JEE Physics Mechanical Properties Of Fluids MCQs Set 14

Answer: (c) 5.0 × 10⁻⁵ m²

Explanation:
1. v² = u² + 2gh = 1 + 2 × 10 × 0.15 = 4, so v = 2 m/s.
2. A1 v1 = A2 v2 gives A2 = 10⁻⁴ × 1 / 2 = 5.0 × 10⁻⁵ m².

Teacher's Note:
1. Find the new speed using the kinematic equation first.
2. Then apply the equation of continuity.

Answer: (a) 10 m/s

Explanation:
1. By Bernoulli's theorem the drop in pressure appears as kinetic energy.
2. ½ ρ v² = 4.5 × 10⁵ − 4 × 10⁵ = 5 × 10⁴, so v² = 100 and v = 10 m/s.

Teacher's Note:
1. With the tap closed, the water is at rest.
2. The pressure difference equals ½ ρ v².

Answer: (a) 0.8 m

Explanation:
1. Velocity head = v² / (2g).
2. = 16 / 20 = 0.8 m.

Teacher's Note:
1. Velocity head is the height of liquid that would give the same pressure as the kinetic energy term.
2. It is ½ρv² divided by ρg.

Answer: (c) \( \dfrac{v}{n^2}\left(\dfrac{R}{r}\right)^{2} \)

Explanation:
1. By continuity π R² v = n² × π r² × v'.
2. v' = v R² / (n² r²) = (v/n²)(R/r)².

Teacher's Note:
1. Total area of the holes is n² π r².
2. The speed is inversely proportional to the total area.

Answer: (d) All of these

Explanation:
1. Torricelli's theorem gives v = √(2gh).
2. The fall height is (H − h), so R = v t = 2√(h(H − h)), which is maximum when h = H/2 and then R_max = H.

Teacher's Note:
1. Time of fall depends on H − h while speed depends on h.
2. The product is largest when the two are equal.

Answer: (c) (7/2) ρ A v²

Explanation:
1. Pressure difference = ½ ρ [(2√2 v)² − v²] = ½ ρ (8v² − v²) = (7/2) ρ v².
2. Lift = pressure difference × area = (7/2) ρ A v².

Teacher's Note:
1. Higher speed on the upper surface means lower pressure above the wing.
2. The net upward force is the pressure difference times the area.

Answer: (a) 2.0

Explanation:
1. In level flight the lift equals the weight, so ΔP × A = m g.
2. ΔP = 5 × 10⁴ × 10 / 250 = 2000 Pa = 2.0 kPa.

Teacher's Note:
1. Level flight means zero net vertical force.
2. Convert pascal to kilopascal at the end.

Answer: (d) Bernoulli's theorem

Explanation:
1. Air blown fast over the top of a narrow tube lowers the pressure there.
2. The higher atmospheric pressure then pushes the liquid up the tube, which is Bernoulli's principle.

Teacher's Note:
1. Sprayers, atomisers and carburettors all use Bernoulli's principle.
2. Faster flow means lower pressure.

Answer: (c) viscous force exerted by air

Explanation:
1. The viscous drag increases with speed.
2. At the terminal velocity the drag (with the small upthrust) balances the weight, so the speed stops increasing.

Teacher's Note:
1. Drag grows with velocity until it balances the weight.
2. Net force is then zero and velocity is constant.

Answer: (d) viscosity

Explanation:
1. Poise is the CGS unit of the coefficient of viscosity.
2. 1 poise = 0.1 Pa s.

Teacher's Note:
1. In SI the unit of viscosity is Pa s (decapoise).
2. Surface tension is measured in N/m.

Answer: (d) 0.0005

Explanation:
1. η = 0.01 poise = 0.001 Pa s.
2. F = η A v / d, so d = 0.001 × 0.01 × 0.1 / 0.002 = 5 × 10⁻⁴ m.

Teacher's Note:
1. Convert poise to SI by multiplying by 0.1.
2. The area is 0.1 × 0.1 = 0.01 m².

Answer: (d) 1 : 4

Explanation:
1. Stokes' law gives F = 6 π η r v.
2. For the same v and the same liquid, F is proportional to r, so the ratio is 1 : 4.

Teacher's Note:
1. Viscous force is proportional to radius and to velocity.
2. Here the velocities are the same.

Answer: (a) almost independent of h

Explanation:
1. A drop falling in air reaches its terminal velocity after a short fall.
2. Beyond that, the speed does not change, so the final velocity is almost independent of the height.

Teacher's Note:
1. Terminal velocity depends on radius, density and viscosity, not on height.
2. This is why raindrops do not hit the ground at very high speed.

Answer: (b) 4^(1/3) cm/s

Explanation:
1. The radius of the combined drop is R = 2^(1/3) r.
2. Terminal velocity is proportional to r², so v' = 2^(2/3) × 1 = 4^(1/3) cm/s.

Teacher's Note:
1. Volumes add, so the radius is multiplied by 2^(1/3).
2. Then use v proportional to r².

Answer: (b) 10 cm/s

Explanation:
1. Terminal speed is proportional to r².
2. Halving the radius gives a quarter of the speed, 40/4 = 10 cm/s.

Teacher's Note:
1. Assume both balls are of the same material and have reached the terminal velocity.
2. The speed is proportional to r².

Answer: (b) 9 : 4

Explanation:
1. Terminal velocity is proportional to r².
2. The ratio is 3² : 2² = 9 : 4.

Teacher's Note:
1. Larger drops fall faster.
2. The ratio is the square of the radius ratio.

Answer: (c) M²/R²

Explanation:
1. Terminal velocity v = 2 R² ρ g / (9 η) with ρ = 3M / (4 π R³), so v is proportional to M/R.
2. Therefore v² is proportional to M²/R².

Teacher's Note:
1. Express the density in terms of the mass and radius.
2. Then square the result.

Answer: (c) 2.5 × 10⁻⁴ m/s

Explanation:
1. v = 2 r² ρ g / (9 η) with r = 1.5 × 10⁻⁶ m.
2. v = 2 × (2.25 × 10⁻¹²) × 10³ × 10 / (9 × 2 × 10⁻⁵) = 2.5 × 10⁻⁴ m/s.

Teacher's Note:
1. Convert mm to m carefully.
2. The formula follows from balancing weight with the Stokes drag.

Answer: (d) 27 : 8

Explanation:
1. Terminal velocity is proportional to r², so r1 : r2 = 3 : 2.
2. Volume is proportional to r³, so the ratio is 27 : 8.

Teacher's Note:
1. Take the square root of the velocity ratio to get the radius ratio.
2. Then cube it for volume.

Answer: (d) decreases with increase in temperature

Explanation:
1. In a liquid, viscosity comes from the cohesive forces between molecules.
2. Heating weakens these forces, so the viscosity falls as temperature rises. (Gases behave in the opposite way.)

Teacher's Note:
1. Honey flows more easily when warm.
2. Viscosity of gases increases with temperature.

Answer: (b) surface tension

Explanation:
1. Surface tension tries to reduce the surface area of the liquid.
2. A sphere has the least area for a given volume, so free drops become spherical.

Teacher's Note:
1. Minimum surface energy means minimum surface area.
2. Small drops are almost perfectly spherical.

Answer: (d) surface tension of liquid lead

Explanation:
1. Molten lead falling from a height forms spherical drops because of surface tension.
2. The drops solidify on the way down to give round shots.

Teacher's Note:
1. Surface tension makes free liquid drops spherical.
2. The same property rounds the drops of molten glass beads.

Answer: (d) R/10

Explanation:
1. Volume is conserved: 1000 × (4/3)π r³ = (4/3)π R³.
2. r³ = R³/1000, so r = R/10.

Teacher's Note:
1. When drops merge, volumes add and not radii.
2. The cube root of 1000 is 10.

Answer: (b) 4π T D²

Explanation:
1. Each tiny drop has radius r = R/5.
2. Increase in area = 4π (125 r² − R²) = 4π R² (5 − 1) = 16 π R², so the energy change = 16 π T R² = 4 π T D².

Teacher's Note:
1. Energy supplied equals surface tension times the increase in area.
2. Use R = D/2 to convert.

Answer: (c) will be one-fourth of that for the smaller droplet

Explanation:
1. The radius of the big drop is 4 times that of a droplet, since 64 = 4³.
2. Excess pressure 2T/R is inversely proportional to radius, so it is one-fourth of that for a smaller droplet.

Teacher's Note:
1. Smaller drops have higher excess pressure.
2. Excess pressure for a drop is 2T/r.