JEE Physics Mechanical Properties Of Fluids MCQs Set 34

Answer: (b) A-3, B-4, C-1, D-2

Explanation:
1. Stokes' law gives the viscous drag (A-3).
2. Turbulence is characterised by a large Reynolds number (B-4).
3. Bernoulli's principle involves pressure energy (C-1).
4. Pascal's law is used in a hydraulic lift (D-2).

Teacher's Note:
1. Link each law to its application.

Answer: (c) Energy

Explanation:
1. Bernoulli's equation is the energy conservation law for an ideal fluid.
2. It balances pressure, kinetic and potential energy per unit volume.

Teacher's Note:
1. Energy conservation.

Answer: (d) \( 0.025\ \text{N m}^{-1} \)

Explanation:
1. The film has two surfaces, so the force is 2TL = W.
2. T = 1.5 × 10⁻²/(2 × 0.3) = 0.025 N/m.

Teacher's Note:
1. Two surfaces double the force.

Answer: (d) \( P_i-P_o=\frac{4s}{r} \)

Explanation:
1. A soap bubble has two free surfaces.
2. The excess pressure is 2 × (2s/r) = 4s/r.

Teacher's Note:
1. Two surfaces double the excess pressure.

Answer: (a) 0.057

Explanation:
1. Small drop radius r = 10⁻² cm = 10⁻⁴ m.
2. ΔA = 4π[10⁶(10⁻⁴)² − (10⁻²)²] = 4π(10⁻² − 10⁻⁴) = 0.1244 m².
3. E = TΔA = 0.46 × 0.1244 = 0.057 J.

Teacher's Note:
1. Energy equals surface tension times the increase in area.

Answer: (a) Both A and R are true, and R is the correct explanation of A.

Explanation:
1. For a floating body the buoyant force equals its weight, so the net force is zero.
2. A balance would read zero, and the reason explains this.

Teacher's Note:
1. Weight is balanced by upthrust.

Answer: (a) \( \frac{VR^2}{nr^2} \)

Explanation:
1. Continuity: πR²V = n πr² v.
2. v = VR²/(nr²).

Teacher's Note:
1. Equation of continuity.

Answer: (a) \( \sqrt{r_1^2+r_2^2} \)

Explanation:
1. In vacuum the pressure inside is 4T/r.
2. Isothermal: P₁V₁ + P₂V₂ = PV gives (4T/r₁)r₁³ + (4T/r₂)r₂³ = (4T/R)R³.
3. So R² = r₁² + r₂².

Teacher's Note:
1. Boyle's law with the bubble pressures.

Answer: (a) 13.6 cm

Explanation:
1. The pressure difference = 10 mm Hg = 1 cm Hg.
2. Equivalent water column = 1 × 13.6 = 13.6 cm.

Teacher's Note:
1. Convert mercury height to water height.

Answer: (c) \( 2\ \text{m s}^{-1} \)

Explanation:
1. Torricelli: v = √(2gh) = √(2 × 10 × 0.2).
2. = 2 m/s.

Teacher's Note:
1. Speed of efflux.

Answer: (d) A is false, but R is true.

Explanation:
1. Pressure acts equally in all directions and is a scalar, so the assertion is false.
2. Force is indeed a vector, so the reason is a true statement.

Teacher's Note:
1. Pressure is a scalar.

Answer: (d) the full length of the tube

Explanation:
1. In a satellite the effective gravity is zero.
2. The surface tension pulls the water along the whole tube.

Teacher's Note:
1. h → ∞ when g_eff = 0.

Answer: (c) 144 erg

Explanation:
1. The film has two surfaces.
2. ΔA = 2 × 10 × 0.1 = 2 cm², W = TΔA = 72 × 2 = 144 erg.

Teacher's Note:
1. Count both sides of the film.

Answer: (c) the area of the bottom surface

Explanation:
1. P = hρg depends on g, h and ρ.
2. It does not depend on the area of the bottom surface.

Teacher's Note:
1. Hydrostatic paradox.

Answer: (a) Both A and R are true, and R is the correct explanation of A.

Explanation:
1. When the drag grows with velocity until it balances the constant weight, the speed stops increasing.
2. The reason gives this general mechanism.

Teacher's Note:
1. Constant force plus rising drag.

Answer: (d) cohesive force between molecules

Explanation:
1. Molecules at the surface are pulled inward by cohesive forces from neighbours.
2. This gives rise to surface tension.

Teacher's Note:
1. Cohesion causes surface tension.

Answer: (a) Terminal velocities of rain drops are proportional to the square of their radii

Explanation:
1. vₜ = 2r²(ρ − σ)g/9η, so vₜ ∝ r².
2. Waterproofing agents increase the angle of contact, detergents lower the surface tension, and hydraulic machines use Pascal's law.

Teacher's Note:
1. Stokes' law gives vₜ ∝ r².

Answer: (d) III and IV only

Explanation:
1. v = √(2gh).
2. It depends on the height h and on g only, not on the hole's area or the liquid's density.

Teacher's Note:
1. Torricelli's theorem.

Answer: (b) \( \frac{2T\cos\theta}{xdg} \)

Explanation:
1. The upward force = 2T cos θ per unit length of the plates (two plate surfaces).
2. The weight raised per unit length = x h d g.
3. So h = 2T cos θ/(xdg).

Teacher's Note:
1. Balance 2T cos θ against the weight of the liquid.

Answer: (b) pressure

Explanation:
1. Pressure = force/area.
2. A smaller area gives a larger pressure for the same force.

Teacher's Note:
1. P = F/A.

Answer: (b) zero

Explanation:
1. At terminal velocity the weight is balanced by buoyancy plus viscous drag.
2. The net force and the acceleration are zero.

Teacher's Note:
1. Forces balance.

Answer: (d) Pascal's law

Explanation:
1. This is the statement of Pascal's law.
2. It follows from the hydrostatic equation P = P₀ + hρg.

Teacher's Note:
1. Same level, same pressure.

Answer: (a) \( \sqrt{\frac{Vg(\rho_1-\rho_2)}{k}} \)

Explanation:
1. At terminal speed kv² = Vρ₁g − Vρ₂g (weight minus upthrust).
2. v = √(Vg(ρ₁ − ρ₂)/k).

Teacher's Note:
1. Net weight equals the drag.

Answer: (a) Bernoulli's theorem

Explanation:
1. A fast air stream lowers the pressure over the paint tube, so paint is drawn up.
2. This uses Bernoulli's theorem.

Teacher's Note:
1. Fast flow, low pressure.

Answer: (c) \( \text{energy}=3VT\left(\frac1r-\frac1R\right)\ \text{is released} \)

Explanation:
1. Number of drops n = R³/r³.
2. Released energy = T(4πr²n − 4πR²) = 4πR³T(1/r − 1/R) = 3VT(1/r − 1/R).

Teacher's Note:
1. Surface area decreases, so energy is released.