Physics · Properties of bulk matter · chapter 9 of 14

Class 11 Physics Chapter 9: Mechanical Properties of Fluids MCQs with answers

10 multiple-choice questions on Mechanical Properties of Fluids, each with its answer and the reason, written from the NCERT chapter. For school exams on the CBSE pattern, JEE Main and NEET, which are set from the same NCERT chapter. Open a question's solution to check your working; then take the free test to find your next chapter.

Question 1
A small flat disc is held at one fixed point under still water, first horizontal, then vertical, then tilted at 45°. The force of the water on one face of the disc is
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Answer: (d) the same in size in every orientation, and always normal to the face

At a point in a fluid at rest the pressure is the same in all directions, so the force on the small face has the same size whatever its orientation. A fluid at rest can push only normal to a surface, never along it.

Question 2
The dimensions of the coefficient of viscosity are
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Answer: (d) [ML⁻¹T⁻¹]

η = Fl/(vA): [MLT⁻²][L]/([LT⁻¹][L²]) = [ML⁻¹T⁻¹]. [ML⁻¹T⁻²] is pressure, [MT⁻²] is surface tension and [MLT⁻¹] is momentum.

Question 3
Three vessels on a table have the same base area: a straight cylinder, a vessel that narrows towards the top, and one that widens towards the top. Each is filled with water to the same height. The pressure of the water at the base is
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Answer: (c) the same in all three

Liquid pressure at a point depends only on the height of the liquid column above it (P = Pa + ρgh), not on the shape of the vessel or the amount of liquid: the hydrostatic paradox. Equal heights give equal pressures.

Question 4
A gauge fixed at the bottom of an underground water sump reads a gauge pressure of 18 kPa. Taking ρ = 1000 kg m⁻³ and g = 10 m s⁻², the depth of water in the sump is
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Answer: (a) 1.8 m

Gauge pressure P − Pa = ρgh, so h = 18 000/(1000 × 10) = 1.8 m. Adding atmospheric pressure (11.9 m) treats the gauge reading as absolute.

Question 5
At a place where a barometer reads 76 cm of mercury, the long barometer tube is tilted so that it makes an angle of 60° with the vertical. The length of the mercury column along the tube is now
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Answer: (a) 152 cm

Pa = ρgh fixes the vertical height of the column at 76 cm. Along a tube at 60° to the vertical, length = 76/cos 60° = 152 cm.

Question 6
One arm of an open-tube mercury manometer is joined to a gas cylinder and the other is open to the air. The mercury in the open arm stands 15 cm higher than in the arm joined to the cylinder. If the atmospheric pressure is 76 cm of mercury, the absolute pressure of the gas is
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Answer: (c) 91 cm of mercury

Pressure is the same at the same level in both arms. At the level of the mercury surface on the cylinder side, the open arm has 15 cm of mercury above it plus the atmosphere: P = 76 + 15 = 91 cm of mercury. 15 cm is the gauge pressure.

Question 7
In an ideal hydraulic lift, which quantity has the same value at the small piston and at the large piston?
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Answer: (c) the pressure of the liquid on the piston

By Pascal's law the pressure applied at the small piston is transmitted undiminished through the liquid, so the pressure is the same at both pistons; the forces differ in the ratio of the areas.

Question 8
A hydraulic jack must raise a load of 1000 kg when a mechanic pushes its small piston, of diameter 2.0 cm, with a force of 250 N. Take g = 10 m s⁻². The smallest diameter the load piston can have is about
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Answer: (d) 12.6 cm

F₂/F₁ = A₂/A₁ = (D₂/D₁)². F₂/F₁ = 10 000/250 = 40, so D₂ = 2.0 × √40 = 12.6 cm. Using the force ratio directly on the diameters gives 80 cm.

Question 9
In a steady flow, two streamlines can never cross because
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Answer: (a) a fluid particle at the crossing point would have two velocities at once

The tangent to a streamline gives the velocity of the fluid at that point. At a crossing there would be two tangents, that is two velocities for the particle there, which is impossible in steady flow.

Question 10
Water moves at 2.0 m s⁻¹ in a fire hose of internal diameter 6.0 cm. For the jet to leave the nozzle at 18 m s⁻¹, the diameter of the nozzle must be
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Answer: (a) 2.0 cm

Continuity: A₁v₁ = A₂v₂, so (D₂/D₁)² = v₁/v₂ = 2/18 = 1/9, D₂ = 6.0/3 = 2.0 cm. Taking the diameters in the ratio of the speeds gives 0.67 cm.

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