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.

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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…

Written & reviewed by the Syllab.in Academic Team (CBSE/NCERT subject experts) · Updated Jul 23, 2026

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Key Questions Covered:

A hydraulic press has a small piston of area 5 cm² and a large piston of area…
Mercury has a density of 13,600 kg/m³. Calculate the pressure exerted by a 76…
A steel ball of radius 2 cm and density 8000 kg/m³ falls through a viscous me…
A soap bubble of radius 5 mm has surface tension 25 × 10⁻³ N/m. Calculate the…
A water droplet of radius 2 mm is formed. If surface tension of water is 0.07…
A glass capillary tube of radius 0.25 mm is dipped in water. If surface tensi…
+ 2 more questions in the full chapter

Solutions Summary:

Question	Status
A hydraulic press has a small piston of area 5 cm² and a …	✓ Solved
Mercury has a density of 13,600 kg/m³. Calculate the pres…	✓ Solved
A steel ball of radius 2 cm and density 8000 kg/m³ falls …	✓ Solved
A soap bubble of radius 5 mm has surface tension 25 × 10⁻…	✓ Solved
A water droplet of radius 2 mm is formed. If surface tens…	✓ Solved
A glass capillary tube of radius 0.25 mm is dipped in wat…	✓ Solved

Showing 6 of 8 questions

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...

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

Showing 6 of 8 questions. Visit the full page for complete solutions.