Mechanical Properties of Fluids — Class 11 Physics NCERT Solutions (Free)
Free step-by-step NCERT solutions for Class 11 Physics chapter "Mechanical Properties of Fluids" — 8 important questions with detailed answers for CBSE board exam preparation.
TL;DR: Free step-by-step NCERT solutions for Class 11 Physics chapter "Mechanical Properties of Fluids" — 8 important questions with detailed answers for CBS…
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NCERT Solutions for Class 11 Physics — Mechanical Properties of Fluids. Step-by-step answers to all 8 textbook questions from this chapter, written for CBSE board preparation and free to use.
Mechanical Properties of Fluids — All 8 Questions Solved
Q1. A hydraulic press has a small piston of area 5 cm² and a large piston of area 50 cm². A force of 100 N is applied on the small piston. Calculate the force exerted by the large piston on the load.
Step 1: In a hydraulic press, pressure is transmitted uniformly throughout the fluid.
P = F/A = constant everywhere
Step 2: Pressure on small piston = 100/(5 × 10⁻⁴) = 2 × 10⁵ Pa
Step 3: Since pressure is same throughout:
P = F_large/A_large
2 × 10⁵ = F_large/(50 × 10⁻⁴)
F_large = 2 × 10⁵ × 50 × 10⁻⁴ = 1000 N
Alternatively: F_small/A_small = F_large/A_large
100/5 = F_large/50
F_large = 100 × 50/5 = 1000 N
Final Answer: Force exerted by large piston = 1000 N or 1 kN
Q2. Mercury has a density of 13,600 kg/m³. Calculate the pressure exerted by a 76 cm column of mercury. Given: g = 10 m/s².
Step 1: Pressure due to a fluid column P = ρgh
where ρ = density, g = acceleration due to gravity, h = height of column
Step 2: Convert height to SI units:
h = 76 cm = 0.76 m
Step 3: Calculate pressure:
P = 13,600 × 10 × 0.76
P = 136,000 × 0.76
P = 103,360 Pa ≈ 1.034 × 10⁵ Pa
Step 4: In atmospheres:
P = 103,360/101,325 ≈ 1.02 atm
Final Answer: Pressure = 1.034 × 10⁵ Pa or 103.4 kPa or ~1 atm
Q3. A steel ball of radius 2 cm and density 8000 kg/m³ falls through a viscous medium. The terminal velocity is 5 m/s. Calculate the coefficient of viscosity of the medium. (Density of medium = 1000 kg/m³)
Step 1: At terminal velocity, net force = 0
Weight = Buoyancy + Viscous drag
mg = m'g + 6πηrv
where η = coefficient of viscosity, r = radius, v = terminal velocity
Step 2: Calculate weight of steel ball:
Volume V = (4/3)πr³ = (4/3)π(0.02)³ = 3.35 × 10⁻⁵ m³
Mass m = ρ_steel × V = 8000 × 3.35 × 10⁻⁵ = 0.268 kg
Weight W = 0.268 × 10 = 2.68 N
Step 3: Buoyancy:
Mass of fluid displaced = 1000 × 3.35 × 10⁻⁵ = 0.0335 kg
Buoyancy F_b = 0.0335 × 10 = 0.335 N
Step 4: Viscous drag at terminal velocity:
F_drag = Weight - Buoyancy = 2.68 - 0.335 = 2.345 N
Step 5: Using Stokes' law: F_drag = 6πηrv
2.345 = 6 × π × η × 0.02 × 5
2.345 = 6π × η × 0.1
η = 2.345/(0.6π) = 1.247 Pa·s
Final Answer: Coefficient of viscosity η ≈ 1.25 Pa·s
Q4. A soap bubble of radius 5 mm has surface tension 25 × 10⁻³ N/m. Calculate the pressure difference between inside and outside the bubble.
Step 1: A soap bubble has TWO surfaces - inner and outer.
Step 2: Pressure difference due to surface tension (Young-Laplace equation):
For a spherical surface: ΔP = 2T/r
where T = surface tension, r = radius
Step 3: Since soap bubble has two surfaces (inner and outer):
Total pressure difference = ΔP_outer + ΔP_inner = 2T/r + 2T/r = 4T/r
Step 4: Calculate:
ΔP = 4 × 25 × 10⁻³ / (5 × 10⁻³)
ΔP = 100 × 10⁻³ / (5 × 10⁻³)
ΔP = 100/5 = 20 Pa
Final Answer: Pressure difference = 20 Pa
Q5. A water droplet of radius 2 mm is formed. If surface tension of water is 0.073 N/m, calculate the energy stored in the droplet due to surface tension.
Step 1: Surface energy is given by: E = T × A
where T = surface tension, A = surface area
Step 2: For a sphere: A = 4πr²
A = 4π(2 × 10⁻³)²
A = 4π × 4 × 10⁻⁶
A = 16π × 10⁻⁶ m²
A = 5.03 × 10⁻⁵ m²
Step 3: Calculate surface energy:
E = 0.073 × 5.03 × 10⁻⁵
E = 3.67 × 10⁻⁶ J
E = 3.67 μJ
Final Answer: Surface energy = 3.67 × 10⁻⁶ J or 3.67 μJ
Q6. A glass capillary tube of radius 0.25 mm is dipped in water. If surface tension of water is 0.075 N/m and density is 1000 kg/m³, calculate the height to which water rises in the capillary. (Assume contact angle = 0°)
Step 1: In capillarity, at equilibrium:
Upward force due to surface tension = Weight of water column
2πr × T × cos θ = ρgπr²h
Step 2: Simplifying:
2T cos θ = ρgrh
h = 2T cos θ / (ρgr)
Step 3: Substitute values (cos 0° = 1):
h = (2 × 0.075 × 1) / (1000 × 10 × 0.25 × 10⁻³)
h = 0.15 / (10,000 × 0.25 × 10⁻³)
h = 0.15 / 2.5
h = 0.06 m = 60 mm
Step 4: Verification:
h = 0.15 / (10 × 0.25 × 10⁻³) = 0.15 / (2.5 × 10⁻³) = 60 mm ✓
Final Answer: Height of water rise = 60 mm or 0.06 m
Q7. Water flows through a horizontal pipe of radius 1 cm with speed 2 m/s. If the pipe narrows to radius 0.5 cm, calculate the speed of water in the narrow section using equation of continuity.
Step 1: Equation of continuity (conservation of mass):
A₁v₁ = A₂v₂
where A = cross-sectional area, v = speed
Step 2: Calculate cross-sectional areas:
A₁ = πr₁² = π(0.01)² = π × 10⁻⁴ m²
A₂ = πr₂² = π(0.005)² = π × 0.25 × 10⁻⁴ m²
Step 3: Apply continuity equation:
π × 10⁻⁴ × 2 = π × 0.25 × 10⁻⁴ × v₂
2 × 10⁻⁴ = 0.25 × 10⁻⁴ × v₂
v₂ = 2 / 0.25 = 8 m/s
Alternatively: v₂/v₁ = A₁/A₂ = r₁²/r₂² = (1)²/(0.5)² = 4
v₂ = 4 × 2 = 8 m/s
Final Answer: Speed in narrow section = 8 m/s
Q8. A U-tube manometer contains mercury with density 13,600 kg/m³. The mercury level on one side is 10 cm higher than the other side. Calculate the pressure difference between the two arms. (g = 10 m/s²)
Step 1: In a U-tube manometer, pressure difference equals the pressure due to height difference of mercury.
Step 2: Height difference h = 10 cm = 0.1 m
Step 3: Pressure difference:
ΔP = ρgh
ΔP = 13,600 × 10 × 0.1
ΔP = 13,600 Pa = 1.36 × 10⁴ Pa
Step 4: In atmospheres:
ΔP = 13,600/101,325 ≈ 0.134 atm
Final Answer: Pressure difference = 1.36 × 10⁴ Pa or 13.6 kPa or 0.134 atm
Mechanical Properties of Fluids — Practice MCQs with Answers
Attempt these 20 multiple-choice questions after working through the solutions above. Each carries the correct option and the reasoning behind it, so a wrong answer tells you which idea to revisit.
MCQ 1. What is the SI unit of pressure?
- A. Newton
- B. Joule
- C. Pascal
- D. Watt
Answer: C. Pascal — Pressure is defined as force per unit area. The SI unit of pressure is Pascal (Pa), which is equivalent to one Newton per square meter (N/m²).
MCQ 2. A hydraulic lift is designed to lift heavy loads. If the small piston has an area of 5 cm² and the large piston has an area of 500 cm², what force must be applied to the small piston to lift a 2000 kg load? (Take g = 10 m/s²)
- A. 20 N
- B. 200 N
- C. 400 N
- D. 2000 N
Answer: B. 200 N — According to Pascal's Law, the pressure applied to the small piston is transmitted equally to the large piston (F₁/A₁ = F₂/A₂). F₂ = 2000 kg * 10 m/s² = 20000 N. So, F₁ = F₂ * (A₁/A₂) = 20000 N * (5/500) = 200 N.
MCQ 3. Which principle states that a body immersed partially or wholly in a fluid experiences an upward buoyant force equal to the weight of the fluid displaced by it?
- A. Pascal's Principle
- B. Bernoulli's Principle
- C. Archimedes' Principle
- D. Torricelli's Law
Answer: C. Archimedes' Principle — Archimedes' Principle describes the buoyant force experienced by an object submerged in a fluid. This force is directed upwards and its magnitude is equal to the weight of the fluid displaced by the object.
MCQ 4. A piece of wood floats in water with 3/5th of its volume submerged. What is the density of the wood? (Density of water = 1000 kg/m³)
- A. 400 kg/m³
- B. 500 kg/m³
- C. 600 kg/m³
- D. 750 kg/m³
Answer: C. 600 kg/m³ — For a floating object, the buoyant force equals the weight of the object. ρ_wood * V_object * g = ρ_water * V_submerged * g. Since V_submerged = (3/5)V_object, ρ_wood = (3/5) * ρ_water = (3/5) * 1000 kg/m³ = 600 kg/m³.
MCQ 5. What is the energy required to increase the radius of a soap bubble from R to 2R, if the surface tension of the soap solution is T?
- A. 4πR²T
- B. 8πR²T
- C. 24πR²T
- D. 12πR²T
Answer: C. 24πR²T — A soap bubble has two free surfaces. Initial area = 2 * 4πR² = 8πR². Final area = 2 * 4π(2R)² = 32πR². Change in area = 32πR² - 8πR² = 24πR². Energy required = T * ΔA = 24πR²T.
MCQ 6. The property of a fluid by virtue of which it opposes the relative motion between its different layers is called:
- A. Surface tension
- B. Viscosity
- C. Elasticity
- D. Buoyancy
Answer: B. Viscosity — Viscosity is the measure of a fluid's resistance to flow. It describes the internal friction of a moving fluid, resisting relative motion between adjacent layers.
MCQ 7. A spherical body of radius R falls through a fluid. Its terminal velocity is proportional to:
- A. R
- B. R²
- C. 1/R
- D. R³
Answer: B. R² — The terminal velocity (v_t) of a spherical body is given by v_t = (2R²g(ρ_body - ρ_fluid))/(9η), where R is the radius. Thus, terminal velocity is proportional to R².
MCQ 8. Water flows through a horizontal pipe of varying cross-section. At a point where the diameter is 2 cm, the speed of water is 1 m/s. What is the speed of water at another point where the diameter is 1 cm?
- A. 0.25 m/s
- B. 0.5 m/s
- C. 2 m/s
- D. 4 m/s
Answer: D. 4 m/s — From the equation of continuity, A₁v₁ = A₂v₂. Since A ∝ D², we have D₁²v₁ = D₂²v₂. Given D₁ = 2 cm, v₁ = 1 m/s, D₂ = 1 cm. So, (2)² * 1 = (1)² * v₂, which gives v₂ = 4 m/s.
MCQ 9. For a liquid that 'wets' a solid surface completely, the angle of contact is typically:
- A. 0°
- B. 90°
- C. 180°
- D. Between 0° and 90°
Answer: A. 0° — When a liquid wets a solid surface completely, it spreads out uniformly. In this case, the adhesive forces are much stronger than cohesive forces, resulting in a concave meniscus and an angle of contact close to 0°.
MCQ 10. Water rises in a capillary tube up to a height h. If the radius of the capillary tube is halved, the height to which water will rise will be:
- A. h/2
- B. h
- C. 2h
- D. 4h
Answer: C. 2h — The height of capillary rise is given by h = (2T cosθ) / (ρrg). This shows that h is inversely proportional to the radius r. If the radius is halved, the height will double.
MCQ 11. The excess pressure inside a soap bubble of radius r is p. If a second soap bubble has a radius 2r, the excess pressure inside it will be:
- A. p/4
- B. p/2
- C. 2p
- D. 4p
Answer: B. p/2 — The excess pressure inside a soap bubble is given by ΔP = 4T/r, where T is the surface tension. This implies ΔP is inversely proportional to r. If the radius doubles (r to 2r), the excess pressure will halve (p to p/2).
MCQ 12. Pressure exerted by a fluid at rest depends on:
- A. Area of the container
- B. Shape of the container
- C. Depth
- D. Fluid velocity
Answer: C. Depth — The pressure inside a fluid at rest (hydrostatic pressure) is given by P = ρgh, which depends on the density of the fluid (ρ), acceleration due to gravity (g), and the depth (h). It does not depend on the container's shape or area.
MCQ 13. According to Bernoulli's principle, for a streamline flow of an incompressible, non-viscous fluid:
- A. Pressure energy is always constant
- B. Kinetic energy is always constant
- C. The sum of pressure energy, kinetic energy, and potential energy per unit volume is constant
- D. Total energy always decreases.
Answer: C. The sum of pressure energy, kinetic energy, and potential energy per unit volume is constant — Bernoulli's principle states that the sum of pressure (P), kinetic energy per unit volume (½ρv²), and potential energy per unit volume (ρgh) remains constant along a streamline for an ideal fluid: P + ½ρv² + ρgh = constant.
MCQ 14. A tank of water has a small hole at a depth H below the free surface. If the tank is open to the atmosphere, what is the speed of efflux of water from the hole?
- A. √(gH)
- B. √(2gH)
- C. 2gH
- D. gH
Answer: B. √(2gH) — Torricelli's Law, derived from Bernoulli's principle, states that the speed of efflux (v) from a hole at depth H below the free surface is v = √(2gH).
MCQ 15. Why do heavy vehicles like trucks and buses use thicker oil for lubrication than smaller vehicles like cars?
- A. Thicker oil reduces friction more effectively
- B. Thicker oil has higher viscosity and provides a better lubricating film under heavy loads
- C. Thicker oil evaporates slower
- D. Thicker oil is cheaper
Answer: B. Thicker oil has higher viscosity and provides a better lubricating film under heavy loads — Thicker oil possesses higher viscosity. Under the high loads and temperatures experienced in heavy vehicles, this higher viscosity helps maintain a robust lubricating film, preventing metal-to-metal contact and ensuring effective lubrication.
MCQ 16. In which type of fluid flow do all particles passing a given point follow the same path, and the velocity of the fluid at any point remains constant with time?
- A. Turbulent flow
- B. Steady flow
- C. Laminar flow
- D. Both B and C
Answer: D. Both B and C — Steady flow (also known as streamline flow) means the fluid velocity at any point is constant over time. Laminar flow is a type of steady flow where fluid particles move in smooth, parallel layers without significant mixing.
MCQ 17. A block of ice floats in a beaker containing water. When the ice melts completely, what happens to the level of water in the beaker?
- A. The level rises
- B. The level falls
- C. The level remains the same
- D. The level first rises then falls
Answer: C. The level remains the same — When ice floats, it displaces a volume of water equal to its own weight. When the ice melts, the volume of water produced is exactly equal to the volume of water it displaced, so the water level in the beaker remains unchanged.
MCQ 18. An air bubble of radius r is at a depth h below the surface of a liquid of density ρ. If atmospheric pressure is P₀, what is the total pressure inside the bubble? (Assume surface tension T).
- A. P₀ + ρgh
- B. P₀ + ρgh + 2T/r
- C. P₀ + ρgh + 4T/r
- D. ρgh + 2T/r
Answer: B. P₀ + ρgh + 2T/r — The pressure at depth h outside the bubble is P_ext = P₀ + ρgh. For an air bubble inside a liquid, the excess pressure due to surface tension across its single curved interface is ΔP = 2T/r. Thus, total pressure inside = P_ext + ΔP = P₀ + ρgh + 2T/r.
MCQ 19. Relative density of a substance is defined as the ratio of the density of the substance to the density of:
- A. Air
- B. Water at 4°C
- C. Any liquid
- D. Vacuum
Answer: B. Water at 4°C — Relative density, also known as specific gravity, compares the density of a substance to the density of a standard reference. For solids and liquids, this reference is typically water at 4°C.
MCQ 20. A Venturi-meter works on the principle of:
- A. Pascal's Law
- B. Archimedes' Principle
- C. Bernoulli's Principle
- D. Stokes' Law
Answer: C. Bernoulli's Principle — A Venturi-meter is used to measure fluid flow speed and operates based on Bernoulli's principle. As fluid velocity increases in the constricted part of the meter, its pressure decreases.
How to Use These Solutions
Attempt each question yourself first and only then compare with the worked answer. Marks in the CBSE board exam are awarded for the METHOD as much as the final result, so reproduce the steps rather than memorising the last line. Where a solution states a law, a formula or a definition, learn that wording — examiners look for it.
Related practice for this chapter: chapter MCQs, previous-year questions, revision notes and sample papers.