Mechanical Properties of Fluids Class 11 Physics NCERT Solutions

Step-by-step solution

1. 1(a) P = hρg — pressure inside a liquid column rises with the height h of the column. The blood column above the feet is taller than that above the brain, so the pressure at the feet is greater.
2. 2(b) Pressure falls because the air density falls with height: at 6 km the density is roughly half the sea-level value, so the pressure there is about half. The upper 100 km of the atmosphere is very thin air and contributes little pressure.
3. 3(c) Pressure acts equally in all directions in a fluid at rest; it has magnitude but no unique direction, so it is a scalar even though it is defined as force/area.

Step-by-step solution

1. 1(a)At the 3-phase contact line, cos θ = (S_sa − S_sl)/S_la. For mercury, adhesion to glass is weak (S_sa < S_la), giving an obtuse angle; for water adhesion is strong (S_sl < S_la), giving an acute angle.
2. 2(b) Mercury molecules attract each other strongly but the glass weakly — they pull together into drops. Water molecules attract each other weakly but the glass strongly — they spread out (wet the glass).
3. 3(c) Surface tension is the force per unit length along a surface; it depends on the molecular interaction of the liquid, not on how large the surface area is.
4. 4(d) A small angle of contact gives fast capillary rise (h ∝ cos θ), which pulls the detergent solution quickly into cloth fibres and cleans effectively.
5. 5(e) Surface tension makes a liquid minimize its surface area; for a fixed volume a sphere has the smallest surface area, so a free drop pulls itself into a sphere.

Step-by-step solution

1. 1(a) Decreases — surface tension falls as temperature rises.
2. 2(b) Increases; decreases — gas viscosity grows with temperature while liquid viscosity falls.
3. 3(c) Shear strain; rate of shear strain — solids resist deformation proportional to strain, fluids to the rate of strain.
4. 4(d) Conservation of mass (the equation of continuity A₁v₁ = A₂v₂).
5. 5(e) Greater — the small model has a smaller Reynolds number, so turbulence sets in at higher speed.

Step-by-step solution

1. 1(a) Blowing over the paper speeds the air above it, lowering pressure there (Bernoulli); the higher pressure beneath lifts the paper up and keeps it horizontal. Blowing under it does the reverse and pushes it down.
2. 2(b) By the equation of continuity, A₁v₁ = A₂v₂: the tiny gaps left between the fingers have a very small area, forcing the water through them at high speed.
3. 3(c) The needle's small area multiplies the flow speed strongly at a given pressure (continuity), so the needle size dominates the flow rate; thumb pressure only adds a small adjustment.
4. 4(d) Expulsion of fluid from the hole (momentum carried by the jet) exerts an equal and opposite reaction on the vessel — the backward thrust (conservation of momentum).
5. 5(e) The spin drags air faster on one side of the ball, lowering pressure there (Bernoulli), so a sideways force curves the path — it is not a plain parabola.

Step-by-step solution

1. 1Force F = mg = 50 × 9.8 = 490 N.
2. 2Area A = πr² = π(0.005)² = 7.85 × 10⁻⁵ m².
3. 3P = F/A = 490/7.85 × 10⁻⁵.
4. 4P = 6.24 × 10⁶ N m⁻².

Step-by-step solution

1. 1Atmospheric pressure supports both columns: ρ₁h₁g = ρ₂h₂g.
2. 2ρ₁ (mercury) = 13.6 × 10³ kg m⁻³, h₁ = 0.76 m, ρ₂ (wine) = 984 kg m⁻³.
3. 3h₂ = ρ₁h₁/ρ₂ = (13.6 × 10³ × 0.76)/984.
4. 4h₂ = 10.5 m.

Step-by-step solution

1. 1Pressure at depth d: P = ρgh = 10³ × 9.8 × 3 × 10³.
2. 2P = 2.94 × 10⁷ Pa.
3. 3This is far less than the maximum withstandable stress 10⁹ Pa.
4. 4Yes — the structure is suitable.

Step-by-step solution

1. 1Load force F = mg = 3000 × 9.8 = 29,400 N; A = 425 × 10⁻⁴ m².
2. 2P = F/A = 29,400/425 × 10⁻⁴.
3. 3P = 6.917 × 10⁵ Pa.
4. 4Pascal's law transmits this pressure to the smaller piston.

Step-by-step solution

1. 1With the mercury levels equal, the pressures on the two mercury surfaces are the same (open to the atmosphere).
2. 2h_water ρ_water g = h_spirit ρ_spirit g.
3. 30.10 ρ_water = 0.125 ρ_spirit.
4. 4ρ_spirit/ρ_water = 0.10/0.125 = 0.8.
5. 5Specific gravity of spirit = 0.8.

Step-by-step solution

1. 1New columns: water h₁ = 10 + 15 = 25 cm; spirit h₂ = 12.5 + 15 = 27.5 cm.
2. 2Pressure difference at the mercury level = h₁ρ₁g − h₂ρ₂g with ρ₁ = 1, ρ₂ = 0.8.
3. 3= 25 × 1 g − 27.5 × 0.8 g = 3g (dyn forces per cm² of mercury column).
4. 4This is balanced by the mercury column of height h: h × 13.6 g = 3 g.
5. 5h = 3/13.6 = 0.22 cm.

Step-by-step solution

1. 1Bernoulli's equation holds for steady, streamline (laminar) flow of a non-viscous fluid.
2. 2A river rapid is violently turbulent — streamlines break up.
3. 3No — Bernoulli's equation cannot be used there.

Step-by-step solution

1. 1Bernoulli's equation is used between two points and involves a difference of pressures.
2. 2Gauge pressure = absolute pressure − atmospheric pressure.
3. 3The constant atmospheric term cancels in the difference, leaving the same result.
4. 4No — it does not matter, provided both pressures are treated consistently.

Step-by-step solution

1. 1Volume flow rate V = M/ρ = 4.0 × 10⁻³/1.3 × 10³ = 3.08 × 10⁻⁶ m³ s⁻¹.
2. 2Poiseuille's formula: V = π p r⁴/(8ηl).
3. 3p = 8ηlV/(πr⁴) = 8 × 0.83 × 1.5 × 3.08 × 10⁻⁶/[π(0.01)⁴].
4. 4p = 9.8 × 10² Pa.
5. 5Laminar check: Re = 4ρV/(πdη) = 4 × 1.3 × 10³ × 3.08 × 10⁻⁶/(π × 0.02 × 0.83) ≈ 0.3.
6. 6Since Re ≪ 2000, the flow is laminar — the assumption is correct.

Step-by-step solution

1. 1By Bernoulli, P₁ + ½ρv₁² = P₂ + ½ρv₂².
2. 2Pressure difference (P₂ − P₁) = ½ρ(v₁² − v₂²).
3. 3Lift = (P₂ − P₁) × A = ½ × 1.3 × (70² − 63²) × 2.5.
4. 4= ½ × 1.3 × 931 × 2.5 = 1512.9 N.
5. 5Lift ≈ 1.51 × 10³ N.

Step-by-step solution

1. 1At a constriction, continuity A₁v₁ = A₂v₂ makes the speed rise.
2. 2By Bernoulli's principle, higher speed means lower pressure.
3. 3In a vertical standpipe the liquid level shows the pressure: the narrower section must show a depressed level.
4. 4Figure (a) shows the level rising at the constriction, which contradicts Bernoulli — figure (a) is incorrect.

Step-by-step solution

1. 1A₁ = 8 × 10⁻⁴ m²; v₁ = 1.5 m min⁻¹ = 0.025 m s⁻¹.
2. 2Total hole area A₂ = 40 × π(0.5 × 10⁻³)² = 31.4 × 10⁻⁶ m².
3. 3Continuity: A₁v₁ = A₂v₂.
4. 4v₂ = (8 × 10⁻⁴ × 0.025)/(31.4 × 10⁻⁶) = 0.637 m s⁻¹ ≈ 0.64 m s⁻¹.

Step-by-step solution

1. 1A soap film has two free surfaces, so the total length of film pulled by the weight is 2l = 0.6 m.
2. 2Surface tension S = W/2l = 1.5 × 10⁻²/0.6.
3. 3S = 2.5 × 10⁻² N m⁻¹.

Step-by-step solution

1. 1For a liquid film with two surfaces, the supported weight is W = 2Sl.
2. 2In all three figures the liquid is the same and the temperature is the same, so S is identical.
3. 3The length of the film's slider is the same (40 cm) in each case.
4. 4Hence each film supports the same weight: 4.5 × 10⁻² N in figures (b) and (c) as well.

Step-by-step solution

1. 1Excess pressure inside a drop = 2S/r = 2 × 4.65 × 10⁻¹/(3 × 10⁻³).
2. 2= 310 Pa.
3. 3Total pressure inside = P₀ + 2S/r = 1.01 × 10⁵ + 310.
4. 4= 1.0131 × 10⁵ Pa ≈ 1.01 × 10⁵ Pa.

Step-by-step solution

1. 1Soap bubble (two surfaces): excess pressure = 4S/r = 4 × 2.5 × 10⁻²/(5 × 10⁻³) = 20 Pa.
2. 2Air bubble in the liquid (one surface): excess pressure = 2S/r = 10 Pa.
3. 3Pressure at depth 0.4 m in the liquid: hρg = 0.4 × 1.2 × 10³ × 9.8 = 4704 Pa.
4. 4Total pressure inside the bubble = 1.01 × 10⁵ + 4704 + 10.
5. 5= 1.057 × 10⁵ Pa ≈ 1.06 × 10⁵ Pa.