoulli's principle and the equation of continuity.

A fully loaded Boeing aircraft has a mass of 3.3 × 10⁵ kg and a total wing area of 500 m². Estimate the pressure difference needed between the lower and upper surfaces of the wings to keep the aircraft in level flight. (g = 9.8 m s⁻²)

In a car lift used in a service garage, compressed air exerts force on a small piston of radius 5 cm. This pressure is transmitted to a second piston of radius 15 cm to lift a car of mass 1350 kg. Calculate the force required on the smaller piston and the air pressure needed. (g = 9.8 m s⁻²)

A swimmer dives to a depth of 10 m in a lake. Calculate the pressure acting on the swimmer and explain why submarines must be specially designed to operate at 1 km depth. (ρ_water = 1000 kg m⁻³, g = 10 m s⁻²)

A capillary tube of radius 0.5 mm is dipped in water. Using the formula for capillary rise, find the height of water that rises in the tube. (S = 0.073 N m⁻¹, ρ = 1000 kg m⁻³, θ ≈ 0°, g = 9.8 m s⁻²)

A copper ball of radius 2 mm falls through oil at 20°C with a terminal velocity of 6.5 cm s⁻¹. Use Stokes' Law to find the viscosity of the oil. (ρ_copper = 8.9 × 10³ kg m⁻³, ρ_oil = 1.5 × 10³ kg m⁻³, g = 9.8 m s⁻²)

A U-tube contains water in one arm and methylated spirit in the other, with the mercury columns at the same level. The water column is 10.0 cm and the spirit column is 12.5 cm. Find the specific gravity of the spirit.

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