Class 11 Physics Chapter 9: Mechanical Properties of Fluids worksheet with answer key

Answer: (d) the same in size in every orientation, and always normal to the face

At a point in a fluid at rest the pressure is the same in all directions, so the force on the small face has the same size whatever its orientation. A fluid at rest can push only normal to a surface, never along it.

Answer: (d) [ML⁻¹T⁻¹]

η = Fl/(vA): [MLT⁻²][L]/([LT⁻¹][L²]) = [ML⁻¹T⁻¹]. [ML⁻¹T⁻²] is pressure, [MT⁻²] is surface tension and [MLT⁻¹] is momentum.

Answer: (a) a fluid particle at the crossing point would have two velocities at once

The tangent to a streamline gives the velocity of the fluid at that point. At a crossing there would be two tangents, that is two velocities for the particle there, which is impossible in steady flow.

Answer: (a) 152 cm

P_a = ρgh fixes the vertical height of the column at 76 cm. Along a tube at 60° to the vertical, length = 76/cos 60° = 152 cm.

Answer: (c) the pressure of the liquid on the piston

By Pascal's law the pressure applied at the small piston is transmitted undiminished through the liquid, so the pressure is the same at both pistons; the forces differ in the ratio of the areas.

Velocity head = v²/2g = 16/(2 × 10) = 0.80 m. Pressure head = P/ρg = 150 000/(1000 × 10) = 15 m. (Dividing Bernoulli's equation by ρg gives P/ρg + v²/2g + h = constant.)

• velocity head v²/2g = 0.80 m (1)
• pressure head P/ρg = 15 m (1)

A liquid sticks to a solid if the solid–liquid surface energy is smaller than the sum of the solid–air and liquid–air surface energies; the liquid is then attracted to the solid. Water wets glass (acute angle of contact); mercury does not (obtuse angle of contact).

• S_sl smaller than S_sa + S_la (1)
• water wets glass; mercury does not (1)

The bubble is lighter than the liquid it displaces, so the upward buoyant force exceeds its weight and it accelerates upward. The viscous drag of the liquid opposes its motion and grows with its speed. When drag plus weight equals the buoyant force, the net force is zero and the bubble moves at a constant (terminal) velocity.

• buoyant force exceeds weight at first; viscous drag grows with speed (1)
• forces balance, net force zero, so constant terminal velocity (1)

(a) P − P_a = ρgh = 1000 × 10 × 0.20 = 2000 Pa. (b) h = 2000/(13 600 × 10) = 0.0147 m = 1.47 cm. (c) Water: for a small pressure difference a low-density liquid gives a large, easily read height difference; mercury suits large pressure differences.

• gauge pressure 2.0 kPa (1)
• mercury height about 1.5 cm (1)
• water, as a low-density liquid suits small pressure differences (1)

Continuity: v₂ = A₁v₁/A₂ = 10 × 2.0/5.0 = 4.0 m s⁻¹. Bernoulli: P₂ = P₁ + ½ρ(v₁² − v₂²) + ρg(h₁ − h₂) = 150 000 + ½ × 1000 × (4 − 16) + 1000 × 10 × 5.0 = 150 000 − 6000 + 50 000 = 194 000 Pa = 1.94 × 10⁵ Pa.

• v₂ = 4.0 m s⁻¹ from continuity (1)
• Bernoulli's equation with both height and speed terms (1)
• P₂ = 1.94 × 10⁵ Pa (1)

Forces: weight downward; buoyant force and viscous drag 6πηav upward. At terminal velocity they balance: 6πηav_t = (4π/3)a³(ρ − σ)g, so v_t = 2a²(ρ − σ)g/(9η) = 2 × (1.0 × 10⁻³)² × 6540 × 10/(9 × 0.83) = 0.0175 m s⁻¹ = 1.75 cm s⁻¹.

• weight, buoyant force and viscous drag (1)
• balance gives v_t = 2a²(ρ − σ)g/(9η) (1)
• v_t ≈ 1.75 cm s⁻¹ (1)