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

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

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A hydraulic lift at a car service station uses Pascal's Law to raise vehicles for inspection and repair. The lift consists of two pistons connected by a fluid-filled pipe. A technician applies a force on the smaller piston of cross-sectional area 10 cm² to generate pressure in the hydraulic fluid. This pressure is transmitted undiminished through the oil to the larger piston of cross-sectional area 500 cm², which lifts the car placed on a platform above it. The technician notices that when the smaller piston is pushed down by 50 cm, the larger piston rises by only 1 cm. The car being lifted has a mass of 1500 kg. The system uses oil as the hydraulic fluid, which is treated as incompressible for this purpose.

Question 1: State Pascal's Law as applied in this hydraulic lift and identify the physical quantity that is transmitted from the smaller piston to the larger piston.

Pascal's Law states that when external pressure is applied to any part of a fluid contained in a vessel, it is transmitted undiminished and equally in all directions throughout the fluid.
The physical quantity transmitted from the smaller piston to the larger piston is pressure (P = F/A), not force; the same pressure acts on both pistons.
This transmission of pressure allows a small force on the small piston to produce a large force on the large piston, giving the lift its mechanical advantage.

Question 2: Calculate the force that must be applied on the smaller piston to lift the car, and determine the mechanical advantage of the hydraulic lift.

Weight of car: F₂ = mg = 1500 × 9.8 = 14700 N; by Pascal's Law F₁/A₁ = F₂/A₂, so F₁ = F₂ × (A₁/A₂) = 14700 × (10/500) = 294 N.
The technician needs to apply only 294 N to lift a car weighing 14700 N, demonstrating the force multiplication effect of the hydraulic lift.
Mechanical advantage = F₂/F₁ = A₂/A₁ = 500/10 = 50; the lift multiplies the applied force by a factor of 50, making it feasible to lift heavy vehicles with modest effort.

Question 3: Verify using the principle of incompressibility that when the smaller piston moves down by 50 cm, the larger piston rises by 1 cm. Also explain why the work done by the technician equals the work done on the car (ignoring friction).

For an incompressible fluid, volume displaced by the small piston equals volume received by the large piston: A₁L₁ = A₂L₂; substituting: 10 × 50 = 500 × L₂, giving L₂ = 500/500 = 1 cm — verified.
Work done by technician: W₁ = F₁ × L₁ = 294 N × 0.50 m = 147 J; work done on car: W₂ = F₂ × L₂ = 14700 N × 0.01 m = 147 J — equal, confirming energy conservation.
The hydraulic system is a force multiplier, not an energy multiplier; the gain in force is exactly compensated by the loss in displacement (the large piston moves 50 times less than the small one), consistent with the work-energy theorem.

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Case Set 2

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During a biology field trip, students observe that water rises to the tops of tall trees through narrow xylem vessels. The teacher explains that this is partly due to capillary action — the same phenomenon that causes water to rise in narrow glass tubes. The xylem vessels in a tree can be modelled as capillary tubes of very small radius. The teacher shows a demonstration: capillary tubes of three different radii (0.1 mm, 0.5 mm, and 1.0 mm) are dipped vertically in water. The students observe that the narrowest tube shows the greatest rise. The surface tension of water at room temperature is 0.073 N m⁻¹, density of water is 1000 kg m⁻³, and the contact angle between water and glass is approximately 0°.

Question 1: Define capillary rise and state the formula for the height h of liquid rise in a capillary tube of radius a.

Capillary rise is the spontaneous rise (or fall) of a liquid in a narrow tube due to surface tension and the pressure difference across the curved liquid meniscus at the liquid-air interface.
The formula for capillary rise is h = 2S cos θ / (ρga), where S is the surface tension, θ is the angle of contact, ρ is the liquid density, g is acceleration due to gravity, and a is the tube radius.
For water in glass θ ≈ 0° so cos θ = 1; the meniscus is concave and pressure below it is less than atmospheric, causing the liquid to rise until the weight of the raised column balances the pressure difference.

Question 2: Calculate the capillary rise for the tube of radius 0.5 mm and explain why the narrowest tube shows the greatest rise.

For a = 0.5 mm = 5 × 10⁻⁴ m, θ = 0°: h = 2 × 0.073 × 1 / (1000 × 9.8 × 5 × 10⁻⁴) = 0.146 / 0.49 ≈ 0.298 m = 2.98 cm.
Since h = 2S cos θ/(ρga) and all quantities are fixed except a, h is inversely proportional to the tube radius — halving the radius doubles the rise.
The narrowest tube (a = 0.1 mm) would give h ≈ 14.9 cm, five times greater than the 0.5 mm tube; this confirms that finer xylem vessels in plants can draw water to much greater heights.

Question 3: If mercury (contact angle 135° with glass, density 13,600 kg m⁻³, surface tension 0.465 N m⁻¹) is used instead of water in the 0.5 mm tube, predict and calculate what happens to the mercury level inside the tube relative to outside. Explain the physical reason.

For mercury, cos 135° = −1/√2 ≈ −0.707; applying h = 2S cos θ/(ρga) = 2 × 0.465 × (−0.707) / (13600 × 9.8 × 5 × 10⁻⁴) = −0.6575 / 66.64 ≈ −0.00987 m ≈ −0.99 cm.
The negative value means mercury is depressed by about 1 cm below the outside level rather than rising; a convex meniscus forms because mercury does not wet glass (obtuse angle of contact).
Physically, for mercury Ssl > Sla, making it energetically costly to create a mercury-glass interface; mercury molecules prefer to stay together rather than spread on glass, so the meniscus curves upward (convex) and the pressure below it is higher than atmospheric, pushing mercury down inside the tube.

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A student conducts an experiment using Torricelli's apparatus to study the speed of efflux. A large cylindrical tank of water (cross-sectional area 0.5 m²) has a small circular hole (area 2 cm²) at its side, at a height of 0.2 m from the bottom. The top surface of water is at a height of 1.4 m from the bottom, and the tank is open to the atmosphere. The student measures the speed of water coming out of the hole and also observes how far from the tank the water stream hits the ground. The ground is at the level of the bottom of the tank. The student notes that as water drains, the efflux speed gradually decreases. The density of water is 1000 kg m⁻³ and g = 9.8 m s⁻².

Question 1: State Torricelli's Law and derive the expression for the speed of efflux when the tank is open to atmosphere.

Torricelli's Law states that the speed of efflux of liquid from a small hole in an open tank at a depth h below the free surface is v = √(2gh), which is identical to the speed of a freely falling body from the same height h.
Applying Bernoulli's equation between the free surface (v₂ ≈ 0, pressure = Pa, height = y₂) and the hole (pressure = Pa, speed = v₁, height = y₁): Pa + ρgy₂ = Pa + ½ρv₁² + ρgy₁.
Simplifying with y₂ − y₁ = h: ½ρv₁² = ρgh, giving v₁ = √(2gh); the atmospheric pressure cancels on both sides since the tank is open and the hole opens to atmosphere.

Question 2: Calculate the speed of efflux when the water level is at 1.4 m from the bottom and the hole is at 0.2 m. Then use the equation of continuity to check if the assumption v₂ ≈ 0 is valid.

Depth of hole below surface: h = 1.4 − 0.2 = 1.2 m; efflux speed: v₁ = √(2gh) = √(2 × 9.8 × 1.2) = √23.52 ≈ 4.85 m s⁻¹.
By equation of continuity: A₁v₁ = A₂v₂; v₂ = A₁v₁/A₂ = (2 × 10⁻⁴ × 4.85) / 0.5 = 9.7 × 10⁻⁴ / 0.5 = 1.94 × 10⁻³ m s⁻¹.
Since v₂ ≈ 1.94 × 10⁻³ m s⁻¹ << v₁ ≈ 4.85 m s⁻¹, the assumption that the free surface is approximately at rest (v₂ ≈ 0) is fully justified for A₂ >> A₁.

Question 3: The water exits the hole horizontally at a height of 0.2 m above the ground. Calculate the horizontal distance from the tank where the jet hits the ground. Also explain why the efflux speed decreases as the tank drains.

The water exits horizontally at height y = 0.2 m above the ground; time to fall: y = ½gt², so t = √(2y/g) = √(2 × 0.2/9.8) = √0.0408 ≈ 0.202 s.
Horizontal range: x = v₁ × t = 4.85 × 0.202 ≈ 0.98 m; the jet hits the ground approximately 0.98 m from the base of the tank.
As the tank drains, the height h of water above the hole decreases; since v₁ = √(2gh) and h is decreasing, the efflux speed decreases as √h — a square-root relationship — until eventually h = 0 and flow stops.

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Case Set 4

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A meteorologist uses a mercury barometer to measure atmospheric pressure. The barometer shows a column height of 76 cm of mercury on a normal day at sea level. The meteorologist notes that the barometric reading has dropped to 74 cm, which historically signals an approaching storm. An intern asks why mercury is used instead of water. The meteorologist explains that the choice of liquid determines the height of the column needed to balance atmospheric pressure, and that denser liquids require shorter columns. The meteorologist also explains the concept of gauge pressure and how blood pressure gauges work on the same principle. Density of mercury = 13,600 kg m⁻³, density of water = 1000 kg m⁻³, g = 9.8 m s⁻².

Question 1: Explain the working principle of the mercury barometer and derive the expression Pa = ρgh.

A long glass tube closed at one end is filled with mercury and inverted into a trough of mercury; the space above the mercury column contains only mercury vapour, whose pressure is negligibly small and is taken as zero.
The atmospheric pressure Pa acts on the open mercury surface in the trough; this pressure is transmitted through the mercury to support the column of height h inside the tube.
At the base of the column (same level as trough surface), pressure inside = pressure outside: 0 + ρgh = Pa, giving Pa = ρgh, where ρ is mercury density and h is the column height.

Question 2: Calculate the height of a water barometer that would balance the same atmospheric pressure of 1.013 × 10⁵ Pa. Explain why water is not practical for this purpose.

Using Pa = ρgh: h = Pa/(ρg) = 1.013 × 10⁵ / (1000 × 9.8) = 101300/9800 ≈ 10.34 m.
A water barometer would need to be over 10 metres tall — impractical for laboratory or field use; mercury, being 13.6 times denser, requires only 76 cm, making it far more convenient.
Pascal himself replicated Torricelli's experiment using French wine (density 984 kg m⁻³) and needed a column height of about 10.3 m, confirming the inverse relationship between density and required column height.

Question 3: Explain gauge pressure with reference to blood pressure measurement. If a blood pressure reading is 120 mm Hg (systolic), what is the corresponding gauge pressure in Pa and the absolute pressure? (ρ_mercury = 13600 kg m⁻³)

Gauge pressure is the excess of actual pressure over atmospheric pressure: Pg = P − Pa = ρgh; a blood pressure gauge (sphygmomanometer) measures gauge pressure, not absolute pressure — a reading of 0 mm Hg on the gauge means pressure equals atmospheric, not zero.
Gauge pressure for 120 mm Hg: Pg = ρgh = 13600 × 9.8 × 0.120 = 15974.4 Pa ≈ 1.6 × 10⁴ Pa ≈ 16 kPa.
Absolute pressure = Pa + Pg = 1.013 × 10⁵ + 1.597 × 10⁴ = 1.173 × 10⁵ Pa ≈ 1.16 atm; blood pressure gauges report gauge pressure because the atmospheric pressure component is the same everywhere and the medically relevant quantity is the excess pressure driving blood flow.

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Case Set 5

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An engineering student studying fluid mechanics observes water flowing through a horizontal pipe that narrows from a diameter of 4 cm to 2 cm. A pressure gauge at the wider section reads 2.5 × 10⁵ Pa, and the flow speed at the wider section is 1.0 m s⁻¹. The student uses Bernoulli's equation and the equation of continuity to find the pressure at the narrow section. The student also notes that in real pipes, there is some energy loss due to viscosity, so the actual pressure at the narrow section will be slightly lower than predicted by ideal Bernoulli's equation. The density of water is 1000 kg m⁻³ and the pipe is horizontal throughout.

Question 1: Using the equation of continuity, find the flow speed at the narrow section of the pipe.

Equation of continuity for incompressible flow: A₁v₁ = A₂v₂, where A₁ and A₂ are cross-sectional areas and v₁, v₂ are flow speeds.
A₁ = π(2 cm)² = 4π cm²; A₂ = π(1 cm)² = π cm²; ratio A₁/A₂ = 4; therefore v₂ = v₁ × (A₁/A₂) = 1.0 × 4 = 4.0 m s⁻¹.
The speed quadruples at the narrow section because the cross-sectional area decreases by a factor of 4 (area is proportional to the square of the radius, and radius halved means area quartered).

Question 2: Calculate the pressure at the narrow section using Bernoulli's equation. The pipe is horizontal.

Bernoulli's equation for horizontal pipe (h₁ = h₂): P₁ + ½ρv₁² = P₂ + ½ρv₂².
Rearranging: P₂ = P₁ + ½ρ(v₁² − v₂²) = 2.5 × 10⁵ + ½ × 1000 × (1² − 4²) = 2.5 × 10⁵ + 500 × (1 − 16) = 2.5 × 10⁵ − 7500.
P₂ = 2.5 × 10⁵ − 0.075 × 10⁵ = 2.425 × 10⁵ Pa; pressure decreases at the narrower section as kinetic energy increases at the expense of pressure energy.

Question 3: The student observes that the actual measured pressure at the narrow section is slightly lower than 2.425 × 10⁵ Pa. Explain why, and state the conditions under which Bernoulli's equation gives exact results.

The actual pressure is lower than predicted because real water has viscosity; viscous forces between fluid layers convert some kinetic energy into heat, causing additional energy loss that Bernoulli's equation does not account for.
Bernoulli's equation gives exact results only for ideal fluids that are: (1) non-viscous (zero viscosity), (2) incompressible (constant density), and (3) in steady (non-turbulent) flow along a streamline.
In practice, as the flow speed increases beyond a critical speed, flow becomes turbulent; in turbulent flow, velocity and pressure fluctuate with time at each point, and Bernoulli's equation completely breaks down — actual pressure drops are significantly larger than the ideal prediction.

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Chapter 9: Mechanical Properties of Fluids | Case Study Questions

Passage

Question 1: State Pascal's Law as applied in this hydraulic lift and identify the physical quantity that is transmitted from the smaller piston to the larger piston.

Question 2: Calculate the force that must be applied on the smaller piston to lift the car, and determine the mechanical advantage of the hydraulic lift.

Question 3: Verify using the principle of incompressibility that when the smaller piston moves down by 50 cm, the larger piston rises by 1 cm. Also explain why the work done by the technician equals the work done on the car (ignoring friction).

Passage

Question 1: Define capillary rise and state the formula for the height h of liquid rise in a capillary tube of radius a.

Question 2: Calculate the capillary rise for the tube of radius 0.5 mm and explain why the narrowest tube shows the greatest rise.

Question 3: If mercury (contact angle 135° with glass, density 13,600 kg m⁻³, surface tension 0.465 N m⁻¹) is used instead of water in the 0.5 mm tube, predict and calculate what happens to the mercury level inside the tube relative to outside. Explain the physical reason.

Passage

Question 1: State Torricelli's Law and derive the expression for the speed of efflux when the tank is open to atmosphere.

Question 2: Calculate the speed of efflux when the water level is at 1.4 m from the bottom and the hole is at 0.2 m. Then use the equation of continuity to check if the assumption v₂ ≈ 0 is valid.

Question 3: The water exits the hole horizontally at a height of 0.2 m above the ground. Calculate the horizontal distance from the tank where the jet hits the ground. Also explain why the efflux speed decreases as the tank drains.

Passage

Question 1: Explain the working principle of the mercury barometer and derive the expression Pa = ρgh.

Question 2: Calculate the height of a water barometer that would balance the same atmospheric pressure of 1.013 × 10⁵ Pa. Explain why water is not practical for this purpose.

Question 3: Explain gauge pressure with reference to blood pressure measurement. If a blood pressure reading is 120 mm Hg (systolic), what is the corresponding gauge pressure in Pa and the absolute pressure? (ρ_mercury = 13600 kg m⁻³)

Passage

Question 1: Using the equation of continuity, find the flow speed at the narrow section of the pipe.

Question 2: Calculate the pressure at the narrow section using Bernoulli's equation. The pipe is horizontal.

Question 3: The student observes that the actual measured pressure at the narrow section is slightly lower than 2.425 × 10⁵ Pa. Explain why, and state the conditions under which Bernoulli's equation gives exact results.

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