JEE Physics Mechanical Properties Of Fluids MCQs Set 16

Answer: (d) 0 ≤ θ1 < θ2 < θ3 < π/2

Explanation:
1. The liquids rise, so each angle is acute.
2. h = 2T cos θ / (r ρ g) is the same for all, so cos θ is proportional to ρ. The largest density has the largest cos θ and the smallest angle, so θ1 < θ2 < θ3.

Teacher's Note:
1. Capillary rise needs an acute contact angle.
2. For the same rise, a denser liquid must have a larger cos θ.

Answer: (c) 928 kg m⁻³

Explanation:
1. The water in the left limb falls by 65 mm and in the right limb rises by 65 mm, so the water surface in the right limb is 130 mm above the oil-water interface on the left.
2. The oil column is 130 + 10 = 140 mm. Pressure balance gives ρ_oil × 140 = 1000 × 130, so ρ_oil ≈ 928 kg/m³.

Teacher's Note:
1. Compare pressures at the level of the oil-water interface.
2. Include the extra 10 mm of oil above the right-limb water level.

Answer: (c) 1 cm

Explanation:
1. The excess pressure inside a soap bubble is 4T/r = 4 × 2.5 × 10⁻² / 10⁻³ = 100 Pa.
2. This equals ρ g Z0, so Z0 = 100 / (10³ × 10) = 0.01 m = 1 cm.

Teacher's Note:
1. A soap bubble has two surfaces and so the excess pressure is 4T/r.
2. Match the excess (gauge) pressure with the water column.

Answer: (a) 12.6 × 10⁻⁶ m³/s

Explanation:
1. Speed of efflux v = √(2 g h) = √40 ≈ 6.32 m/s.
2. Rate = A v = 2 × 10⁻⁶ × 6.32 ≈ 12.6 × 10⁻⁶ m³/s.

Teacher's Note:
1. Convert mm² to m² by multiplying by 10⁻⁶.
2. This is Torricelli's theorem.

Answer: (b) 750 kg/m³

Explanation:
1. The pressures at the interface level are equal.
2. ρ_oil × 20 = 1000 × 15, so ρ_oil = 750 kg/m³.

Teacher's Note:
1. The oil column is longer, so it is lighter.
2. Equal pressure at the same horizontal level in the same connected liquid.

Answer: (a) \( \dfrac{\rho - \rho_2}{\rho_1 - \rho_2} \)

Explanation:
1. Let V1 be the volume in the upper liquid. Equilibrium: ρ V = ρ1 V1 + ρ2 (V − V1).
2. So V1/V = (ρ2 − ρ)/(ρ2 − ρ1) = (ρ − ρ2)/(ρ1 − ρ2).

Teacher's Note:
1. The weight is balanced by the sum of the two buoyant forces.
2. Both forms of the fraction are equal.

Answer: (b) (5/4) ρ

Explanation:
1. Weight = buoyancy from the two layers.
2. d L = (3L/4)ρ + (L/4)(2ρ) = (5L/4) ρ, so d = (5/4) ρ.

Teacher's Note:
1. Add the contributions of both liquids to the upthrust.
2. The block's density lies between ρ and 2ρ.

Answer: (a) 26 ℓ

Explanation:
1. At the bottom the pressure is (ℓ + h) and radius r; at the top it is ℓ and radius 3r.
2. Boyle's law (ignoring surface tension): (ℓ + h) r³ = ℓ (27 r³), so h = 26 ℓ.

Teacher's Note:
1. Use pressure in the same unit, metres of water.
2. Volume goes as the cube of the radius.

Answer: (a) 2

Explanation:
1. Force at the base = ρ g L × L² = ρ g L³.
2. Force on one side wall = ½ ρ g L × L² = ½ ρ g L³, so the ratio is 2.

Teacher's Note:
1. The mean pressure on a vertical wall is half of the pressure at the base.
2. The wall and the base have the same area.

Answer: (b) 300 g

Explanation:
1. The extra weight m was balanced by extra buoyancy from a 3 cm deeper immersion.
2. m = ρ × (area × 0.03) = 1000 × 0.01 × 0.03 = 0.3 kg = 300 g.

Teacher's Note:
1. The extra displaced volume equals area times the rise.
2. Neglect the change in the water level of the large container.

Answer: (b) 12.50 cm

Explanation:
1. Water falls 25 cm on the oil side and rises 25 cm on the other side, so the water levels differ by 50 cm.
2. Oil column height = 50 × 1/0.8 = 62.5 cm above the interface, so it stands 62.5 − 50 = 12.5 cm above the water level of the other limb.

Teacher's Note:
1. Balance pressure at the level of the oil-water interface.
2. The oil column is longer than the water column it balances.

Answer: (c) 915 kg/m³

Explanation:
1. The oil column above the interface level is ℓ + d = 147.5 mm.
2. Pressure balance gives ρ_oil × 147.5 = 1000 × 135, so ρ_oil ≈ 915 kg/m³.

Teacher's Note:
1. Include the extra height d in the oil column.
2. Oil is lighter than water, so its density is below 1000.

Answer: (a) 37.12 N

Explanation:
1. Volume of the block = 4 / 10⁴ = 4 × 10⁻⁴ m³.
2. Buoyant force = 720 × 4 × 10⁻⁴ × 10 = 2.88 N, so T = 40 − 2.88 = 37.12 N.

Teacher's Note:
1. Tension = weight − buoyant force.
2. Relative density gives the density in units of 1000 kg/m³.

Answer: (a) decreases

Explanation:
1. The bubble expands as more air is pushed in, so its radius increases.
2. The excess pressure 4T/r decreases with increasing radius, so the pressure falls.

Teacher's Note:
1. Smaller bubbles have higher pressure.
2. This is why a small bubble connected to a large one empties into the large one.

Answer: (b) air from end 1 flows towards end 2 and the volume of the soap bubble at end 1 decreases

Explanation:
1. A sub-hemispherical bubble has a larger radius of curvature, so its excess pressure 4T/R is smaller.
2. Air flows from the higher pressure (end 1) to the lower pressure (end 2), so the bubble at end 1 shrinks.

Teacher's Note:
1. The bubble with the smaller radius has the higher pressure.
2. Flow is always from high pressure to low pressure.

Answer: (c) 1130 Pa

Explanation:
1. Pressure due to the liquid depth = ρ g h = 1000 × 9.8 × 0.1 = 980 Pa.
2. Excess pressure due to the single surface = 2T/r = 2 × 0.075 / 10⁻³ = 150 Pa, so the total is 1130 Pa.

Teacher's Note:
1. An air bubble in a liquid has one surface, so 2T/r.
2. Add the hydrostatic pressure to the surface tension excess.

Answer: (a) 117.6

Explanation:
1. Pressure difference = 6 × 10⁻³ × 1000 × 9.8 = 58.8 Pa.
2. A soap bubble has two surfaces, so ΔP = 4T/r gives T = 58.8 × 8 × 10⁻³ / 4 = 0.1176 N/m = 117.6 dyne/cm.

Teacher's Note:
1. 1 N/m equals 1000 dyne/cm.
2. Use 4T/r for a soap bubble.

Answer: (a) √400 m/s

Explanation:
1. Only the gauge pressure drives the flow: 3 atm − 1 atm = 2 × 10⁵ Pa.
2. v = √(2 ΔP / ρ) = √(2 × 2 × 10⁵ / 1000) = √400 m/s.

Teacher's Note:
1. The outside of the hole is at atmospheric pressure.
2. Use the excess over atmospheric pressure in Torricelli's relation.

Answer: (b) a curve starting from zero and rising with decreasing slope until it becomes constant at the terminal speed

Explanation:
1. The ball starts from rest and its acceleration is (weight − upthrust − drag)/mass.
2. The drag grows with speed, so the acceleration decreases to zero and the speed levels off at the terminal value.

Teacher's Note:
1. Acceleration decreases as drag increases.
2. The graph approaches a horizontal line asymptotically.

Answer: (a) 0.4 m/s

Explanation:
1. Terminal velocity is proportional to (ρ − σ) for the same size.
2. v_silver = 0.8 × (10.5 − 1.5)/(19.5 − 1.5) = 0.8 × 9/18 = 0.4 m/s.

Teacher's Note:
1. Subtract the density of the liquid from the density of the sphere.
2. Same radius means the other factors cancel.

Answer: (d) 1 : 32

Explanation:
1. Terminal velocity is proportional to r² and mass is proportional to r³.
2. Momentum m v is proportional to r⁵, so the ratio is 1 : 32.

Teacher's Note:
1. Both mass and speed depend on the radius.
2. Multiply their power laws together.

Answer: (a) \( \left(\dfrac{2T}{r}\right)\cos\theta \)

Explanation:
1. The radius of curvature of the meniscus is R = r / cos θ.
2. The pressure difference across the meniscus is 2T/R = (2T/r) cos θ.

Teacher's Note:
1. The pressure is greater on the concave side of the meniscus.
2. The angle of contact reduces the curvature of the meniscus.

Answer: (d) 0.84 cm

Explanation:
1. h = 2T cos θ / (ρ g) × (1/r1 − 1/r2) with r1 = 1 mm and r2 = 2.5 mm.
2. h = 2 × 0.07 / (10⁴) × (1000 − 400) = 8.4 × 10⁻³ m = 0.84 cm.

Teacher's Note:
1. The narrower limb has the higher level.
2. The difference in capillary rises gives the level difference.

Answer: (c) A is false but B is true

Explanation:
1. The viscosity of a liquid decreases with increasing temperature, so A is false.
2. Surface tension decreases as temperature rises, so B is true.

Teacher's Note:
1. Liquids become thinner and less sticky on heating.
2. Surface tension becomes zero at the critical temperature.