How to Use This Page
Attempt each NCERT exercise question yourself first, then compare with the detailed solutions below. For numericals, every step — formula, substitution, calculation, and final answer with units — is shown explicitly. Watch for unit conversions, especially cm to m and mm to m.
Key Concepts (Quick Recap)
- Pressure P = F/A; in a fluid column P = ρgh
- Pascal's law: Pressure applied to an enclosed fluid is transmitted equally throughout
- Bernoulli's equation: P + ½ρv² + ρgh = constant (for incompressible, non-viscous, steady flow)
- Viscosity: F = ηA(dv/dx); coefficient of viscosity η (unit: Pa s or poise)
- Stokes' law: F = 6πηrv (viscous drag on a sphere)
- Terminal velocity: v_t = 2r²(ρ−σ)g / (9η)
- Surface tension: T = F/l (unit: N m⁻¹); energy per unit area
- Capillarity (Jurin's law): h = 2T cos θ / (rρg)
Intext Questions
Page 257
Q1. Why does a fluid exert pressure on the walls of its container?
Q2. The pressure at the bottom of a tank of water is 3 atm. What is the depth of water? Given 1 atm = 1.013 × 10⁵ Pa.
Gauge pressure = ρgh
3 atm − 1 atm = 2 atm gauge = 2 × 1.013 × 10⁵ = 2.026 × 10⁵ Pa
h = P/(ρg) = 2.026 × 10⁵ / (1000 × 9.8) = 20.67 m
Page 263
Q3. A 50 kg girl on heels of area 1 cm² each, and a 500 kg elephant on footpads of area 1000 cm² each — who exerts more pressure?
Girl: A = 2 × 1 × 10⁻⁴ = 2 × 10⁻⁴ m², P = 50 × 9.8 / (2 × 10⁻⁴) = 2.45 × 10⁶ Pa
Elephant: A = 4 × 1000 × 10⁻⁴ = 0.4 m², P = 500 × 9.8 / 0.4 = 1.225 × 10⁴ Pa
Page 268
Q4. Show that Bernoulli's equation is a statement of conservation of energy.
Q5. Why does a spinning cricket ball curve in its path? (Magnus effect)
Chapter-End Exercise
Q1. Explain why: (a) The blood pressure in humans is greater at the feet than at the brain. (b) Atmospheric pressure at a height of about 6 km decreases to nearly half of its value at sea level. (c) Hydrostatic pressure is a scalar quantity.
Q2. Explain why: (a) To keep a piece of paper horizontal, you blow over it. (b) A spinning ball follows a curved path. (c) A flag flutters in a breeze.
Q3. Fill in the blanks using the word(s): Bernoulli's principle, surface tension, Pascal's law. (a) The working of a hydraulic lift is based on ___. (b) The shape of a liquid drop is governed by ___. (c) The rise of mercury in a barometer is due to ___.
Q4. A U-tube contains water and methylated spirit separated by mercury. The mercury columns in the two arms are at the same level. If the water column is 10 cm, what is the height of the spirit column? ρ_water = 1 g cm⁻³, ρ_spirit = 0.8 g cm⁻³.
Since mercury levels are equal, the pressure due to each liquid column is the same:
ρ_w g h_w = ρ_s g h_s
h_s = (ρ_w × h_w) / ρ_s = (1 × 10) / 0.8 = 12.5 cm
Q5. In the previous question, if 10 cm of water is replaced by 10 cm of a liquid of density 1.5 g cm⁻³, what is the difference in mercury levels?
Pressure difference = (ρ_liquid − ρ_spirit) × g × h
= (1.5 − 0.8) × g × 10 = 7g cm⁻² × g
This excess pressure = ρ_Hg × g × Δh
Δh = 7 × g × 10 / (13.6 × g) = 70/13.6 ≈ 0.515 cm
Q6. A tank 10 m deep has water up to the top. Find the pressure at: (a) the surface (b) at a depth of 5 m (c) at the bottom.
Q7. A vertical off-shore structure is built in sea water. Find the total force on a vertical wall of height 20 m and width 30 m. ρ_sea = 1.03 × 10³ kg m⁻³.
Pressure at depth h: P = ρgh
Average pressure on wall = ½ × ρgH = ½ × 1.03 × 10³ × 9.8 × 20 = 1.009 × 10⁵ Pa
Force = average P × area = 1.009 × 10⁵ × (20 × 30) = 6.056 × 10⁷ N
Q8. A 50 kg block is placed on a hydraulic press with a piston area 0.02 m². If the larger piston has an area of 2 m², what load can it support?
By Pascal's law: P₁ = P₂
F₁/A₁ = F₂/A₂
F₂ = F₁ × A₂/A₁ = (50 × 9.8) × 2/0.02 = 490 × 100 = 49000 N
Load = 49000/9.8 = 5000 kg
Q9. A rain drop of radius 0.3 mm falls through air with a terminal velocity of 1 m/s. Calculate the viscosity of air. ρ_water = 1000 kg m⁻³, ρ_air = 1.2 kg m⁻³.
v_t = 2r²(ρ − σ)g / (9η)
η = 2r²(ρ − σ)g / (9 v_t)
r = 0.3 × 10⁻³ m
η = 2 × (0.3 × 10⁻³)² × (1000 − 1.2) × 9.8 / (9 × 1)
= 2 × 9 × 10⁻⁸ × 998.8 × 9.8 / 9
= 2 × 9 × 10⁻⁸ × 9798.24 / 9
= 2 × 9 × 10⁻⁸ × 1088.69
= 1.96 × 10⁻⁴ Pa s ≈ 1.96 × 10⁻⁵ Pa s
Q10. The terminal velocity of a copper ball of radius 2 mm falling through a tank of oil at 20°C is 6.5 cm/s. Find the viscosity of oil. ρ_copper = 8.9 × 10³ kg m⁻³, ρ_oil = 1.5 × 10³ kg m⁻³.
r = 2 × 10⁻³ m, v_t = 6.5 × 10⁻² m/s
η = 2r²(ρ − σ)g / (9 v_t)
= 2 × (2 × 10⁻³)² × (8900 − 1500) × 9.8 / (9 × 6.5 × 10⁻²)
= 2 × 4 × 10⁻⁶ × 7400 × 9.8 / (5.85 × 10⁻¹)
= 8 × 10⁻⁶ × 72520 / 0.585
= 580.16 × 10⁻⁶ / 0.585
= 9.92 × 10⁻⁴ ≈ 0.99 Pa s
Q11. Calculate the force required to separate two glass plates of area 0.01 m² each with a water film of thickness 0.01 mm between them. T_water = 0.075 N m⁻¹.
The water film has two surfaces. Excess pressure due to surface tension in the film:
ΔP = 2T/d where d = thickness = 0.01 × 10⁻³ m
F = ΔP × A = (2T/d) × A = (2 × 0.075 / 10⁻⁵) × 0.01
= 1.5 × 10⁴ × 0.01 = 150 N
Q12. Mercury has an angle of contact of 120° with glass. The depression of mercury in a capillary tube of radius 0.5 mm is: T = 0.465 N m⁻¹, ρ = 13.6 × 10³ kg m⁻³.
h = 2T cos θ / (rρg)
cos 120° = −0.5
h = 2 × 0.465 × (−0.5) / (0.5 × 10⁻³ × 13600 × 9.8)
= −0.465 / 66.64 = −6.98 × 10⁻³ m = −6.98 mm
Q13. Find the excess pressure inside a soap bubble of radius 5 mm. T_soap = 0.025 N m⁻¹.
Soap bubble has two surfaces:
ΔP = 4T/r = 4 × 0.025 / (5 × 10⁻³) = 0.1/0.005 = 20 Pa
Q14. What is the work done in blowing a soap bubble of radius 3 cm? T = 0.025 N m⁻¹.
A soap bubble has two surfaces. Surface area = 2 × 4πr²
r = 3 × 10⁻² m
W = T × ΔA = T × 2 × 4πr² = 0.025 × 2 × 4π × (3 × 10⁻²)²
= 0.025 × 8π × 9 × 10⁻⁴ = 0.025 × 226.2 × 10⁻⁴
= 5.655 × 10⁻⁴ J ≈ 5.65 × 10⁻⁴ J
Q15. A capillary tube of radius 0.5 mm is dipped vertically in water. Find the height to which water rises. T = 0.075 N m⁻¹, ρ = 1000 kg m⁻³, θ ≈ 0°.
h = 2T cos θ / (rρg)
= 2 × 0.075 × 1 / (0.5 × 10⁻³ × 1000 × 9.8)
= 0.15 / 4.9 = 0.0306 m = 3.06 cm
Q16. The velocity of water in a river is 2 m/s at a point where the cross-section is 10 m². If the cross-section downstream is 20 m², find the velocity there.
By the equation of continuity: A₁v₁ = A₂v₂
v₂ = A₁v₁/A₂ = 10 × 2 / 20 = 1 m/s
Q17. Water flows through a horizontal pipe of varying cross-section. The pressure at a point where the area is 10⁻³ m² and velocity is 2 m/s is 1.5 × 10⁵ Pa. Find the pressure at a point where the area is 5 × 10⁻⁴ m².
By continuity: A₁v₁ = A₂v₂
v₂ = A₁v₁/A₂ = (10⁻³ × 2) / (5 × 10⁻⁴) = 4 m/s
By Bernoulli (horizontal pipe, h same):
P₁ + ½ρv₁² = P₂ + ½ρv₂²
P₂ = P₁ + ½ρ(v₁² − v₂²)
= 1.5 × 10⁵ + ½ × 1000 × (4 − 16)
= 1.5 × 10⁵ − 6000 = 1.44 × 10⁵ Pa
Q18. Explain why a drop of liquid under no external force is spherical.
Q19. State and prove Pascal's law.
Q20. What is Reynold's number? What is its significance?
Q21. What is the effect of temperature on viscosity of liquids and gases?
Internal Links
- [Chapter 10 Guide — Mechanical Properties of Fluids](/school/cbse/class/11/physics/cbse-mechanical-properties-of-fluids)
- [All Class 11 Physics Solutions](/school/cbse/class/11/physics/solutions)
- [Class 11 Physics Subject Guide](/school/cbse/class/11/physics)
Frequently Asked Questions
Pascal's law states that a pressure change applied to an enclosed incompressible fluid is transmitted equally and undiminished to every portion of the fluid and the walls of the container.