NEET Physics: Gravitation and Properties of Matter · Chapter 3

Mechanical Properties of Fluids

What to remember

Pressure at depth h in a liquid of density ρ is P = P₀ + ρgh. Pascal's law: a pressure change applied to an enclosed fluid is transmitted undiminished to every part of the fluid and to the walls.
Equation of continuity: A₁v₁ = A₂v₂. Bernoulli's principle: P + ½ρv² + ρgh is constant along a streamline of a steady, non-viscous, incompressible flow. Where the speed is high, the pressure is low.
Stokes' law: viscous drag on a sphere is F = 6πηrv. Terminal velocity is v_t = 2r²(ρ − σ)g/9η. Flow is streamline for a small Reynolds number and turbulent for a large one.
Surface tension S = force per unit length = surface energy per unit area. Excess pressure is 2S/r in a drop and 4S/r in a soap bubble. Capillary rise is h = 2S cos θ/(ρrg).

Pressure, Pascal's law and its applications

A fluid is a substance that can flow: a liquid or a gas. A fluid at rest exerts a force perpendicular to any surface in contact with it.

Pressure is the normal force per unit area. P = F/A. The SI unit is the pascal (Pa), 1 Pa = 1 N m⁻². Pressure is a scalar. Other units: 1 atmosphere (atm) = 1.013 × 10⁵ Pa, 1 bar = 10⁵ Pa. Density is ρ = m/V, in kg m⁻³. The density of water is 1000 kg m⁻³ and that of mercury is about 13,600 kg m⁻³.

Pascal's law. The pressure applied to an enclosed fluid is transmitted undiminished to every point of the fluid and to the walls of the container.

Hydraulic lift. Two cylinders, one narrow with piston area A₁ and one wide with piston area A₂, are joined and filled with a liquid. A small force F₁ on the narrow piston makes a pressure F₁/A₁. The same pressure acts on the wide piston and gives

F₂ = F₁ (A₂/A₁)

A small force can lift a heavy load. The force is multiplied, but the work is not (ideal case): the small piston moves farther by the factor A₂/A₁.

Hydraulic brakes. The force on the brake pedal is transmitted through the brake oil to the pistons at all four wheels, and presses the brake shoes equally on each wheel.

Variation of pressure with depth, atmospheric pressure and gauge pressure

For a liquid of constant density ρ, the pressure at depth h below the free surface is

P = P₀ + ρ g h

where P₀ is the pressure at the surface (atmospheric pressure for an open vessel). The pressure depends only on the depth, not on the shape of the vessel. All points at the same level in a connected liquid at rest have the same pressure.

Atmospheric pressure is the pressure due to the weight of the air column above. At sea level, P₀ = 1.013 × 10⁵ Pa = 1 atm.

Absolute and gauge pressure. The excess of the absolute pressure over the atmospheric pressure is the gauge pressure.

Gauge pressure = P − P₀ = ρ g h

Mercury barometer. A long glass tube closed at one end is filled with mercury and inverted in a dish of mercury. The mercury column falls and stands at a height of about 76 cm, leaving a vacuum (Torricelli vacuum) above it. The atmospheric pressure is P₀ = ρ g h = 13,600 × 9.8 × 0.76 ≈ 1.013 × 10⁵ Pa. A pressure of 1 mm of mercury is called 1 torr; 1 atm = 760 mm of Hg = 760 torr.

Open-tube manometer. A U-shaped tube, partly filled with a liquid, has one arm connected to the vessel and the other open to the air. The difference in the liquid levels in the two arms is proportional to the gauge pressure of the gas in the vessel: P − P₀ = ρ g h, with h being the height difference.

Instrument	Measures	Working relation
Mercury barometer	Atmospheric pressure	P₀ = ρ g h (h about 76 cm)
Open-tube manometer	Gauge pressure of a gas	P − P₀ = ρ g h
Hydraulic lift	Multiplies force	F₂ = F₁ A₂/A₁

Streamline flow, equation of continuity and Bernoulli's principle

In streamline (steady) flow, the velocity of the fluid at any fixed point does not change with time. The path of a fluid particle is a streamline, and Flow is streamline at low speed. Above a critical speed, the flow becomes turbulent: the velocity at a point changes irregularly and eddies form.

Equation of continuity. For an incompressible fluid in steady flow through a pipe of varying cross-section, the mass entering per second equals the mass leaving per second:

A₁ v₁ = A₂ v₂ (A v = constant)

The fluid moves faster where the pipe is narrow and slower where it is wide. The product A v is the volume flow rate, in m³ s⁻¹.

Bernoulli's principle. For a steady, non-viscous, incompressible flow, along a streamline:

P + ½ ρ v² + ρ g h = constant

This is the statement of conservation of energy per unit volume. For a horizontal flow, P + ½ρv² is constant, so where the speed is high the pressure is low. Bernoulli's principle does not hold for viscous or turbulent flow.

Applications.

1. Speed of efflux (Torricelli's law). A liquid flows out of a small hole at a depth h below the open surface of a large tank. Both the top of the liquid and the jet are at atmospheric pressure, and the speed at the top is almost zero. So v = √(2 g h).
2. Venturi meter. A pipe has a narrow throat. A fluid of density ρ flows with speed v₁ in the wide section (area A₁) and v₂ in the narrow section (area A₂). By continuity, v₂ > v₁. By Bernoulli's principle, the pressure in the throat is lower: P₁ − P₂ = ½ ρ (v₂² − v₁²). The pressure difference is measured, and the speed of flow is found from it.
3. Blood flow. If an artery is narrowed (by plaque), the blood speeds up at the constriction and its pressure falls, which can make the artery collapse temporarily.
4. Dynamic lift (aerofoil). The wing of an aircraft has a shape such that the air moves faster above the wing than below it. The pressure above is lower than the pressure below, so there is a net upward force. This is the dynamic lift.
5. Magnus effect. A spinning ball drags air round with it. On one side the spin adds to the speed of the air, and on the other side it reduces the air speed. The pressure is lower on the side where the air moves faster. The net sideways force makes the ball curve. This is the Magnus effect, and it explains the swing of a spinning cricket or tennis ball.

Viscosity, Stokes' law and Reynolds number

Viscosity is the internal friction of a fluid that opposes the relative motion between its layers. Consider a fluid between a fixed plate and a parallel plate of area A at distance l moving with speed v. The force needed to keep the top plate moving is

F = η A v / l

η is the coefficient of viscosity. Its SI unit is the pascal second (Pa s), also called the poiseuille. Its dimensions are [M L⁻¹ T⁻¹]. The CGS unit is the poise: 1 poise = 0.1 Pa s. For water at 20 °C, η is about 1 × 10⁻³ Pa s. The viscosity of liquids decreases as the temperature rises. The viscosity of gases increases as the temperature rises.

Stokes' law. A sphere of radius r moving slowly with speed v through a fluid of viscosity η experiences a viscous drag

F = 6 π η r v

Terminal velocity. A small sphere of density ρ falling in a fluid of density σ is acted on by weight (4/3)πr³ρg downwards, buoyancy (4/3)πr³σg upwards and viscous drag 6πηrv upwards. The speed increases until the net force is zero. Then the sphere falls with constant speed, the terminal velocity:

v_t = 2 r² (ρ − σ) g / (9 η)

The terminal velocity is proportional to r², is larger for a denser sphere, and is smaller for a more viscous fluid. If ρ is less than σ, the sphere rises with a terminal velocity (an air bubble in a liquid).

Reynolds number. It decides whether the flow is streamline or turbulent.

R_e = ρ v d / η

Here d is the diameter of the pipe, or a typical dimension of the flow. R_e has no unit. For flow in a pipe, the flow is streamline if R_e is below about 1000 and turbulent if R_e is above about 2000. Between these values the flow is unsteady. The critical speed is the speed above which the flow becomes turbulent. It is v_c = R_e η / (ρ d).

Surface tension, surface energy and angle of contact

The surface of a liquid behaves like a stretched elastic membrane.

Surface tension (S) is the force per unit length acting perpendicular to a line drawn on the surface, tangent to the surface:

S = F / l

Unit: N m⁻¹. Dimensions [M T⁻²]. For water at 20 °C, S is about 0.073 N m⁻¹ (air–water interface). Surface tension decreases as the temperature rises.

Surface energy. To increase the area of a liquid surface by ΔA, work is done against the surface tension:

W = S ΔA

So surface tension is also equal to the surface energy per unit area (unit J m⁻², which is the same as N m⁻¹). A soap film has two surfaces, so its surface energy is 2 S A.

Angle of contact (θ). It is the angle between the tangent to the liquid surface at the point of contact and the solid surface, measured inside the liquid.

Liquid and solid	Angle of contact	Surface shape in a tube	Behaviour
Water and clean glass	acute (less than 90°)	concave	Wets the glass; rises in a capillary
Mercury and glass	obtuse (more than 90°)	convex	Does not wet the glass; falls in a capillary

Excess pressure, capillary rise and detergents

The pressure on the concave side of a curved liquid surface is greater than the pressure on the convex side. The difference is called the excess pressure.

Liquid drop of radius r (one free surface): ΔP = 2S / r
Soap bubble of radius r in air (two free surfaces, inside and outside): ΔP = 4S / r
Air bubble of radius r inside a liquid (one surface): ΔP = 2S / r

The excess pressure is larger in a smaller drop or bubble. If two soap bubbles of different sizes are joined by a tube, air flows from the smaller bubble to the larger bubble.

Capillary rise. In a narrow tube (capillary) of radius r dipped in a liquid of density ρ and surface tension S, the liquid rises or falls by

h = 2 S cos θ / (ρ r g)

For water in glass, θ is acute, cos θ is positive, and the water rises. For mercury in glass, θ is obtuse, cos θ is negative, and the mercury is depressed. The rise is greater in a narrower tube (h ∝ 1/r).

Detergents and surface tension. Pure water has a high surface tension, so it does not wet or penetrate dirty cloth well. A detergent or soap lowers the surface tension of water. The solution then spreads easily, enters the fibres, and carries away oil and dirt.

Worked examples

Example 1 (hydraulic lift). The areas of the two pistons of a hydraulic lift are 10 cm² and 500 cm². A force of 200 N is applied on the small piston. The force on the large piston is F₂ = 200 × (500/10) = 10,000 N.

Example 2 (continuity and Bernoulli). Water flows at 2 m s⁻¹ in a horizontal pipe of cross-section 10 cm². It enters a section of 5 cm². The speed in the narrow section is 2 × (10/5) = 4 m s⁻¹. The pressure drop is ΔP = ½ρ(v₂² − v₁²) = ½ × 1000 × (16 − 4) = 6000 Pa.

Example 3 (terminal velocity). A small steel ball has radius 1 mm and density 8000 kg m⁻³. It falls through a liquid of density 1000 kg m⁻³ and viscosity 0.7 Pa s. Take g = 10 m s⁻². v_t = 2r²(ρ − σ)g/(9η) = 2 × 10⁻⁶ × 7000 × 10/(9 × 0.7) = 0.14/6.3 ≈ 0.022 m s⁻¹.

Example 4 (capillary rise). Water (S = 0.07 N m⁻¹, θ = 0°) rises in a capillary tube of radius 0.5 mm. Take ρ = 1000 kg m⁻³ and g = 10 m s⁻². h = 2 × 0.07/(1000 × 0.5 × 10⁻³ × 10) = 0.14/5 = 0.028 m = 2.8 cm.

Formula summary

Quantity	Formula	Unit
Pressure	P = F/A	Pa
Pressure at depth	P = P₀ + ρgh	Pa
Gauge pressure	ρgh	Pa
Hydraulic lift	F₂ = F₁ (A₂/A₁)	N
Continuity	A₁v₁ = A₂v₂	m³ s⁻¹
Bernoulli	P + ½ρv² + ρgh = constant	Pa
Speed of efflux	v = √(2gh)	m s⁻¹
Venturi pressure drop	P₁ − P₂ = ½ρ(v₂² − v₁²)	Pa
Viscous force	F = ηAv/l	N
Stokes' drag	F = 6πηrv	N
Terminal velocity	v_t = 2r²(ρ − σ)g/(9η)	m s⁻¹
Reynolds number	R_e = ρvd/η	none
Surface tension	S = F/l = W/ΔA	N m⁻¹
Excess pressure	drop 2S/r; bubble 4S/r	Pa
Capillary rise	h = 2S cos θ/(ρrg)	m

NEET traps

1. Pascal's law concerns the transmission of a pressure change; the pressure itself still increases with depth.
2. In a hydraulic lift the force is magnified but the work is not: the work done on both pistons is equal.
3. Pressure in a liquid depends on the vertical depth and the density, not on the shape of the vessel or the amount of liquid.
4. Gauge pressure is the absolute pressure minus the atmospheric pressure; do not add P₀ again.
5. Where the speed of a fluid is high, the pressure is low. In a Venturi meter the pressure is lowest in the narrow throat.
6. Terminal velocity depends on r², so doubling the radius makes it four times larger, not two times.
7. Viscosity of a liquid falls and viscosity of a gas rises when the temperature is increased.
8. A soap bubble has two surfaces (4S/r), and a liquid drop or an air bubble inside a liquid has one surface (2S/r).
9. Mercury shows capillary depression because its angle of contact is obtuse; water shows capillary rise.

One-liners

1. Pressure = force per unit area, in pascal; 1 atm = 1.013 × 10⁵ Pa.
2. The pressure at depth h is P₀ + ρgh.
3. A mercury barometer reads a column of about 76 cm at sea level.
4. The open-tube manometer measures gauge pressure.
5. A hydraulic lift works on Pascal's law.
6. The equation of continuity expresses conservation of mass: Av = constant.
7. Bernoulli's equation expresses conservation of energy for a flowing fluid.
8. The speed of efflux from a hole at depth h is √(2gh).
9. A Venturi meter measures the speed of flow from the pressure difference between the wide part and the throat.
10. The lift on an aircraft wing is due to lower pressure above the wing.
11. The unit of the coefficient of viscosity is Pa s, and 1 poise = 0.1 Pa s.
12. Stokes' drag on a sphere is 6πηrv.
13. Flow in a pipe is streamline below R_e of about 1000 and turbulent above about 2000.
14. Surface tension is force per unit length and also energy per unit area.
15. Excess pressure is 2S/r in a drop and 4S/r in a soap bubble.

Practice questions

A hydraulic lift works on the principle of
1. Stokes' law
2. Archimedes' principle
3. Pascal's law
4. Bernoulli's principle
Answer
C. Pascal's law
Pressure applied to an enclosed liquid is transmitted undiminished, so F₂ = F₁A₂/A₁.
The pistons of a hydraulic lift have areas 4 cm² and 400 cm². The minimum force needed on the smaller piston to lift a load of 5000 N on the larger piston is
1. 5 × 10⁵ N
2. 50 N
3. 500 N
4. 5 N
Answer
B. 50 N
F₁ = F₂ A₁/A₂ = 5000 × 4/400 = 50 N. Check: pressure = 5000/0.04 = 1.25 × 10⁵ Pa and 1.25 × 10⁵ × 4 × 10⁻⁴ = 50 N.
Take g = 10 m s⁻². The gauge pressure at a depth of 5 m in a liquid of density 800 kg m⁻³ is
1. 8 × 10⁴ Pa
2. 4 × 10⁵ Pa
3. 4 × 10³ Pa
4. 4 × 10⁴ Pa
Answer
D. 4 × 10⁴ Pa
Gauge pressure = ρgh = 800 × 10 × 5 = 4 × 10⁴ Pa.
A pressure of 1 atm is equal to a mercury column of
1. 760 mm
2. 7.6 mm
3. 76 mm
4. 7600 mm
Answer
A. 760 mm
1 atm = 760 mm of Hg = 760 torr = 1.013 × 10⁵ Pa.
Take g = 10 m s⁻² and atmospheric pressure 1 × 10⁵ Pa. If a barometer were made with water (density 1000 kg m⁻³) instead of mercury, the height of the water column would be
1. 100 m
2. 1.0 m
3. 0.76 m
4. 10 m
Answer
D. 10 m
h = P₀/ρg = 10⁵/(1000 × 10) = 10 m. Check: mercury is 13.6 times denser, so 76 cm × 13.6 ≈ 10.3 m.
Water flows at 1.5 m s⁻¹ in a pipe of cross-section 8 cm². It then enters a section of cross-section 2 cm². The speed of water in the narrow section is
1. 0.375 m s⁻¹
2. 24 m s⁻¹
3. 6 m s⁻¹
4. 1.5 m s⁻¹
Answer
C. 6 m s⁻¹
A₁v₁ = A₂v₂, so v₂ = 8 × 1.5/2 = 6 m s⁻¹.
Take g = 10 m s⁻². A small hole is made in the side of a large open tank at a depth of 1.25 m below the water surface. The speed with which water leaves the hole is
1. 25 m s⁻¹
2. 5 m s⁻¹
3. 3.5 m s⁻¹
4. 12.5 m s⁻¹
Answer
B. 5 m s⁻¹
v = √(2gh) = √(2 × 10 × 1.25) = √25 = 5 m s⁻¹. Check by energy: ½v² = gh = 12.5, so v = 5.
The SI unit of the coefficient of viscosity is
1. Pa s
2. N m⁻¹
3. N s
4. Pa
Answer
A. Pa s
From F = ηAv/l, η = Fl/(Av) = N m⁻² s = Pa s, dimensions [M L⁻¹ T⁻¹].
Two identical-material spheres fall through the same liquid. The radius of the second is twice that of the first. The ratio of the terminal velocity of the second to that of the first is
1. 4
2. 2
3. 8
4. 16
Answer
A. 4
v_t ∝ r², so the ratio is 2² = 4.
Flow of a liquid through a pipe is expected to be turbulent when the Reynolds number is
1. below about 1000
2. exactly 1
3. above about 2000
4. below about 100
Answer
C. above about 2000
Flow is streamline below about 1000 and turbulent above about 2000.
A soap film (two free surfaces) is stretched in a wire frame so that the area of the frame covered by the film increases by 20 cm². The surface tension of the soap solution is 0.03 N m⁻¹. The work done against surface tension is
1. 1.2 × 10⁻³ J
2. 2.4 × 10⁻⁴ J
3. 6 × 10⁻⁵ J
4. 1.2 × 10⁻⁴ J
Answer
D. 1.2 × 10⁻⁴ J
Total new surface = 2 × 20 cm² = 40 × 10⁻⁴ m². W = SΔA = 0.03 × 4 × 10⁻³ = 1.2 × 10⁻⁴ J. Using one surface gives 6 × 10⁻⁵ J.
A soap bubble of radius 2 cm is made with a solution of surface tension 0.04 N m⁻¹. The excess pressure inside the bubble is
1. 4 Pa
2. 8 Pa
3. 2 Pa
4. 16 Pa
Answer
B. 8 Pa
ΔP = 4S/r = 4 × 0.04/0.02 = 8 Pa. Using 2S/r (one surface) would give 4 Pa.
Water has surface tension 0.07 N m⁻¹ and zero angle of contact with glass. Take ρ = 1000 kg m⁻³ and g = 10 m s⁻². In a glass capillary of radius 0.35 mm, water rises to a height of
1. 8 cm
2. 2 cm
3. 0.4 cm
4. 4 cm
Answer
D. 4 cm
h = 2S cosθ/(ρrg) = 2 × 0.07/(1000 × 0.35 × 10⁻³ × 10) = 0.14/3.5 = 0.04 m. Check: for r = 0.5 mm it is 2.8 cm and h ∝ 1/r.
The angle of contact between mercury and clean glass is
1. zero
2. obtuse
3. acute
4. exactly 90° for every glass
Answer
B. obtuse
Mercury does not wet glass, so its angle of contact is greater than 90°, and its level is depressed in a capillary tube.
When a detergent is dissolved in water, the surface tension of the water
1. increases
2. does not change
3. decreases
4. becomes infinite
Answer
C. decreases
A detergent lowers the surface tension, so the solution spreads and enters the cloth fibres.
An air bubble of radius 1 mm is inside water of surface tension 0.07 N m⁻¹. The excess pressure of the air inside the bubble over the water outside is
1. 140 Pa
2. 280 Pa
3. 14 Pa
4. 70 Pa
Answer
A. 140 Pa
A bubble inside a liquid has one surface, so ΔP = 2S/r = 2 × 0.07/10⁻³ = 140 Pa.
Given below are two statements. Statement I: Pressure is a scalar quantity. Statement II: The SI unit of pressure is the pascal. Choose the correct option.
1. Both Statement I and Statement II are correct
2. Both Statement I and Statement II are incorrect
3. Statement I is correct but Statement II is incorrect
4. Statement I is incorrect but Statement II is correct
Answer
A. Both Statement I and Statement II are correct
Pressure is a scalar (force per area) and its SI unit is Pa = N m⁻².
Given below are two statements. Statement I: The pressure at a given depth in a liquid depends on the shape of the vessel. Statement II: In a connected liquid at rest, all points at the same level have the same pressure. Choose the correct option.
1. Both Statement I and Statement II are correct
2. Both Statement I and Statement II are incorrect
3. Statement I is correct but Statement II is incorrect
4. Statement I is incorrect but Statement II is correct
Answer
D. Statement I is incorrect but Statement II is correct
Pressure depends only on depth and density, not on vessel shape. I is incorrect, II is correct.
Given below are two statements. Statement I: In a hydraulic lift a small force can lift a large load. Statement II: The work done on the larger piston is greater than the work done on the smaller piston. Choose the correct option.
1. Both Statement I and Statement II are correct
2. Both Statement I and Statement II are incorrect
3. Statement I is correct but Statement II is incorrect
4. Statement I is incorrect but Statement II is correct
Answer
C. Statement I is correct but Statement II is incorrect
Force is multiplied, but in the ideal case the work is the same on both pistons. II is incorrect.
Which of the following statements are correct? 1. Gauge pressure is the absolute pressure minus the atmospheric pressure. 2. An open-tube manometer measures the gauge pressure of a gas. 3. A mercury barometer measures the gauge pressure of a gas.
1. 1 and 2 only
2. 2 and 3 only
3. 1 and 3 only
4. 1, 2 and 3
Answer
A. 1 and 2 only
A barometer measures the atmospheric pressure, so 3 is wrong.
Given below are two statements. Statement I: Bernoulli's principle holds for viscous and turbulent flow. Statement II: In a horizontal steady flow of a non-viscous liquid, the pressure is low where the speed is high. Choose the correct option.
1. Both Statement I and Statement II are correct
2. Both Statement I and Statement II are incorrect
3. Statement I is correct but Statement II is incorrect
4. Statement I is incorrect but Statement II is correct
Answer
D. Statement I is incorrect but Statement II is correct
Bernoulli's principle needs steady, non-viscous, incompressible flow. I is incorrect, II is correct.
Which of the following statements are correct about the speed of efflux from a small hole in a large open tank? 1. The speed is √(2gh), where h is the depth of the hole below the free surface. 2. The speed depends on the density of the liquid. 3. The speed increases when the hole is made deeper.
1. 1 and 2 only
2. 2 and 3 only
3. 1 and 3 only
4. 1, 2 and 3
Answer
C. 1 and 3 only
v = √(2gh) is independent of density, so 2 is wrong. It increases with h.
Given below are two statements. Statement I: The viscosity of a liquid increases when its temperature rises. Statement II: The viscosity of a gas increases when its temperature rises. Choose the correct option.
1. Both Statement I and Statement II are correct
2. Both Statement I and Statement II are incorrect
3. Statement I is correct but Statement II is incorrect
4. Statement I is incorrect but Statement II is correct
Answer
D. Statement I is incorrect but Statement II is correct
Liquid viscosity decreases with temperature; gas viscosity increases. I is incorrect, II is correct.
Which of the following statements are correct? 1. The terminal velocity of a falling sphere is proportional to its radius. 2. The terminal velocity is smaller in a more viscous fluid. 3. An air bubble in a liquid rises with a terminal velocity.
1. 1 and 2 only
2. 2 and 3 only
3. 1 and 3 only
4. 1, 2 and 3
Answer
B. 2 and 3 only
v_t ∝ r², so 1 is wrong. v_t ∝ 1/η, and for ρ < σ the sphere rises, so 2 and 3 are correct.
Given below are two statements. Statement I: The excess pressure inside a soap bubble of radius r is 4S/r. Statement II: When two soap bubbles of different sizes are joined by a tube, air flows from the larger bubble to the smaller bubble. Choose the correct option.
1. Both Statement I and Statement II are correct
2. Both Statement I and Statement II are incorrect
3. Statement I is correct but Statement II is incorrect
4. Statement I is incorrect but Statement II is correct
Answer
C. Statement I is correct but Statement II is incorrect
I is correct. The smaller bubble has the higher pressure, so air flows from the smaller to the larger bubble; II is incorrect.
Given below are two statements. Statement I: The height of capillary rise is directly proportional to the radius of the tube. Statement II: The height of capillary rise depends on the length of the tube. Choose the correct option.
1. Both Statement I and Statement II are correct
2. Both Statement I and Statement II are incorrect
3. Statement I is correct but Statement II is incorrect
4. Statement I is incorrect but Statement II is correct
Answer
B. Both Statement I and Statement II are incorrect
h = 2S cosθ/(ρrg) is inversely proportional to r and independent of the tube length. Both are incorrect.
Assertion (A): Water rises in a glass capillary tube. Reason (R): The angle of contact between water and glass is acute.
1. Both A and R are true and R is the correct explanation of A
2. Both A and R are true but R is not the correct explanation of A
3. A is true but R is false
4. A is false but R is true
Answer
A. Both A and R are true and R is the correct explanation of A
An acute angle gives cosθ > 0 in h = 2S cosθ/(ρrg), so water rises. R explains A.
Assertion (A): Mercury is depressed in a glass capillary tube. Reason (R): The density of mercury is large.
1. Both A and R are true and R is the correct explanation of A
2. Both A and R are true but R is not the correct explanation of A
3. A is true but R is false
4. A is false but R is true
Answer
B. Both A and R are true but R is not the correct explanation of A
Both are true, but the depression is caused by the obtuse angle of contact, not by the large density.
Assertion (A): The terminal velocity of a small sphere falling in a fluid increases with its radius. Reason (R): The viscous drag on the sphere is independent of its speed.
1. Both A and R are true and R is the correct explanation of A
2. Both A and R are true but R is not the correct explanation of A
3. A is true but R is false
4. A is false but R is true
Answer
C. A is true but R is false
v_t ∝ r², so A is true. Stokes' drag 6πηrv depends on speed, so R is false.
Assertion (A): In a Venturi meter the pressure in the narrow throat is higher than in the wide part. Reason (R): The speed of the liquid is greater in the narrow section.
1. Both A and R are true and R is the correct explanation of A
2. Both A and R are true but R is not the correct explanation of A
3. A is true but R is false
4. A is false but R is true
Answer
D. A is false but R is true
A higher speed means lower pressure (Bernoulli), so A is false. R is true by continuity.
Assertion (A): A spinning ball moving through air curves from its straight path. Reason (R): The spin makes the air speed on the two sides of the ball different, so the pressures differ.
1. Both A and R are true and R is the correct explanation of A
2. Both A and R are true but R is not the correct explanation of A
3. A is true but R is false
4. A is false but R is true
Answer
A. Both A and R are true and R is the correct explanation of A
This is the Magnus effect. The pressure difference gives the sideways force, so R explains A.
Assertion (A): The excess pressure inside a small liquid drop is smaller than that inside a large drop of the same liquid. Reason (R): The excess pressure inside a drop is 2S/r.
1. Both A and R are true and R is the correct explanation of A
2. Both A and R are true but R is not the correct explanation of A
3. A is true but R is false
4. A is false but R is true
Answer
D. A is false but R is true
ΔP = 2S/r is larger for a smaller r, so A is false. R is true.
Match List I with List II. List I: a. Hydraulic brake b. Mercury barometer c. Open-tube manometer d. Venturi meter. List II: i. Gauge pressure of a gas ii. Atmospheric pressure iii. Pascal's law iv. Speed of flow from a pressure difference
1. a-iii, b-i, c-ii, d-iv
2. a-iii, b-ii, c-i, d-iv
3. a-iii, b-ii, c-iv, d-i
4. a-ii, b-iii, c-i, d-iv
Answer
B. a-iii, b-ii, c-i, d-iv
Brake - Pascal's law (iii); barometer - atmospheric pressure (ii); manometer - gauge pressure (i); Venturi meter - flow speed (iv).
Match List I with List II. List I: a. Reynolds number b. Surface tension c. Pressure d. Coefficient of viscosity. List II: i. Pa ii. N m⁻¹ iii. Pa s iv. No unit
1. a-ii, b-iv, c-i, d-iii
2. a-iv, b-ii, c-i, d-iii
3. a-i, b-ii, c-iv, d-iii
4. a-iv, b-iii, c-i, d-ii
Answer
B. a-iv, b-ii, c-i, d-iii
Re is a pure number (iv); S is in N m⁻¹ (ii); pressure in Pa (i); η in Pa s (iii).
Match List I with List II. List I: a. Speed of efflux b. Stokes' drag c. Terminal velocity d. Capillary rise. List II: i. 6πηrv ii. 2r²(ρ − σ)g/9η iii. √(2gh) iv. 2S cosθ/(ρrg)
1. a-i, b-iii, c-ii, d-iv
2. a-iii, b-i, c-iv, d-ii
3. a-iii, b-i, c-ii, d-iv
4. a-ii, b-i, c-iii, d-iv
Answer
C. a-iii, b-i, c-ii, d-iv
Efflux speed √(2gh) (iii); drag 6πηrv (i); terminal velocity (ii); capillary rise (iv).
Match List I with List II. List I: a. Streamline flow b. Turbulent flow c. Equation of continuity d. Bernoulli's principle. List II: i. Conservation of energy ii. Conservation of mass iii. Velocity at a point does not change with time iv. Eddies and irregular velocity
1. a-iii, b-iv, c-ii, d-i
2. a-iii, b-iv, c-i, d-ii
3. a-iv, b-iii, c-ii, d-i
4. a-iii, b-ii, c-iv, d-i
Answer
A. a-iii, b-iv, c-ii, d-i
Streamline: steady velocity (iii); turbulent: eddies (iv); continuity: mass (ii); Bernoulli: energy (i).
Match List I with List II. List I: a. Dynamic lift of an aerofoil b. Magnus effect c. Blood flow past a narrowed artery d. Speed of efflux. List II: i. Curving of a spinning ball ii. Pressure falls at the constriction iii. Lower pressure above the wing iv. v = √(2gh)
1. a-iii, b-i, c-iv, d-ii
2. a-ii, b-i, c-iii, d-iv
3. a-i, b-iii, c-ii, d-iv
4. a-iii, b-i, c-ii, d-iv
Answer
D. a-iii, b-i, c-ii, d-iv
Aerofoil - lower pressure above (iii); Magnus - spinning ball (i); narrowed artery - pressure falls (ii); efflux - √(2gh) (iv).
Match List I with List II. List I: a. Coefficient of viscosity b. Surface tension c. Pressure d. Reynolds number. List II: i. [M T⁻²] ii. [M L⁻¹ T⁻¹] iii. [M⁰ L⁰ T⁰] iv. [M L⁻¹ T⁻²]
1. a-ii, b-i, c-iv, d-iii
2. a-ii, b-iv, c-i, d-iii
3. a-iv, b-i, c-ii, d-iii
4. a-ii, b-i, c-iii, d-iv
Answer
A. a-ii, b-i, c-iv, d-iii
η = [M L⁻¹ T⁻¹]; S = force/length = [M T⁻²]; P = [M L⁻¹ T⁻²]; Re is dimensionless.
Water flows at 1 m s⁻¹ through the wide part of a horizontal pipe. The throat has one quarter of the cross-sectional area of the wide part. Take ρ = 1000 kg m⁻³. The fall in pressure at the throat is
1. 8000 Pa
2. 7500 Pa
3. 4500 Pa
4. 15,000 Pa
Answer
B. 7500 Pa
v₂ = 4 m s⁻¹. ΔP = ½ρ(v₂² − v₁²) = 500 × (16 − 1) = 7500 Pa. Using (v₂ − v₁)² gives 4500 Pa, which is wrong.
Water leaves a hose pipe of cross-section 4 cm² with a speed of 5 m s⁻¹. The volume of water delivered per second is
1. 2 × 10⁻² m³ s⁻¹
2. 2 × 10⁻⁴ m³ s⁻¹
3. 2 × 10⁻³ m³ s⁻¹
4. 8 × 10⁻⁵ m³ s⁻¹
Answer
C. 2 × 10⁻³ m³ s⁻¹
Q = Av = 4 × 10⁻⁴ × 5 = 2 × 10⁻³ m³ s⁻¹ (2 litre per second).
A small sphere of density 2000 kg m⁻³ falls with terminal velocity v₁ in a liquid of density 1000 kg m⁻³, and with terminal velocity v₂ in another liquid of density 500 kg m⁻³. Both liquids have the same viscosity. The ratio v₁ : v₂ is
1. 1 : 2
2. 3 : 2
3. 2 : 3
4. 1 : 1
Answer
C. 2 : 3
v_t ∝ (ρ − σ). v₁ : v₂ = (2000 − 1000) : (2000 − 500) = 1000 : 1500 = 2 : 3.
Water (density 1000 kg m⁻³, viscosity 1 × 10⁻³ Pa s) flows in a pipe of diameter 2 cm. Taking the critical Reynolds number as 2000, the critical speed is
1. 0.1 m s⁻¹
2. 1.0 m s⁻¹
3. 0.01 m s⁻¹
4. 0.2 m s⁻¹
Answer
A. 0.1 m s⁻¹
v_c = R_e η/(ρd) = 2000 × 10⁻³/(1000 × 0.02) = 0.1 m s⁻¹. Check: R_e = 1000 × 0.1 × 0.02/10⁻³ = 2000.
Eight identical water drops, each of radius 1 mm, combine to form one big drop. The surface tension of water is 0.07 N m⁻¹. The surface energy released is
1. 1.68π × 10⁻⁶ J
2. 2.24π × 10⁻⁶ J
3. 1.96π × 10⁻⁶ J
4. 1.12π × 10⁻⁶ J
Answer
D. 1.12π × 10⁻⁶ J
Big drop radius = 2 mm. Area before = 8 × 4π(10⁻³)² = 32π × 10⁻⁶ m²; after = 4π(2 × 10⁻³)² = 16π × 10⁻⁶. Released = 0.07 × 16π × 10⁻⁶ = 1.12π × 10⁻⁶ J.
A gas is connected to an open-tube manometer containing water. The water level in the open arm is 50 cm higher than in the arm joined to the gas. Take atmospheric pressure 1 × 10⁵ Pa, ρ = 1000 kg m⁻³ and g = 10 m s⁻². The absolute pressure of the gas is
1. 9.5 × 10⁴ Pa
2. 1.05 × 10⁵ Pa
3. 1.5 × 10⁵ Pa
4. 5 × 10³ Pa
Answer
B. 1.05 × 10⁵ Pa
The gas pushes the liquid down on its side, so the gas pressure is higher than atmospheric. Gauge pressure = ρgh = 1000 × 10 × 0.5 = 5000 Pa. P = 10⁵ + 5000 = 1.05 × 10⁵ Pa.
A liquid of surface tension 0.07 N m⁻¹ and density 1000 kg m⁻³ makes an angle of contact of 60° with the wall of a capillary of radius 0.5 mm. Take cos 60° = 0.5 and g = 10 m s⁻². The height of the rise in the tube is
1. 2.8 cm
2. 0.7 cm
3. 1.2 cm
4. 1.4 cm
Answer
D. 1.4 cm
h = 2S cosθ/(ρrg) = 2 × 0.07 × 0.5/(1000 × 0.5 × 10⁻³ × 10) = 0.07/5 = 0.014 m. Check: half of the 2.8 cm obtained for θ = 0°.

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