Chapter 9 – Mechanical Properties of Fluids

Q1

The SI unit of pressure is Pascal (Pa). Which of the following is equivalent to 1 Pa? Easy

(A)1 N m²

(B)1 N m⁻²

(C)1 kg m⁻¹s⁻²

(D)Both (B) and (C)

Solution

✓ Correct: (D)

P = F/A. 1 Pa = 1 N m⁻². Since 1 N = 1 kg·m·s⁻², also 1 Pa = 1 kg m⁻¹s⁻². Both (B) and (C) are correct.

Q2

Pascal's law: pressure applied to an enclosed fluid is transmitted: Easy

(A)Only to the bottom

(B)Undiminished in all directions

(C)Proportional to depth

(D)Only to vertical walls

Solution

✓ Correct: (B)

Pascal's Law: any pressure change is transmitted undiminished and equally in all directions throughout an enclosed fluid and to its container walls.

Q3

Pressure at depth h in a liquid (density ρ) open to atmosphere Pₐ is: Easy

(A)P = ρgh

(B)P = Pₐ − ρgh

(C)P = Pₐ + ρgh

(D)P = Pₐ × ρgh

Solution

✓ Correct: (C)

P = Pₐ + ρgh

Absolute pressure = atmospheric + gauge pressure (ρgh). The term ρgh is the gauge pressure at depth h.

Q4

Hydraulic press: A₁ = 10 cm², A₂ = 100 cm², F₁ = 50 N. Find F₂. Moderate

(A)50 N

(B)500 N

(C)5 N

(D)5000 N

Solution

✓ Correct: (B)

F₂ = F₁ × (A₂/A₁) = 50 × (100/10) = 500 N

By Pascal's Law: P = F₁/A₁ = F₂/A₂. Mechanical advantage = A₂/A₁ = 10.

Q5

Bernoulli's equation is applicable to: Easy

(A)Viscous, compressible flow

(B)Non-viscous, incompressible steady flow

(C)Turbulent flow only

(D)Fluid at rest only

Solution

✓ Correct: (B)

Bernoulli's equation (energy conservation) applies to non-viscous, incompressible, steady (streamline) flow. It does NOT hold for turbulent or viscous flow.

Q6

The correct form of Bernoulli's equation is: Easy

(A)P + ρv² + ρgh = C

(B)P + ½ρv² + ρgh = C

(C)P − ½ρv² + ρgh = C

(D)P + ½ρv² − ρgh = C

Solution

✓ Correct: (B)

P + ½ρv² + ρgh = constant

P = static pressure · ½ρv² = KE per unit volume · ρgh = PE per unit volume. Their sum is conserved along a streamline.

Q7

The equation of continuity (Av = constant) is a consequence of: Easy

(A)Conservation of energy

(B)Conservation of momentum

(C)Conservation of mass

(D)Newton's Second Law

Solution

✓ Correct: (C)

A₁v₁ = A₂v₂ follows from conservation of mass — no mass is created or destroyed in steady flow of incompressible fluid.

Q8

When fluid flows through a narrower section of pipe, velocity and pressure respectively: Easy

(A)Decreases, increases

(B)Increases, increases

(C)Increases, decreases

(D)Both remain unchanged

Solution

✓ Correct: (C)

Continuity: narrow area → higher velocity. Bernoulli: higher velocity → lower pressure. This Venturi effect is used in carburettors and flow meters.

Q9

Torricelli's theorem gives speed of efflux from an open tank at depth h as: Moderate

(A)v = √(gh)

(B)v = √(2gh)

(C)v = 2√(gh)

(D)v = √(gh/2)

Solution

✓ Correct: (B)

v = √(2gh)

Applying Bernoulli between surface and hole: v = √(2gh) — identical to speed of a freely falling body from height h.

Q10

The SI unit of coefficient of viscosity η is: Easy

(A)N m⁻²

(B)Pa·s (Poiseuille)

(C)N m⁻¹

(D)kg m⁻² s⁻¹

Solution

✓ Correct: (B)

SI unit = Pa·s (Pascal-second) = Poiseuille (Pl). Dimensions: [ML⁻¹T⁻¹]. Equivalent to N·s·m⁻².

Q11

Stokes' law for viscous drag on a sphere (radius a, velocity v, viscosity η) is: Moderate

(A)F = 4πηav

(B)F = 6πηa²v

(C)F = 6πηav

(D)F = 2πηav

Solution

✓ Correct: (C)

F = 6πηav

Viscous drag ∝ velocity (linear). Used to derive terminal velocity. Radius a appears to first power, not squared.

Q12

At terminal velocity, a falling sphere experiences: Moderate

(A)Zero net force, zero acceleration

(B)Net downward force

(C)Net upward force

(D)Increasing acceleration

Solution

✓ Correct: (A)

At terminal velocity: Weight = Buoyant Force + Viscous Drag. Net force = 0 → acceleration = 0 → constant velocity.

Q13

Surface tension is defined as: Easy

(A)Force per unit area of liquid surface

(B)Force per unit length along liquid surface

(C)Pressure difference across surface

(D)Energy per unit volume of liquid

Solution

✓ Correct: (B)

S = F/l (force per unit length). Also = surface energy per unit area. SI unit: N/m or J/m². Dimensions: [MT⁻²].

Q14

Excess pressure inside a spherical soap bubble (radius r, surface tension S) is: Moderate

(A)2S/r

(B)4S/r

(C)S/r

(D)8S/r

Solution

✓ Correct: (B)

Soap bubble (2 surfaces): ΔP = 4S/r Liquid drop (1 surface): ΔP = 2S/r

A soap bubble has two surfaces (inner + outer), hence ΔP = 4S/r.

Q15

Capillary rise h in tube of radius a (contact angle θ) is: Moderate

(A)h = 2S cosθ / (ρga)

(B)h = S cosθ / (ρga)

(C)h = 4S cosθ / (ρga)

(D)h = 2S / (ρga)

Solution

✓ Correct: (A)

h = 2S cosθ / (ρga)

Water-glass (acute θ): liquid rises. Mercury-glass (obtuse θ, cosθ < 0): liquid is depressed. Rise is larger for smaller radius.

Q16

Mercury has an obtuse angle of contact with glass. Mercury in a glass capillary: Moderate

(A)Rises in capillary

(B)Is depressed in capillary

(C)Neither rises nor falls

(D)Wets the glass

Solution

✓ Correct: (B)

θ > 90° → cosθ < 0 → h < 0 (depression). Mercury forms a convex meniscus and its level is lower inside the capillary.

Q17

With increase in temperature, viscosity of a liquid: Easy

(A)Increases

(B)Decreases

(C)First increases then decreases

(D)Remains unchanged

Solution

✓ Correct: (B)

Higher temperature weakens cohesive forces → viscosity of liquids decreases. (Opposite for gases: viscosity increases with temperature.)

Q18

Terminal velocity of a sphere of radius a is proportional to: Moderate

(A)a

(B)a²

(C)a³

(D)1/a

Solution

✓ Correct: (B)

vₜ = 2a²(ρ − σ)g / (9η)

vₜ ∝ a². Double the radius → 4× terminal velocity. Large raindrops fall much faster than drizzle droplets.

Q19

Magnus effect (deviation of spinning cricket ball) is caused by: Moderate

(A)Gravity on the ball

(B)Pressure difference from unequal air speeds

(C)Surface roughness friction

(D)Wind speed alone

Solution

✓ Correct: (B)

Spinning ball drags air: one side has higher speed → lower pressure (Bernoulli) → net lift/deflection force. Explains swing in cricket, top-spin in tennis.

Q20

In streamline flow, two streamlines: Easy

(A)Can intersect at right angles

(B)Never intersect each other

(C)Always run parallel

(D)Can intersect at any angle

Solution

✓ Correct: (B)

Streamlines cannot intersect in steady flow — intersection would imply a particle has two velocities simultaneously, which is impossible.

Q21

Normal atmospheric pressure at sea level is approximately: Easy

(A)10⁵ Pa

(B)1.013 × 10⁵ Pa

(C)76 Pa

(D)13.6 × 10³ Pa

Solution

✓ Correct: (B)

1 atm = 1.013 × 10⁵ Pa = 101.3 kPa = 760 mm Hg. Measured by Torricelli's mercury barometer.

Q22

Hydrostatic paradox — liquid pressure at given depth depends on: Moderate

(A)Shape of the container

(B)Base cross-sectional area

(C)Only vertical height of liquid

(D)Both shape and depth

Solution

✓ Correct: (C)

P = ρgh — depends only on vertical depth h, not on shape or cross-section. Three different-shaped vessels filled to same height exert same pressure at base level.

Q23

Surface tension arises due to: Moderate

(A)Gravity on surface molecules

(B)Extra potential energy of surface molecules

(C)Kinetic energy of surface molecules

(D)Atmospheric pressure on surface

Solution

✓ Correct: (B)

Surface molecules are only half-surrounded by liquid — they have extra potential energy vs interior molecules. The liquid minimises this by minimising surface area.

Q24

Which is NOT an application of Bernoulli's principle? Moderate

(A)Lift on aircraft wings

(B)Working of a hydraulic lift

(C)Flow through Venturi meter

(D)Magnus effect on spinning balls

Solution

✓ Correct: (B)

Hydraulic lift is based on Pascal's Law (pressure transmission), NOT Bernoulli's. Aircraft lift, Venturi meter, and Magnus effect all use Bernoulli's principle.

Q25

A liquid drop tends to be spherical because: Moderate

(A)Sphere has maximum volume for given area

(B)Sphere has minimum surface area for given volume

(C)Gravity pulls drop to sphere

(D)Atmosphere compresses drop to sphere

Solution

✓ Correct: (B)

A liquid minimises surface energy by minimising surface area. For a given volume, a sphere has the minimum surface area — hence free drops are spherical when gravity is negligible.

N1

A 50 kg girl in high-heel shoes balances on one heel of diameter 1.0 cm. Calculate the pressure on the floor. (g = 10 m/s²) Moderate

Solution

Given: m = 50 kg, d = 1 cm → r = 0.005 m, g = 10 m/s²

A = πr² = π × (0.005)² = 7.854 × 10⁻⁵ m² F = mg = 500 N P = F/A = 500 / 7.854 × 10⁻⁵ P ≈ 6.37 × 10⁶ Pa ≈ 63.7 atm

~63× atmospheric pressure — explains why high heels damage floors!

N2

Find pressure on a swimmer 10 m below a lake surface. (ρ = 1000 kg/m³, g = 10 m/s², Pₐ = 1.01 × 10⁵ Pa) Moderate

Solution

P = Pₐ + ρgh = 1.01 × 10⁵ + 1000 × 10 × 10 = 1.01 × 10⁵ + 1.0 × 10⁵ P = 2.01 × 10⁵ Pa ≈ 2 atm

Pressure doubles at 10 m. At 1 km depth: ~100 atm — why submarines need reinforcement.

N3

Hydraulic lift: A₁ = 5 cm², A₂ = 75 cm², F₁ = 20 N. Find (a) F₂, (b) mechanical advantage. Moderate

Solution

(a) F₂ = F₁ × (A₂/A₁) = 20 × (75/5) = 300 N (b) MA = A₂/A₁ = 75/5 = 15

20 N input → 300 N output (15× amplification). Principle of hydraulic brakes and car lifts.

N4

Water flows in pipe (A₁ = 4 cm², v₁ = 5 m/s) narrowing to A₂ = 2 cm². Find v₂ and pressure difference. (ρ = 1000 kg/m³) Moderate

Solution

Continuity: v₂ = (A₁/A₂)v₁ = (4/2)×5 = 10 m/s Bernoulli (horizontal): ΔP = ½ρ(v₂²−v₁²) = ½×1000×(100−25) = 500 × 75 = 3.75 × 10⁴ Pa

N5

Open tank has a hole at depth 3.2 m. Find speed of efflux. (g = 10 m/s²) Moderate

Solution

Torricelli's Law: v = √(2gh) = √(2 × 10 × 3.2) = √64 v = 8 m/s

Same as speed of free fall from h = 3.2 m — Torricelli's Law (special case of Bernoulli).

N6

Find terminal velocity of a steel ball (r = 1 mm) in glycerine. ρ_steel = 8000, ρ_gly = 1260 kg/m³, η = 1.5 Pa·s, g = 9.8 m/s². Moderate

Solution

vₜ = 2a²(ρ−σ)g / (9η) = [2×(10⁻³)²×(8000−1260)×9.8] / (9×1.5) = [2×10⁻⁶×6740×9.8] / 13.5 = 0.1321 / 13.5 vₜ ≈ 9.78 × 10⁻³ m/s ≈ 9.78 mm/s

N7

U-tube: water column 12 cm, unknown liquid column 15 cm in equilibrium. Find density of liquid. Moderate

Solution

ρ_water × h_water = ρ_liq × h_liq ρ_liq = (1000 × 12) / 15 ρ_liq = 800 kg/m³ (Relative density = 0.8)

N8

Find excess pressure inside a soap bubble of radius 2 cm. Surface tension = 0.03 N/m. Moderate

Solution

ΔP = 4S/r = (4 × 0.03) / 0.02 ΔP = 0.12 / 0.02 = 6 Pa (liquid drop: ΔP = 2S/r = 3 Pa)

N9

Find capillary rise for water (r = 0.1 mm). S = 0.073 N/m, θ = 0°, ρ = 1000 kg/m³, g = 9.8 m/s². Moderate

Solution

h = 2S cosθ / (ρga) = (2 × 0.073 × 1) / (1000 × 9.8 × 10⁻⁴) = 0.146 / 0.98 h ≈ 14.9 cm

This 15 cm rise explains capillary action in plant stems and oil wicks.

N10

At depth 800 m in ocean find (a) absolute pressure, (b) gauge pressure. ρ = 1.03 × 10³ kg/m³, g = 10 m/s². Moderate

Solution

(b) Gauge: P_g = ρgh = 1.03×10³×10×800 = 82.4×10⁵ Pa ≈ 82.4 atm (a) Absolute: P = Pₐ + P_g = (1.01 + 82.4)×10⁵ = 83.41×10⁵ Pa ≈ 83.4 atm

N11

Flow speeds on upper/lower wing surfaces: 70 m/s and 63 m/s. Wing area = 2.5 m². Find lift force. (ρ_air = 1.3 kg/m³) Moderate

Solution

ΔP = ½ρ(v_top²−v_bot²) = ½×1.3×(4900−3969) = 0.65×931 = 605.15 Pa Lift = ΔP×A = 605.15×2.5 ≈ 1513 N ≈ 1.5×10³ N

N12

Metal block (A = 0.10 m²) connected to 10 g mass via frictionless pulley moves at constant 0.085 m/s. Film thickness = 0.30 mm. Find coefficient of viscosity. Hard

Solution

F = mg = 0.010×9.8 = 9.8×10⁻² N Shear stress = F/A = 9.8×10⁻²/0.10 = 0.98 N/m² Strain rate = v/l = 0.085/(3×10⁻⁴) = 283.3 s⁻¹ η = stress/strain rate = 0.98/283.3 η ≈ 3.46×10⁻³ Pa·s

N13

ρ_air = 1.29 kg/m³ at sea level (constant). Estimate the height of the atmosphere. (g = 9.8 m/s², Pₐ = 1.01 × 10⁵ Pa) Moderate

Solution

h = P/(ρg) = 1.01×10⁵ / (1.29×9.8) h = 1.01×10⁵ / 12.642 h ≈ 7991 m ≈ 8 km

Rough estimate (actual ~100 km since density decreases with altitude).

N14

Two syringes (d₁ = 1 cm, d₂ = 3 cm) connected by rubber tube. (a) F₂ when F₁ = 10 N. (b) If smaller piston moves 6 cm, how far does larger move? Moderate

Solution

(a) A₁=π(0.5)²=0.785 cm², A₂=π(1.5)²=7.069 cm² F₂ = F₁×(A₂/A₁) = 10×(7.069/0.785) = 10×9 = 90 N (b) Volume conservation: A₁L₁ = A₂L₂ L₂ = (A₁/A₂)×L₁ = (1/9)×6 ≈ 0.67 cm

N15

Spray pump tube (A = 8 cm²) has 40 holes (d = 1 mm each). Liquid flows at 1.5 m/min. Find ejection speed through holes. Moderate

Solution

v_tube = 1.5 m/min = 0.025 m/s a_hole = π(0.5×10⁻³)² = 7.854×10⁻⁷ m² A_total = 40 × 7.854×10⁻⁷ = 3.14×10⁻⁵ m² Continuity: A_tube×v_tube = A_total×v_holes v_holes = (8×10⁻⁴×0.025) / (3.14×10⁻⁵) v_holes ≈ 0.637 m/s ≈ 38.2 m/min

Ejection speed is ~25× tube speed — mass conservation across 40 small holes.