JEE Physics Mechanical Properties Of Fluids MCQs Set 13

Answer: (b) 4/3 g/cm³

Explanation:
1. For mass m of each, total mass = 2m and total volume = m/1 + m/2 = 3m/2.
2. Density = 2m / (3m/2) = 4/3 g/cm³.

Teacher's Note:
1. When equal masses are mixed, the density is the harmonic mean, not the arithmetic mean.
2. Always add the volumes, not the densities.

Answer: (a) pressure

Explanation:
1. One torr is the pressure exerted by a column of mercury 1 mm high.
2. It is therefore a unit of pressure (1 torr is about 133.3 Pa).

Teacher's Note:
1. Torr is named after Torricelli, who invented the mercury barometer.
2. 760 torr equals one atmosphere.

Answer: (d) 4 hdg

Explanation:
1. Pressure on the wall increases linearly from 0 at the top of the liquid to ρ g H at the bottom.
2. Mean pressure = ½ ρ g H = ½ × 2d × g × 4h = 4 h d g.

Teacher's Note:
1. Pressure varies linearly with depth, so the mean is half of the bottom value.
2. Use the liquid density 2d, not d.

Answer: (c) 1/3

Explanation:
1. P_A = P0 + ρ g (10) = 10⁵ + 10⁵ = 2 × 10⁵ Pa.
2. P_B = P_A + (2ρ) g (20) = 2 × 10⁵ + 4 × 10⁵ = 6 × 10⁵ Pa, so P_A / P_B = 1/3.

Teacher's Note:
1. Add the pressure contribution of each layer separately.
2. The denser layer contributes twice as much per metre of depth.

Answer: (b) 0.8 × 10⁵ Pa

Explanation:
1. ρ g h = 10⁵ Pa.
2. The pressure at a height h/5 above the open surface is 10⁵ − ρ g (h/5) = 10⁵ − 0.2 × 10⁵ = 0.8 × 10⁵ Pa.

Teacher's Note:
1. Pressure falls by ρ g per unit height as you go up in the column.
2. At the top of the column (vacuum) the pressure is zero.

Answer: (c) Hydraulic lift

Explanation:
1. A hydraulic lift transmits a pressure applied to a small piston equally to a large piston.
2. The sprayer and venturimeter work on Bernoulli's principle.

Teacher's Note:
1. Pascal's law: pressure applied to an enclosed fluid is transmitted undiminished.
2. Hydraulic brakes, jacks and presses are applications.

Answer: (a) Pascal's law

Explanation:
1. Pascal's law says a change in pressure applied to an enclosed fluid is transmitted undiminished to every part of the fluid and the walls.
2. The other laws deal with gases.

Teacher's Note:
1. Gay-Lussac, Dalton and Boyle all deal with gases.
2. Pascal's law applies to fluids at rest.

Answer: (c) Pascal's law

Explanation:
1. A small force on a small piston creates a pressure that is transmitted equally to a large piston.
2. This multiplies the force, which is Pascal's law.

Teacher's Note:
1. F1/A1 = F2/A2 in a hydraulic press.
2. The force gain equals the area ratio.

Answer: (b) 1.1

Explanation:
1. Take the level of the A-B interface in the right limb.
2. The left limb has a column of B, 2 cm high, above this level and the right limb has a column of A, also 2 cm high. Equal pressures give ρB × 2 = 1.1 × 2, so the specific gravity of B is 1.1.

Teacher's Note:
1. Pressures at the same level in the same liquid are equal.
2. Equal heights above the level then mean equal densities.

Answer: (c) float completely immersed in water

Explanation:
1. The density of the block equals that of water.
2. Buoyant force equals weight when it is completely immersed, so it floats just fully submerged.

Teacher's Note:
1. Equal density means neutral buoyancy.
2. The block stays at rest at any depth.

Answer: (a) 1/2

Explanation:
1. In the accelerating frame, effective gravity becomes g' = 4g/3 for both the weight and the buoyancy.
2. The condition ρ_block V g' = ρ_liquid V_in g' does not contain g', so the fraction remains 1/2.

Teacher's Note:
1. A change in effective gravity affects weight and buoyancy equally.
2. The submerged fraction depends only on the ratio of the densities.

Answer: (a) 4/9 g/cc

Explanation:
1. In water, 1/3 of the volume is immersed, so the density of the sphere is 1/3 g/cc.
2. In the liquid, 3/4 is immersed, so ρ_liquid × 3/4 = 1/3, giving ρ_liquid = 4/9 g/cc.

Teacher's Note:
1. Density of a floating body = (fraction immersed) × liquid density.
2. Compute the sphere's density first.

Answer: (c) the balance A will read less than 2 kg and B will read more than 5 kg

Explanation:
1. The liquid exerts an upward buoyant force on the block, so balance A reads less than 2 kg.
2. By Newton's third law the block pushes down on the liquid with an equal force, so balance B reads more than 5 kg.

Teacher's Note:
1. The buoyant force acts on the block and the reaction acts on the liquid.
2. The sum of the two readings stays 7 kg.

Answer: (a) constant acceleration

Explanation:
1. The net force is the buoyant force minus the weight, and both are constant while the block is fully under water.
2. So the acceleration is constant until it reaches the surface.

Teacher's Note:
1. Fully immersed means constant buoyant force.
2. The acceleration changes only when the block breaks the surface.

Answer: (a) 3

Explanation:
1. Loss of weight in water = 60 − 40 = 20 N.
2. Relative density = weight in air / loss of weight = 60/20 = 3.

Teacher's Note:
1. Relative density = weight in air divided by the loss of weight in water.
2. This is Archimedes' method for relative density.

Answer: (c) 1.6 × 10³ kg/m³

Explanation:
1. Let the outer volume be V. The material occupies V/2.
2. Floating gives mass = (4/5) V × 1000, so the density of material = 0.8 V × 1000 / (V/2) = 1.6 × 10³ kg/m³.

Teacher's Note:
1. Use the volume of the material, not the outer volume, for its density.
2. The mass equals the mass of the displaced water.

Answer: (c) 72 kg

Explanation:
1. Extra displaced water = 12 × 2 × 0.003 = 0.072 m³.
2. Mass = 0.072 × 1000 = 72 kg.

Teacher's Note:
1. Extra weight equals the weight of extra water displaced.
2. Convert 0.3 cm to 0.003 m.

Answer: (c) \( \dfrac{d - d_0}{d} \)

Explanation:
1. The immersed fraction is d0/d.
2. The fraction above the surface is 1 − d0/d = (d − d0)/d.

Teacher's Note:
1. Immersed fraction plus emerged fraction equals one.
2. A lighter object floats higher.

Answer: (c) remains unchanged

Explanation:
1. The floating ice displaces water equal to its own weight.
2. Melted ice produces water of the same weight and the same volume, so the level is unchanged.

Teacher's Note:
1. Floating ice displaces its own weight of water.
2. It would rise only if the ice were resting on something above the water.

Answer: (b) 2.5

Explanation:
1. By the equation of continuity A1 v1 = A2 v2.
2. v2 = (2 × 5)/4 = 2.5 cm/s.

Teacher's Note:
1. A wider section means slower flow for an incompressible liquid.
2. Equation of continuity expresses conservation of mass.

Answer: (a) 4 times that of Q

Explanation:
1. A1 v1 = A2 v2 and A is proportional to d².
2. v_P / v_Q = (d_Q/d_P)² = 4.

Teacher's Note:
1. Velocity is inversely proportional to the area, which goes as the square of the diameter.
2. Doubling the diameter reduces the speed to a quarter.

Answer: (a) 2v

Explanation:
1. A_A v = A_B v_B, with area proportional to r².
2. v_B = v × (√2 R)²/R² = 2v.

Teacher's Note:
1. Area ratio is the square of the radius ratio.
2. The velocity ratio is the reverse of the area ratio.

Answer: (a) area decreases

Explanation:
1. The water speeds up as it falls under gravity.
2. By continuity A v is constant, so the cross-sectional area of the stream decreases.

Teacher's Note:
1. Higher speed with the same flow rate means a thinner stream.
2. This is why a stream from a tap narrows as it falls.

Answer: (d) the height of the liquid in the tubes A and C is the same

Explanation:
1. At A and C the pipe has the same cross-section, so the speed and hence the pressure are the same, giving equal heights.
2. At B the speed is higher and the pressure is lower, so the liquid in B stands lower.

Teacher's Note:
1. Faster flow means lower pressure (Bernoulli).
2. The height in each tube indicates the static pressure at that point.

Answer: (a) high in the wide part, falls through the narrowing section and stays lower in the narrow part

Explanation:
1. The velocity is higher in the narrow part, so the pressure there is lower by Bernoulli's theorem.
2. The pressure therefore falls through the narrowing section and stays low in the narrow part.

Teacher's Note:
1. In a horizontal pipe, pressure is greater where the speed is smaller.
2. The pressure only recovers if the duct widens again.