Chapter 10: Mechanical Properties of Fluids

Streamline Flow

Concepts

• Streamline flow: Path of a fluid particle in steady flow; tangent at any point aligns with fluid velocity.
• Laminar flow: Fluid flows in layers with different velocities; layers do not mix.
• Turbulent flow: Occurs when fluid velocity exceeds critical velocity; flow becomes irregular and chaotic.
• Critical velocity: Maximum velocity for streamline flow; above this, flow turns turbulent.

Solved Examples

• Ex 1: Identify flow type if fluid velocity is below critical velocity.
  • Answer: Laminar flow (steady layers, no mixing).
• Ex 2: What happens if velocity exceeds critical velocity?
  • Answer: Flow becomes turbulent (irregular motion).

Practice Set

• Q1: What defines a streamline?
  • A) Curve tangent to fluid velocity
  • B) Random fluid path
  • C) Path of fastest particle
  • D) Path perpendicular to velocity
• Q2: Laminar flow is characterized by:
  • A) Mixing layers
  • B) Steady layers with no mixing
  • C) Irregular motion
  • D) Velocity above critical
• Q3: Critical velocity is:
  • A) Velocity below which flow is turbulent
  • B) Velocity above which flow is laminar
  • C) Velocity up to which flow is streamline
  • D) Velocity causing fluid to stop

Answer Key

• Q1: A - Streamline is a curve tangent to fluid velocity at every point.
• Q2: B - Laminar flow has steady layers that do not mix.
• Q3: C - Critical velocity is the max velocity for streamline flow.

Memory Aids

• Streamline = tangent to velocity
• Laminar = Layers, no mixing
• Turbulent = Velocity > critical, irregular
• Critical velocity = boundary between laminar and turbulent

Equation of Continuity

Concepts

• For incompressible fluid, mass flow rate is constant.
• Formula: A₁ v₁ = A₂ v₂ where A = cross-sectional area, v = velocity.
• Velocity increases if area decreases, and vice versa.

Solved Examples

• Ex 1: Fluid flows from area 0.5 m² at 2 m/s to 0.25 m². Find velocity at smaller area.
  • Use A₁ v₁ = A₂ v₂ : 0.5 × 2 = 0.25 × v₂ ⇒ v₂ = 4 m/s.
• Ex 2: If velocity at narrow end is 6 m/s and area is 0.1 m², find velocity at wider end with area 0.3 m².
  • A₁ v₁ = A₂ v₂ ⇒ 0.3 × v₂ = 0.1 × 6 ⇒ v₂ = 2 m/s.

Practice Set

• Q1: If area halves, velocity:
  • A) Halves
  • B) Doubles
  • C) Stays same
  • D) Quadruples
• Q2: For incompressible fluid, A₁ = 3 m² , v₁ = 4 m/s , A₂ = 1 m² . Find v₂ .
  • A) 12 m/s
  • B) 1.33 m/s
  • C) 7 m/s
  • D) 4 m/s
• Q3: Fluid velocity at narrow pipe is 8 m/s, area 0.2 m²; wider pipe area 0.5 m². Find velocity at wider pipe.
  • A) 3.2 m/s
  • B) 20 m/s
  • C) 8 m/s
  • D) 0.32 m/s

Answer Key

• Q1: B - Velocity doubles if area halves (Av = constant).
• Q2: A - v₂ = A₁ v₁/A₂ = 3 × 4/1 = 12 m/s.
• Q3: A - v₂ = A₁ v₁/A₂ = 0.2 × 8/0.5 = 3.2 m/s.

Memory Aids

• Continuity: A v = constant
• Smaller area → higher velocity
• Use A₁ v₁ = A₂ v₂ for calculations

Concepts

• For incompressible fluid, mass flow rate is constant.
• Formula: A₁ v₁ = A₂ v₂ where A = cross-sectional area, v = velocity.
• Velocity increases if area decreases, and vice versa.

Solved Examples

• Ex 1: Fluid flows from area 0.5 m² at 2 m/s to 0.25 m². Find velocity at smaller area.
  • Use A₁ v₁ = A₂ v₂: 0.5 × 2 = 0.25 × v₂ ⇒ v₂ = 4 m/s.
• Ex 2: If velocity at narrow end is 6 m/s and area is 0.1 m², find velocity at wider end with area 0.3 m².
  • A₁ v₁ = A₂ v₂ ⇒ 0.3 × v₂ = 0.1 × 6 ⇒ v₂ = 2 m/s.

Practice Set

• Q1: If area halves, velocity:
  • A) Halves
  • B) Doubles
  • C) Stays same
  • D) Quadruples
• Q2: For incompressible fluid, A₁ = 3 m², v₁ = 4 m/s, A₂ = 1 m². Find v₂.
  • A) 12 m/s
  • B) 1.33 m/s
  • C) 7 m/s
  • D) 4 m/s
• Q3: Fluid velocity at narrow pipe is 8 m/s, area 0.2 m²; wider pipe area 0.5 m². Find velocity at wider pipe.
  • A) 3.2 m/s
  • B) 20 m/s
  • C) 8 m/s
  • D) 0.32 m/s

Answer Key

• Q1: B - Velocity doubles if area halves (Av = constant).
• Q2: A - v₂ = A₁ v₁/A₂ = 3 × 4/1 = 12 m/s.
• Q3: A - v₂ = A₁ v₁/A₂ = 0.2 × 8/0.5 = 3.2 m/s.

Memory Aids

• Continuity: A v = constant
• Smaller area → higher velocity
• Use A₁ v₁ = A₂ v₂ for calculations

Concepts

• For incompressible fluid, mass flow rate is constant.
• Formula: A₁ v₁ = A₂ v₂ where A = cross-sectional area, v = velocity.
• Velocity increases if area decreases, and vice versa.

Solved Examples

• Ex 1: Fluid flows from area 0.5 m² at 2 m/s to 0.25 m². Find velocity at smaller area.
  • Use A₁ v₁ = A₂ v₂: 0.5 × 2 = 0.25 × v₂ ⇒ v₂ = 4 m/s.
• Ex 2: If velocity at narrow end is 6 m/s and area is 0.1 m², find velocity at wider end with area 0.3 m².
  • A₁ v₁ = A₂ v₂ ⇒ 0.3 × v₂ = 0.1 × 6 ⇒ v₂ = 2 m/s.

Practice Set

• Q1: If area halves, velocity:
  • A) Halves
  • B) Doubles
  • C) Stays same
  • D) Quadruples
• Q2: For incompressible fluid, A₁ = 3 m², v₁ = 4 m/s, A₂ = 1 m². Find v₂.
  • A) 12 m/s
  • B) 1.33 m/s
  • C) 7 m/s
  • D) 4 m/s
• Q3: Fluid velocity at narrow pipe is 8 m/s, area 0.2 m²; wider pipe area 0.5 m². Find velocity at wider pipe.
  • A) 3.2 m/s
  • B) 20 m/s
  • C) 8 m/s
  • D) 0.32 m/s

Answer Key

• Q1: B - Velocity doubles if area halves (Av = constant).
• Q2: A - v₂ = A₁ v₁/A₂ = 3 × 4/1 = 12 m/s.
• Q3: A - v₂ = A₁ v₁/A₂ = 0.2 × 8/0.5 = 3.2 m/s.

Memory Aids

• Continuity: A v = constant
• Smaller area → higher velocity
• Use A₁ v₁ = A₂ v₂ for calculations

Concepts

• For incompressible fluid, mass flow rate is constant.
• Formula: A₁ v₁ = A₂ v₂ where A = cross-sectional area, v = velocity.
• Velocity increases if area decreases, and vice versa.

Solved Examples

• Ex 1: Fluid flows from area 0.5 m² at 2 m/s to 0.25 m². Find velocity at smaller area.
  • Use A₁ v₁ = A₂ v₂: 0.5 × 2 = 0.25 × v₂ ⇒ v₂ = 4 m/s.
• Ex 2: If velocity at narrow end is 6 m/s and area is 0.1 m², find velocity at wider end with area 0.3 m².
  • A₁ v₁ = A₂ v₂ ⇒ 0.3 × v₂ = 0.1 × 6 ⇒ v₂ = 2 m/s.

Practice Set

• Q1: If area halves, velocity:
  • A) Halves
  • B) Doubles
  • C) Stays same
  • D) Quadruples
• Q2: For incompressible fluid, A₁ = 3 m², v₁ = 4 m/s, A₂ = 1 m². Find v₂.
  • A) 12 m/s
  • B) 1.33 m/s
  • C) 7 m/s
  • D) 4 m/s
• Q3: Fluid velocity at narrow pipe is 8 m/s, area 0.2 m²; wider pipe area 0.5 m². Find velocity at wider pipe.
  • A) 3.2 m/s
  • B) 20 m/s
  • C) 8 m/s
  • D) 0.32 m/s

Answer Key

• Q1: B - Velocity doubles if area halves (Av = constant).
• Q2: A - v₂ = A₁ v₁/A₂ = 3 × 4/1 = 12 m/s.
• Q3: A - v₂ = A₁ v₁/A₂ = 0.2 × 8/0.5 = 3.2 m/s.

Memory Aids

• Continuity: A v = constant
• Smaller area → higher velocity
• Use A₁ v₁ = A₂ v₂ for calculations

Concepts

• For incompressible fluid, mass flow rate is constant.
• Formula: A₁ v₁ = A₂ v₂ where A = cross-sectional area, v = velocity.
• Velocity increases if area decreases, and vice versa.

Solved Examples

• Ex 1: Fluid flows from area 0.5 m² at 2 m/s to 0.25 m². Find velocity at smaller area.
  • Use A₁ v₁ = A₂ v₂: 0.5 × 2 = 0.25 × v₂ ⇒ v₂ = 4 m/s.
• Ex 2: If velocity at narrow end is 6 m/s and area is 0.1 m², find velocity at wider end with area 0.3 m².
  • A₁ v₁ = A₂ v₂ ⇒ 0.3 × v₂ = 0.1 × 6 ⇒ v₂ = 2 m/s.

Practice Set

• Q1: If area halves, velocity:
  • A) Halves
  • B) Doubles
  • C) Stays same
  • D) Quadruples
• Q2: For incompressible fluid, A₁ = 3 m² , v₁ = 4 m/s , A₂ = 1 m² . Find v₂ .
  • A) 12 m/s
  • B) 1.33 m/s
  • C) 7 m/s
  • D) 4 m/s
• Q3: Fluid velocity at narrow pipe is 8 m/s, area 0.2 m²; wider pipe area 0.5 m². Find velocity at wider pipe.
  • A) 3.2 m/s
  • B) 20 m/s
  • C) 8 m/s
  • D) 0.32 m/s

Answer Key

• Q1: B - Velocity doubles if area halves (Av = constant).
• Q2: A - v₂ = A₁ v₁/A₂ = 3 × 4/1 = 12 m/s.
• Q3: A - v₂ = A₁ v₁/A₂ = 0.2 × 8/0.5 = 3.2 m/s.

Memory Aids

• Continuity: A v = constant
• Smaller area → higher velocity
• Use A₁ v₁ = A₂ v₂ for calculations

Concepts

• For incompressible fluid, mass flow rate is constant.
• Formula: A₁ v₁ = A₂ v₂ where A = cross-sectional area, v = velocity.
• Velocity increases if area decreases, and vice versa.

Solved Examples

• Ex 1: Fluid flows from area 0.5 m² at 2 m/s to 0.25 m². Find velocity at smaller area.
  • Use A₁ v₁ = A₂ v₂: 0.5 × 2 = 0.25 × v₂ ⇒ v₂ = 4 m/s.
• Ex 2: If velocity at narrow end is 6 m/s and area is 0.1 m², find velocity at wider end with area 0.3 m².
  • A₁ v₁ = A₂ v₂ ⇒ 0.3 × v₂ = 0.1 × 6 ⇒ v₂ = 2 m/s.

Practice Set

• Q1: If area halves, velocity:
  • A) Halves
  • B) Doubles
  • C) Stays same
  • D) Quadruples
• Q2: For incompressible fluid, A₁ = 3 m², v₁ = 4 m/s, A₂ = 1 m². Find v₂.
  • A) 12 m/s
  • B) 1.33 m/s
  • C) 7 m/s
  • D) 4 m/s
• Q3: Fluid velocity at narrow pipe is 8 m/s, area 0.2 m²; wider pipe area 0.5 m². Find velocity at wider pipe.
  • A) 3.2 m/s
  • B) 20 m/s
  • C) 8 m/s
  • D) 0.32 m/s

Answer Key

• Q1: B - Velocity doubles if area halves (Av = constant).
• Q2: A - v₂ = A₁ v₁/A₂ = 3 × 4/1 = 12 m/s.
• Q3: A - v₂ = A₁ v₁/A₂ = 0.2 × 8/0.5 = 3.2 m/s.

Memory Aids

• Continuity: A v = constant
• Smaller area → higher velocity
• Use A₁ v₁ = A₂ v₂ for calculations

Concepts

• For incompressible fluid, mass flow rate is constant.
• Formula: A₁ v₁ = A₂ v₂ where A = cross-sectional area, v = velocity.
• Velocity increases if area decreases, and vice versa.

Solved Examples

• Ex 1: Fluid flows from area 0.5 m² at 2 m/s to 0.25 m². Find velocity at smaller area.
  • Use A₁ v₁ = A₂ v₂ : 0.5 × 2 = 0.25 × v₂ ⇒ v₂ = 4 m/s.
• Ex 2: If velocity at narrow end is 6 m/s and area is 0.1 m², find velocity at wider end with area 0.3 m².
  • A₁ v₁ = A₂ v₂ ⇒ 0.3 × v₂ = 0.1 × 6 ⇒ v₂ = 2 m/s.

Practice Set

• Q1: If area halves, velocity:
  • A) Halves
  • B) Doubles
  • C) Stays same
  • D) Quadruples
• Q2: For incompressible fluid, A₁ = 3 m² , v₁ = 4 m/s , A₂ = 1 m² . Find v₂ .
  • A) 12 m/s
  • B) 1.33 m/s
  • C) 7 m/s
  • D) 4 m/s
• Q3: Fluid velocity at narrow pipe is 8 m/s, area 0.2 m²; wider pipe area 0.5 m². Find velocity at wider pipe.
  • A) 3.2 m/s
  • B) 20 m/s
  • C) 8 m/s
  • D) 0.32 m/s

Answer Key

• Q1: B - Velocity doubles if area halves (Av = constant).
• Q2: A - v₂ = A₁ v₁/A₂ = 3 × 4/1 = 12 m/s.
• Q3: A - v₂ = A₁ v₁/A₂ = 0.2 × 8/0.5 = 3.2 m/s.

Memory Aids

• Continuity: A v = constant
• Smaller area → higher velocity
• Use A₁ v₁ = A₂ v₂ for calculations

Concepts

• For incompressible fluid, mass flow rate is constant.
• Formula: A₁ v₁ = A₂ v₂ where A = cross-sectional area, v = velocity.
• Velocity increases if area decreases, and vice versa.

Solved Examples

• Ex 1: Fluid flows from area 0.5 m² at 2 m/s to 0.25 m². Find velocity at smaller area.
  • Use A₁ v₁ = A₂ v₂: 0.5 × 2 = 0.25 × v₂ ⇒ v₂ = 4 m/s.
• Ex 2: If velocity at narrow end is 6 m/s and area is 0.1 m², find velocity at wider end with area 0.3 m².
  • A₁ v₁ = A₂ v₂ ⇒ 0.3 × v₂ = 0.1 × 6 ⇒ v₂ = 2 m/s.

Practice Set

• Q1: If area halves, velocity:
  • A) Halves
  • B) Doubles
  • C) Stays same
  • D) Quadruples
• Q2: For incompressible fluid, A₁ = 3 m², v₁ = 4 m/s, A₂ = 1 m². Find v₂.
  • A) 12 m/s
  • B) 1.33 m/s
  • C) 7 m/s
  • D) 4 m/s
• Q3: Fluid velocity at narrow pipe is 8 m/s, area 0.2 m²; wider pipe area 0.5 m². Find velocity at wider pipe.
  • A) 3.2 m/s
  • B) 20 m/s
  • C) 8 m/s
  • D) 0.32 m/s

Answer Key

• Q1: B - Velocity doubles if area halves (Av = constant).
• Q2: A - v₂ = A₁ v₁/A₂ = 3 × 4/1 = 12 m/s.
• Q3: A - v₂ = A₁ v₁/A₂ = 0.2 × 8/0.5 = 3.2 m/s.

Memory Aids

• Continuity: A v = constant
• Smaller area → higher velocity
• Use A₁ v₁ = A₂ v₂ for calculations

Concepts

• For incompressible fluid, mass flow rate is constant.
• Formula: A₁ v₁ = A₂ v₂ where A = cross-sectional area, v = velocity.
• Velocity increases if area decreases, and vice versa.

Solved Examples

• Ex 1: Fluid flows from area 0.5 m² at 2 m/s to 0.25 m². Find velocity at smaller area.
  • Use A₁ v₁ = A₂ v₂: 0.5 × 2 = 0.25 × v₂ ⇒ v₂ = 4 m/s.
• Ex 2: If velocity at narrow end is 6 m/s and area is 0.1 m², find velocity at wider end with area 0.3 m².
  • A₁ v₁ = A₂ v₂ ⇒ 0.3 × v₂ = 0.1 × 6 ⇒ v₂ = 2 m/s.

Practice Set

• Q1: If area halves, velocity:
  • A) Halves
  • B) Doubles
  • C) Stays same
  • D) Quadruples
• Q2: For incompressible fluid, A₁ = 3 m², v₁ = 4 m/s, A₂ = 1 m². Find v₂.
  • A) 12 m/s
  • B) 1.33 m/s
  • C) 7 m/s
  • D) 4 m/s
• Q3: Fluid velocity at narrow pipe is 8 m/s, area 0.2 m²; wider pipe area 0.5 m². Find velocity at wider pipe.
  • A) 3.2 m/s
  • B) 20 m/s
  • C) 8 m/s
  • D) 0.32 m/s

Answer Key

• Q1: B - Velocity doubles if area halves (Av = constant).
• Q2: A - v₂ = A₁ v₁/A₂ = 3 × 4/1 = 12 m/s.
• Q3: A - v₂ = A₁ v₁/A₂ = 0.2 × 8/0.5 = 3.2 m/s.

Memory Aids

• Continuity: A v = constant
• Smaller area → higher velocity
• Use A₁ v₁ = A₂ v₂ for calculations

Concepts

• For incompressible fluid, mass flow rate is constant.
• Formula: A₁ v₁ = A₂ v₂ where A = cross-sectional area, v = velocity.
• Velocity increases if area decreases, and vice versa.

Solved Examples

• Ex 1: Fluid flows from area 0.5 m² at 2 m/s to 0.25 m². Find velocity at smaller area.
  • Use A₁ v₁ = A₂ v₂: 0.5 × 2 = 0.25 × v₂ ⇒ v₂ = 4 m/s.
• Ex 2: If velocity at narrow end is 6 m/s and area is 0.1 m², find velocity at wider end with area 0.3 m².
  • A₁ v₁ = A₂ v₂ ⇒ 0.3 × v₂ = 0.1 × 6 ⇒ v₂ = 2 m/s.

Practice Set

• Q1: If area halves, velocity:
  • A) Halves
  • B) Doubles
  • C) Stays same
  • D) Quadruples
• Q2: For incompressible fluid, A₁ = 3 m², v₁ = 4 m/s, A₂ = 1 m². Find v₂.
  • A) 12 m/s
  • B) 1.33 m/s
  • C) 7 m/s
  • D) 4 m/s
• Q3: Fluid velocity at narrow pipe is 8 m/s, area 0.2 m²; wider pipe area 0.5 m². Find velocity at wider pipe.
  • A) 3.2 m/s
  • B) 20 m/s
  • C) 8 m/s
  • D) 0.32 m/s

Answer Key

• Q1: B - Velocity doubles if area halves (Av = constant).
• Q2: A - v₂ = A₁ v₁/A₂ = 3 × 4/1 = 12 m/s.
• Q3: A - v₂ = A₁ v₁/A₂ = 0.2 × 8/0.5 = 3.2 m/s.

Memory Aids

• Continuity: A v = constant
• Smaller area → higher velocity
• Use A₁ v₁ = A₂ v₂ for calculations

Concepts

• For incompressible fluid, mass flow rate is constant.
• Formula: A₁ v₁ = A₂ v₂ where A = cross-sectional area, v = velocity.
• Velocity increases if area decreases, and vice versa.

Solved Examples

• Ex 1: Fluid flows from area 0.5 m² at 2 m/s to 0.25 m². Find velocity at smaller area.
  • Use A₁ v₁ = A₂ v₂ : 0.5 × 2 = 0.25 × v₂ ⇒ v₂ = 4 m/s.
• Ex 2: If velocity at narrow end is 6 m/s and area is 0.1 m², find velocity at wider end with area 0.3 m².
  • A₁ v₁ = A₂ v₂ ⇒ 0.3 × v₂ = 0.1 × 6 ⇒ v₂ = 2 m/s.

Practice Set

• Q1: If area halves, velocity:
  • A) Halves
  • B) Doubles
  • C) Stays same
  • D) Quadruples
• Q2: For incompressible fluid, A₁ = 3 m² , v₁ = 4 m/s , A₂ = 1 m² . Find v₂ .
  • A) 12 m/s
  • B) 1.33 m/s
  • C) 7 m/s
  • D) 4 m/s
• Q3: Fluid velocity at narrow pipe is 8 m/s, area 0.2 m²; wider pipe area 0.5 m². Find velocity at wider pipe.
  • A) 3.2 m/s
  • B) 20 m/s
  • C) 8 m/s
  • D) 0.32 m/s

Answer Key

• Q1: B - Velocity doubles if area halves (Av = constant).
• Q2: A - v₂ = A₁ v₁/A₂ = 3 × 4/1 = 12 m/s.
• Q3: A - v₂ = A₁ v₁/A₂ = 0.2 × 8/0.5 = 3.2 m/s.

Memory Aids

• Continuity: A v = constant
• Smaller area → higher velocity
• Use A₁ v₁ = A₂ v₂ for calculations

Concepts

• For incompressible fluid, mass flow rate is constant.
• Formula: A₁ v₁ = A₂ v₂ where A = cross-sectional area, v = velocity.
• Velocity increases if area decreases, and vice versa.

Solved Examples

• Ex 1: Fluid flows from area 0.5 m² at 2 m/s to 0.25 m². Find velocity at smaller area.
  • Use A₁ v₁ = A₂ v₂: 0.5 × 2 = 0.25 × v₂ ⇒ v₂ = 4 m/s.
• Ex 2: If velocity at narrow end is 6 m/s and area is 0.1 m², find velocity at wider end with area 0.3 m².
  • A₁ v₁ = A₂ v₂ ⇒ 0.3 × v₂ = 0.1 × 6 ⇒ v₂ = 2 m/s.

Practice Set

• Q1: If area halves, velocity:
  • A) Halves
  • B) Doubles
  • C) Stays same
  • D) Quadruples
• Q2: For incompressible fluid, A₁ = 3 m², v₁ = 4 m/s, A₂ = 1 m². Find v₂.
  • A) 12 m/s
  • B) 1.33 m/s
  • C) 7 m/s
  • D) 4 m/s
• Q3: Fluid velocity at narrow pipe is 8 m/s, area 0.2 m²; wider pipe area 0.5 m². Find velocity at wider pipe.
  • A) 3.2 m/s
  • B) 20 m/s
  • C) 8 m/s
  • D) 0.32 m/s

Answer Key

• Q1: B - Velocity doubles if area halves (Av = constant).
• Q2: A - v₂ = A₁ v₁/A₂ = 3 × 4/1 = 12 m/s.
• Q3: A - v₂ = A₁ v₁/A₂ = 0.2 × 8/0.5 = 3.2 m/s.

Memory Aids

• Continuity: A v = constant
• Smaller area → higher velocity
• Use A₁ v₁ = A₂ v₂ for calculations

Concepts

• For incompressible fluid, mass flow rate is constant.
• Formula: A₁ v₁ = A₂ v₂ where A = cross-sectional area, v = velocity.
• Velocity increases if area decreases, and vice versa.

Solved Examples

• Ex 1: Fluid flows from area 0.5 m² at 2 m/s to 0.25 m². Find velocity at smaller area.
  • Use A₁ v₁ = A₂ v₂ : 0.5 × 2 = 0.25 × v₂ ⇒ v₂ = 4 m/s.
• Ex 2: If velocity at narrow end is 6 m/s and area is 0.1 m², find velocity at wider end with area 0.3 m².
  • A₁ v₁ = A₂ v₂ ⇒ 0.3 × v₂ = 0.1 × 6 ⇒ v₂ = 2 m/s.

Practice Set

• Q1: If area halves, velocity:
  • A) Halves
  • B) Doubles
  • C) Stays same
  • D) Quadruples
• Q2: For incompressible fluid, A₁ = 3 m² , v₁ = 4 m/s , A₂ = 1 m² . Find v₂ .
  • A) 12 m/s
  • B) 1.33 m/s
  • C) 7 m/s
  • D) 4 m/s
• Q3: Fluid velocity at narrow pipe is 8 m/s, area 0.2 m²; wider pipe area 0.5 m². Find velocity at wider pipe.
  • A) 3.2 m/s
  • B) 20 m/s
  • C) 8 m/s
  • D) 0.32 m/s

Answer Key

• Q1: B - Velocity doubles if area halves (Av = constant).
• Q2: A - v₂ = A₁ v₁/A₂ = 3 × 4/1 = 12 m/s.
• Q3: A - v₂ = A₁ v₁/A₂ = 0.2 × 8/0.5 = 3.2 m/s.

Memory Aids

• Continuity: A v = constant
• Smaller area → higher velocity
• Use A₁ v₁ = A₂ v₂ for calculations

Concepts

• For incompressible fluid, mass flow rate is constant.
• Formula: \( A_1 v_1 = A_2 v_2 \) where \( A \) = cross-sectional area, \( v \) = velocity.
• Velocity increases if area decreases, and vice versa.

Solved Examples

• Ex 1: Fluid flows from area 0.5 m² at 2 m/s to 0.25 m². Find velocity at smaller area.
  • Use \( A_1 v_1 = A_2 v_2 \): \( 0.5 \times 2 = 0.25 \times v_2 \) ⇒ \( v_2 = 4 \) m/s.
• Ex 2: If velocity at narrow end is 6 m/s and area is 0.1 m², find velocity at wider end with area 0.3 m².
  • \( A_1 v_1 = A_2 v_2 \) ⇒ \( 0.3 \times v_2 = 0.1 \times 6 \) ⇒ \( v_2 = 2 \) m/s.

Practice Set

• Q1: If area halves, velocity:
  • A) Halves
  • B) Doubles
  • C) Stays same
  • D) Quadruples
• Q2: For incompressible fluid, \( A_1 = 3 m^2 \), \( v_1 = 4 m/s \), \( A_2 = 1 m^2 \). Find \( v_2 \).
  • A) 12 m/s
  • B) 1.33 m/s
  • C) 7 m/s
  • D) 4 m/s
• Q3: Fluid velocity at narrow pipe is 8 m/s, area 0.2 m²; wider pipe area 0.5 m². Find velocity at wider pipe.
  • A) 3.2 m/s
  • B) 20 m/s
  • C) 8 m/s
  • D) 0.32 m/s

Answer Key

• Q1: B - Velocity doubles if area halves (Av = constant).
• Q2: A - \( v_2 = \frac{A_1 v_1}{A_2} = \frac{3 \times 4}{1} = 12 \) m/s.
• Q3: A - \( v_2 = \frac{A_1 v_1}{A_2} = \frac{0.2 \times 8}{0.5} = 3.2 \) m/s.

Memory Aids

• Continuity: \( A v = \) constant
• Smaller area → higher velocity
• Use \( A_1 v_1 = A_2 v_2 \) for calculations

Bernoulli’s Principle

Explanation

• For steady flow of incompressible, non-viscous fluid, sum of pressure energy, kinetic energy per unit volume, and potential energy per unit volume is constant.
• Work done on fluid at ends A and B:
  W₁ = P₁ A₁ v₁ Δt = P₁ ΔV,
  W₂ = P₂ A₂ v₂ Δt = P₂ ΔV
• Total work done: W₁ − W₂ = (P₁ − P₂) ΔV
• Change in potential energy: ΔU = ρgΔV (h₂ − h₁)
• Change in kinetic energy: ΔK = 1/2 ρΔV (v₂² − v₁²)
• From work-energy theorem: (P₁ − P₂) ΔV = ΔK + ΔU
• Bernoulli’s equation:
  P + 1/2 ρv² + ρgh = constant

Solved Examples

• Ex 1: Calculate pressure difference if velocity changes from 2 m/s to 4 m/s, height difference 3 m, fluid density 1000 kg/m³.
  Use Bernoulli: P₁ − P₂ = 1/2 ρ(v₂² − v₁²) + ρg(h₂ − h₁)
  = 0.5×1000×(16−4) + 1000×9.8×3 = 6000 + 29400 = 35400 Pa

Practice Set

• Q1: Bernoulli’s equation applies to:
  A) Compressible fluids
  B) Non-viscous, incompressible fluids
  C) Viscous fluids
  D) Gases only
• Q2: If fluid velocity increases, pressure:
  A) Increases
  B) Decreases
  C) Remains same
  D) Depends on temperature
• Q3: For fluid flowing from height 5 m to 2 m with velocity change from 1 m/s to 3 m/s, density 1000 kg/m³, find pressure difference:
  A) 14700 Pa
  B) 9800 Pa
  C) 19600 Pa
  D) 24500 Pa

Answer Key

• Q1: B - Bernoulli’s principle is valid for steady, incompressible, non-viscous fluids.
• Q2: B - Pressure decreases as velocity increases (inverse relation).
• Q3: A - P₁ − P₂ = 0.5×1000×(9−1) + 1000×9.8×(2−5) = 4000 − 29400 = -25400 Pa (magnitude 14700 Pa considering direction).

Memory Aids

• Bernoulli’s equation: Pressure + Kinetic + Potential energy constant.
• Higher velocity → lower pressure.
• Use work-energy theorem for pressure-velocity-height relations.

Explanation

• For steady flow of incompressible, non-viscous fluid, sum of pressure energy, kinetic energy per unit volume, and potential energy per unit volume is constant.
• Work done on fluid at ends A and B:
  W₁ = P₁ A₁ v₁ Δt = P₁ ΔV,
  W₂ = P₂ A₂ v₂ Δt = P₂ ΔV
• Total work done: W₁ − W₂ = (P₁ − P₂) ΔV
• Change in potential energy: ΔU = ρgΔV (h₂ − h₁)
• Change in kinetic energy: ΔK = 1/2 ρΔV (v₂² − v₁²)
• From work-energy theorem: (P₁ − P₂) ΔV = ΔK + ΔU
• Bernoulli’s equation:
  P + 1/2 ρv² + ρgh = constant

Solved Examples

• Ex 1: Calculate pressure difference if velocity changes from 2 m/s to 4 m/s, height difference 3 m, fluid density 1000 kg/m³.
  Use Bernoulli: P₁ − P₂ = 1/2 ρ(v₂² − v₁²) + ρg(h₂ − h₁)
  = 0.5×1000×(16−4) + 1000×9.8×3 = 6000 + 29400 = 35400 Pa

Practice Set

• Q1: Bernoulli’s equation applies to:
  A) Compressible fluids
  B) Non-viscous, incompressible fluids
  C) Viscous fluids
  D) Gases only
• Q2: If fluid velocity increases, pressure:
  A) Increases
  B) Decreases
  C) Remains same
  D) Depends on temperature
• Q3: For fluid flowing from height 5 m to 2 m with velocity change from 1 m/s to 3 m/s, density 1000 kg/m³, find pressure difference:
  A) 14700 Pa
  B) 9800 Pa
  C) 19600 Pa
  D) 24500 Pa

Answer Key

• Q1: B - Bernoulli’s principle is valid for steady, incompressible, non-viscous fluids.
• Q2: B - Pressure decreases as velocity increases (inverse relation).
• Q3: A - P₁ − P₂ = 0.5×1000×(9−1) + 1000×9.8×(2−5) = 4000 − 29400 = -25400 Pa (magnitude 14700 Pa considering direction).

Memory Aids

• Bernoulli’s equation: Pressure + Kinetic + Potential energy constant.
• Higher velocity → lower pressure.
• Use work-energy theorem for pressure-velocity-height relations.

Explanation

• For steady flow of incompressible, non-viscous fluid, sum of pressure energy, kinetic energy per unit volume, and potential energy per unit volume is constant.
• Work done on fluid at ends A and B:
  W₁ = P₁ A₁ v₁ Δt = P₁ ΔV,
  W₂ = P₂ A₂ v₂ Δt = P₂ ΔV
• Total work done: W₁ − W₂ = (P₁ − P₂) ΔV
• Change in potential energy: ΔU = ρgΔV (h₂ − h₁)
• Change in kinetic energy: ΔK = 1/2 ρΔV (v₂² − v₁²)
• From work-energy theorem: (P₁ − P₂) ΔV = ΔK + ΔU
• Bernoulli’s equation:
  P + 1/2 ρv² + ρgh = constant

Solved Examples

• Ex 1: Calculate pressure difference if velocity changes from 2 m/s to 4 m/s, height difference 3 m, fluid density 1000 kg/m³.
  Use Bernoulli: P₁ − P₂ = 1/2 ρ(v₂² − v₁²) + ρg(h₂ − h₁)
  = 0.5×1000×(16−4) + 1000×9.8×3 = 6000 + 29400 = 35400 Pa

Practice Set

• Q1: Bernoulli’s equation applies to:
  A) Compressible fluids
  B) Non-viscous, incompressible fluids
  C) Viscous fluids
  D) Gases only
• Q2: If fluid velocity increases, pressure:
  A) Increases
  B) Decreases
  C) Remains same
  D) Depends on temperature
• Q3: For fluid flowing from height 5 m to 2 m with velocity change from 1 m/s to 3 m/s, density 1000 kg/m³, find pressure difference:
  A) 14700 Pa
  B) 9800 Pa
  C) 19600 Pa
  D) 24500 Pa

Answer Key

• Q1: B - Bernoulli’s principle is valid for steady, incompressible, non-viscous fluids.
• Q2: B - Pressure decreases as velocity increases (inverse relation).
• Q3: A - P₁ − P₂ = 0.5×1000×(9−1) + 1000×9.8×(2−5) = 4000 − 29400 = -25400 Pa (magnitude 14700 Pa considering direction).

Memory Aids

• Bernoulli’s equation: Pressure + Kinetic + Potential energy constant.
• Higher velocity → lower pressure.
• Use work-energy theorem for pressure-velocity-height relations.

Explanation

• For steady flow of incompressible, non-viscous fluid, sum of pressure energy, kinetic energy per unit volume, and potential energy per unit volume is constant.
• Work done on fluid at ends A and B:
  W₁ = P₁ A₁ v₁ Δt = P₁ ΔV,
  W₂ = P₂ A₂ v₂ Δt = P₂ ΔV
• Total work done: W₁ − W₂ = (P₁ − P₂) ΔV
• Change in potential energy: ΔU = ρgΔV (h₂ − h₁)
• Change in kinetic energy: ΔK = 1/2 ρΔV (v₂² − v₁²)
• From work-energy theorem: (P₁ − P₂) ΔV = ΔK + ΔU
• Bernoulli’s equation:
  P + 1/2 ρv² + ρgh = constant

Solved Examples

• Ex 1: Calculate pressure difference if velocity changes from 2 m/s to 4 m/s, height difference 3 m, fluid density 1000 kg/m³.
  Use Bernoulli: P₁ − P₂ = 1/2 ρ(v₂² − v₁²) + ρg(h₂ − h₁)
  = 0.5×1000×(16−4) + 1000×9.8×3 = 6000 + 29400 = 35400 Pa

Practice Set

• Q1: Bernoulli’s equation applies to:
  A) Compressible fluids
  B) Non-viscous, incompressible fluids
  C) Viscous fluids
  D) Gases only
• Q2: If fluid velocity increases, pressure:
  A) Increases
  B) Decreases
  C) Remains same
  D) Depends on temperature
• Q3: For fluid flowing from height 5 m to 2 m with velocity change from 1 m/s to 3 m/s, density 1000 kg/m³, find pressure difference:
  A) 14700 Pa
  B) 9800 Pa
  C) 19600 Pa
  D) 24500 Pa

Answer Key

• Q1: B - Bernoulli’s principle is valid for steady, incompressible, non-viscous fluids.
• Q2: B - Pressure decreases as velocity increases (inverse relation).
• Q3: A - P₁ − P₂ = 0.5×1000×(9−1) + 1000×9.8×(2−5) = 4000 − 29400 = -25400 Pa (magnitude 14700 Pa considering direction).

Memory Aids

• Bernoulli’s equation: Pressure + Kinetic + Potential energy constant.
• Higher velocity → lower pressure.
• Use work-energy theorem for pressure-velocity-height relations.

Explanation

• For steady flow of incompressible, non-viscous fluid, sum of pressure energy, kinetic energy per unit volume, and potential energy per unit volume is constant.
• Work done on fluid at ends A and B:
  W₁ = P₁ A₁ v₁ Δt = P₁ ΔV,
  W₂ = P₂ A₂ v₂ Δt = P₂ ΔV
• Total work done: W₁ − W₂ = (P₁ − P₂) ΔV
• Change in potential energy: ΔU = ρgΔV (h₂ − h₁)
• Change in kinetic energy: ΔK = 1/2 ρΔV (v₂² − v₁²)
• From work-energy theorem: (P₁ − P₂) ΔV = ΔK + ΔU
• Bernoulli’s equation:
  P + 1/2 ρv² + ρgh = constant

Solved Examples

• Ex 1: Calculate pressure difference if velocity changes from 2 m/s to 4 m/s, height difference 3 m, fluid density 1000 kg/m³.
  Use Bernoulli: P₁ − P₂ = 1/2 ρ(v₂² − v₁²) + ρg(h₂ − h₁)
  = 0.5×1000×(16−4) + 1000×9.8×3 = 6000 + 29400 = 35400 Pa

Practice Set

• Q1: Bernoulli’s equation applies to:
  A) Compressible fluids
  B) Non-viscous, incompressible fluids
  C) Viscous fluids
  D) Gases only
• Q2: If fluid velocity increases, pressure:
  A) Increases
  B) Decreases
  C) Remains same
  D) Depends on temperature
• Q3: For fluid flowing from height 5 m to 2 m with velocity change from 1 m/s to 3 m/s, density 1000 kg/m³, find pressure difference:
  A) 14700 Pa
  B) 9800 Pa
  C) 19600 Pa
  D) 24500 Pa

Answer Key

• Q1: B - Bernoulli’s principle is valid for steady, incompressible, non-viscous fluids.
• Q2: B - Pressure decreases as velocity increases (inverse relation).
• Q3: A - P₁ − P₂ = 0.5×1000×(9−1) + 1000×9.8×(2−5) = 4000 − 29400 = -25400 Pa (magnitude 14700 Pa considering direction).

Memory Aids

• Bernoulli’s equation: Pressure + Kinetic + Potential energy constant.
• Higher velocity → lower pressure.
• Use work-energy theorem for pressure-velocity-height relations.

Explanation

• For steady flow of incompressible, non-viscous fluid, sum of pressure energy, kinetic energy per unit volume, and potential energy per unit volume is constant.
• Work done on fluid at ends A and B:
  W₁ = P₁ A₁ v₁ Δt = P₁ ΔV,
  W₂ = P₂ A₂ v₂ Δt = P₂ ΔV
• Total work done: W₁ − W₂ = (P₁ − P₂) ΔV
• Change in potential energy: ΔU = ρgΔV (h₂ − h₁)
• Change in kinetic energy: ΔK = 1/2 ρΔV (v₂² − v₁²)
• From work-energy theorem: (P₁ − P₂) ΔV = ΔK + ΔU
• Bernoulli’s equation:
  P + 1/2 ρv² + ρgh = constant

Solved Examples

• Ex 1: Calculate pressure difference if velocity changes from 2 m/s to 4 m/s, height difference 3 m, fluid density 1000 kg/m³.
  Use Bernoulli: P₁ − P₂ = 1/2 ρ(v₂² − v₁²) + ρg(h₂ − h₁)
  = 0.5×1000×(16−4) + 1000×9.8×3 = 6000 + 29400 = 35400 Pa

Practice Set

• Q1: Bernoulli’s equation applies to:
  A) Compressible fluids
  B) Non-viscous, incompressible fluids
  C) Viscous fluids
  D) Gases only
• Q2: If fluid velocity increases, pressure:
  A) Increases
  B) Decreases
  C) Remains same
  D) Depends on temperature
• Q3: For fluid flowing from height 5 m to 2 m with velocity change from 1 m/s to 3 m/s, density 1000 kg/m³, find pressure difference:
  A) 14700 Pa
  B) 9800 Pa
  C) 19600 Pa
  D) 24500 Pa

Answer Key

• Q1: B - Bernoulli’s principle is valid for steady, incompressible, non-viscous fluids.
• Q2: B - Pressure decreases as velocity increases (inverse relation).
• Q3: A - P₁ − P₂ = 0.5×1000×(9−1) + 1000×9.8×(2−5) = 4000 − 29400 = -25400 Pa (magnitude 14700 Pa considering direction).

Memory Aids

• Bernoulli’s equation: Pressure + Kinetic + Potential energy constant.
• Higher velocity → lower pressure.
• Use work-energy theorem for pressure-velocity-height relations.

Explanation

• For steady flow of incompressible, non-viscous fluid, sum of pressure energy, kinetic energy per unit volume, and potential energy per unit volume is constant.
• Work done on fluid at ends A and B:
  W₁ = P₁ A₁ v₁ Δt = P₁ ΔV,
  W₂ = P₂ A₂ v₂ Δt = P₂ ΔV
• Total work done: W₁ − W₂ = (P₁ − P₂) ΔV
• Change in potential energy: ΔU = ρgΔV (h₂ − h₁)
• Change in kinetic energy: ΔK = 1/2 ρΔV (v₂² − v₁²)
• From work-energy theorem: (P₁ − P₂) ΔV = ΔK + ΔU
• Bernoulli’s equation:
  P + 1/2 ρv² + ρgh = constant

Solved Examples

• Ex 1: Calculate pressure difference if velocity changes from 2 m/s to 4 m/s, height difference 3 m, fluid density 1000 kg/m³.
  Use Bernoulli: P₁ − P₂ = 1/2 ρ(v₂² − v₁²) + ρg(h₂ − h₁)
  = 0.5×1000×(16−4) + 1000×9.8×3 = 6000 + 29400 = 35400 Pa

Practice Set

• Q1: Bernoulli’s equation applies to:
  A) Compressible fluids
  B) Non-viscous, incompressible fluids
  C) Viscous fluids
  D) Gases only
• Q2: If fluid velocity increases, pressure:
  A) Increases
  B) Decreases
  C) Remains same
  D) Depends on temperature
• Q3: For fluid flowing from height 5 m to 2 m with velocity change from 1 m/s to 3 m/s, density 1000 kg/m³, find pressure difference:
  A) 14700 Pa
  B) 9800 Pa
  C) 19600 Pa
  D) 24500 Pa

Answer Key

• Q1: B - Bernoulli’s principle is valid for steady, incompressible, non-viscous fluids.
• Q2: B - Pressure decreases as velocity increases (inverse relation).
• Q3: A - P₁ − P₂ = 0.5×1000×(9−1) + 1000×9.8×(2−5) = 4000 − 29400 = -25400 Pa (magnitude 14700 Pa considering direction).

Memory Aids

• Bernoulli’s equation: Pressure + Kinetic + Potential energy constant.
• Higher velocity → lower pressure.
• Use work-energy theorem for pressure-velocity-height relations.

Explanation

• For steady flow of incompressible, non-viscous fluid, sum of pressure energy, kinetic energy per unit volume, and potential energy per unit volume is constant.
• Work done on fluid at ends A and B:
  W₁ = P₁ A₁ v₁ Δt = P₁ ΔV,
  W₂ = P₂ A₂ v₂ Δt = P₂ ΔV
• Total work done: W₁ − W₂ = (P₁ − P₂) ΔV
• Change in potential energy: ΔU = ρgΔV (h₂ − h₁)
• Change in kinetic energy: ΔK = 1/2 ρΔV (v₂² − v₁²)
• From work-energy theorem: (P₁ − P₂) ΔV = ΔK + ΔU
• Bernoulli’s equation:
  P + 1/2 ρv² + ρgh = constant

Solved Examples

• Ex 1: Calculate pressure difference if velocity changes from 2 m/s to 4 m/s, height difference 3 m, fluid density 1000 kg/m³.
  Use Bernoulli: P₁ − P₂ = 1/2 ρ(v₂² − v₁²) + ρg(h₂ − h₁)
  = 0.5×1000×(16−4) + 1000×9.8×3 = 6000 + 29400 = 35400 Pa

Practice Set

• Q1: Bernoulli’s equation applies to:
  A) Compressible fluids
  B) Non-viscous, incompressible fluids
  C) Viscous fluids
  D) Gases only
• Q2: If fluid velocity increases, pressure:
  A) Increases
  B) Decreases
  C) Remains same
  D) Depends on temperature
• Q3: For fluid flowing from height 5 m to 2 m with velocity change from 1 m/s to 3 m/s, density 1000 kg/m³, find pressure difference:
  A) 14700 Pa
  B) 9800 Pa
  C) 19600 Pa
  D) 24500 Pa

Answer Key

• Q1: B - Bernoulli’s principle is valid for steady, incompressible, non-viscous fluids.
• Q2: B - Pressure decreases as velocity increases (inverse relation).
• Q3: A - P₁ − P₂ = 0.5×1000×(9−1) + 1000×9.8×(2−5) = 4000 − 29400 = -25400 Pa (magnitude 14700 Pa considering direction).

Memory Aids

• Bernoulli’s equation: Pressure + Kinetic + Potential energy constant.
• Higher velocity → lower pressure.
• Use work-energy theorem for pressure-velocity-height relations.

Explanation

• For steady flow of incompressible, non-viscous fluid, sum of pressure energy, kinetic energy per unit volume, and potential energy per unit volume is constant.
• Work done on fluid at ends A and B:
  W₁ = P₁ A₁ v₁ Δt = P₁ ΔV,
  W₂ = P₂ A₂ v₂ Δt = P₂ ΔV
• Total work done: W₁ − W₂ = (P₁ − P₂) ΔV
• Change in potential energy: ΔU = ρgΔV (h₂ − h₁)
• Change in kinetic energy: ΔK = 1/2 ρΔV (v₂² − v₁²)
• From work-energy theorem: (P₁ − P₂) ΔV = ΔK + ΔU
• Bernoulli’s equation:
  P + 1/2 ρv² + ρgh = constant

Solved Examples

• Ex 1: Calculate pressure difference if velocity changes from 2 m/s to 4 m/s, height difference 3 m, fluid density 1000 kg/m³.
  Use Bernoulli: P₁ − P₂ = 1/2 ρ(v₂² − v₁²) + ρg(h₂ − h₁)
  = 0.5×1000×(16−4) + 1000×9.8×3 = 6000 + 29400 = 35400 Pa

Practice Set

• Q1: Bernoulli’s equation applies to:
  A) Compressible fluids
  B) Non-viscous, incompressible fluids
  C) Viscous fluids
  D) Gases only
• Q2: If fluid velocity increases, pressure:
  A) Increases
  B) Decreases
  C) Remains same
  D) Depends on temperature
• Q3: For fluid flowing from height 5 m to 2 m with velocity change from 1 m/s to 3 m/s, density 1000 kg/m³, find pressure difference:
  A) 14700 Pa
  B) 9800 Pa
  C) 19600 Pa
  D) 24500 Pa

Answer Key

• Q1: B - Bernoulli’s principle is valid for steady, incompressible, non-viscous fluids.
• Q2: B - Pressure decreases as velocity increases (inverse relation).
• Q3: A - P₁ − P₂ = 0.5×1000×(9−1) + 1000×9.8×(2−5) = 4000 − 29400 = -25400 Pa (magnitude 14700 Pa considering direction).

Memory Aids

• Bernoulli’s equation: Pressure + Kinetic + Potential energy constant.
• Higher velocity → lower pressure.
• Use work-energy theorem for pressure-velocity-height relations.

Explanation

• For steady flow of incompressible, non-viscous fluid, sum of pressure energy, kinetic energy per unit volume, and potential energy per unit volume is constant.
• Work done on fluid at ends A and B:
  W₁ = P₁ A₁ v₁ Δt = P₁ ΔV,
  W₂ = P₂ A₂ v₂ Δt = P₂ ΔV
• Total work done: W₁ − W₂ = (P₁ − P₂) ΔV
• Change in potential energy: ΔU = ρgΔV (h₂ − h₁)
• Change in kinetic energy: ΔK = 1/2 ρΔV (v₂² − v₁²)
• From work-energy theorem: (P₁ − P₂) ΔV = ΔK + ΔU
• Bernoulli’s equation:
  P + 1/2 ρv² + ρgh = constant

Solved Examples

• Ex 1: Calculate pressure difference if velocity changes from 2 m/s to 4 m/s, height difference 3 m, fluid density 1000 kg/m³.
  Use Bernoulli: P₁ − P₂ = 1/2 ρ(v₂² − v₁²) + ρg(h₂ − h₁)
  = 0.5×1000×(16−4) + 1000×9.8×3 = 6000 + 29400 = 35400 Pa

Practice Set

• Q1: Bernoulli’s equation applies to:
  A) Compressible fluids
  B) Non-viscous, incompressible fluids
  C) Viscous fluids
  D) Gases only
• Q2: If fluid velocity increases, pressure:
  A) Increases
  B) Decreases
  C) Remains same
  D) Depends on temperature
• Q3: For fluid flowing from height 5 m to 2 m with velocity change from 1 m/s to 3 m/s, density 1000 kg/m³, find pressure difference:
  A) 14700 Pa
  B) 9800 Pa
  C) 19600 Pa
  D) 24500 Pa

Answer Key

• Q1: B - Bernoulli’s principle is valid for steady, incompressible, non-viscous fluids.
• Q2: B - Pressure decreases as velocity increases (inverse relation).
• Q3: A - P₁ − P₂ = 0.5×1000×(9−1) + 1000×9.8×(2−5) = 4000 − 29400 = -25400 Pa (magnitude 14700 Pa considering direction).

Memory Aids

• Bernoulli’s equation: Pressure + Kinetic + Potential energy constant.
• Higher velocity → lower pressure.
• Use work-energy theorem for pressure-velocity-height relations.

Explanation

• For steady flow of incompressible, non-viscous fluid, sum of pressure energy, kinetic energy per unit volume, and potential energy per unit volume is constant.
• Work done on fluid at ends A and B:
  W₁ = P₁ A₁ v₁ Δt = P₁ ΔV,
  W₂ = P₂ A₂ v₂ Δt = P₂ ΔV
• Total work done: W₁ − W₂ = (P₁ − P₂) ΔV
• Change in potential energy: ΔU = ρgΔV (h₂ − h₁)
• Change in kinetic energy: ΔK = 1/2 ρΔV (v₂² − v₁²)
• From work-energy theorem: (P₁ − P₂) ΔV = ΔK + ΔU
• Bernoulli’s equation:
  P + 1/2 ρv² + ρgh = constant

Solved Examples

• Ex 1: Calculate pressure difference if velocity changes from 2 m/s to 4 m/s, height difference 3 m, fluid density 1000 kg/m³.
  Use Bernoulli: P₁ − P₂ = 1/2 ρ(v₂² − v₁²) + ρg(h₂ − h₁)
  = 0.5×1000×(16−4) + 1000×9.8×3 = 6000 + 29400 = 35400 Pa

Practice Set

• Q1: Bernoulli’s equation applies to:
  A) Compressible fluids
  B) Non-viscous, incompressible fluids
  C) Viscous fluids
  D) Gases only
• Q2: If fluid velocity increases, pressure:
  A) Increases
  B) Decreases
  C) Remains same
  D) Depends on temperature
• Q3: For fluid flowing from height 5 m to 2 m with velocity change from 1 m/s to 3 m/s, density 1000 kg/m³, find pressure difference:
  A) 14700 Pa
  B) 9800 Pa
  C) 19600 Pa
  D) 24500 Pa

Answer Key

• Q1: B - Bernoulli’s principle is valid for steady, incompressible, non-viscous fluids.
• Q2: B - Pressure decreases as velocity increases (inverse relation).
• Q3: A - P₁ − P₂ = 0.5×1000×(9−1) + 1000×9.8×(2−5) = 4000 − 29400 = -25400 Pa (magnitude 14700 Pa considering direction).

Memory Aids

• Bernoulli’s equation: Pressure + Kinetic + Potential energy constant.
• Higher velocity → lower pressure.
• Use work-energy theorem for pressure-velocity-height relations.

Explanation

• For steady flow of incompressible, non-viscous fluid, sum of pressure energy, kinetic energy per unit volume, and potential energy per unit volume is constant.
• Work done on fluid at ends A and B:
  W₁ = P₁ A₁ v₁ Δt = P₁ ΔV,
  W₂ = P₂ A₂ v₂ Δt = P₂ ΔV
• Total work done: W₁ − W₂ = (P₁ − P₂) ΔV
• Change in potential energy: ΔU = ρgΔV (h₂ − h₁)
• Change in kinetic energy: ΔK = 1/2 ρΔV (v₂² − v₁²)
• From work-energy theorem: (P₁ − P₂) ΔV = ΔK + ΔU
• Bernoulli’s equation:
  P + 1/2 ρv² + ρgh = constant

Solved Examples

• Ex 1: Calculate pressure difference if velocity changes from 2 m/s to 4 m/s, height difference 3 m, fluid density 1000 kg/m³.
  Use Bernoulli: P₁ − P₂ = 1/2 ρ(v₂² − v₁²) + ρg(h₂ − h₁)
  = 0.5×1000×(16−4) + 1000×9.8×3 = 6000 + 29400 = 35400 Pa

Practice Set

• Q1: Bernoulli’s equation applies to:
  A) Compressible fluids
  B) Non-viscous, incompressible fluids
  C) Viscous fluids
  D) Gases only
• Q2: If fluid velocity increases, pressure:
  A) Increases
  B) Decreases
  C) Remains same
  D) Depends on temperature
• Q3: For fluid flowing from height 5 m to 2 m with velocity change from 1 m/s to 3 m/s, density 1000 kg/m³, find pressure difference:
  A) 14700 Pa
  B) 9800 Pa
  C) 19600 Pa
  D) 24500 Pa

Answer Key

• Q1: B - Bernoulli’s principle is valid for steady, incompressible, non-viscous fluids.
• Q2: B - Pressure decreases as velocity increases (inverse relation).
• Q3: A - P₁ − P₂ = 0.5×1000×(9−1) + 1000×9.8×(2−5) = 4000 − 29400 = -25400 Pa (magnitude 14700 Pa considering direction).

Memory Aids

• Bernoulli’s equation: Pressure + Kinetic + Potential energy constant.
• Higher velocity → lower pressure.
• Use work-energy theorem for pressure-velocity-height relations.

Explanation

• For steady flow of incompressible, non-viscous fluid, sum of pressure energy, kinetic energy per unit volume, and potential energy per unit volume is constant.
• Work done on fluid at ends A and B:
  W₁ = P₁ A₁ v₁ Δt = P₁ ΔV,
  W₂ = P₂ A₂ v₂ Δt = P₂ ΔV
• Total work done: W₁ − W₂ = (P₁ − P₂) ΔV
• Change in potential energy: ΔU = ρgΔV (h₂ − h₁)
• Change in kinetic energy: ΔK = 1/2 ρΔV (v₂² − v₁²)
• From work-energy theorem: (P₁ − P₂) ΔV = ΔK + ΔU
• Bernoulli’s equation:
  P + 1/2 ρv² + ρgh = constant

Solved Examples

• Ex 1: Calculate pressure difference if velocity changes from 2 m/s to 4 m/s, height difference 3 m, fluid density 1000 kg/m³.
  Use Bernoulli: P₁ − P₂ = 1/2 ρ(v₂² − v₁²) + ρg(h₂ − h₁)
  = 0.5×1000×(16−4) + 1000×9.8×3 = 6000 + 29400 = 35400 Pa

Practice Set

• Q1: Bernoulli’s equation applies to:
  A) Compressible fluids
  B) Non-viscous, incompressible fluids
  C) Viscous fluids
  D) Gases only
• Q2: If fluid velocity increases, pressure:
  A) Increases
  B) Decreases
  C) Remains same
  D) Depends on temperature
• Q3: For fluid flowing from height 5 m to 2 m with velocity change from 1 m/s to 3 m/s, density 1000 kg/m³, find pressure difference:
  A) 14700 Pa
  B) 9800 Pa
  C) 19600 Pa
  D) 24500 Pa

Answer Key

• Q1: B - Bernoulli’s principle is valid for steady, incompressible, non-viscous fluids.
• Q2: B - Pressure decreases as velocity increases (inverse relation).
• Q3: A - P₁ − P₂ = 0.5×1000×(9−1) + 1000×9.8×(2−5) = 4000 − 29400 = -25400 Pa (magnitude 14700 Pa considering direction).

Memory Aids

• Bernoulli’s equation: Pressure + Kinetic + Potential energy constant.
• Higher velocity → lower pressure.
• Use work-energy theorem for pressure-velocity-height relations.

Torricelli’s Law

Explanation

• Speed of efflux from open container:
  v₁ = √2gh
• Horizontal range of efflux jet:
  R = 2√h(H - h)
• Maximum range when h = H/2, max range R_max = H
• General speed of outflow:
  v₁ = √(2gh + 2(P - P_a)/ρ)

Solved Examples

• Ex 1: Calculate speed of water efflux from tank with height 4 m, atmospheric pressure at top.
  v₁ = √2gh = √(2 × 9.8 × 4) = 8.85 m/s

Practice Set

• Q1: Speed of efflux depends on:
  A) Height of fluid column
  B) Fluid density
  C) Atmospheric pressure only
  D) Temperature
• Q2: Maximum horizontal range occurs at:
  A) h = H
  B) h = 0
  C) h = H/2
  D) h = 2H
• Q3: For H = 8 m, h = 2 m, find horizontal range:
  A) 8 m
  B) 6 m
  C) 4 m
  D) 10 m

Answer Key

• Q1: A - Speed depends on height of fluid column (potential energy).
• Q2: C - Maximum range at half the height.
• Q3: B - R = 2√2 × (8-2) = 2√12 = 6.93 ≈ 6 m

Memory Aids

• Efflux speed v = √2gh
• Max range at half height
• Pressure difference affects speed

Explanation

• Speed of efflux from open container:
  v₁ = √2gh
• Horizontal range of efflux jet:
  R = 2√h(H - h)
• Maximum range when h = H/2, max range R_max = H
• General speed of outflow:
  v₁ = √(2gh + 2(P - P_a)/ρ)

Solved Examples

• Ex 1: Calculate speed of water efflux from tank with height 4 m, atmospheric pressure at top.
  v₁ = √2gh = √(2 × 9.8 × 4) = 8.85 m/s

Practice Set

• Q1: Speed of efflux depends on:
  A) Height of fluid column
  B) Fluid density
  C) Atmospheric pressure only
  D) Temperature
• Q2: Maximum horizontal range occurs at:
  A) h = H
  B) h = 0
  C) h = H/2
  D) h = 2H
• Q3: For H = 8 m, h = 2 m, find horizontal range:
  A) 8 m
  B) 6 m
  C) 4 m
  D) 10 m

Answer Key

• Q1: A - Speed depends on height of fluid column (potential energy).
• Q2: C - Maximum range at half the height.
• Q3: B - R = 2√2 × (8-2) = 2√12 = 6.93 ≈ 6 m

Memory Aids

• Efflux speed v = √2gh
• Max range at half height
• Pressure difference affects speed

Explanation

• Speed of efflux from open container:
  v₁ = √2gh
• Horizontal range of efflux jet:
  R = 2√h(H - h)
• Maximum range when h = H/2, max range R_max = H
• General speed of outflow:
  v₁ = √(2gh + 2(P - P_a)/ρ)

Solved Examples

• Ex 1: Calculate speed of water efflux from tank with height 4 m, atmospheric pressure at top.
  v₁ = √2gh = √(2 × 9.8 × 4) = 8.85 m/s

Practice Set

• Q1: Speed of efflux depends on:
  A) Height of fluid column
  B) Fluid density
  C) Atmospheric pressure only
  D) Temperature
• Q2: Maximum horizontal range occurs at:
  A) h = H
  B) h = 0
  C) h = H/2
  D) h = 2H
• Q3: For H = 8 m, h = 2 m, find horizontal range:
  A) 8 m
  B) 6 m
  C) 4 m
  D) 10 m

Answer Key

• Q1: A - Speed depends on height of fluid column (potential energy).
• Q2: C - Maximum range at half the height.
• Q3: B - R = 2√2 × (8-2) = 2√12 = 6.93 ≈ 6 m

Memory Aids

• Efflux speed v = √2gh
• Max range at half height
• Pressure difference affects speed

Explanation

• Speed of efflux from open container:
  v₁ = √2gh
• Horizontal range of efflux jet:
  R = 2√h(H - h)
• Maximum range when h = H/2, max range R_max = H
• General speed of outflow:
  v₁ = √(2gh + 2(P - P_a)/ρ)

Solved Examples

• Ex 1: Calculate speed of water efflux from tank with height 4 m, atmospheric pressure at top.
  v₁ = √2gh = √(2 × 9.8 × 4) = 8.85 m/s

Practice Set

• Q1: Speed of efflux depends on:
  A) Height of fluid column
  B) Fluid density
  C) Atmospheric pressure only
  D) Temperature
• Q2: Maximum horizontal range occurs at:
  A) h = H
  B) h = 0
  C) h = H/2
  D) h = 2H
• Q3: For H = 8 m, h = 2 m, find horizontal range:
  A) 8 m
  B) 6 m
  C) 4 m
  D) 10 m

Answer Key

• Q1: A - Speed depends on height of fluid column (potential energy).
• Q2: C - Maximum range at half the height.
• Q3: B - R = 2√2 × (8-2) = 2√12 = 6.93 ≈ 6 m

Memory Aids

• Efflux speed v = √2gh
• Max range at half height
• Pressure difference affects speed

Explanation

• Speed of efflux from open container:
  v₁ = √2gh
• Horizontal range of efflux jet:
  R = 2√h(H - h)
• Maximum range when h = H/2, max range R_max = H
• General speed of outflow:
  v₁ = √(2gh + 2(P - P_a)/ρ)

Solved Examples

• Ex 1: Calculate speed of water efflux from tank with height 4 m, atmospheric pressure at top.
  v₁ = √2gh = √(2 × 9.8 × 4) = 8.85 m/s

Practice Set

• Q1: Speed of efflux depends on:
  A) Height of fluid column
  B) Fluid density
  C) Atmospheric pressure only
  D) Temperature
• Q2: Maximum horizontal range occurs at:
  A) h = H
  B) h = 0
  C) h = H/2
  D) h = 2H
• Q3: For H = 8 m, h = 2 m, find horizontal range:
  A) 8 m
  B) 6 m
  C) 4 m
  D) 10 m

Answer Key

• Q1: A - Speed depends on height of fluid column (potential energy).
• Q2: C - Maximum range at half the height.
• Q3: B - R = 2√2 × (8-2) = 2√12 = 6.93 ≈ 6 m

Memory Aids

• Efflux speed v = √2gh
• Max range at half height
• Pressure difference affects speed

Explanation

• Speed of efflux from open container:
  v₁ = √2gh
• Horizontal range of efflux jet:
  R = 2√h(H - h)
• Maximum range when h = H/2, max range R_max = H
• General speed of outflow:
  v₁ = √(2gh + 2(P - P_a)/ρ)

Solved Examples

• Ex 1: Calculate speed of water efflux from tank with height 4 m, atmospheric pressure at top.
  v₁ = √2gh = √(2 × 9.8 × 4) = 8.85 m/s

Practice Set

• Q1: Speed of efflux depends on:
  A) Height of fluid column
  B) Fluid density
  C) Atmospheric pressure only
  D) Temperature
• Q2: Maximum horizontal range occurs at:
  A) h = H
  B) h = 0
  C) h = H/2
  D) h = 2H
• Q3: For H = 8 m, h = 2 m, find horizontal range:
  A) 8 m
  B) 6 m
  C) 4 m
  D) 10 m

Answer Key

• Q1: A - Speed depends on height of fluid column (potential energy).
• Q2: C - Maximum range at half the height.
• Q3: B - R = 2√2 × (8-2) = 2√12 = 6.93 ≈ 6 m

Memory Aids

• Efflux speed v = √2gh
• Max range at half height
• Pressure difference affects speed

Explanation

• Speed of efflux from open container:
  v₁ = √2gh
• Horizontal range of efflux jet:
  R = 2√h(H - h)
• Maximum range when h = H/2, max range R_max = H
• General speed of outflow:
  v₁ = √(2gh + 2(P - P_a)/ρ)

Solved Examples

• Ex 1: Calculate speed of water efflux from tank with height 4 m, atmospheric pressure at top.
  v₁ = √2gh = √(2 × 9.8 × 4) = 8.85 m/s

Practice Set

• Q1: Speed of efflux depends on:
  A) Height of fluid column
  B) Fluid density
  C) Atmospheric pressure only
  D) Temperature
• Q2: Maximum horizontal range occurs at:
  A) h = H
  B) h = 0
  C) h = H/2
  D) h = 2H
• Q3: For H = 8 m, h = 2 m, find horizontal range:
  A) 8 m
  B) 6 m
  C) 4 m
  D) 10 m

Answer Key

• Q1: A - Speed depends on height of fluid column (potential energy).
• Q2: C - Maximum range at half the height.
• Q3: B - R = 2√2 × (8-2) = 2√12 = 6.93 ≈ 6 m

Memory Aids

• Efflux speed v = √2gh
• Max range at half height
• Pressure difference affects speed

Explanation

• Speed of efflux from open container:
  v₁ = √2gh
• Horizontal range of efflux jet:
  R = 2√h(H - h)
• Maximum range when h = H/2, max range R_max = H
• General speed of outflow:
  v₁ = √(2gh + 2(P - P_a)/ρ)

Solved Examples

• Ex 1: Calculate speed of water efflux from tank with height 4 m, atmospheric pressure at top.
  v₁ = √2gh = √(2 × 9.8 × 4) = 8.85 m/s

Practice Set

• Q1: Speed of efflux depends on:
  A) Height of fluid column
  B) Fluid density
  C) Atmospheric pressure only
  D) Temperature
• Q2: Maximum horizontal range occurs at:
  A) h = H
  B) h = 0
  C) h = H/2
  D) h = 2H
• Q3: For H = 8 m, h = 2 m, find horizontal range:
  A) 8 m
  B) 6 m
  C) 4 m
  D) 10 m

Answer Key

• Q1: A - Speed depends on height of fluid column (potential energy).
• Q2: C - Maximum range at half the height.
• Q3: B - R = 2√2 × (8-2) = 2√12 = 6.93 ≈ 6 m

Memory Aids

• Efflux speed v = √2gh
• Max range at half height
• Pressure difference affects speed

Explanation

• Speed of efflux from open container:
  v₁ = √2gh
• Horizontal range of efflux jet:
  R = 2√h(H - h)
• Maximum range when h = H/2, max range R_max = H
• General speed of outflow:
  v₁ = √(2gh + 2(P - P_a)/ρ)

Solved Examples

• Ex 1: Calculate speed of water efflux from tank with height 4 m, atmospheric pressure at top.
  v₁ = √2gh = √(2 × 9.8 × 4) = 8.85 m/s

Practice Set

• Q1: Speed of efflux depends on:
  A) Height of fluid column
  B) Fluid density
  C) Atmospheric pressure only
  D) Temperature
• Q2: Maximum horizontal range occurs at:
  A) h = H
  B) h = 0
  C) h = H/2
  D) h = 2H
• Q3: For H = 8 m, h = 2 m, find horizontal range:
  A) 8 m
  B) 6 m
  C) 4 m
  D) 10 m

Answer Key

• Q1: A - Speed depends on height of fluid column (potential energy).
• Q2: C - Maximum range at half the height.
• Q3: B - R = 2√2 × (8-2) = 2√12 = 6.93 ≈ 6 m

Memory Aids

• Efflux speed v = √2gh
• Max range at half height
• Pressure difference affects speed

Explanation

• Speed of efflux from open container:
  v₁ = √2gh
• Horizontal range of efflux jet:
  R = 2√h(H - h)
• Maximum range when h = H/2, max range R_max = H
• General speed of outflow:
  v₁ = √(2gh + 2(P - P_a)/ρ)

Solved Examples

• Ex 1: Calculate speed of water efflux from tank with height 4 m, atmospheric pressure at top.
  v₁ = √2gh = √(2 × 9.8 × 4) = 8.85 m/s

Practice Set

• Q1: Speed of efflux depends on:
  A) Height of fluid column
  B) Fluid density
  C) Atmospheric pressure only
  D) Temperature
• Q2: Maximum horizontal range occurs at:
  A) h = H
  B) h = 0
  C) h = H/2
  D) h = 2H
• Q3: For H = 8 m, h = 2 m, find horizontal range:
  A) 8 m
  B) 6 m
  C) 4 m
  D) 10 m

Answer Key

• Q1: A - Speed depends on height of fluid column (potential energy).
• Q2: C - Maximum range at half the height.
• Q3: B - R = 2√2 × (8-2) = 2√12 = 6.93 ≈ 6 m

Memory Aids

• Efflux speed v = √2gh
• Max range at half height
• Pressure difference affects speed

Explanation

• Speed of efflux from open container:
  v₁ = √2gh
• Horizontal range of efflux jet:
  R = 2√h(H - h)
• Maximum range when h = H/2, max range R_max = H
• General speed of outflow:
  v₁ = √(2gh + 2(P - P_a)/ρ)

Solved Examples

• Ex 1: Calculate speed of water efflux from tank with height 4 m, atmospheric pressure at top.
  v₁ = √2gh = √(2 × 9.8 × 4) = 8.85 m/s

Practice Set

• Q1: Speed of efflux depends on:
  A) Height of fluid column
  B) Fluid density
  C) Atmospheric pressure only
  D) Temperature
• Q2: Maximum horizontal range occurs at:
  A) h = H
  B) h = 0
  C) h = H/2
  D) h = 2H
• Q3: For H = 8 m, h = 2 m, find horizontal range:
  A) 8 m
  B) 6 m
  C) 4 m
  D) 10 m

Answer Key

• Q1: A - Speed depends on height of fluid column (potential energy).
• Q2: C - Maximum range at half the height.
• Q3: B - R = 2√2 × (8-2) = 2√12 = 6.93 ≈ 6 m

Memory Aids

• Efflux speed v = √2gh
• Max range at half height
• Pressure difference affects speed

Explanation

• Speed of efflux from open container:
  v₁ = √2gh
• Horizontal range of efflux jet:
  R = 2√h(H - h)
• Maximum range when h = H/2, max range R_max = H
• General speed of outflow:
  v₁ = √(2gh + 2(P - P_a)/ρ)

Solved Examples

• Ex 1: Calculate speed of water efflux from tank with height 4 m, atmospheric pressure at top.
  v₁ = √2gh = √(2 × 9.8 × 4) = 8.85 m/s

Practice Set

• Q1: Speed of efflux depends on:
  A) Height of fluid column
  B) Fluid density
  C) Atmospheric pressure only
  D) Temperature
• Q2: Maximum horizontal range occurs at:
  A) h = H
  B) h = 0
  C) h = H/2
  D) h = 2H
• Q3: For H = 8 m, h = 2 m, find horizontal range:
  A) 8 m
  B) 6 m
  C) 4 m
  D) 10 m

Answer Key

• Q1: A - Speed depends on height of fluid column (potential energy).
• Q2: C - Maximum range at half the height.
• Q3: B - R = 2√2 × (8-2) = 2√12 = 6.93 ≈ 6 m

Memory Aids

• Efflux speed v = √2gh
• Max range at half height
• Pressure difference affects speed

Explanation

• Speed of efflux from open container:
  v₁ = √2gh
• Horizontal range of efflux jet:
  R = 2√h(H - h)
• Maximum range when h = H/2, max range R_max = H
• General speed of outflow:
  v₁ = √(2gh + 2(P - P_a)/ρ)

Solved Examples

• Ex 1: Calculate speed of water efflux from tank with height 4 m, atmospheric pressure at top.
  v₁ = √2gh = √(2 × 9.8 × 4) = 8.85 m/s

Practice Set

• Q1: Speed of efflux depends on:
  A) Height of fluid column
  B) Fluid density
  C) Atmospheric pressure only
  D) Temperature
• Q2: Maximum horizontal range occurs at:
  A) h = H
  B) h = 0
  C) h = H/2
  D) h = 2H
• Q3: For H = 8 m, h = 2 m, find horizontal range:
  A) 8 m
  B) 6 m
  C) 4 m
  D) 10 m

Answer Key

• Q1: A - Speed depends on height of fluid column (potential energy).
• Q2: C - Maximum range at half the height.
• Q3: B - R = 2√2 × (8-2) = 2√12 = 6.93 ≈ 6 m

Memory Aids

• Efflux speed v = √2gh
• Max range at half height
• Pressure difference affects speed

Explanation

• Speed of efflux from open container:
  \( v_1 = \sqrt{2gh} \)
• Horizontal range of efflux jet:
  \( R = 2\sqrt{h(H - h)} \)
• Maximum range when \( h = H/2 \), max range \( R_{max} = H \)
• General speed of outflow:
  \( v_1 = \sqrt{2gh + \frac{2(P - P_a)}{\rho}} \)

Solved Examples

• Ex 1: Calculate speed of water efflux from tank with height 4 m, atmospheric pressure at top.
  \( v_1 = \sqrt{2gh} = \sqrt{2 \times 9.8 \times 4} = 8.85 \text{ m/s} \)

Practice Set

• Q1: Speed of efflux depends on:
  A) Height of fluid column
  B) Fluid density
  C) Atmospheric pressure only
  D) Temperature
• Q2: Maximum horizontal range occurs at:
  A) \( h = H \)
  B) \( h = 0 \)
  C) \( h = H/2 \)
  D) \( h = 2H \)
• Q3: For \( H = 8 \text{ m} \), \( h = 2 \text{ m} \), find horizontal range:
  A) 8 m
  B) 6 m
  C) 4 m
  D) 10 m

Answer Key

• Q1: A - Speed depends on height of fluid column (potential energy).
• Q2: C - Maximum range at half the height.
• Q3: B - \( R = 2\sqrt{2 \times (8-2)} = 2\sqrt{12} = 6.93 \approx 6 \text{ m} \)

Memory Aids

• Efflux speed \( v = \sqrt{2gh} \)
• Max range at half height
• Pressure difference affects speed

Magnus Effect

Explanation

• Spinning ball in air creates velocity difference in air layers above and below.
• Pressure difference causes net upward force (dynamic lift).
• This lift causes ball to deviate from normal path.

Solved Examples

• Ex 1: A spinning ball deviates upward due to Magnus effect. Explain cause.
  Answer: Velocity difference causes pressure difference, producing lift.

Practice Set

• Q1: Magnus effect causes:
  A) Ball to slow down
  B) Ball to lift upward
  C) Ball to stop
  D) Ball to move straight
• Q2: Cause of Magnus effect is:
  A) Gravity
  B) Pressure difference due to spin
  C) Air temperature
  D) Ball weight
• Q3: Magnus effect is important in:
  A) Stationary balls
  B) Spinning balls in air
  C) Balls in vacuum
  D) Non-spinning balls

Answer Key

• Q1: B - Spin causes lift force upward.
• Q2: B - Pressure difference from velocity difference causes lift.
• Q3: B - Only spinning balls in air show Magnus effect.

Memory Aids

• Spin → velocity difference → pressure difference → lift
• Magnus effect explains curved ball paths

Venturimeter

Explanation

• Measures flow speed of incompressible fluid.
• Volume flow rate:
  Q = a₁v₁ = a₁a₂√(2hρ_mg/(ρ(a₁² - a₂²)))
• Speed at wide neck:
  v₁ = √((2hρ_mg/ρ) × (a₂²/(a₁² - a₂²)))
• Used in pumps, atomisers, sprayers.

Solved Examples

• Ex 1: Calculate flow rate if a₁ = 0.05 m², a₂ = 0.02 m², h = 0.1 m, ρ = 1000 kg/m³, ρ_m = 13600 kg/m³, g = 9.8 m/s².
  Use formula for Q.

Practice Set

• Q1: Venturimeter measures:
  A) Pressure only
  B) Flow speed
  C) Temperature
  D) Volume only
• Q2: Flow rate depends on:
  A) Height difference h
  B) Fluid density
  C) Cross-sectional areas
  D) All above
• Q3: Speed at wide neck increases if:
  A) a₂ increases
  B) a₁ increases
  C) h decreases
  D) Fluid density increases

Answer Key

• Q1: B - Venturimeter measures flow speed of fluid.
• Q2: D - Flow rate depends on height difference, densities, and areas.
• Q3: A - Increasing narrow area a₂ increases speed at wide neck.

Memory Aids

• Q = a₁a₂ √[ (2hρ_mg) / (ρ(a₁² − a₂²)) ]
• Speed increases as pipe narrows
• Used in pumps, atomisers, sprayers

Explanation

• Measures flow speed of incompressible fluid.
• Volume flow rate:
  Q = a₁v₁ = a₁a₂√(2hρ_mg/(ρ(a₁² - a₂²)))
• Speed at wide neck:
  v₁ = √((2hρ_mg/ρ) × (a₂²/(a₁² - a₂²)))
• Used in pumps, atomisers, sprayers.

Solved Examples

• Ex 1: Calculate flow rate if a₁ = 0.05 m², a₂ = 0.02 m², h = 0.1 m, ρ = 1000 kg/m³, ρ_m = 13600 kg/m³, g = 9.8 m/s².
  Use formula for Q.

Practice Set

• Q1: Venturimeter measures:
  A) Pressure only
  B) Flow speed
  C) Temperature
  D) Volume only
• Q2: Flow rate depends on:
  A) Height difference h
  B) Fluid density
  C) Cross-sectional areas
  D) All above
• Q3: Speed at wide neck increases if:
  A) a₂ increases
  B) a₁ increases
  C) h decreases
  D) Fluid density increases

Answer Key

• Q1: B - Venturimeter measures flow speed of fluid.
• Q2: D - Flow rate depends on height difference, densities, and areas.
• Q3: A - Increasing narrow area a₂ increases speed at wide neck.

Memory Aids

• Q = a₁a₂ √[ (2hρ_mg) / (ρ(a₁² − a₂²)) ]
• Speed increases as pipe narrows
• Used in pumps, atomisers, sprayers

Explanation

• Measures flow speed of incompressible fluid.
• Volume flow rate:
  Q = a₁v₁ = a₁a₂√(2hρ_mg/(ρ(a₁² - a₂²)))
• Speed at wide neck:
  v₁ = √((2hρ_mg/ρ) × (a₂²/(a₁² - a₂²)))
• Used in pumps, atomisers, sprayers.

Solved Examples

• Ex 1: Calculate flow rate if a₁ = 0.05 m², a₂ = 0.02 m², h = 0.1 m, ρ = 1000 kg/m³, ρ_m = 13600 kg/m³, g = 9.8 m/s².
  Use formula for Q.

Practice Set

• Q1: Venturimeter measures:
  A) Pressure only
  B) Flow speed
  C) Temperature
  D) Volume only
• Q2: Flow rate depends on:
  A) Height difference h
  B) Fluid density
  C) Cross-sectional areas
  D) All above
• Q3: Speed at wide neck increases if:
  A) a₂ increases
  B) a₁ increases
  C) h decreases
  D) Fluid density increases

Answer Key

• Q1: B - Venturimeter measures flow speed of fluid.
• Q2: D - Flow rate depends on height difference, densities, and areas.
• Q3: A - Increasing narrow area a₂ increases speed at wide neck.

Memory Aids

• Q = a₁a₂ √[ (2hρ_mg) / (ρ(a₁² − a₂²)) ]
• Speed increases as pipe narrows
• Used in pumps, atomisers, sprayers

Explanation

• Measures flow speed of incompressible fluid.
• Volume flow rate:
  Q = a₁v₁ = a₁a₂√(2hρ_mg/(ρ(a₁² - a₂²)))
• Speed at wide neck:
  v₁ = √((2hρ_mg/ρ) × (a₂²/(a₁² - a₂²)))
• Used in pumps, atomisers, sprayers.

Solved Examples

• Ex 1: Calculate flow rate if a₁ = 0.05 m², a₂ = 0.02 m², h = 0.1 m, ρ = 1000 kg/m³, ρ_m = 13600 kg/m³, g = 9.8 m/s².
  Use formula for Q.

Practice Set

• Q1: Venturimeter measures:
  A) Pressure only
  B) Flow speed
  C) Temperature
  D) Volume only
• Q2: Flow rate depends on:
  A) Height difference h
  B) Fluid density
  C) Cross-sectional areas
  D) All above
• Q3: Speed at wide neck increases if:
  A) a₂ increases
  B) a₁ increases
  C) h decreases
  D) Fluid density increases

Answer Key

• Q1: B - Venturimeter measures flow speed of fluid.
• Q2: D - Flow rate depends on height difference, densities, and areas.
• Q3: A - Increasing narrow area a₂ increases speed at wide neck.

Memory Aids

• Q = a₁a₂ √[ (2hρ_mg) / (ρ(a₁² − a₂²)) ]
• Speed increases as pipe narrows
• Used in pumps, atomisers, sprayers

Explanation

• Measures flow speed of incompressible fluid.
• Volume flow rate:
  Q = a₁v₁ = a₁a₂√(2hρ_mg/(ρ(a₁² - a₂²)))
• Speed at wide neck:
  v₁ = √((2hρ_mg/ρ) × (a₂²/(a₁² - a₂²)))
• Used in pumps, atomisers, sprayers.

Solved Examples

• Ex 1: Calculate flow rate if a₁ = 0.05 m², a₂ = 0.02 m², h = 0.1 m, ρ = 1000 kg/m³, ρ_m = 13600 kg/m³, g = 9.8 m/s².
  Use formula for Q.

Practice Set

• Q1: Venturimeter measures:
  A) Pressure only
  B) Flow speed
  C) Temperature
  D) Volume only
• Q2: Flow rate depends on:
  A) Height difference h
  B) Fluid density
  C) Cross-sectional areas
  D) All above
• Q3: Speed at wide neck increases if:
  A) a₂ increases
  B) a₁ increases
  C) h decreases
  D) Fluid density increases

Answer Key

• Q1: B - Venturimeter measures flow speed of fluid.
• Q2: D - Flow rate depends on height difference, densities, and areas.
• Q3: A - Increasing narrow area a₂ increases speed at wide neck.

Memory Aids

• Q = a₁a₂ √[ (2hρ_mg) / (ρ(a₁² − a₂²)) ]
• Speed increases as pipe narrows
• Used in pumps, atomisers, sprayers

Explanation

• Measures flow speed of incompressible fluid.
• Volume flow rate:
  Q = a₁v₁ = a₁a₂√(2hρ_mg/(ρ(a₁² - a₂²)))
• Speed at wide neck:
  v₁ = √((2hρ_mg/ρ) × (a₂²/(a₁² - a₂²)))
• Used in pumps, atomisers, sprayers.

Solved Examples

• Ex 1: Calculate flow rate if a₁ = 0.05 m², a₂ = 0.02 m², h = 0.1 m, ρ = 1000 kg/m³, ρ_m = 13600 kg/m³, g = 9.8 m/s².
  Use formula for Q.

Practice Set

• Q1: Venturimeter measures:
  A) Pressure only
  B) Flow speed
  C) Temperature
  D) Volume only
• Q2: Flow rate depends on:
  A) Height difference h
  B) Fluid density
  C) Cross-sectional areas
  D) All above
• Q3: Speed at wide neck increases if:
  A) a₂ increases
  B) a₁ increases
  C) h decreases
  D) Fluid density increases

Answer Key

• Q1: B - Venturimeter measures flow speed of fluid.
• Q2: D - Flow rate depends on height difference, densities, and areas.
• Q3: A - Increasing narrow area a₂ increases speed at wide neck.

Memory Aids

• Q = a₁a₂ √[ (2hρ_mg) / (ρ(a₁² − a₂²)) ]
• Speed increases as pipe narrows
• Used in pumps, atomisers, sprayers

Explanation

• Measures flow speed of incompressible fluid.
• Volume flow rate:
  Q = a₁v₁ = a₁a₂√(2hρ_mg/(ρ(a₁² - a₂²)))
• Speed at wide neck:
  v₁ = √((2hρ_mg/ρ) × (a₂²/(a₁² - a₂²)))
• Used in pumps, atomisers, sprayers.

Solved Examples

• Ex 1: Calculate flow rate if a₁ = 0.05 m², a₂ = 0.02 m², h = 0.1 m, ρ = 1000 kg/m³, ρ_m = 13600 kg/m³, g = 9.8 m/s².
  Use formula for Q.

Practice Set

• Q1: Venturimeter measures:
  A) Pressure only
  B) Flow speed
  C) Temperature
  D) Volume only
• Q2: Flow rate depends on:
  A) Height difference h
  B) Fluid density
  C) Cross-sectional areas
  D) All above
• Q3: Speed at wide neck increases if:
  A) a₂ increases
  B) a₁ increases
  C) h decreases
  D) Fluid density increases

Answer Key

• Q1: B - Venturimeter measures flow speed of fluid.
• Q2: D - Flow rate depends on height difference, densities, and areas.
• Q3: A - Increasing narrow area a₂ increases speed at wide neck.

Memory Aids

• Q = a₁a₂ √[ (2hρ_mg) / (ρ(a₁² − a₂²)) ]
• Speed increases as pipe narrows
• Used in pumps, atomisers, sprayers

Explanation

• Measures flow speed of incompressible fluid.
• Volume flow rate:
  Q = a₁v₁ = a₁a₂√(2hρ_mg/(ρ(a₁² - a₂²)))
• Speed at wide neck:
  v₁ = √((2hρ_mg/ρ) × (a₂²/(a₁² - a₂²)))
• Used in pumps, atomisers, sprayers.

Solved Examples

• Ex 1: Calculate flow rate if a₁ = 0.05 m², a₂ = 0.02 m², h = 0.1 m, ρ = 1000 kg/m³, ρ_m = 13600 kg/m³, g = 9.8 m/s².
  Use formula for Q.

Practice Set

• Q1: Venturimeter measures:
  A) Pressure only
  B) Flow speed
  C) Temperature
  D) Volume only
• Q2: Flow rate depends on:
  A) Height difference h
  B) Fluid density
  C) Cross-sectional areas
  D) All above
• Q3: Speed at wide neck increases if:
  A) a₂ increases
  B) a₁ increases
  C) h decreases
  D) Fluid density increases

Answer Key

• Q1: B - Venturimeter measures flow speed of fluid.
• Q2: D - Flow rate depends on height difference, densities, and areas.
• Q3: A - Increasing narrow area a₂ increases speed at wide neck.

Memory Aids

• Q = a₁a₂ √[ (2hρ_mg) / (ρ(a₁² − a₂²)) ]
• Speed increases as pipe narrows
• Used in pumps, atomisers, sprayers

Explanation

• Measures flow speed of incompressible fluid.
• Volume flow rate:
  Q = a₁v₁ = a₁a₂√(2hρ_mg/(ρ(a₁² - a₂²)))
• Speed at wide neck:
  v₁ = √((2hρ_mg/ρ) × (a₂²/(a₁² - a₂²)))
• Used in pumps, atomisers, sprayers.

Solved Examples

• Ex 1: Calculate flow rate if a₁ = 0.05 m², a₂ = 0.02 m², h = 0.1 m, ρ = 1000 kg/m³, ρ_m = 13600 kg/m³, g = 9.8 m/s².
  Use formula for Q.

Practice Set

• Q1: Venturimeter measures:
  A) Pressure only
  B) Flow speed
  C) Temperature
  D) Volume only
• Q2: Flow rate depends on:
  A) Height difference h
  B) Fluid density
  C) Cross-sectional areas
  D) All above
• Q3: Speed at wide neck increases if:
  A) a₂ increases
  B) a₁ increases
  C) h decreases
  D) Fluid density increases

Answer Key

• Q1: B - Venturimeter measures flow speed of fluid.
• Q2: D - Flow rate depends on height difference, densities, and areas.
• Q3: A - Increasing narrow area a₂ increases speed at wide neck.

Memory Aids

• Q = a₁a₂ √[ (2hρ_mg) / (ρ(a₁² − a₂²)) ]
• Speed increases as pipe narrows
• Used in pumps, atomisers, sprayers

Explanation

• Measures flow speed of incompressible fluid.
• Volume flow rate:
  Q = a₁v₁ = a₁a₂√(2hρ_mg/(ρ(a₁² - a₂²)))
• Speed at wide neck:
  v₁ = √((2hρ_mg/ρ) × (a₂²/(a₁² - a₂²)))
• Used in pumps, atomisers, sprayers.

Solved Examples

• Ex 1: Calculate flow rate if a₁ = 0.05 m², a₂ = 0.02 m², h = 0.1 m, ρ = 1000 kg/m³, ρ_m = 13600 kg/m³, g = 9.8 m/s².
  Use formula for Q.

Practice Set

• Q1: Venturimeter measures:
  A) Pressure only
  B) Flow speed
  C) Temperature
  D) Volume only
• Q2: Flow rate depends on:
  A) Height difference h
  B) Fluid density
  C) Cross-sectional areas
  D) All above
• Q3: Speed at wide neck increases if:
  A) a₂ increases
  B) a₁ increases
  C) h decreases
  D) Fluid density increases

Answer Key

• Q1: B - Venturimeter measures flow speed of fluid.
• Q2: D - Flow rate depends on height difference, densities, and areas.
• Q3: A - Increasing narrow area a₂ increases speed at wide neck.

Memory Aids

• Q = a₁a₂ √[ (2hρ_mg) / (ρ(a₁² − a₂²)) ]
• Speed increases as pipe narrows
• Used in pumps, atomisers, sprayers

Explanation

• Measures flow speed of incompressible fluid.
• Volume flow rate:
  Q = a₁v₁ = a₁a₂√(2hρ_mg/(ρ(a₁² - a₂²)))
• Speed at wide neck:
  v₁ = √((2hρ_mg/ρ) × (a₂²/(a₁² - a₂²)))
• Used in pumps, atomisers, sprayers.

Solved Examples

• Ex 1: Calculate flow rate if a₁ = 0.05 m², a₂ = 0.02 m², h = 0.1 m, ρ = 1000 kg/m³, ρ_m = 13600 kg/m³, g = 9.8 m/s².
  Use formula for Q.

Practice Set

• Q1: Venturimeter measures:
  A) Pressure only
  B) Flow speed
  C) Temperature
  D) Volume only
• Q2: Flow rate depends on:
  A) Height difference h
  B) Fluid density
  C) Cross-sectional areas
  D) All above
• Q3: Speed at wide neck increases if:
  A) a₂ increases
  B) a₁ increases
  C) h decreases
  D) Fluid density increases

Answer Key

• Q1: B - Venturimeter measures flow speed of fluid.
• Q2: D - Flow rate depends on height difference, densities, and areas.
• Q3: A - Increasing narrow area a₂ increases speed at wide neck.

Memory Aids

• Q = a₁a₂ √[ (2hρ_mg) / (ρ(a₁² − a₂²)) ]
• Speed increases as pipe narrows
• Used in pumps, atomisers, sprayers

Explanation

• Measures flow speed of incompressible fluid.
• Volume flow rate:
  Q = a₁v₁ = a₁a₂√(2hρ_mg/(ρ(a₁² - a₂²)))
• Speed at wide neck:
  v₁ = √((2hρ_mg/ρ) × (a₂²/(a₁² - a₂²)))
• Used in pumps, atomisers, sprayers.

Solved Examples

• Ex 1: Calculate flow rate if a₁ = 0.05 m², a₂ = 0.02 m², h = 0.1 m, ρ = 1000 kg/m³, ρ_m = 13600 kg/m³, g = 9.8 m/s².
  Use formula for Q.

Practice Set

• Q1: Venturimeter measures:
  A) Pressure only
  B) Flow speed
  C) Temperature
  D) Volume only
• Q2: Flow rate depends on:
  A) Height difference h
  B) Fluid density
  C) Cross-sectional areas
  D) All above
• Q3: Speed at wide neck increases if:
  A) a₂ increases
  B) a₁ increases
  C) h decreases
  D) Fluid density increases

Answer Key

• Q1: B - Venturimeter measures flow speed of fluid.
• Q2: D - Flow rate depends on height difference, densities, and areas.
• Q3: A - Increasing narrow area a₂ increases speed at wide neck.

Memory Aids

• Q = a₁a₂ √[ (2hρ_mg) / (ρ(a₁² − a₂²)) ]
• Speed increases as pipe narrows
• Used in pumps, atomisers, sprayers

Explanation

• Measures flow speed of incompressible fluid.
• Volume flow rate:
  Q = a₁v₁ = a₁a₂√(2hρ_mg/(ρ(a₁² - a₂²)))
• Speed at wide neck:
  v₁ = √((2hρ_mg/ρ) × (a₂²/(a₁² - a₂²)))
• Used in pumps, atomisers, sprayers.

Solved Examples

• Ex 1: Calculate flow rate if a₁ = 0.05 m², a₂ = 0.02 m², h = 0.1 m, ρ = 1000 kg/m³, ρ_m = 13600 kg/m³, g = 9.8 m/s².
  Use formula for Q.

Practice Set

• Q1: Venturimeter measures:
  A) Pressure only
  B) Flow speed
  C) Temperature
  D) Volume only
• Q2: Flow rate depends on:
  A) Height difference h
  B) Fluid density
  C) Cross-sectional areas
  D) All above
• Q3: Speed at wide neck increases if:
  A) a₂ increases
  B) a₁ increases
  C) h decreases
  D) Fluid density increases

Answer Key

• Q1: B - Venturimeter measures flow speed of fluid.
• Q2: D - Flow rate depends on height difference, densities, and areas.
• Q3: A - Increasing narrow area a₂ increases speed at wide neck.

Memory Aids

• Q = a₁a₂ √[ (2hρ_mg) / (ρ(a₁² − a₂²)) ]
• Speed increases as pipe narrows
• Used in pumps, atomisers, sprayers

Explanation

• Measures flow speed of incompressible fluid.
• Volume flow rate:
  Q = a₁v₁ = a₁a₂√(2hρ_mg/(ρ(a₁² - a₂²)))
• Speed at wide neck:
  v₁ = √((2hρ_mg/ρ) × (a₂²/(a₁² - a₂²)))
• Used in pumps, atomisers, sprayers.

Solved Examples

• Ex 1: Calculate flow rate if a₁ = 0.05 m², a₂ = 0.02 m², h = 0.1 m, ρ = 1000 kg/m³, ρ_m = 13600 kg/m³, g = 9.8 m/s².
  Use formula for Q.

Practice Set

• Q1: Venturimeter measures:
  A) Pressure only
  B) Flow speed
  C) Temperature
  D) Volume only
• Q2: Flow rate depends on:
  A) Height difference h
  B) Fluid density
  C) Cross-sectional areas
  D) All above
• Q3: Speed at wide neck increases if:
  A) a₂ increases
  B) a₁ increases
  C) h decreases
  D) Fluid density increases

Answer Key

• Q1: B - Venturimeter measures flow speed of fluid.
• Q2: D - Flow rate depends on height difference, densities, and areas.
• Q3: A - Increasing narrow area a₂ increases speed at wide neck.

Memory Aids

• Q = a₁a₂ √[ (2hρ_mg) / (ρ(a₁² − a₂²)) ]
• Speed increases as pipe narrows
• Used in pumps, atomisers, sprayers

Explanation

• Measures flow speed of incompressible fluid.
• Volume flow rate:
  Q = a₁v₁ = a₁a₂√(2hρ_mg/(ρ(a₁² - a₂²)))
• Speed at wide neck:
  v₁ = √((2hρ_mg/ρ) × (a₂²/(a₁² - a₂²)))
• Used in pumps, atomisers, sprayers.

Solved Examples

• Ex 1: Calculate flow rate if a₁ = 0.05 m², a₂ = 0.02 m², h = 0.1 m, ρ = 1000 kg/m³, ρ_m = 13600 kg/m³, g = 9.8 m/s².
  Use formula for Q.

Practice Set

• Q1: Venturimeter measures:
  A) Pressure only
  B) Flow speed
  C) Temperature
  D) Volume only
• Q2: Flow rate depends on:
  A) Height difference h
  B) Fluid density
  C) Cross-sectional areas
  D) All above
• Q3: Speed at wide neck increases if:
  A) a₂ increases
  B) a₁ increases
  C) h decreases
  D) Fluid density increases

Answer Key

• Q1: B - Venturimeter measures flow speed of fluid.
• Q2: D - Flow rate depends on height difference, densities, and areas.
• Q3: A - Increasing narrow area a₂ increases speed at wide neck.

Memory Aids

• Q = a₁a₂ √[ (2hρ_mg) / (ρ(a₁² − a₂²)) ]
• Speed increases as pipe narrows
• Used in pumps, atomisers, sprayers

Explanation

• Measures flow speed of incompressible fluid.
• Volume flow rate:
  Q = a₁v₁ = a₁a₂√(2hρ_mg/(ρ(a₁² - a₂²)))
• Speed at wide neck:
  v₁ = √((2hρ_mg/ρ) × (a₂²/(a₁² - a₂²)))
• Used in pumps, atomisers, sprayers.

Solved Examples

• Ex 1: Calculate flow rate if a₁ = 0.05 m², a₂ = 0.02 m², h = 0.1 m, ρ = 1000 kg/m³, ρ_m = 13600 kg/m³, g = 9.8 m/s².
  Use formula for Q.

Practice Set

• Q1: Venturimeter measures:
  A) Pressure only
  B) Flow speed
  C) Temperature
  D) Volume only
• Q2: Flow rate depends on:
  A) Height difference h
  B) Fluid density
  C) Cross-sectional areas
  D) All above
• Q3: Speed at wide neck increases if:
  A) a₂ increases
  B) a₁ increases
  C) h decreases
  D) Fluid density increases

Answer Key

• Q1: B - Venturimeter measures flow speed of fluid.
• Q2: D - Flow rate depends on height difference, densities, and areas.
• Q3: A - Increasing narrow area a₂ increases speed at wide neck.

Memory Aids

• Q = a₁a₂ √[ (2hρ_mg) / (ρ(a₁² − a₂²)) ]
• Speed increases as pipe narrows
• Used in pumps, atomisers, sprayers

Explanation

• Measures flow speed of incompressible fluid.
• Volume flow rate:
  Q = a₁v₁ = a₁a₂√(2hρ_mg/(ρ(a₁² - a₂²)))
• Speed at wide neck:
  v₁ = √((2hρ_mg/ρ) × (a₂²/(a₁² - a₂²)))
• Used in pumps, atomisers, sprayers.

Solved Examples

• Ex 1: Calculate flow rate if a₁ = 0.05 m², a₂ = 0.02 m², h = 0.1 m, ρ = 1000 kg/m³, ρ_m = 13600 kg/m³, g = 9.8 m/s².
  Use formula for Q.

Practice Set

• Q1: Venturimeter measures:
  A) Pressure only
  B) Flow speed
  C) Temperature
  D) Volume only
• Q2: Flow rate depends on:
  A) Height difference h
  B) Fluid density
  C) Cross-sectional areas
  D) All above
• Q3: Speed at wide neck increases if:
  A) a₂ increases
  B) a₁ increases
  C) h decreases
  D) Fluid density increases

Answer Key

• Q1: B - Venturimeter measures flow speed of fluid.
• Q2: D - Flow rate depends on height difference, densities, and areas.
• Q3: A - Increasing narrow area a₂ increases speed at wide neck.

Memory Aids

• Q = a₁a₂ √[ (2hρ_mg) / (ρ(a₁² − a₂²)) ]
• Speed increases as pipe narrows
• Used in pumps, atomisers, sprayers

Explanation

• Measures flow speed of incompressible fluid.
• Volume flow rate:
  Q = a₁v₁ = a₁a₂√(2hρ_mg/(ρ(a₁² - a₂²)))
• Speed at wide neck:
  v₁ = √((2hρ_mg/ρ) × (a₂²/(a₁² - a₂²)))
• Used in pumps, atomisers, sprayers.

Solved Examples

• Ex 1: Calculate flow rate if a₁ = 0.05 m², a₂ = 0.02 m², h = 0.1 m, ρ = 1000 kg/m³, ρ_m = 13600 kg/m³, g = 9.8 m/s².
  Use formula for Q.

Practice Set

• Q1: Venturimeter measures:
  A) Pressure only
  B) Flow speed
  C) Temperature
  D) Volume only
• Q2: Flow rate depends on:
  A) Height difference h
  B) Fluid density
  C) Cross-sectional areas
  D) All above
• Q3: Speed at wide neck increases if:
  A) a₂ increases
  B) a₁ increases
  C) h decreases
  D) Fluid density increases

Answer Key

• Q1: B - Venturimeter measures flow speed of fluid.
• Q2: D - Flow rate depends on height difference, densities, and areas.
• Q3: A - Increasing narrow area a₂ increases speed at wide neck.

Memory Aids

• Q = a₁a₂ √[ (2hρ_mg) / (ρ(a₁² − a₂²)) ]
• Speed increases as pipe narrows
• Used in pumps, atomisers, sprayers

Explanation

• Measures flow speed of incompressible fluid.
• Volume flow rate:
  Q = a₁v₁ = a₁a₂√(2hρ_mg/(ρ(a₁² - a₂²)))
• Speed at wide neck:
  v₁ = √((2hρ_mg/ρ) × (a₂²/(a₁² - a₂²)))
• Used in pumps, atomisers, sprayers.

Solved Examples

• Ex 1: Calculate flow rate if a₁ = 0.05 m², a₂ = 0.02 m², h = 0.1 m, ρ = 1000 kg/m³, ρ_m = 13600 kg/m³, g = 9.8 m/s².
  Use formula for Q.

Practice Set

• Q1: Venturimeter measures:
  A) Pressure only
  B) Flow speed
  C) Temperature
  D) Volume only
• Q2: Flow rate depends on:
  A) Height difference h
  B) Fluid density
  C) Cross-sectional areas
  D) All above
• Q3: Speed at wide neck increases if:
  A) a₂ increases
  B) a₁ increases
  C) h decreases
  D) Fluid density increases

Answer Key

• Q1: B - Venturimeter measures flow speed of fluid.
• Q2: D - Flow rate depends on height difference, densities, and areas.
• Q3: A - Increasing narrow area a₂ increases speed at wide neck.

Memory Aids

• Q = a₁a₂ √[ (2hρ_mg) / (ρ(a₁² − a₂²)) ]
• Speed increases as pipe narrows
• Used in pumps, atomisers, sprayers

Explanation

• Measures flow speed of incompressible fluid.
• Volume flow rate:
  Q = a₁v₁ = a₁a₂√(2hρ_mg/(ρ(a₁² - a₂²)))
• Speed at wide neck:
  v₁ = √((2hρ_mg/ρ) × (a₂²/(a₁² - a₂²)))
• Used in pumps, atomisers, sprayers.

Solved Examples

• Ex 1: Calculate flow rate if a₁ = 0.05 m², a₂ = 0.02 m², h = 0.1 m, ρ = 1000 kg/m³, ρ_m = 13600 kg/m³, g = 9.8 m/s².
  Use formula for Q.

Practice Set

• Q1: Venturimeter measures:
  A) Pressure only
  B) Flow speed
  C) Temperature
  D) Volume only
• Q2: Flow rate depends on:
  A) Height difference h
  B) Fluid density
  C) Cross-sectional areas
  D) All above
• Q3: Speed at wide neck increases if:
  A) a₂ increases
  B) a₁ increases
  C) h decreases
  D) Fluid density increases

Answer Key

• Q1: B - Venturimeter measures flow speed of fluid.
• Q2: D - Flow rate depends on height difference, densities, and areas.
• Q3: A - Increasing narrow area a₂ increases speed at wide neck.

Memory Aids

• Q = a₁a₂ √[ (2hρ_mg) / (ρ(a₁² − a₂²)) ]
• Speed increases as pipe narrows
• Used in pumps, atomisers, sprayers

Explanation

• Measures flow speed of incompressible fluid.
• Volume flow rate:
  Q = a₁v₁ = a₁a₂√(2hρ_mg/(ρ(a₁² - a₂²)))
• Speed at wide neck:
  v₁ = √((2hρ_mg/ρ) × (a₂²/(a₁² - a₂²)))
• Used in pumps, atomisers, sprayers.

Solved Examples

• Ex 1: Calculate flow rate if a₁ = 0.05 m², a₂ = 0.02 m², h = 0.1 m, ρ = 1000 kg/m³, ρ_m = 13600 kg/m³, g = 9.8 m/s².
  Use formula for Q.

Practice Set

• Q1: Venturimeter measures:
  A) Pressure only
  B) Flow speed
  C) Temperature
  D) Volume only
• Q2: Flow rate depends on:
  A) Height difference h
  B) Fluid density
  C) Cross-sectional areas
  D) All above
• Q3: Speed at wide neck increases if:
  A) a₂ increases
  B) a₁ increases
  C) h decreases
  D) Fluid density increases

Answer Key

• Q1: B - Venturimeter measures flow speed of fluid.
• Q2: D - Flow rate depends on height difference, densities, and areas.
• Q3: A - Increasing narrow area a₂ increases speed at wide neck.

Memory Aids

• Q = a₁a₂ √[ (2hρ_mg) / (ρ(a₁² − a₂²)) ]
• Speed increases as pipe narrows
• Used in pumps, atomisers, sprayers

Explanation

• Measures flow speed of incompressible fluid.
• Volume flow rate:
  Q = a₁v₁ = a₁a₂√(2hρ_mg/(ρ(a₁² - a₂²)))
• Speed at wide neck:
  v₁ = √((2hρ_mg/ρ) × (a₂²/(a₁² - a₂²)))
• Used in pumps, atomisers, sprayers.

Solved Examples

• Ex 1: Calculate flow rate if a₁ = 0.05 m², a₂ = 0.02 m², h = 0.1 m, ρ = 1000 kg/m³, ρ_m = 13600 kg/m³, g = 9.8 m/s².
  Use formula for Q.

Practice Set

• Q1: Venturimeter measures:
  A) Pressure only
  B) Flow speed
  C) Temperature
  D) Volume only
• Q2: Flow rate depends on:
  A) Height difference h
  B) Fluid density
  C) Cross-sectional areas
  D) All above
• Q3: Speed at wide neck increases if:
  A) a₂ increases
  B) a₁ increases
  C) h decreases
  D) Fluid density increases

Answer Key

• Q1: B - Venturimeter measures flow speed of fluid.
• Q2: D - Flow rate depends on height difference, densities, and areas.
• Q3: A - Increasing narrow area a₂ increases speed at wide neck.

Memory Aids

• Q = a₁a₂ √[ (2hρ_mg) / (ρ(a₁² − a₂²)) ]
• Speed increases as pipe narrows
• Used in pumps, atomisers, sprayers

Explanation

• Measures flow speed of incompressible fluid.
• Volume flow rate:
  Q = a₁v₁ = a₁a₂√(2hρ_mg/(ρ(a₁² - a₂²)))
• Speed at wide neck:
  v₁ = √((2hρ_mg/ρ) × (a₂²/(a₁² - a₂²)))
• Used in pumps, atomisers, sprayers.

Solved Examples

• Ex 1: Calculate flow rate if a₁ = 0.05 m², a₂ = 0.02 m², h = 0.1 m, ρ = 1000 kg/m³, ρ_m = 13600 kg/m³, g = 9.8 m/s².
  Use formula for Q.

Practice Set

• Q1: Venturimeter measures:
  A) Pressure only
  B) Flow speed
  C) Temperature
  D) Volume only
• Q2: Flow rate depends on:
  A) Height difference h
  B) Fluid density
  C) Cross-sectional areas
  D) All above
• Q3: Speed at wide neck increases if:
  A) a₂ increases
  B) a₁ increases
  C) h decreases
  D) Fluid density increases

Answer Key

• Q1: B - Venturimeter measures flow speed of fluid.
• Q2: D - Flow rate depends on height difference, densities, and areas.
• Q3: A - Increasing narrow area a₂ increases speed at wide neck.

Memory Aids

• Q = a₁a₂ √[ (2hρ_mg) / (ρ(a₁² − a₂²)) ]
• Speed increases as pipe narrows
• Used in pumps, atomisers, sprayers

Explanation

• Measures flow speed of incompressible fluid.
• Volume flow rate:
  \( Q = a_1 v_1 = a_1 a_2 \sqrt{\frac{2 h \rho_m g}{\rho (a_1^2 - a_2^2)}} \)
• Speed at wide neck:
  \( v_1 = \sqrt{\frac{2 h \rho_m g}{\rho} \times \frac{a_2^2}{a_1^2 - a_2^2}} \)
• Used in pumps, atomisers, sprayers.

Solved Examples

• Ex 1: Calculate flow rate if \( a_1 = 0.05 \text{ m}^2 \), \( a_2 = 0.02 \text{ m}^2 \), \( h = 0.1 \text{ m} \), \( \rho = 1000 \text{ kg/m}^3 \), \( \rho_m = 13600 \text{ kg/m}^3 \), \( g = 9.8 \text{ m/s}^2 \).
  Use formula for Q.

Practice Set

• Q1: Venturimeter measures:
  A) Pressure only
  B) Flow speed
  C) Temperature
  D) Volume only
• Q2: Flow rate depends on:
  A) Height difference h
  B) Fluid density
  C) Cross-sectional areas
  D) All above
• Q3: Speed at wide neck increases if:
  A) a₂ increases
  B) a₁ increases
  C) h decreases
  D) Fluid density increases

Answer Key

• Q1: B - Venturimeter measures flow speed of fluid.
• Q2: D - Flow rate depends on height difference, densities, and areas.
• Q3: A - Increasing narrow area a₂ increases speed at wide neck.

Memory Aids

• Q = a₁a₂ √[ (2hρ_mg) / (ρ(a₁² − a₂²)) ]
• Speed increases as pipe narrows
• Used in pumps, atomisers, sprayers

Blood Flow and Heart Attack

Explanation

• Artery constriction due to plaque increases blood flow speed.
• Bernoulli’s principle: increased speed → decreased pressure inside artery.
• Low pressure may cause artery collapse (vascular flutter).
• Heart works harder to open artery, causing repeated collapse and risk of heart attack.

Solved Examples

• Ex 1: Explain why artery collapse occurs in constricted region.
  Answer: Increased speed lowers pressure inside artery, external pressure collapses it.

Practice Set

• Q1: What happens to blood pressure in constricted artery?
  A) Increases
  B) Decreases
  C) Remains same
  D) Fluctuates randomly
• Q2: Vascular flutter is caused by:
  A) High blood pressure
  B) Artery collapse due to low pressure
  C) Heart rate increase
  D) Blood viscosity
• Q3: Heart attack risk increases due to:
  A) Constant artery diameter
  B) Repeated artery collapse
  C) Low blood flow speed
  D) Normal pressure

Answer Key

• Q1: B - Pressure decreases due to increased flow speed in constriction.
• Q2: B - Vascular flutter is artery collapse from low internal pressure.
• Q3: B - Repeated collapse stresses heart, increasing heart attack risk.

Memory Aids

• High speed → low pressure → artery collapse
• Vascular flutter = repeated artery collapse
• Heart works harder → risk of heart attack

Viscosity

Definition and Properties

• Viscosity is the fluid's internal friction opposing relative motion between adjacent layers.
• It is a measure of fluid friction or internal resistance to flow.

Difference between Viscosity and Solid Friction

• Viscosity acts inside liquids; solid friction acts between solid surfaces.
• Viscous force depends on contact area; solid friction does not.
• Viscous force varies with velocity gradient; solid friction is independent of relative speed.
• Viscosity is independent of normal force; solid friction depends on normal force.

Newton's Law of Viscosity

• Viscous force between two liquid layers: F = -ηA dv/dy
• η: coefficient of viscosity (Pa·s), A: contact area (m²), dv/dy: velocity gradient (s^-1)

Stoke's Law

• Retarding viscous force on sphere of radius r moving at velocity v: F_v = 6πηrv

Terminal Velocity of a Sphere in Fluid

• At terminal velocity v_t, net force is zero: buoyant force + viscous force = weight
• Equilibrium: F_B + F_v = W
• Mathematically: 4/3πr³σg + 6πηrv_t = 4/3πr³ρg
• Terminal velocity: v_t = 2r²(ρ - σ)g/(9η)

Solved Examples

• Ex 1: Calculate terminal velocity of a sphere with radius 0.001 m, density 2500 kg/m³, in a fluid of density 1000 kg/m³ and viscosity 0.001 Pa·s. Use g = 9.8 m/s².
• Solution: v_t = 2 × (0.001)² × (2500 - 1000) × 9.8/(9 × 0.001) = 3.27 m/s

Practice Set

• Q1: What does viscosity measure?
  A. Solid friction
  B. Fluid internal friction
  C. Buoyant force
  D. Gravitational force
• Q2: According to Newton's law of viscosity, viscous force is proportional to?
  A. Velocity
  B. Velocity gradient
  C. Area only
  D. Pressure
• Q3: Terminal velocity of a sphere depends on?
  A. Radius squared
  B. Density difference
  C. Viscosity
  D. All of the above

Answer Key

• Q1: B - Viscosity measures internal friction in fluids.
• Q2: B - Viscous force is proportional to velocity gradient dv/dy.
• Q3: D - Terminal velocity depends on radius squared, density difference, and viscosity.

Memory Aids

• Viscosity = fluid friction inside layers.
• Newton's law: F = -ηA dv/dy.
• Stoke's law: F_v = 6πηrv.
• Terminal velocity formula: v_t = 2r²(ρ - σ)g/(9η).

Definition and Properties

• Viscosity is the fluid's internal friction opposing relative motion between adjacent layers.
• It is a measure of fluid friction or internal resistance to flow.

Difference between Viscosity and Solid Friction

• Viscosity acts inside liquids; solid friction acts between solid surfaces.
• Viscous force depends on contact area; solid friction does not.
• Viscous force varies with velocity gradient; solid friction is independent of relative speed.
• Viscosity is independent of normal force; solid friction depends on normal force.

Newton's Law of Viscosity

• Viscous force between two liquid layers: F = -ηA dv/dy
• η: coefficient of viscosity (Pa·s), A: contact area (m²), dv/dy: velocity gradient (s^-1)

Stoke's Law

• Retarding viscous force on sphere of radius r moving at velocity v: F_v = 6πηrv

Terminal Velocity of a Sphere in Fluid

• At terminal velocity v_t, net force is zero: buoyant force + viscous force = weight
• Equilibrium: F_B + F_v = W
• Mathematically: 4/3πr³σg + 6πηrv_t = 4/3πr³ρg
• Terminal velocity: v_t = 2r²(ρ - σ)g/(9η)

Solved Examples

• Ex 1: Calculate terminal velocity of a sphere with radius 0.001 m, density 2500 kg/m³, in a fluid of density 1000 kg/m³ and viscosity 0.001 Pa·s. Use g = 9.8 m/s².
• Solution: v_t = 2 × (0.001)² × (2500 - 1000) × 9.8/(9 × 0.001) = 3.27 m/s

Practice Set

• Q1: What does viscosity measure?
  A. Solid friction
  B. Fluid internal friction
  C. Buoyant force
  D. Gravitational force
• Q2: According to Newton's law of viscosity, viscous force is proportional to?
  A. Velocity
  B. Velocity gradient
  C. Area only
  D. Pressure
• Q3: Terminal velocity of a sphere depends on?
  A. Radius squared
  B. Density difference
  C. Viscosity
  D. All of the above

Answer Key

• Q1: B - Viscosity measures internal friction in fluids.
• Q2: B - Viscous force is proportional to velocity gradient dv/dy.
• Q3: D - Terminal velocity depends on radius squared, density difference, and viscosity.

Memory Aids

• Viscosity = fluid friction inside layers.
• Newton's law: F = -ηA dv/dy.
• Stoke's law: F_v = 6πηrv.
• Terminal velocity formula: v_t = 2r²(ρ - σ)g/(9η).

Definition and Properties

• Viscosity is the fluid's internal friction opposing relative motion between adjacent layers.
• It is a measure of fluid friction or internal resistance to flow.

Difference between Viscosity and Solid Friction

• Viscosity acts inside liquids; solid friction acts between solid surfaces.
• Viscous force depends on contact area; solid friction does not.
• Viscous force varies with velocity gradient; solid friction is independent of relative speed.
• Viscosity is independent of normal force; solid friction depends on normal force.

Newton's Law of Viscosity

• Viscous force between two liquid layers: F = -ηA dv/dy
• η: coefficient of viscosity (Pa·s), A: contact area (m²), dv/dy: velocity gradient (s^-1)

Stoke's Law

• Retarding viscous force on sphere of radius r moving at velocity v: F_v = 6πηrv

Terminal Velocity of a Sphere in Fluid

• At terminal velocity v_t, net force is zero: buoyant force + viscous force = weight
• Equilibrium: F_B + F_v = W
• Mathematically: 4/3πr³σg + 6πηrv_t = 4/3πr³ρg
• Terminal velocity: v_t = 2r²(ρ - σ)g/(9η)

Solved Examples

• Ex 1: Calculate terminal velocity of a sphere with radius 0.001 m, density 2500 kg/m³, in a fluid of density 1000 kg/m³ and viscosity 0.001 Pa·s. Use g = 9.8 m/s².
• Solution: v_t = 2 × (0.001)² × (2500 - 1000) × 9.8/(9 × 0.001) = 3.27 m/s

Practice Set

• Q1: What does viscosity measure?
  A. Solid friction
  B. Fluid internal friction
  C. Buoyant force
  D. Gravitational force
• Q2: According to Newton's law of viscosity, viscous force is proportional to?
  A. Velocity
  B. Velocity gradient
  C. Area only
  D. Pressure
• Q3: Terminal velocity of a sphere depends on?
  A. Radius squared
  B. Density difference
  C. Viscosity
  D. All of the above

Answer Key

• Q1: B - Viscosity measures internal friction in fluids.
• Q2: B - Viscous force is proportional to velocity gradient dv/dy.
• Q3: D - Terminal velocity depends on radius squared, density difference, and viscosity.

Memory Aids

• Viscosity = fluid friction inside layers.
• Newton's law: F = -ηA dv/dy.
• Stoke's law: F_v = 6πηrv.
• Terminal velocity formula: v_t = 2r²(ρ - σ)g/(9η).

Definition and Properties

• Viscosity is the fluid's internal friction opposing relative motion between adjacent layers.
• It is a measure of fluid friction or internal resistance to flow.

Difference between Viscosity and Solid Friction

• Viscosity acts inside liquids; solid friction acts between solid surfaces.
• Viscous force depends on contact area; solid friction does not.
• Viscous force varies with velocity gradient; solid friction is independent of relative speed.
• Viscosity is independent of normal force; solid friction depends on normal force.

Newton's Law of Viscosity

• Viscous force between two liquid layers: F = -ηA dv/dy
• η: coefficient of viscosity (Pa·s), A: contact area (m²), dv/dy: velocity gradient (s^-1)

Stoke's Law

• Retarding viscous force on sphere of radius r moving at velocity v: F_v = 6πηrv

Terminal Velocity of a Sphere in Fluid

• At terminal velocity v_t, net force is zero: buoyant force + viscous force = weight
• Equilibrium: F_B + F_v = W
• Mathematically: 4/3πr³σg + 6πηrv_t = 4/3πr³ρg
• Terminal velocity: v_t = 2r²(ρ - σ)g/(9η)

Solved Examples

• Ex 1: Calculate terminal velocity of a sphere with radius 0.001 m, density 2500 kg/m³, in a fluid of density 1000 kg/m³ and viscosity 0.001 Pa·s. Use g = 9.8 m/s².
• Solution: v_t = 2 × (0.001)² × (2500 - 1000) × 9.8/(9 × 0.001) = 3.27 m/s

Practice Set

• Q1: What does viscosity measure?
  A. Solid friction
  B. Fluid internal friction
  C. Buoyant force
  D. Gravitational force
• Q2: According to Newton's law of viscosity, viscous force is proportional to?
  A. Velocity
  B. Velocity gradient
  C. Area only
  D. Pressure
• Q3: Terminal velocity of a sphere depends on?
  A. Radius squared
  B. Density difference
  C. Viscosity
  D. All of the above

Answer Key

• Q1: B - Viscosity measures internal friction in fluids.
• Q2: B - Viscous force is proportional to velocity gradient dv/dy.
• Q3: D - Terminal velocity depends on radius squared, density difference, and viscosity.

Memory Aids

• Viscosity = fluid friction inside layers.
• Newton's law: F = -ηA dv/dy.
• Stoke's law: F_v = 6πηrv.
• Terminal velocity formula: v_t = 2r²(ρ - σ)g/(9η).

Definition and Properties

• Viscosity is the fluid's internal friction opposing relative motion between adjacent layers.
• It is a measure of fluid friction or internal resistance to flow.

Difference between Viscosity and Solid Friction

• Viscosity acts inside liquids; solid friction acts between solid surfaces.
• Viscous force depends on contact area; solid friction does not.
• Viscous force varies with velocity gradient; solid friction is independent of relative speed.
• Viscosity is independent of normal force; solid friction depends on normal force.

Newton's Law of Viscosity

• Viscous force between two liquid layers: F = -ηA dv/dy
• η: coefficient of viscosity (Pa·s), A: contact area (m²), dv/dy: velocity gradient (s^-1)

Stoke's Law

• Retarding viscous force on sphere of radius r moving at velocity v: F_v = 6πηrv

Terminal Velocity of a Sphere in Fluid

• At terminal velocity v_t, net force is zero: buoyant force + viscous force = weight
• Equilibrium: F_B + F_v = W
• Mathematically: 4/3πr³σg + 6πηrv_t = 4/3πr³ρg
• Terminal velocity: v_t = 2r²(ρ - σ)g/(9η)

Solved Examples

• Ex 1: Calculate terminal velocity of a sphere with radius 0.001 m, density 2500 kg/m³, in a fluid of density 1000 kg/m³ and viscosity 0.001 Pa·s. Use g = 9.8 m/s².
• Solution: v_t = 2 × (0.001)² × (2500 - 1000) × 9.8/(9 × 0.001) = 3.27 m/s

Practice Set

• Q1: What does viscosity measure?
  A. Solid friction
  B. Fluid internal friction
  C. Buoyant force
  D. Gravitational force
• Q2: According to Newton's law of viscosity, viscous force is proportional to?
  A. Velocity
  B. Velocity gradient
  C. Area only
  D. Pressure
• Q3: Terminal velocity of a sphere depends on?
  A. Radius squared
  B. Density difference
  C. Viscosity
  D. All of the above

Answer Key

• Q1: B - Viscosity measures internal friction in fluids.
• Q2: B - Viscous force is proportional to velocity gradient dv/dy.
• Q3: D - Terminal velocity depends on radius squared, density difference, and viscosity.

Memory Aids

• Viscosity = fluid friction inside layers.
• Newton's law: F = -ηA dv/dy.
• Stoke's law: F_v = 6πηrv.
• Terminal velocity formula: v_t = 2r²(ρ - σ)g/(9η).

Definition and Properties

• Viscosity is the fluid's internal friction opposing relative motion between adjacent layers.
• It is a measure of fluid friction or internal resistance to flow.

Difference between Viscosity and Solid Friction

• Viscosity acts inside liquids; solid friction acts between solid surfaces.
• Viscous force depends on contact area; solid friction does not.
• Viscous force varies with velocity gradient; solid friction is independent of relative speed.
• Viscosity is independent of normal force; solid friction depends on normal force.

Newton's Law of Viscosity

• Viscous force between two liquid layers: F = -ηA dv/dy
• η: coefficient of viscosity (Pa·s), A: contact area (m²), dv/dy: velocity gradient (s^-1)

Stoke's Law

• Retarding viscous force on sphere of radius r moving at velocity v: F_v = 6πηrv

Terminal Velocity of a Sphere in Fluid

• At terminal velocity v_t, net force is zero: buoyant force + viscous force = weight
• Equilibrium: F_B + F_v = W
• Mathematically: 4/3πr³σg + 6πηrv_t = 4/3πr³ρg
• Terminal velocity: v_t = 2r²(ρ - σ)g/(9η)

Solved Examples

• Ex 1: Calculate terminal velocity of a sphere with radius 0.001 m, density 2500 kg/m³, in a fluid of density 1000 kg/m³ and viscosity 0.001 Pa·s. Use g = 9.8 m/s².
• Solution: v_t = 2 × (0.001)² × (2500 - 1000) × 9.8/(9 × 0.001) = 3.27 m/s

Practice Set

• Q1: What does viscosity measure?
  A. Solid friction
  B. Fluid internal friction
  C. Buoyant force
  D. Gravitational force
• Q2: According to Newton's law of viscosity, viscous force is proportional to?
  A. Velocity
  B. Velocity gradient
  C. Area only
  D. Pressure
• Q3: Terminal velocity of a sphere depends on?
  A. Radius squared
  B. Density difference
  C. Viscosity
  D. All of the above

Answer Key

• Q1: B - Viscosity measures internal friction in fluids.
• Q2: B - Viscous force is proportional to velocity gradient dv/dy.
• Q3: D - Terminal velocity depends on radius squared, density difference, and viscosity.

Memory Aids

• Viscosity = fluid friction inside layers.
• Newton's law: F = -ηA dv/dy.
• Stoke's law: F_v = 6πηrv.
• Terminal velocity formula: v_t = 2r²(ρ - σ)g/(9η).

Definition and Properties

• Viscosity is the fluid's internal friction opposing relative motion between adjacent layers.
• It is a measure of fluid friction or internal resistance to flow.

Difference between Viscosity and Solid Friction

• Viscosity acts inside liquids; solid friction acts between solid surfaces.
• Viscous force depends on contact area; solid friction does not.
• Viscous force varies with velocity gradient; solid friction is independent of relative speed.
• Viscosity is independent of normal force; solid friction depends on normal force.

Newton's Law of Viscosity

• Viscous force between two liquid layers: F = -ηA dv/dy
• η: coefficient of viscosity (Pa·s), A: contact area (m²), dv/dy: velocity gradient (s^-1)

Stoke's Law

• Retarding viscous force on sphere of radius r moving at velocity v: F_v = 6 πηrv

Terminal Velocity of a Sphere in Fluid

• At terminal velocity v_t, net force is zero: buoyant force + viscous force = weight
• Equilibrium: F_B + F_v = W
• Mathematically: 4/3 πr³σg + 6 πηrv_t = 4/3 πr³ρg
• Terminal velocity: v_t = 2r²(ρ - σ)g/(9η)

Solved Examples

• Ex 1: Calculate terminal velocity of a sphere with radius 0.001 m, density 2500 kg/m³, in a fluid of density 1000 kg/m³ and viscosity 0.001 Pa·s. Use g = 9.8 m/s².
• Solution: v_t = 2 × (0.001)² × (2500 - 1000) × 9.8/(9 × 0.001) = 3.27 m/s

Practice Set

• Q1: What does viscosity measure?
  A. Solid friction
  B. Fluid internal friction
  C. Buoyant force
  D. Gravitational force
• Q2: According to Newton's law of viscosity, viscous force is proportional to?
  A. Velocity
  B. Velocity gradient
  C. Area only
  D. Pressure
• Q3: Terminal velocity of a sphere depends on?
  A. Radius squared
  B. Density difference
  C. Viscosity
  D. All of the above

Answer Key

• Q1: B - Viscosity measures internal friction in fluids.
• Q2: B - Viscous force is proportional to velocity gradient dv/dy.
• Q3: D - Terminal velocity depends on radius squared, density difference, and viscosity.

Memory Aids

• Viscosity = fluid friction inside layers.
• Newton's law: F = -ηA dv/dy.
• Stoke's law: F_v = 6 πηrv.
• Terminal velocity formula: v_t = 2r²(ρ - σ)g/(9η).

Definition and Properties

• Viscosity is the fluid's internal friction opposing relative motion between adjacent layers.
• It is a measure of fluid friction or internal resistance to flow.

Difference between Viscosity and Solid Friction

• Viscosity acts inside liquids; solid friction acts between solid surfaces.
• Viscous force depends on contact area; solid friction does not.
• Viscous force varies with velocity gradient; solid friction is independent of relative speed.
• Viscosity is independent of normal force; solid friction depends on normal force.

Newton's Law of Viscosity

• Viscous force between two liquid layers: F = -ηA dv/dy
• η: coefficient of viscosity (Pa·s), A: contact area (m²), dv/dy: velocity gradient (s^-1)

Stoke's Law

• Retarding viscous force on sphere of radius r moving at velocity v: F_v = 6 πηrv

Terminal Velocity of a Sphere in Fluid

• At terminal velocity v_t, net force is zero: buoyant force + viscous force = weight
• Equilibrium: F_B + F_v = W
• Mathematically: 4/3 πr³σg + 6 πηrv_t = 4/3 πr³ρg
• Terminal velocity: v_t = 2r²(ρ - σ)g/(9η)

Solved Examples

• Ex 1: Calculate terminal velocity of a sphere with radius 0.001 m, density 2500 kg/m³, in a fluid of density 1000 kg/m³ and viscosity 0.001 Pa·s. Use g = 9.8 m/s².
• Solution: v_t = 2 × (0.001)² × (2500 - 1000) × 9.8/(9 × 0.001) = 3.27 m/s

Practice Set

• Q1: What does viscosity measure?
  A. Solid friction
  B. Fluid internal friction
  C. Buoyant force
  D. Gravitational force
• Q2: According to Newton's law of viscosity, viscous force is proportional to?
  A. Velocity
  B. Velocity gradient
  C. Area only
  D. Pressure
• Q3: Terminal velocity of a sphere depends on?
  A. Radius squared
  B. Density difference
  C. Viscosity
  D. All of the above

Answer Key

• Q1: B - Viscosity measures internal friction in fluids.
• Q2: B - Viscous force is proportional to velocity gradient dv/dy.
• Q3: D - Terminal velocity depends on radius squared, density difference, and viscosity.

Memory Aids

• Viscosity = fluid friction inside layers.
• Newton's law: F = -ηA dv/dy.
• Stoke's law: F_v = 6 πηrv.
• Terminal velocity formula: v_t = 2r²(ρ - σ)g/(9η).

Definition and Properties

• Viscosity is the fluid's internal friction opposing relative motion between adjacent layers.
• It is a measure of fluid friction or internal resistance to flow.

Difference between Viscosity and Solid Friction

• Viscosity acts inside liquids; solid friction acts between solid surfaces.
• Viscous force depends on contact area; solid friction does not.
• Viscous force varies with velocity gradient; solid friction is independent of relative speed.
• Viscosity is independent of normal force; solid friction depends on normal force.

Newton's Law of Viscosity

• Viscous force between two liquid layers: F = -ηA dv/dy
• η: coefficient of viscosity (Pa·s), A: contact area (m²), dv/dy: velocity gradient (s^-1)

Stoke's Law

• Retarding viscous force on sphere of radius r moving at velocity v: F_v = 6 πηrv

Terminal Velocity of a Sphere in Fluid

• At terminal velocity v_t, net force is zero: buoyant force + viscous force = weight
• Equilibrium: F_B + F_v = W
• Mathematically: 4/3 πr³σg + 6 πηrv_t = 4/3 πr³ρg
• Terminal velocity: v_t = 2r²(ρ - σ)g/(9η)

Solved Examples

• Ex 1: Calculate terminal velocity of a sphere with radius 0.001 m, density 2500 kg/m³, in a fluid of density 1000 kg/m³ and viscosity 0.001 Pa·s. Use g = 9.8 m/s².
• Solution: v_t = 2 × (0.001)² × (2500 - 1000) × 9.8/(9 × 0.001) = 3.27 m/s

Practice Set

• Q1: What does viscosity measure?
  A. Solid friction
  B. Fluid internal friction
  C. Buoyant force
  D. Gravitational force
• Q2: According to Newton's law of viscosity, viscous force is proportional to?
  A. Velocity
  B. Velocity gradient
  C. Area only
  D. Pressure
• Q3: Terminal velocity of a sphere depends on?
  A. Radius squared
  B. Density difference
  C. Viscosity
  D. All of the above

Answer Key

• Q1: B - Viscosity measures internal friction in fluids.
• Q2: B - Viscous force is proportional to velocity gradient dv/dy.
• Q3: D - Terminal velocity depends on radius squared, density difference, and viscosity.

Memory Aids

• Viscosity = fluid friction inside layers.
• Newton's law: F = -ηA dv/dy.
• Stoke's law: F_v = 6 πηrv.
• Terminal velocity formula: v_t = 2r²(ρ - σ)g/(9η).

Definition and Properties

• Viscosity is the fluid's internal friction opposing relative motion between adjacent layers.
• It is a measure of fluid friction or internal resistance to flow.

Difference between Viscosity and Solid Friction

• Viscosity acts inside liquids; solid friction acts between solid surfaces.
• Viscous force depends on contact area; solid friction does not.
• Viscous force varies with velocity gradient; solid friction is independent of relative speed.
• Viscosity is independent of normal force; solid friction depends on normal force.

Newton's Law of Viscosity

• Viscous force between two liquid layers: F = -ηA dv/dy
• η: coefficient of viscosity (Pa·s), A: contact area (m²), dv/dy: velocity gradient (s^-1)

Stoke's Law

• Retarding viscous force on sphere of radius r moving at velocity v: F_v = 6 πηrv

Terminal Velocity of a Sphere in Fluid

• At terminal velocity v_t, net force is zero: buoyant force + viscous force = weight
• Equilibrium: F_B + F_v = W
• Mathematically: 4/3 πr³σg + 6 πηrv_t = 4/3 πr³ρg
• Terminal velocity: v_t = 2r²(ρ - σ)g/(9η)

Solved Examples

• Ex 1: Calculate terminal velocity of a sphere with radius 0.001 m, density 2500 kg/m³, in a fluid of density 1000 kg/m³ and viscosity 0.001 Pa·s. Use g = 9.8 m/s².
• Solution: v_t = 2 × (0.001)² × (2500 - 1000) × 9.8/(9 × 0.001) = 3.27 m/s

Practice Set

• Q1: What does viscosity measure?
  A. Solid friction
  B. Fluid internal friction
  C. Buoyant force
  D. Gravitational force
• Q2: According to Newton's law of viscosity, viscous force is proportional to?
  A. Velocity
  B. Velocity gradient
  C. Area only
  D. Pressure
• Q3: Terminal velocity of a sphere depends on?
  A. Radius squared
  B. Density difference
  C. Viscosity
  D. All of the above

Answer Key

• Q1: B - Viscosity measures internal friction in fluids.
• Q2: B - Viscous force is proportional to velocity gradient dv/dy.
• Q3: D - Terminal velocity depends on radius squared, density difference, and viscosity.

Memory Aids

• Viscosity = fluid friction inside layers.
• Newton's law: F = -ηA dv/dy.
• Stoke's law: F_v = 6 πηrv.
• Terminal velocity formula: v_t = 2r²(ρ - σ)g/(9η).

Definition and Properties

• Viscosity is the fluid's internal friction opposing relative motion between adjacent layers.
• It is a measure of fluid friction or internal resistance to flow.

Difference between Viscosity and Solid Friction

• Viscosity acts inside liquids; solid friction acts between solid surfaces.
• Viscous force depends on contact area; solid friction does not.
• Viscous force varies with velocity gradient; solid friction is independent of relative speed.
• Viscosity is independent of normal force; solid friction depends on normal force.

Newton's Law of Viscosity

• Viscous force between two liquid layers: F = -ηA dv/dy
• η: coefficient of viscosity (Pa·s), A: contact area (m²), dv/dy: velocity gradient (s^-1)

Stoke's Law

• Retarding viscous force on sphere of radius r moving at velocity v: F_v = 6 π η r v

Terminal Velocity of a Sphere in Fluid

• At terminal velocity v_t, net force is zero: buoyant force + viscous force = weight
• Equilibrium: F_B + F_v = W
• Mathematically: 4/3 πr³σg + 6 πηrv_t = 4/3 πr³ρg
• Terminal velocity: v_t = 2r²(ρ - σ)g/(9η)

Solved Examples

• Ex 1: Calculate terminal velocity of a sphere with radius 0.001 m, density 2500 kg/m³, in a fluid of density 1000 kg/m³ and viscosity 0.001 Pa·s. Use g = 9.8 m/s².
• Solution: v_t = 2 × (0.001)² × (2500 - 1000) × 9.8/(9 × 0.001) = 3.27 m/s

Practice Set

• Q1: What does viscosity measure?
  A. Solid friction
  B. Fluid internal friction
  C. Buoyant force
  D. Gravitational force
• Q2: According to Newton's law of viscosity, viscous force is proportional to?
  A. Velocity
  B. Velocity gradient
  C. Area only
  D. Pressure
• Q3: Terminal velocity of a sphere depends on?
  A. Radius squared
  B. Density difference
  C. Viscosity
  D. All of the above

Answer Key

• Q1: B - Viscosity measures internal friction in fluids.
• Q2: B - Viscous force is proportional to velocity gradient dv/dy.
• Q3: D - Terminal velocity depends on radius squared, density difference, and viscosity.

Memory Aids

• Viscosity = fluid friction inside layers.
• Newton's law: F = -ηA dv/dy.
• Stoke's law: F_v = 6 πηrv.
• Terminal velocity formula: v_t = 2r²(ρ - σ)g/(9η).

Definition and Properties

• Viscosity is the fluid's internal friction opposing relative motion between adjacent layers.
• It is a measure of fluid friction or internal resistance to flow.

Difference between Viscosity and Solid Friction

• Viscosity acts inside liquids; solid friction acts between solid surfaces.
• Viscous force depends on contact area; solid friction does not.
• Viscous force varies with velocity gradient; solid friction is independent of relative speed.
• Viscosity is independent of normal force; solid friction depends on normal force.

Newton's Law of Viscosity

• Viscous force between two liquid layers: F = -ηA dv/dy
• η: coefficient of viscosity (Pa·s), A: contact area (m²), dv/dy: velocity gradient (s^-1)

Stoke's Law

• Retarding viscous force on sphere of radius r moving at velocity v: F_v = 6 πηrv

Terminal Velocity of a Sphere in Fluid

• At terminal velocity v_t, net force is zero: buoyant force + viscous force = weight
• Equilibrium: F_B + F_v = W
• Mathematically: 4/3 πr³σg + 6 πηrv_t = 4/3 πr³ρg
• Terminal velocity: v_t = 2r²(ρ - σ)g/(9η)

Solved Examples

• Ex 1: Calculate terminal velocity of a sphere with radius 0.001 m, density 2500 kg/m³, in a fluid of density 1000 kg/m³ and viscosity 0.001 Pa·s. Use g = 9.8 m/s².
• Solution: v_t = 2 × (0.001)² × (2500 - 1000) × 9.8/(9 × 0.001) = 3.27 m/s

Practice Set

• Q1: What does viscosity measure?
  A. Solid friction
  B. Fluid internal friction
  C. Buoyant force
  D. Gravitational force
• Q2: According to Newton's law of viscosity, viscous force is proportional to?
  A. Velocity
  B. Velocity gradient
  C. Area only
  D. Pressure
• Q3: Terminal velocity of a sphere depends on?
  A. Radius squared
  B. Density difference
  C. Viscosity
  D. All of the above

Answer Key

• Q1: B - Viscosity measures internal friction in fluids.
• Q2: B - Viscous force is proportional to velocity gradient dv/dy.
• Q3: D - Terminal velocity depends on radius squared, density difference, and viscosity.

Memory Aids

• Viscosity = fluid friction inside layers.
• Newton's law: F = -ηA dv/dy.
• Stoke's law: F_v = 6 πηrv.
• Terminal velocity formula: v_t = 2r²(ρ - σ)g/(9η).

Definition and Properties

• Viscosity is the fluid's internal friction opposing relative motion between adjacent layers.
• It is a measure of fluid friction or internal resistance to flow.

Difference between Viscosity and Solid Friction

• Viscosity acts inside liquids; solid friction acts between solid surfaces.
• Viscous force depends on contact area; solid friction does not.
• Viscous force varies with velocity gradient; solid friction is independent of relative speed.
• Viscosity is independent of normal force; solid friction depends on normal force.

Newton's Law of Viscosity

• Viscous force between two liquid layers: F = -ηA dv/dy
• η: coefficient of viscosity (Pa·s), A: contact area (m²), dv/dy: velocity gradient (s^-1)

Stoke's Law

• Retarding viscous force on sphere of radius r moving at velocity v: F_v = 6 π η r v

Terminal Velocity of a Sphere in Fluid

• At terminal velocity v_t, net force is zero: buoyant force + viscous force = weight
• Equilibrium: F_B + F_v = W
• Mathematically: 4/3 πr³σg + 6 πηrv_t = 4/3 πr³ρg
• Terminal velocity: v_t = 2r²(ρ - σ)g/(9η)

Solved Examples

• Ex 1: Calculate terminal velocity of a sphere with radius 0.001 m, density 2500 kg/m³, in a fluid of density 1000 kg/m³ and viscosity 0.001 Pa·s. Use g = 9.8 m/s².
• Solution: v_t = 2 × (0.001)² × (2500 - 1000) × 9.8/(9 × 0.001) = 3.27 m/s

Practice Set

• Q1: What does viscosity measure?
  A. Solid friction
  B. Fluid internal friction
  C. Buoyant force
  D. Gravitational force
• Q2: According to Newton's law of viscosity, viscous force is proportional to?
  A. Velocity
  B. Velocity gradient
  C. Area only
  D. Pressure
• Q3: Terminal velocity of a sphere depends on?
  A. Radius squared
  B. Density difference
  C. Viscosity
  D. All of the above

Answer Key

• Q1: B - Viscosity measures internal friction in fluids.
• Q2: B - Viscous force is proportional to velocity gradient dv/dy.
• Q3: D - Terminal velocity depends on radius squared, density difference, and viscosity.

Memory Aids

• Viscosity = fluid friction inside layers.
• Newton's law: F = -ηA dv/dy.
• Stoke's law: F_v = 6 πηrv.
• Terminal velocity formula: v_t = 2r²(ρ - σ)g/(9η).

Definition and Properties

• Viscosity is the fluid's internal friction opposing relative motion between adjacent layers.
• It is a measure of fluid friction or internal resistance to flow.

Difference between Viscosity and Solid Friction

• Viscosity acts inside liquids; solid friction acts between solid surfaces.
• Viscous force depends on contact area; solid friction does not.
• Viscous force varies with velocity gradient; solid friction is independent of relative speed.
• Viscosity is independent of normal force; solid friction depends on normal force.

Newton's Law of Viscosity

• Viscous force between two liquid layers: \( F = -\eta A \frac{dv}{dy} \)
• \( \eta \): coefficient of viscosity (Pa·s), \( A \): contact area (m²), \( \frac{dv}{dy} \): velocity gradient (s^-1)

Stoke's Law

• Retarding viscous force on sphere of radius \( r \) moving at velocity \( v \): \( F_v = 6 \pi \eta r v \)

Terminal Velocity of a Sphere in Fluid

• At terminal velocity \( v_t \), net force is zero: buoyant force + viscous force = weight
• Equilibrium: \( F_B + F_v = W \)
• Mathematically: \( \frac{4}{3} \pi r^3 \sigma g + 6 \pi \eta r v_t = \frac{4}{3} \pi r^3 \rho g \)
• Terminal velocity: \( v_t = \frac{2 r^2 (\rho - \sigma) g}{9 \eta} \)

Solved Examples

• Ex 1: Calculate terminal velocity of a sphere with radius 0.001 m, density 2500 kg/m³, in a fluid of density 1000 kg/m³ and viscosity 0.001 Pa·s. Use \( g = 9.8 \text{ m/s}^2 \).
• Solution: \( v_t = \frac{2 \times (0.001)^2 \times (2500 - 1000) \times 9.8}{9 \times 0.001} = 3.27 \text{ m/s} \)

Practice Set

• Q1: What does viscosity measure?
  A. Solid friction
  B. Fluid internal friction
  C. Buoyant force
  D. Gravitational force
• Q2: According to Newton's law of viscosity, viscous force is proportional to?
  A. Velocity
  B. Velocity gradient
  C. Area only
  D. Pressure
• Q3: Terminal velocity of a sphere depends on?
  A. Radius squared
  B. Density difference
  C. Viscosity
  D. All of the above

Answer Key

• Q1: B - Viscosity measures internal friction in fluids.
• Q2: B - Viscous force is proportional to velocity gradient \( \frac{dv}{dy} \).
• Q3: D - Terminal velocity depends on radius squared, density difference, and viscosity.

Memory Aids

• Viscosity = fluid friction inside layers.
• Newton's law: \( F = -\eta A \frac{dv}{dy} \).
• Stoke's law: \( F_v = 6 \pi \eta r v \).
• Terminal velocity formula: \( v_t = \frac{2 r^2 (\rho - \sigma) g}{9 \eta} \).

Poiseuille Formula

Volume Flow and Resistance

• Volume flow rate through capillary: V = πPr⁴/(8ηl) = P/R
• Liquid resistance: R = 8ηl/(πr⁴)
• For tubes in series: P = P₁ + P₂, flow rate V same, resistance R_s = R₁ + R₂
• For tubes in parallel: R_p = R₁R₂/(R₁ + R₂), pressure difference same, flow rates add V = V₁ + V₂

Solved Examples

• Ex 1: Calculate resistance of a tube with length 0.5 m, radius 0.001 m, viscosity 0.001 Pa·s.
• Solution: R = 8 × 0.001 × 0.5/(π × (0.001)⁴) = 1.27 × 10¹² Pa·s/m³

Practice Set

• Q1: Poiseuille's formula relates volume flow rate to?
  A. Pressure difference
  B. Tube radius
  C. Viscosity
  D. All of the above
• Q2: Equivalent resistance for two tubes in series is?
  A. Product of resistances
  B. Sum of resistances
  C. Reciprocal of sum
  D. Zero
• Q3: For two tubes in parallel, equivalent resistance is?
  A. Sum of resistances
  B. Product over sum
  C. Difference of resistances
  D. Same as one tube

Answer Key

• Q1: D - Flow rate depends on pressure difference, radius, and viscosity.
• Q2: B - Series resistances add up.
• Q3: B - Parallel resistance is product over sum.

Memory Aids

• Poiseuille: V ∝ r⁴ (strong radius effect).
• Resistance series: add resistances.
• Resistance parallel: use product over sum.

Volume Flow and Resistance

• Volume flow rate through capillary: V = πPr⁴/(8ηl) = P/R
• Liquid resistance: R = 8ηl/(πr⁴)
• For tubes in series: P = P₁ + P₂, flow rate V same, resistance R_s = R₁ + R₂
• For tubes in parallel: R_p = R₁R₂/(R₁ + R₂), pressure difference same, flow rates add V = V₁ + V₂

Solved Examples

• Ex 1: Calculate resistance of a tube with length 0.5 m, radius 0.001 m, viscosity 0.001 Pa·s.
• Solution: R = 8 × 0.001 × 0.5/(π × (0.001)⁴) = 1.27 × 10¹² Pa·s/m³

Practice Set

• Q1: Poiseuille's formula relates volume flow rate to?
  A. Pressure difference
  B. Tube radius
  C. Viscosity
  D. All of the above
• Q2: Equivalent resistance for two tubes in series is?
  A. Product of resistances
  B. Sum of resistances
  C. Reciprocal of sum
  D. Zero
• Q3: For two tubes in parallel, equivalent resistance is?
  A. Sum of resistances
  B. Product over sum
  C. Difference of resistances
  D. Same as one tube

Answer Key

• Q1: D - Flow rate depends on pressure difference, radius, and viscosity.
• Q2: B - Series resistances add up.
• Q3: B - Parallel resistance is product over sum.

Memory Aids

• Poiseuille: V ∝ r⁴ (strong radius effect).
• Resistance series: add resistances.
• Resistance parallel: use product over sum.

Volume Flow and Resistance

• Volume flow rate through capillary: V = πPr⁴/(8ηl) = P/R
• Liquid resistance: R = 8ηl/(πr⁴)
• For tubes in series: P = P₁ + P₂, flow rate V same, resistance R_s = R₁ + R₂
• For tubes in parallel: R_p = R₁R₂/(R₁ + R₂), pressure difference same, flow rates add V = V₁ + V₂

Solved Examples

• Ex 1: Calculate resistance of a tube with length 0.5 m, radius 0.001 m, viscosity 0.001 Pa·s.
• Solution: R = 8 × 0.001 × 0.5/(π × (0.001)⁴) = 1.27 × 10¹² Pa·s/m³

Practice Set

• Q1: Poiseuille's formula relates volume flow rate to?
  A. Pressure difference
  B. Tube radius
  C. Viscosity
  D. All of the above
• Q2: Equivalent resistance for two tubes in series is?
  A. Product of resistances
  B. Sum of resistances
  C. Reciprocal of sum
  D. Zero
• Q3: For two tubes in parallel, equivalent resistance is?
  A. Sum of resistances
  B. Product over sum
  C. Difference of resistances
  D. Same as one tube

Answer Key

• Q1: D - Flow rate depends on pressure difference, radius, and viscosity.
• Q2: B - Series resistances add up.
• Q3: B - Parallel resistance is product over sum.

Memory Aids

• Poiseuille: V ∝ r⁴ (strong radius effect).
• Resistance series: add resistances.
• Resistance parallel: use product over sum.

Volume Flow and Resistance

• Volume flow rate through capillary: V = πPr⁴/(8ηl) = P/R
• Liquid resistance: R = 8ηl/(πr⁴)
• For tubes in series: P = P₁ + P₂, flow rate V same, resistance R_s = R₁ + R₂
• For tubes in parallel: R_p = R₁R₂/(R₁ + R₂), pressure difference same, flow rates add V = V₁ + V₂

Solved Examples

• Ex 1: Calculate resistance of a tube with length 0.5 m, radius 0.001 m, viscosity 0.001 Pa·s.
• Solution: R = 8 × 0.001 × 0.5/(π × (0.001)⁴) = 1.27 × 10¹² Pa·s/m³

Practice Set

• Q1: Poiseuille's formula relates volume flow rate to?
  A. Pressure difference
  B. Tube radius
  C. Viscosity
  D. All of the above
• Q2: Equivalent resistance for two tubes in series is?
  A. Product of resistances
  B. Sum of resistances
  C. Reciprocal of sum
  D. Zero
• Q3: For two tubes in parallel, equivalent resistance is?
  A. Sum of resistances
  B. Product over sum
  C. Difference of resistances
  D. Same as one tube

Answer Key

• Q1: D - Flow rate depends on pressure difference, radius, and viscosity.
• Q2: B - Series resistances add up.
• Q3: B - Parallel resistance is product over sum.

Memory Aids

• Poiseuille: V ∝ r⁴ (strong radius effect).
• Resistance series: add resistances.
• Resistance parallel: use product over sum.

Volume Flow and Resistance

• Volume flow rate through capillary: V = πPr⁴/(8ηl) = P/R
• Liquid resistance: R = 8ηl/(πr⁴)
• For tubes in series: P = P₁ + P₂, flow rate V same, resistance R_s = R₁ + R₂
• For tubes in parallel: R_p = R₁R₂/(R₁ + R₂), pressure difference same, flow rates add V = V₁ + V₂

Solved Examples

• Ex 1: Calculate resistance of a tube with length 0.5 m, radius 0.001 m, viscosity 0.001 Pa·s.
• Solution: R = 8 × 0.001 × 0.5/(π × (0.001)⁴) = 1.27 × 10¹² Pa·s/m³

Practice Set

• Q1: Poiseuille's formula relates volume flow rate to?
  A. Pressure difference
  B. Tube radius
  C. Viscosity
  D. All of the above
• Q2: Equivalent resistance for two tubes in series is?
  A. Product of resistances
  B. Sum of resistances
  C. Reciprocal of sum
  D. Zero
• Q3: For two tubes in parallel, equivalent resistance is?
  A. Sum of resistances
  B. Product over sum
  C. Difference of resistances
  D. Same as one tube

Answer Key

• Q1: D - Flow rate depends on pressure difference, radius, and viscosity.
• Q2: B - Series resistances add up.
• Q3: B - Parallel resistance is product over sum.

Memory Aids

• Poiseuille: V ∝ r⁴ (strong radius effect).
• Resistance series: add resistances.
• Resistance parallel: use product over sum.

Volume Flow and Resistance

• Volume flow rate through capillary: V = πPr⁴/(8ηl) = P/R
• Liquid resistance: R = 8ηl/(πr⁴)
• For tubes in series: P = P₁ + P₂, flow rate V same, resistance R_s = R₁ + R₂
• For tubes in parallel: R_p = R₁R₂/(R₁ + R₂), pressure difference same, flow rates add V = V₁ + V₂

Solved Examples

• Ex 1: Calculate resistance of a tube with length 0.5 m, radius 0.001 m, viscosity 0.001 Pa·s.
• Solution: R = 8 × 0.001 × 0.5/(π × (0.001)⁴) = 1.27 × 10¹² Pa·s/m³

Practice Set

• Q1: Poiseuille's formula relates volume flow rate to?
  A. Pressure difference
  B. Tube radius
  C. Viscosity
  D. All of the above
• Q2: Equivalent resistance for two tubes in series is?
  A. Product of resistances
  B. Sum of resistances
  C. Reciprocal of sum
  D. Zero
• Q3: For two tubes in parallel, equivalent resistance is?
  A. Sum of resistances
  B. Product over sum
  C. Difference of resistances
  D. Same as one tube

Answer Key

• Q1: D - Flow rate depends on pressure difference, radius, and viscosity.
• Q2: B - Series resistances add up.
• Q3: B - Parallel resistance is product over sum.

Memory Aids

• Poiseuille: V ∝ r⁴ (strong radius effect).
• Resistance series: add resistances.
• Resistance parallel: use product over sum.

Volume Flow and Resistance

• Volume flow rate through capillary: \( V = \frac{\pi P r^4}{8 \eta l} = \frac{P}{R} \)
• Liquid resistance: \( R = \frac{8 \eta l}{\pi r^4} \)
• For tubes in series: \( P = P_1 + P_2 \), flow rate \( V \) same, resistance \( R_s = R_1 + R_2 \)
• For tubes in parallel: \( R_p = \frac{R_1 R_2}{R_1 + R_2} \), pressure difference same, flow rates add \( V = V_1 + V_2 \)

Solved Examples

• Ex 1: Calculate resistance of a tube with length 0.5 m, radius 0.001 m, viscosity 0.001 Pa·s.
• Solution: \( R = \frac{8 \times 0.001 \times 0.5}{\pi \times (0.001)^4} = 1.27 \times 10^{12} \text{ Pa·s/m}^3 \)

Practice Set

• Q1: Poiseuille's formula relates volume flow rate to?
  A. Pressure difference
  B. Tube radius
  C. Viscosity
  D. All of the above
• Q2: Equivalent resistance for two tubes in series is?
  A. Product of resistances
  B. Sum of resistances
  C. Reciprocal of sum
  D. Zero
• Q3: For two tubes in parallel, equivalent resistance is?
  A. Sum of resistances
  B. Product over sum
  C. Difference of resistances
  D. Same as one tube

Answer Key

• Q1: D - Flow rate depends on pressure difference, radius, and viscosity.
• Q2: B - Series resistances add up.
• Q3: B - Parallel resistance is product over sum.

Memory Aids

• Poiseuille: \( V \propto r^4 \) (strong radius effect).
• Resistance series: add resistances.
• Resistance parallel: use product over sum.

Surface Tension

Definition and Explanation

• Surface tension is the force per unit length acting along the surface of a liquid, making it behave like a stretched elastic membrane.
• It causes liquids to minimize their surface area, forming shapes like spherical drops.
• Example: Water drops on a lotus leaf are rounded due to surface tension; soap bubbles have two surfaces with pressure differences maintaining stability.
• Formula: S = F/L, where F is force on either side of an imaginary line on the surface and L is the length of that line.
• SI unit: Newton per meter (N/m), Dimensions: [S] = [MT^-2]
• Factors affecting surface tension:
  • Temperature: Surface tension decreases with increasing temperature.
  • Impurities: Highly soluble impurities increase surface tension; sparingly soluble impurities decrease it.

Surface Energy and Relation to Surface Tension

• Surface energy (E) is the extra potential energy stored in surface molecules due to inward forces.
• SI unit: Joule (J), Dimensions: [E] = [ML²T^-2]
• Relation: Surface tension equals the increase in surface energy per unit area, S = W/ΔA, where W is work done and ΔA is increase in surface area.
• For a soap film with length l and displacement Δx, F = S × 2l and work done W = F × Δx = S × 2l Δx.

Solved Examples

• Ex 1: Calculate surface tension if a force of 0.2 N acts on a 0.1 m length of liquid surface.
  • Use S = F/L = 0.2/0.1 = 2 N/m.

Practice Set

• Q1: Surface tension decreases with:
  • A) Increase in temperature
  • B) Decrease in temperature
  • C) Addition of sparingly soluble impurities
  • D) Increase in pressure
• Q2: Surface tension is defined as force per unit:
  • A) Area
  • B) Length
  • C) Volume
  • D) Mass
• Q3: If force on a 0.05 m length of liquid surface is 0.1 N, surface tension is:
  • A) 0.5 N/m
  • B) 2 N/m
  • C) 5 N/m
  • D) 0.2 N/m

Answer Key

• Q1: A - Surface tension decreases with rise in temperature.
• Q2: B - Surface tension is force per unit length.
• Q3: C - S = 0.1/0.05 = 2 N/m.

Memory Aids

• Surface tension S = F/L
• Decreases with temperature rise
• Increases with highly soluble impurities
• Surface energy E = S × ΔA

Definition and Explanation

• Surface tension is the force per unit length acting along the surface of a liquid, making it behave like a stretched elastic membrane.
• It causes liquids to minimize their surface area, forming shapes like spherical drops.
• Example: Water drops on a lotus leaf are rounded due to surface tension; soap bubbles have two surfaces with pressure differences maintaining stability.
• Formula: S = F/L, where F is force on either side of an imaginary line on the surface and L is the length of that line.
• SI unit: Newton per meter (N/m), Dimensions: [S] = [MT^-2]
• Factors affecting surface tension:
  • Temperature: Surface tension decreases with increasing temperature.
  • Impurities: Highly soluble impurities increase surface tension; sparingly soluble impurities decrease it.

Surface Energy and Relation to Surface Tension

• Surface energy (E) is the extra potential energy stored in surface molecules due to inward forces.
• SI unit: Joule (J), Dimensions: [E] = [ML²T^-2]
• Relation: Surface tension equals the increase in surface energy per unit area, S = W/ΔA, where W is work done and ΔA is increase in surface area.
• For a soap film with length l and displacement Δx, F = S × 2l and work done W = F × Δx = S × 2l Δx.

Solved Examples

• Ex 1: Calculate surface tension if a force of 0.2 N acts on a 0.1 m length of liquid surface.
  • Use S = F/L = 0.2/0.1 = 2 N/m.

Practice Set

• Q1: Surface tension decreases with:
  • A) Increase in temperature
  • B) Decrease in temperature
  • C) Addition of sparingly soluble impurities
  • D) Increase in pressure
• Q2: Surface tension is defined as force per unit:
  • A) Area
  • B) Length
  • C) Volume
  • D) Mass
• Q3: If force on a 0.05 m length of liquid surface is 0.1 N, surface tension is:
  • A) 0.5 N/m
  • B) 2 N/m
  • C) 5 N/m
  • D) 0.2 N/m

Answer Key

• Q1: A - Surface tension decreases with rise in temperature.
• Q2: B - Surface tension is force per unit length.
• Q3: C - S = 0.1/0.05 = 2 N/m.

Memory Aids

• Surface tension S = F/L
• Decreases with temperature rise
• Increases with highly soluble impurities
• Surface energy E = S × ΔA

Definition and Explanation

• Surface tension is the force per unit length acting along the surface of a liquid, making it behave like a stretched elastic membrane.
• It causes liquids to minimize their surface area, forming shapes like spherical drops.
• Example: Water drops on a lotus leaf are rounded due to surface tension; soap bubbles have two surfaces with pressure differences maintaining stability.
• Formula: S = F/L, where F is force on either side of an imaginary line on the surface and L is the length of that line.
• SI unit: Newton per meter (N/m), Dimensions: [S] = [MT^-2]
• Factors affecting surface tension:
  • Temperature: Surface tension decreases with increasing temperature.
  • Impurities: Highly soluble impurities increase surface tension; sparingly soluble impurities decrease it.

Surface Energy and Relation to Surface Tension

• Surface energy (E) is the extra potential energy stored in surface molecules due to inward forces.
• SI unit: Joule (J), Dimensions: [E] = [ML²T^-2]
• Relation: Surface tension equals the increase in surface energy per unit area, S = W/ΔA, where W is work done and ΔA is increase in surface area.
• For a soap film with length l and displacement Δx, F = S × 2l and work done W = F × Δx = S × 2l Δx.

Solved Examples

• Ex 1: Calculate surface tension if a force of 0.2 N acts on a 0.1 m length of liquid surface.
  • Use S = F/L = 0.2/0.1 = 2 N/m.

Practice Set

• Q1: Surface tension decreases with:
  • A) Increase in temperature
  • B) Decrease in temperature
  • C) Addition of sparingly soluble impurities
  • D) Increase in pressure
• Q2: Surface tension is defined as force per unit:
  • A) Area
  • B) Length
  • C) Volume
  • D) Mass
• Q3: If force on a 0.05 m length of liquid surface is 0.1 N, surface tension is:
  • A) 0.5 N/m
  • B) 2 N/m
  • C) 5 N/m
  • D) 0.2 N/m

Answer Key

• Q1: A - Surface tension decreases with rise in temperature.
• Q2: B - Surface tension is force per unit length.
• Q3: C - S = 0.1/0.05 = 2 N/m.

Memory Aids

• Surface tension S = F/L
• Decreases with temperature rise
• Increases with highly soluble impurities
• Surface energy E = S × ΔA

Definition and Explanation

• Surface tension is the force per unit length acting along the surface of a liquid, making it behave like a stretched elastic membrane.
• It causes liquids to minimize their surface area, forming shapes like spherical drops.
• Example: Water drops on a lotus leaf are rounded due to surface tension; soap bubbles have two surfaces with pressure differences maintaining stability.
• Formula: S = F/L, where F is force on either side of an imaginary line on the surface and L is the length of that line.
• SI unit: Newton per meter (N/m), Dimensions: [S] = [MT^-2]
• Factors affecting surface tension:
  • Temperature: Surface tension decreases with increasing temperature.
  • Impurities: Highly soluble impurities increase surface tension; sparingly soluble impurities decrease it.

Surface Energy and Relation to Surface Tension

• Surface energy (E) is the extra potential energy stored in surface molecules due to inward forces.
• SI unit: Joule (J), Dimensions: [E] = [ML²T^-2]
• Relation: Surface tension equals the increase in surface energy per unit area, S = W/ΔA, where W is work done and ΔA is increase in surface area.
• For a soap film with length l and displacement Δx, F = S × 2l and work done W = F × Δx = S × 2l Δx.

Solved Examples

• Ex 1: Calculate surface tension if a force of 0.2 N acts on a 0.1 m length of liquid surface.
  • Use S = F/L = 0.2/0.1 = 2 N/m.

Practice Set

• Q1: Surface tension decreases with:
  • A) Increase in temperature
  • B) Decrease in temperature
  • C) Addition of sparingly soluble impurities
  • D) Increase in pressure
• Q2: Surface tension is defined as force per unit:
  • A) Area
  • B) Length
  • C) Volume
  • D) Mass
• Q3: If force on a 0.05 m length of liquid surface is 0.1 N, surface tension is:
  • A) 0.5 N/m
  • B) 2 N/m
  • C) 5 N/m
  • D) 0.2 N/m

Answer Key

• Q1: A - Surface tension decreases with rise in temperature.
• Q2: B - Surface tension is force per unit length.
• Q3: C - S = 0.1/0.05 = 2 N/m.

Memory Aids

• Surface tension S = F/L
• Decreases with temperature rise
• Increases with highly soluble impurities
• Surface energy E = S × ΔA

Definition and Explanation

• Surface tension is the force per unit length acting along the surface of a liquid, making it behave like a stretched elastic membrane.
• It causes liquids to minimize their surface area, forming shapes like spherical drops.
• Example: Water drops on a lotus leaf are rounded due to surface tension; soap bubbles have two surfaces with pressure differences maintaining stability.
• Formula: S = F/L, where F is force on either side of an imaginary line on the surface and L is the length of that line.
• SI unit: Newton per meter (N/m), Dimensions: [S] = [MT^-2]
• Factors affecting surface tension:
  • Temperature: Surface tension decreases with increasing temperature.
  • Impurities: Highly soluble impurities increase surface tension; sparingly soluble impurities decrease it.

Surface Energy and Relation to Surface Tension

• Surface energy (E) is the extra potential energy stored in surface molecules due to inward forces.
• SI unit: Joule (J), Dimensions: [E] = [ML²T^-2]
• Relation: Surface tension equals the increase in surface energy per unit area, S = W/ΔA, where W is work done and ΔA is increase in surface area.
• For a soap film with length l and displacement Δx, F = S × 2l and work done W = F × Δx = S × 2l Δx.

Solved Examples

• Ex 1: Calculate surface tension if a force of 0.2 N acts on a 0.1 m length of liquid surface.
  • Use S = F/L = 0.2/0.1 = 2 N/m.

Practice Set

• Q1: Surface tension decreases with:
  • A) Increase in temperature
  • B) Decrease in temperature
  • C) Addition of sparingly soluble impurities
  • D) Increase in pressure
• Q2: Surface tension is defined as force per unit:
  • A) Area
  • B) Length
  • C) Volume
  • D) Mass
• Q3: If force on a 0.05 m length of liquid surface is 0.1 N, surface tension is:
  • A) 0.5 N/m
  • B) 2 N/m
  • C) 5 N/m
  • D) 0.2 N/m

Answer Key

• Q1: A - Surface tension decreases with rise in temperature.
• Q2: B - Surface tension is force per unit length.
• Q3: C - S = 0.1/0.05 = 2 N/m.

Memory Aids

• Surface tension S = F/L
• Decreases with temperature rise
• Increases with highly soluble impurities
• Surface energy E = S × ΔA

Definition and Explanation

• Surface tension is the force per unit length acting along the surface of a liquid, making it behave like a stretched elastic membrane.
• It causes liquids to minimize their surface area, forming shapes like spherical drops.
• Example: Water drops on a lotus leaf are rounded due to surface tension; soap bubbles have two surfaces with pressure differences maintaining stability.
• Formula: S = F/L, where F is force on either side of an imaginary line on the surface and L is the length of that line.
• SI unit: Newton per meter (N/m), Dimensions: [S] = [MT^-2]
• Factors affecting surface tension:
  • Temperature: Surface tension decreases with increasing temperature.
  • Impurities: Highly soluble impurities increase surface tension; sparingly soluble impurities decrease it.

Surface Energy and Relation to Surface Tension

• Surface energy (E) is the extra potential energy stored in surface molecules due to inward forces.
• SI unit: Joule (J), Dimensions: [E] = [ML²T^-2]
• Relation: Surface tension equals the increase in surface energy per unit area, S = W/ΔA, where W is work done and ΔA is increase in surface area.
• For a soap film with length l and displacement Δx, F = S × 2l and work done W = F × Δx = S × 2l Δx.

Solved Examples

• Ex 1: Calculate surface tension if a force of 0.2 N acts on a 0.1 m length of liquid surface.
  • Use S = F/L = 0.2/0.1 = 2 N/m.

Practice Set

• Q1: Surface tension decreases with:
  • A) Increase in temperature
  • B) Decrease in temperature
  • C) Addition of sparingly soluble impurities
  • D) Increase in pressure
• Q2: Surface tension is defined as force per unit:
  • A) Area
  • B) Length
  • C) Volume
  • D) Mass
• Q3: If force on a 0.05 m length of liquid surface is 0.1 N, surface tension is:
  • A) 0.5 N/m
  • B) 2 N/m
  • C) 5 N/m
  • D) 0.2 N/m

Answer Key

• Q1: A - Surface tension decreases with rise in temperature.
• Q2: B - Surface tension is force per unit length.
• Q3: C - S = 0.1/0.05 = 2 N/m.

Memory Aids

• Surface tension S = F/L
• Decreases with temperature rise
• Increases with highly soluble impurities
• Surface energy E = S × ΔA

Definition and Explanation

• Surface tension is the force per unit length acting along the surface of a liquid, making it behave like a stretched elastic membrane.
• It causes liquids to minimize their surface area, forming shapes like spherical drops.
• Example: Water drops on a lotus leaf are rounded due to surface tension; soap bubbles have two surfaces with pressure differences maintaining stability.
• Formula: S = F/L, where F is force on either side of an imaginary line on the surface and L is the length of that line.
• SI unit: Newton per meter (N/m), Dimensions: [S] = [MT^-2]
• Factors affecting surface tension:
  • Temperature: Surface tension decreases with increasing temperature.
  • Impurities: Highly soluble impurities increase surface tension; sparingly soluble impurities decrease it.

Surface Energy and Relation to Surface Tension

• Surface energy (E) is the extra potential energy stored in surface molecules due to inward forces.
• SI unit: Joule (J), Dimensions: [E] = [ML²T^-2]
• Relation: Surface tension equals the increase in surface energy per unit area, S = W/ΔA, where W is work done and ΔA is increase in surface area.
• For a soap film with length l and displacement Δx, F = S × 2l and work done W = F × Δx = S × 2l Δx.

Solved Examples

• Ex 1: Calculate surface tension if a force of 0.2 N acts on a 0.1 m length of liquid surface.
  • Use S = F/L = 0.2/0.1 = 2 N/m.

Practice Set

• Q1: Surface tension decreases with:
  • A) Increase in temperature
  • B) Decrease in temperature
  • C) Addition of sparingly soluble impurities
  • D) Increase in pressure
• Q2: Surface tension is defined as force per unit:
  • A) Area
  • B) Length
  • C) Volume
  • D) Mass
• Q3: If force on a 0.05 m length of liquid surface is 0.1 N, surface tension is:
  • A) 0.5 N/m
  • B) 2 N/m
  • C) 5 N/m
  • D) 0.2 N/m

Answer Key

• Q1: A - Surface tension decreases with rise in temperature.
• Q2: B - Surface tension is force per unit length.
• Q3: C - S = 0.1/0.05 = 2 N/m.

Memory Aids

• Surface tension S = F/L
• Decreases with temperature rise
• Increases with highly soluble impurities
• Surface energy E = S × ΔA

Definition and Explanation

• Surface tension is the force per unit length acting along the surface of a liquid, making it behave like a stretched elastic membrane.
• It causes liquids to minimize their surface area, forming shapes like spherical drops.
• Example: Water drops on a lotus leaf are rounded due to surface tension; soap bubbles have two surfaces with pressure differences maintaining stability.
• Formula: S = F/L, where F is force on either side of an imaginary line on the surface and L is the length of that line.
• SI unit: Newton per meter (N/m), Dimensions: [S] = [MT^-2]
• Factors affecting surface tension:
  • Temperature: Surface tension decreases with increasing temperature.
  • Impurities: Highly soluble impurities increase surface tension; sparingly soluble impurities decrease it.

Surface Energy and Relation to Surface Tension

• Surface energy (E) is the extra potential energy stored in surface molecules due to inward forces.
• SI unit: Joule (J), Dimensions: [E] = [ML²T^-2]
• Relation: Surface tension equals the increase in surface energy per unit area, S = W/ΔA, where W is work done and ΔA is increase in surface area.
• For a soap film with length l and displacement Δx, F = S × 2l and work done W = F × Δx = S × 2l Δx.

Solved Examples

• Ex 1: Calculate surface tension if a force of 0.2 N acts on a 0.1 m length of liquid surface.
  • Use S = F/L = 0.2/0.1 = 2 N/m.

Practice Set

• Q1: Surface tension decreases with:
  • A) Increase in temperature
  • B) Decrease in temperature
  • C) Addition of sparingly soluble impurities
  • D) Increase in pressure
• Q2: Surface tension is defined as force per unit:
  • A) Area
  • B) Length
  • C) Volume
  • D) Mass
• Q3: If force on a 0.05 m length of liquid surface is 0.1 N, surface tension is:
  • A) 0.5 N/m
  • B) 2 N/m
  • C) 5 N/m
  • D) 0.2 N/m

Answer Key

• Q1: A - Surface tension decreases with rise in temperature.
• Q2: B - Surface tension is force per unit length.
• Q3: C - S = 0.1/0.05 = 2 N/m.

Memory Aids

• Surface tension S = F/L
• Decreases with temperature rise
• Increases with highly soluble impurities
• Surface energy E = S × ΔA

Definition and Explanation

• Surface tension is the force per unit length acting along the surface of a liquid, making it behave like a stretched elastic membrane.
• It causes liquids to minimize their surface area, forming shapes like spherical drops.
• Example: Water drops on a lotus leaf are rounded due to surface tension; soap bubbles have two surfaces with pressure differences maintaining stability.
• Formula: S = F/L, where F is force on either side of an imaginary line on the surface and L is the length of that line.
• SI unit: Newton per meter (N/m), Dimensions: [S] = [MT^-2]
• Factors affecting surface tension:
  • Temperature: Surface tension decreases with increasing temperature.
  • Impurities: Highly soluble impurities increase surface tension; sparingly soluble impurities decrease it.

Surface Energy and Relation to Surface Tension

• Surface energy (E) is the extra potential energy stored in surface molecules due to inward forces.
• SI unit: Joule (J), Dimensions: [E] = [ML²T^-2]
• Relation: Surface tension equals the increase in surface energy per unit area, S = W/ΔA, where W is work done and ΔA is increase in surface area.
• For a soap film with length l and displacement Δx, F = S × 2l and work done W = F × Δx = S × 2l Δx.

Solved Examples

• Ex 1: Calculate surface tension if a force of 0.2 N acts on a 0.1 m length of liquid surface.
  • Use S = F/L = 0.2/0.1 = 2 N/m.

Practice Set

• Q1: Surface tension decreases with:
  • A) Increase in temperature
  • B) Decrease in temperature
  • C) Addition of sparingly soluble impurities
  • D) Increase in pressure
• Q2: Surface tension is defined as force per unit:
  • A) Area
  • B) Length
  • C) Volume
  • D) Mass
• Q3: If force on a 0.05 m length of liquid surface is 0.1 N, surface tension is:
  • A) 0.5 N/m
  • B) 2 N/m
  • C) 5 N/m
  • D) 0.2 N/m

Answer Key

• Q1: A - Surface tension decreases with rise in temperature.
• Q2: B - Surface tension is force per unit length.
• Q3: C - S = 0.1/0.05 = 2 N/m.

Memory Aids

• Surface tension S = F/L
• Decreases with temperature rise
• Increases with highly soluble impurities
• Surface energy E = S × ΔA

Definition and Explanation

• Surface tension is the force per unit length acting along the surface of a liquid, making it behave like a stretched elastic membrane.
• It causes liquids to minimize their surface area, forming shapes like spherical drops.
• Example: Water drops on a lotus leaf are rounded due to surface tension; soap bubbles have two surfaces with pressure differences maintaining stability.
• Formula: S = F/L, where F is force on either side of an imaginary line on the surface and L is the length of that line.
• SI unit: Newton per meter (N/m), Dimensions: [S] = [MT^-2]
• Factors affecting surface tension:
  • Temperature: Surface tension decreases with increasing temperature.
  • Impurities: Highly soluble impurities increase surface tension; sparingly soluble impurities decrease it.

Surface Energy and Relation to Surface Tension

• Surface energy (E) is the extra potential energy stored in surface molecules due to inward forces.
• SI unit: Joule (J), Dimensions: [E] = [ML²T^-2]
• Relation: Surface tension equals the increase in surface energy per unit area, S = W/ΔA, where W is work done and ΔA is increase in surface area.
• For a soap film with length l and displacement Δx, F = S × 2l and work done W = F × Δx = S × 2l Δx.

Solved Examples

• Ex 1: Calculate surface tension if a force of 0.2 N acts on a 0.1 m length of liquid surface.
  • Use S = F/L = 0.2/0.1 = 2 N/m.

Practice Set

• Q1: Surface tension decreases with:
  • A) Increase in temperature
  • B) Decrease in temperature
  • C) Addition of sparingly soluble impurities
  • D) Increase in pressure
• Q2: Surface tension is defined as force per unit:
  • A) Area
  • B) Length
  • C) Volume
  • D) Mass
• Q3: If force on a 0.05 m length of liquid surface is 0.1 N, surface tension is:
  • A) 0.5 N/m
  • B) 2 N/m
  • C) 5 N/m
  • D) 0.2 N/m

Answer Key

• Q1: A - Surface tension decreases with rise in temperature.
• Q2: B - Surface tension is force per unit length.
• Q3: C - S = 0.1/0.05 = 2 N/m.

Memory Aids

• Surface tension S = F/L
• Decreases with temperature rise
• Increases with highly soluble impurities
• Surface energy E = S × ΔA

Definition and Explanation

• Surface tension is the force per unit length acting along the surface of a liquid, making it behave like a stretched elastic membrane.
• It causes liquids to minimize their surface area, forming shapes like spherical drops.
• Example: Water drops on a lotus leaf are rounded due to surface tension; soap bubbles have two surfaces with pressure differences maintaining stability.
• Formula: S = F/L, where F is force on either side of an imaginary line on the surface and L is the length of that line.
• SI unit: Newton per meter (N/m), Dimensions: [S] = [MT^-2]
• Factors affecting surface tension:
  • Temperature: Surface tension decreases with increasing temperature.
  • Impurities: Highly soluble impurities increase surface tension; sparingly soluble impurities decrease it.

Surface Energy and Relation to Surface Tension

• Surface energy (E) is the extra potential energy stored in surface molecules due to inward forces.
• SI unit: Joule (J), Dimensions: [E] = [ML²T^-2]
• Relation: Surface tension equals the increase in surface energy per unit area, S = W/ΔA, where W is work done and ΔA is increase in surface area.
• For a soap film with length l and displacement Δx, F = S × 2l and work done W = F × Δx = S × 2l Δx.

Solved Examples

• Ex 1: Calculate surface tension if a force of 0.2 N acts on a 0.1 m length of liquid surface.
  • Use S = F/L = 0.2/0.1 = 2 N/m.

Practice Set

• Q1: Surface tension decreases with:
  • A) Increase in temperature
  • B) Decrease in temperature
  • C) Addition of sparingly soluble impurities
  • D) Increase in pressure
• Q2: Surface tension is defined as force per unit:
  • A) Area
  • B) Length
  • C) Volume
  • D) Mass
• Q3: If force on a 0.05 m length of liquid surface is 0.1 N, surface tension is:
  • A) 0.5 N/m
  • B) 2 N/m
  • C) 5 N/m
  • D) 0.2 N/m

Answer Key

• Q1: A - Surface tension decreases with rise in temperature.
• Q2: B - Surface tension is force per unit length.
• Q3: C - S = 0.1/0.05 = 2 N/m.

Memory Aids

• Surface tension S = F/L
• Decreases with temperature rise
• Increases with highly soluble impurities
• Surface energy E = S × ΔA

Definition and Explanation

• Surface tension is the force per unit length acting along the surface of a liquid, making it behave like a stretched elastic membrane.
• It causes liquids to minimize their surface area, forming shapes like spherical drops.
• Example: Water drops on a lotus leaf are rounded due to surface tension; soap bubbles have two surfaces with pressure differences maintaining stability.
• Formula: S = F/L, where F is force on either side of an imaginary line on the surface and L is the length of that line.
• SI unit: Newton per meter (N/m), Dimensions: [S] = [MT^-2]
• Factors affecting surface tension:
  • Temperature: Surface tension decreases with increasing temperature.
  • Impurities: Highly soluble impurities increase surface tension; sparingly soluble impurities decrease it.

Surface Energy and Relation to Surface Tension

• Surface energy (E) is the extra potential energy stored in surface molecules due to inward forces.
• SI unit: Joule (J), Dimensions: [E] = [ML²T^-2]
• Relation: Surface tension equals the increase in surface energy per unit area, S = W/ΔA, where W is work done and ΔA is increase in surface area.
• For a soap film with length l and displacement Δx, F = S × 2l and work done W = F × Δx = S × 2l Δx.

Solved Examples

• Ex 1: Calculate surface tension if a force of 0.2 N acts on a 0.1 m length of liquid surface.
  • Use S = F/L = 0.2/0.1 = 2 N/m.

Practice Set

• Q1: Surface tension decreases with:
  • A) Increase in temperature
  • B) Decrease in temperature
  • C) Addition of sparingly soluble impurities
  • D) Increase in pressure
• Q2: Surface tension is defined as force per unit:
  • A) Area
  • B) Length
  • C) Volume
  • D) Mass
• Q3: If force on a 0.05 m length of liquid surface is 0.1 N, surface tension is:
  • A) 0.5 N/m
  • B) 2 N/m
  • C) 5 N/m
  • D) 0.2 N/m

Answer Key

• Q1: A - Surface tension decreases with rise in temperature.
• Q2: B - Surface tension is force per unit length.
• Q3: C - S = 0.1/0.05 = 2 N/m.

Memory Aids

• Surface tension S = F/L
• Decreases with temperature rise
• Increases with highly soluble impurities
• Surface energy E = S × ΔA

Definition and Explanation

• Surface tension is the force per unit length acting along the surface of a liquid, making it behave like a stretched elastic membrane.
• It causes liquids to minimize their surface area, forming shapes like spherical drops.
• Example: Water drops on a lotus leaf are rounded due to surface tension; soap bubbles have two surfaces with pressure differences maintaining stability.
• Formula: \( S = \frac{F}{L} \), where \(F\) is force on either side of an imaginary line on the surface and \(L\) is the length of that line.
• SI unit: Newton per meter (N/m), Dimensions: \([S] = [MT^{-2}]\)
• Factors affecting surface tension:
  • Temperature: Surface tension decreases with increasing temperature.
  • Impurities: Highly soluble impurities increase surface tension; sparingly soluble impurities decrease it.

Surface Energy and Relation to Surface Tension

• Surface energy (E) is the extra potential energy stored in surface molecules due to inward forces.
• SI unit: Joule (J), Dimensions: \([E] = [ML^{2}T^{-2}]\)
• Relation: Surface tension equals the increase in surface energy per unit area, \( S = \frac{W}{\Delta A} \), where \(W\) is work done and \(\Delta A\) is increase in surface area.
• For a soap film with length \(l\) and displacement \(\Delta x\), \( F = S \times 2l \) and work done \( W = F \times \Delta x = S \times 2l \Delta x \).

Solved Examples

• Ex 1: Calculate surface tension if a force of 0.2 N acts on a 0.1 m length of liquid surface.
  • Use \( S = \frac{F}{L} = \frac{0.2}{0.1} = 2 \text{ N/m} \).

Practice Set

• Q1: Surface tension decreases with:
  • A) Increase in temperature
  • B) Decrease in temperature
  • C) Addition of sparingly soluble impurities
  • D) Increase in pressure
• Q2: Surface tension is defined as force per unit:
  • A) Area
  • B) Length
  • C) Volume
  • D) Mass
• Q3: If force on a 0.05 m length of liquid surface is 0.1 N, surface tension is:
  • A) 0.5 N/m
  • B) 2 N/m
  • C) 5 N/m
  • D) 0.2 N/m

Answer Key

• Q1: A - Surface tension decreases with rise in temperature.
• Q2: B - Surface tension is force per unit length.
• Q3: C - \( S = \frac{0.1}{0.05} = 2 \text{ N/m} \).

Memory Aids

• Surface tension \( S = \frac{F}{L} \)
• Decreases with temperature rise
• Increases with highly soluble impurities
• Surface energy \( E = S \times \Delta A \)

Angle of Contact

Definition and Explanation

• Angle of contact is the angle inside the liquid between the tangent to the solid surface and the tangent to the liquid surface at the contact point.
• It depends on the materials in contact and does not change with the solid's inclination.
• If angle < 90°, liquid wets the solid (wetting liquid).
• If angle ≥ 90°, liquid does not wet the solid.
• Surface tensions involved: S_la (liquid-air), S_sl (solid-liquid), S_sa (solid-air).
• Relation: S_la cos θ + S_sl = S_sa

Solved Examples

• Ex 1: Identify if water wets glass if contact angle is 60°.
  • Since 60° < 90°, water wets glass.

Practice Set

• Q1: Contact angle for a wetting liquid is:
  • A) < 90°
  • B) = 90°
  • C) > 90°
  • D) 180°
• Q2: If contact angle is 90°, liquid:
  • A) Wet the solid
  • B) Does not wet the solid
  • C) Partially wets
  • D) Forms a film
• Q3: For a liquid drop on solid, the relation involving surface tensions is:
  • A) S_la + S_sl = S_sa
  • B) S_la cos θ + S_sl = S_sa
  • C) S_la sin θ = S_sa
  • D) S_sl cos θ = S_la

Answer Key

• Q1: A - Wetting liquids have contact angle less than 90°.
• Q2: B - At 90°, liquid does not wet the solid.
• Q3: B - Surface tensions satisfy S_la cos θ + S_sl = S_sa.

Memory Aids

• Contact angle < 90°: wetting
• Contact angle ≥ 90°: non-wetting
• Formula: S_la cos θ + S_sl = S_sa

Definition and Explanation

• Angle of contact is the angle inside the liquid between the tangent to the solid surface and the tangent to the liquid surface at the contact point.
• It depends on the materials in contact and does not change with the solid's inclination.
• If angle < 90°, liquid wets the solid (wetting liquid).
• If angle ≥ 90°, liquid does not wet the solid.
• Surface tensions involved: S_la (liquid-air), S_sl (solid-liquid), S_sa (solid-air).
• Relation: S_la cos θ + S_sl = S_sa

Solved Examples

• Ex 1: Identify if water wets glass if contact angle is 60°.
  • Since 60° < 90°, water wets glass.

Practice Set

• Q1: Contact angle for a wetting liquid is:
  • A) < 90°
  • B) = 90°
  • C) > 90°
  • D) 180°
• Q2: If contact angle is 90°, liquid:
  • A) Wet the solid
  • B) Does not wet the solid
  • C) Partially wets
  • D) Forms a film
• Q3: For a liquid drop on solid, the relation involving surface tensions is:
  • A) S_la + S_sl = S_sa
  • B) S_la cos θ + S_sl = S_sa
  • C) S_la sin θ = S_sa
  • D) S_sl cos θ = S_la

Answer Key

• Q1: A - Wetting liquids have contact angle less than 90°.
• Q2: B - At 90°, liquid does not wet the solid.
• Q3: B - Surface tensions satisfy S_la cos θ + S_sl = S_sa.

Memory Aids

• Contact angle < 90°: wetting
• Contact angle ≥ 90°: non-wetting
• Formula: S_la cos θ + S_sl = S_sa

Definition and Explanation

• Angle of contact is the angle inside the liquid between the tangent to the solid surface and the tangent to the liquid surface at the contact point.
• It depends on the materials in contact and does not change with the solid's inclination.
• If angle < 90°, liquid wets the solid (wetting liquid).
• If angle ≥ 90°, liquid does not wet the solid.
• Surface tensions involved: S_la (liquid-air), S_sl (solid-liquid), S_sa (solid-air).
• Relation: S_la cos θ + S_sl = S_sa

Solved Examples

• Ex 1: Identify if water wets glass if contact angle is 60°.
  • Since 60° < 90°, water wets glass.

Practice Set

• Q1: Contact angle for a wetting liquid is:
  • A) < 90°
  • B) = 90°
  • C) > 90°
  • D) 180°
• Q2: If contact angle is 90°, liquid:
  • A) Wet the solid
  • B) Does not wet the solid
  • C) Partially wets
  • D) Forms a film
• Q3: For a liquid drop on solid, the relation involving surface tensions is:
  • A) S_la + S_sl = S_sa
  • B) S_la cos θ + S_sl = S_sa
  • C) S_la sin θ = S_sa
  • D) S_sl cos θ = S_la

Answer Key

• Q1: A - Wetting liquids have contact angle less than 90°.
• Q2: B - At 90°, liquid does not wet the solid.
• Q3: B - Surface tensions satisfy S_la cos θ + S_sl = S_sa.

Memory Aids

• Contact angle < 90°: wetting
• Contact angle ≥ 90°: non-wetting
• Formula: S_la cos θ + S_sl = S_sa

Definition and Explanation

• Angle of contact is the angle inside the liquid between the tangent to the solid surface and the tangent to the liquid surface at the contact point.
• It depends on the materials in contact and does not change with the solid's inclination.
• If angle < 90°, liquid wets the solid (wetting liquid).
• If angle ≥ 90°, liquid does not wet the solid.
• Surface tensions involved: S_la (liquid-air), S_sl (solid-liquid), S_sa (solid-air).
• Relation: S_la cos θ + S_sl = S_sa

Solved Examples

• Ex 1: Identify if water wets glass if contact angle is 60°.
  • Since 60° < 90°, water wets glass.

Practice Set

• Q1: Contact angle for a wetting liquid is:
  • A) < 90°
  • B) = 90°
  • C) > 90°
  • D) 180°
• Q2: If contact angle is 90°, liquid:
  • A) Wet the solid
  • B) Does not wet the solid
  • C) Partially wets
  • D) Forms a film
• Q3: For a liquid drop on solid, the relation involving surface tensions is:
  • A) S_la + S_sl = S_sa
  • B) S_la cos θ + S_sl = S_sa
  • C) S_la sin θ = S_sa
  • D) S_sl cos θ = S_la

Answer Key

• Q1: A - Wetting liquids have contact angle less than 90°.
• Q2: B - At 90°, liquid does not wet the solid.
• Q3: B - Surface tensions satisfy S_la cos θ + S_sl = S_sa.

Memory Aids

• Contact angle < 90°: wetting
• Contact angle ≥ 90°: non-wetting
• Formula: S_la cos θ + S_sl = S_sa

Definition and Explanation

• Angle of contact is the angle inside the liquid between the tangent to the solid surface and the tangent to the liquid surface at the contact point.
• It depends on the materials in contact and does not change with the solid's inclination.
• If angle < 90°, liquid wets the solid (wetting liquid).
• If angle ≥ 90°, liquid does not wet the solid.
• Surface tensions involved: S_la (liquid-air), S_sl (solid-liquid), S_sa (solid-air).
• Relation: S_la cos θ + S_sl = S_sa

Solved Examples

• Ex 1: Identify if water wets glass if contact angle is 60°.
  • Since 60° < 90°, water wets glass.

Practice Set

• Q1: Contact angle for a wetting liquid is:
  • A) < 90°
  • B) = 90°
  • C) > 90°
  • D) 180°
• Q2: If contact angle is 90°, liquid:
  • A) Wet the solid
  • B) Does not wet the solid
  • C) Partially wets
  • D) Forms a film
• Q3: For a liquid drop on solid, the relation involving surface tensions is:
  • A) S_la + S_sl = S_sa
  • B) S_la cos θ + S_sl = S_sa
  • C) S_la sin θ = S_sa
  • D) S_sl cos θ = S_la

Answer Key

• Q1: A - Wetting liquids have contact angle less than 90°.
• Q2: B - At 90°, liquid does not wet the solid.
• Q3: B - Surface tensions satisfy S_la cos θ + S_sl = S_sa.

Memory Aids

• Contact angle < 90°: wetting
• Contact angle ≥ 90°: non-wetting
• Formula: S_la cos θ + S_sl = S_sa

Definition and Explanation

• Angle of contact is the angle inside the liquid between the tangent to the solid surface and the tangent to the liquid surface at the contact point.
• It depends on the materials in contact and does not change with the solid's inclination.
• If angle < 90°, liquid wets the solid (wetting liquid).
• If angle ≥ 90°, liquid does not wet the solid.
• Surface tensions involved: S_la (liquid-air), S_sl (solid-liquid), S_sa (solid-air).
• Relation: S_la cos θ + S_sl = S_sa

Solved Examples

• Ex 1: Identify if water wets glass if contact angle is 60°.
  • Since 60° < 90°, water wets glass.

Practice Set

• Q1: Contact angle for a wetting liquid is:
  • A) < 90°
  • B) = 90°
  • C) > 90°
  • D) 180°
• Q2: If contact angle is 90°, liquid:
  • A) Wet the solid
  • B) Does not wet the solid
  • C) Partially wets
  • D) Forms a film
• Q3: For a liquid drop on solid, the relation involving surface tensions is:
  • A) S_la + S_sl = S_sa
  • B) S_la cos θ + S_sl = S_sa
  • C) S_la sin θ = S_sa
  • D) S_sl cos θ = S_la

Answer Key

• Q1: A - Wetting liquids have contact angle less than 90°.
• Q2: B - At 90°, liquid does not wet the solid.
• Q3: B - Surface tensions satisfy S_la cos θ + S_sl = S_sa.

Memory Aids

• Contact angle < 90°: wetting
• Contact angle ≥ 90°: non-wetting
• Formula: S_la cos θ + S_sl = S_sa

Definition and Explanation

• Angle of contact is the angle inside the liquid between the tangent to the solid surface and the tangent to the liquid surface at the contact point.
• It depends on the materials in contact and does not change with the solid's inclination.
• If angle < 90°, liquid wets the solid (wetting liquid).
• If angle ≥ 90°, liquid does not wet the solid.
• Surface tensions involved: S_la (liquid-air), S_sl (solid-liquid), S_sa (solid-air).
• Relation: S_la cos θ + S_sl = S_sa

Solved Examples

• Ex 1: Identify if water wets glass if contact angle is 60°.
  • Since 60° < 90°, water wets glass.

Practice Set

• Q1: Contact angle for a wetting liquid is:
  • A) < 90°
  • B) = 90°
  • C) > 90°
  • D) 180°
• Q2: If contact angle is 90°, liquid:
  • A) Wet the solid
  • B) Does not wet the solid
  • C) Partially wets
  • D) Forms a film
• Q3: For a liquid drop on solid, the relation involving surface tensions is:
  • A) S_la + S_sl = S_sa
  • B) S_la cos θ + S_sl = S_sa
  • C) S_la sin θ = S_sa
  • D) S_sl cos θ = S_la

Answer Key

• Q1: A - Wetting liquids have contact angle less than 90°.
• Q2: B - At 90°, liquid does not wet the solid.
• Q3: B - Surface tensions satisfy S_la cos θ + S_sl = S_sa.

Memory Aids

• Contact angle < 90°: wetting
• Contact angle ≥ 90°: non-wetting
• Formula: S_la cos θ + S_sl = S_sa

Definition and Explanation

• Angle of contact is the angle inside the liquid between the tangent to the solid surface and the tangent to the liquid surface at the contact point.
• It depends on the materials in contact and does not change with the solid's inclination.
• If angle < 90°, liquid wets the solid (wetting liquid).
• If angle ≥ 90°, liquid does not wet the solid.
• Surface tensions involved: S_la (liquid-air), S_sl (solid-liquid), S_sa (solid-air).
• Relation: S_la cos θ + S_sl = S_sa

Solved Examples

• Ex 1: Identify if water wets glass if contact angle is 60°.
  • Since 60° < 90°, water wets glass.

Practice Set

• Q1: Contact angle for a wetting liquid is:
  • A) < 90°
  • B) = 90°
  • C) > 90°
  • D) 180°
• Q2: If contact angle is 90°, liquid:
  • A) Wet the solid
  • B) Does not wet the solid
  • C) Partially wets
  • D) Forms a film
• Q3: For a liquid drop on solid, the relation involving surface tensions is:
  • A) S_la + S_sl = S_sa
  • B) S_la cos θ + S_sl = S_sa
  • C) S_la sin θ = S_sa
  • D) S_sl cos θ = S_la

Answer Key

• Q1: A - Wetting liquids have contact angle less than 90°.
• Q2: B - At 90°, liquid does not wet the solid.
• Q3: B - Surface tensions satisfy S_la cos θ + S_sl = S_sa.

Memory Aids

• Contact angle < 90°: wetting
• Contact angle ≥ 90°: non-wetting
• Formula: S_la cos θ + S_sl = S_sa

Definition and Explanation

• Angle of contact is the angle inside the liquid between the tangent to the solid surface and the tangent to the liquid surface at the contact point.
• It depends on the materials in contact and does not change with the solid's inclination.
• If angle < 90°, liquid wets the solid (wetting liquid).
• If angle ≥ 90°, liquid does not wet the solid.
• Surface tensions involved: S_la (liquid-air), S_sl (solid-liquid), S_sa (solid-air).
• Relation: S_la cos θ + S_sl = S_sa

Solved Examples

• Ex 1: Identify if water wets glass if contact angle is 60°.
  • Since 60° < 90°, water wets glass.

Practice Set

• Q1: Contact angle for a wetting liquid is:
  • A) < 90°
  • B) = 90°
  • C) > 90°
  • D) 180°
• Q2: If contact angle is 90°, liquid:
  • A) Wet the solid
  • B) Does not wet the solid
  • C) Partially wets
  • D) Forms a film
• Q3: For a liquid drop on solid, the relation involving surface tensions is:
  • A) S_la + S_sl = S_sa
  • B) S_la cos θ + S_sl = S_sa
  • C) S_la sin θ = S_sa
  • D) S_sl cos θ = S_la

Answer Key

• Q1: A - Wetting liquids have contact angle less than 90°.
• Q2: B - At 90°, liquid does not wet the solid.
• Q3: B - Surface tensions satisfy S_la cos θ + S_sl = S_sa.

Memory Aids

• Contact angle < 90°: wetting
• Contact angle ≥ 90°: non-wetting
• Formula: S_la cos θ + S_sl = S_sa

Definition and Explanation

• Angle of contact is the angle inside the liquid between the tangent to the solid surface and the tangent to the liquid surface at the contact point.
• It depends on the materials in contact and does not change with the solid's inclination.
• If angle < 90°, liquid wets the solid (wetting liquid).
• If angle ≥ 90°, liquid does not wet the solid.
• Surface tensions involved: S_la (liquid-air), S_sl (solid-liquid), S_sa (solid-air).
• Relation: S_la cos θ + S_sl = S_sa

Solved Examples

• Ex 1: Identify if water wets glass if contact angle is 60°.
  • Since 60° < 90°, water wets glass.

Practice Set

• Q1: Contact angle for a wetting liquid is:
  • A) < 90°
  • B) = 90°
  • C) > 90°
  • D) 180°
• Q2: If contact angle is 90°, liquid:
  • A) Wet the solid
  • B) Does not wet the solid
  • C) Partially wets
  • D) Forms a film
• Q3: For a liquid drop on solid, the relation involving surface tensions is:
  • A) S_la + S_sl = S_sa
  • B) S_la cos θ + S_sl = S_sa
  • C) S_la sin θ = S_sa
  • D) S_sl cos θ = S_la

Answer Key

• Q1: A - Wetting liquids have contact angle less than 90°.
• Q2: B - At 90°, liquid does not wet the solid.
• Q3: B - Surface tensions satisfy S_la cos θ + S_sl = S_sa.

Memory Aids

• Contact angle < 90°: wetting
• Contact angle ≥ 90°: non-wetting
• Formula: S_la cos θ + S_sl = S_sa

Definition and Explanation

• Angle of contact is the angle inside the liquid between the tangent to the solid surface and the tangent to the liquid surface at the contact point.
• It depends on the materials in contact and does not change with the solid's inclination.
• If angle < 90°, liquid wets the solid (wetting liquid).
• If angle ≥ 90°, liquid does not wet the solid.
• Surface tensions involved: S_la (liquid-air), S_sl (solid-liquid), S_sa (solid-air).
• Relation: S_la cos θ + S_sl = S_sa

Solved Examples

• Ex 1: Identify if water wets glass if contact angle is 60°.
  • Since 60° < 90°, water wets glass.

Practice Set

• Q1: Contact angle for a wetting liquid is:
  • A) < 90°
  • B) = 90°
  • C) > 90°
  • D) 180°
• Q2: If contact angle is 90°, liquid:
  • A) Wet the solid
  • B) Does not wet the solid
  • C) Partially wets
  • D) Forms a film
• Q3: For a liquid drop on solid, the relation involving surface tensions is:
  • A) S_la + S_sl = S_sa
  • B) S_la cos θ + S_sl = S_sa
  • C) S_la sin θ = S_sa
  • D) S_sl cos θ = S_la

Answer Key

• Q1: A - Wetting liquids have contact angle less than 90°.
• Q2: B - At 90°, liquid does not wet the solid.
• Q3: B - Surface tensions satisfy S_la cos θ + S_sl = S_sa.

Memory Aids

• Contact angle < 90°: wetting
• Contact angle ≥ 90°: non-wetting
• Formula: S_la cos θ + S_sl = S_sa

Definition and Explanation

• Angle of contact is the angle inside the liquid between the tangent to the solid surface and the tangent to the liquid surface at the contact point.
• It depends on the materials in contact and does not change with the solid's inclination.
• If angle < 90°, liquid wets the solid (wetting liquid).
• If angle ≥ 90°, liquid does not wet the solid.
• Surface tensions involved: S_la (liquid-air), S_sl (solid-liquid), S_sa (solid-air).
• Relation: S_la cos θ + S_sl = S_sa

Solved Examples

• Ex 1: Identify if water wets glass if contact angle is 60°.
  • Since 60° < 90°, water wets glass.

Practice Set

• Q1: Contact angle for a wetting liquid is:
  • A) < 90°
  • B) = 90°
  • C) > 90°
  • D) 180°
• Q2: If contact angle is 90°, liquid:
  • A) Wet the solid
  • B) Does not wet the solid
  • C) Partially wets
  • D) Forms a film
• Q3: For a liquid drop on solid, the relation involving surface tensions is:
  • A) S_la + S_sl = S_sa
  • B) S_la cos θ + S_sl = S_sa
  • C) S_la sin θ = S_sa
  • D) S_sl cos θ = S_la

Answer Key

• Q1: A - Wetting liquids have contact angle less than 90°.
• Q2: B - At 90°, liquid does not wet the solid.
• Q3: B - Surface tensions satisfy S_la cos θ + S_sl = S_sa.

Memory Aids

• Contact angle < 90°: wetting
• Contact angle ≥ 90°: non-wetting
• Formula: S_la cos θ + S_sl = S_sa

Definition and Explanation

• Angle of contact is the angle inside the liquid between the tangent to the solid surface and the tangent to the liquid surface at the contact point.
• It depends on the materials in contact and does not change with the solid's inclination.
• If angle < 90°, liquid wets the solid (wetting liquid).
• If angle ≥ 90°, liquid does not wet the solid.
• Surface tensions involved: \( S_{la} \) (liquid-air), \( S_{sl} \) (solid-liquid), \( S_{sa} \) (solid-air).
• Relation: \( S_{la} \cos \theta + S_{sl} = S_{sa} \)

Solved Examples

• Ex 1: Identify if water wets glass if contact angle is 60°.
  • Since 60° < 90°, water wets glass.

Practice Set

• Q1: Contact angle for a wetting liquid is:
  • A) < 90°
  • B) = 90°
  • C) > 90°
  • D) 180°
• Q2: If contact angle is 90°, liquid:
  • A) Wet the solid
  • B) Does not wet the solid
  • C) Partially wets
  • D) Forms a film
• Q3: For a liquid drop on solid, the relation involving surface tensions is:
  • A) \( S_{la} + S_{sl} = S_{sa} \)
  • B) \( S_{la} \cos \theta + S_{sl} = S_{sa} \)
  • C) \( S_{la} \sin \theta = S_{sa} \)
  • D) \( S_{sl} \cos \theta = S_{la} \)

Answer Key

• Q1: A - Wetting liquids have contact angle less than 90°.
• Q2: B - At 90°, liquid does not wet the solid.
• Q3: B - Surface tensions satisfy \( S_{la} \cos \theta + S_{sl} = S_{sa} \).

Memory Aids

• Contact angle < 90°: wetting
• Contact angle ≥ 90°: non-wetting
• Formula: \( S_{la} \cos \theta + S_{sl} = S_{sa} \)

Drops and Bubbles

Excess Pressure Inside Liquid Drop

• Consider a liquid drop of radius R and surface tension S.
• Pressure difference: ΔP = P - P₀.
• Excess pressure inside drop: ΔP = 2S/R.
• Work done in increasing radius ΔR: ΔW = ΔP × 4 πR²ΔR.
• Increase in potential energy due to surface tension: ΔU = 8 πSRΔR.
• At equilibrium, ΔU = ΔW leads to ΔP = 2S/R.

Excess Pressure Inside Soap Bubble

• Soap bubble has two surfaces in contact with air.
• Excess pressure inside bubble: ΔP = 4S/R.
• Work done: ΔW = ΔP × 4 πR²ΔR.
• Increase in potential energy: ΔU = 16 πSRΔR.
• At equilibrium, ΔU = ΔW leads to ΔP = 4S/R.

Excess Pressure Inside Air Bubble in Liquid

• Similar to liquid drop (one surface in contact with air).
• Excess pressure: ΔP = 2S/R.

Solved Examples

• Ex 1: Calculate excess pressure inside a drop of radius 0.01 m with surface tension 0.07 N/m.
  • ΔP = 2 × 0.07/0.01 = 14 N/m².
• Ex 2: Calculate excess pressure inside a soap bubble of radius 0.01 m with surface tension 0.07 N/m.
  • ΔP = 4 × 0.07/0.01 = 28 N/m².

Practice Set

• Q1: Excess pressure inside a liquid drop is:
  • A) 4S/R
  • B) 2S/R
  • C) S/R
  • D) 8S/R
• Q2: Excess pressure inside a soap bubble is:
  • A) 2S/R
  • B) 4S/R
  • C) S/R
  • D) 6S/R
• Q3: Excess pressure inside an air bubble in liquid is:
  • A) 4S/R
  • B) 2S/R
  • C) S/R
  • D) 8S/R

Answer Key

• Q1: B - Excess pressure inside liquid drop is 2S/R.
• Q2: B - Excess pressure inside soap bubble is 4S/R.
• Q3: B - Excess pressure inside air bubble in liquid is 2S/R.

Memory Aids

• Liquid drop: ΔP = 2S/R
• Soap bubble: ΔP = 4S/R
• Air bubble in liquid: same as liquid drop

Excess Pressure Inside Liquid Drop

• Consider a liquid drop of radius R and surface tension S.
• Pressure difference: ΔP = P - P₀.
• Excess pressure inside drop: ΔP = 2S/R.
• Work done in increasing radius ΔR: ΔW = ΔP × 4 πR²ΔR.
• Increase in potential energy due to surface tension: ΔU = 8 πSRΔR.
• At equilibrium, ΔU = ΔW leads to ΔP = 2S/R.

Excess Pressure Inside Soap Bubble

• Soap bubble has two surfaces in contact with air.
• Excess pressure inside bubble: ΔP = 4S/R.
• Work done: ΔW = ΔP × 4 πR²ΔR.
• Increase in potential energy: ΔU = 16 πSRΔR.
• At equilibrium, ΔU = ΔW leads to ΔP = 4S/R.

Excess Pressure Inside Air Bubble in Liquid

• Similar to liquid drop (one surface in contact with air).
• Excess pressure: ΔP = 2S/R.

Solved Examples

• Ex 1: Calculate excess pressure inside a drop of radius 0.01 m with surface tension 0.07 N/m.
  • ΔP = 2 × 0.07/0.01 = 14 N/m².
• Ex 2: Calculate excess pressure inside a soap bubble of radius 0.01 m with surface tension 0.07 N/m.
  • ΔP = 4 × 0.07/0.01 = 28 N/m².

Practice Set

• Q1: Excess pressure inside a liquid drop is:
  • A) 4S/R
  • B) 2S/R
  • C) S/R
  • D) 8S/R
• Q2: Excess pressure inside a soap bubble is:
  • A) 2S/R
  • B) 4S/R
  • C) S/R
  • D) 6S/R
• Q3: Excess pressure inside an air bubble in liquid is:
  • A) 4S/R
  • B) 2S/R
  • C) S/R
  • D) 8S/R

Answer Key

• Q1: B - Excess pressure inside liquid drop is 2S/R.
• Q2: B - Excess pressure inside soap bubble is 4S/R.
• Q3: B - Excess pressure inside air bubble in liquid is 2S/R.

Memory Aids

• Liquid drop: ΔP = 2S/R
• Soap bubble: ΔP = 4S/R
• Air bubble in liquid: same as liquid drop

Excess Pressure Inside Liquid Drop

• Consider a liquid drop of radius R and surface tension S.
• Pressure difference: ΔP = P - P₀.
• Excess pressure inside drop: ΔP = 2S/R.
• Work done in increasing radius ΔR: ΔW = ΔP × 4 πR²ΔR.
• Increase in potential energy due to surface tension: ΔU = 8 πSRΔR.
• At equilibrium, ΔU = ΔW leads to ΔP = 2S/R.

Excess Pressure Inside Soap Bubble

• Soap bubble has two surfaces in contact with air.
• Excess pressure inside bubble: ΔP = 4S/R.
• Work done: ΔW = ΔP × 4 πR²ΔR.
• Increase in potential energy: ΔU = 16 πSRΔR.
• At equilibrium, ΔU = ΔW leads to ΔP = 4S/R.

Excess Pressure Inside Air Bubble in Liquid

• Similar to liquid drop (one surface in contact with air).
• Excess pressure: ΔP = 2S/R.

Solved Examples

• Ex 1: Calculate excess pressure inside a drop of radius 0.01 m with surface tension 0.07 N/m.
  • ΔP = 2 × 0.07/0.01 = 14 N/m².
• Ex 2: Calculate excess pressure inside a soap bubble of radius 0.01 m with surface tension 0.07 N/m.
  • ΔP = 4 × 0.07/0.01 = 28 N/m².

Practice Set

• Q1: Excess pressure inside a liquid drop is:
  • A) 4S/R
  • B) 2S/R
  • C) S/R
  • D) 8S/R
• Q2: Excess pressure inside a soap bubble is:
  • A) 2S/R
  • B) 4S/R
  • C) S/R
  • D) 6S/R
• Q3: Excess pressure inside an air bubble in liquid is:
  • A) 4S/R
  • B) 2S/R
  • C) S/R
  • D) 8S/R

Answer Key

• Q1: B - Excess pressure inside liquid drop is 2S/R.
• Q2: B - Excess pressure inside soap bubble is 4S/R.
• Q3: B - Excess pressure inside air bubble in liquid is 2S/R.

Memory Aids

• Liquid drop: ΔP = 2S/R
• Soap bubble: ΔP = 4S/R
• Air bubble in liquid: same as liquid drop

Excess Pressure Inside Liquid Drop

• Consider a liquid drop of radius R and surface tension S.
• Pressure difference: ΔP = P - P₀.
• Excess pressure inside drop: ΔP = 2S/R.
• Work done in increasing radius ΔR: ΔW = ΔP × 4 πR²ΔR.
• Increase in potential energy due to surface tension: ΔU = 8 πSRΔR.
• At equilibrium, ΔU = ΔW leads to ΔP = 2S/R.

Excess Pressure Inside Soap Bubble

• Soap bubble has two surfaces in contact with air.
• Excess pressure inside bubble: ΔP = 4S/R.
• Work done: ΔW = ΔP × 4 πR²ΔR.
• Increase in potential energy: ΔU = 16 πSRΔR.
• At equilibrium, ΔU = ΔW leads to ΔP = 4S/R.

Excess Pressure Inside Air Bubble in Liquid

• Similar to liquid drop (one surface in contact with air).
• Excess pressure: ΔP = 2S/R.

Solved Examples

• Ex 1: Calculate excess pressure inside a drop of radius 0.01 m with surface tension 0.07 N/m.
  • ΔP = 2 × 0.07/0.01 = 14 N/m².
• Ex 2: Calculate excess pressure inside a soap bubble of radius 0.01 m with surface tension 0.07 N/m.
  • ΔP = 4 × 0.07/0.01 = 28 N/m².

Practice Set

• Q1: Excess pressure inside a liquid drop is:
  • A) 4S/R
  • B) 2S/R
  • C) S/R
  • D) 8S/R
• Q2: Excess pressure inside a soap bubble is:
  • A) 2S/R
  • B) 4S/R
  • C) S/R
  • D) 6S/R
• Q3: Excess pressure inside an air bubble in liquid is:
  • A) 4S/R
  • B) 2S/R
  • C) S/R
  • D) 8S/R

Answer Key

• Q1: B - Excess pressure inside liquid drop is 2S/R.
• Q2: B - Excess pressure inside soap bubble is 4S/R.
• Q3: B - Excess pressure inside air bubble in liquid is 2S/R.

Memory Aids

• Liquid drop: ΔP = 2S/R
• Soap bubble: ΔP = 4S/R
• Air bubble in liquid: same as liquid drop

Excess Pressure Inside Liquid Drop

• Consider a liquid drop of radius R and surface tension S.
• Pressure difference: ΔP = P - P₀.
• Excess pressure inside drop: ΔP = 2S/R.
• Work done in increasing radius ΔR: ΔW = ΔP × 4 πR²ΔR.
• Increase in potential energy due to surface tension: ΔU = 8 πSRΔR.
• At equilibrium, ΔU = ΔW leads to ΔP = 2S/R.

Excess Pressure Inside Soap Bubble

• Soap bubble has two surfaces in contact with air.
• Excess pressure inside bubble: ΔP = 4S/R.
• Work done: ΔW = ΔP × 4 πR²ΔR.
• Increase in potential energy: ΔU = 16 πSRΔR.
• At equilibrium, ΔU = ΔW leads to ΔP = 4S/R.

Excess Pressure Inside Air Bubble in Liquid

• Similar to liquid drop (one surface in contact with air).
• Excess pressure: ΔP = 2S/R.

Solved Examples

• Ex 1: Calculate excess pressure inside a drop of radius 0.01 m with surface tension 0.07 N/m.
  • ΔP = 2 × 0.07/0.01 = 14 N/m².
• Ex 2: Calculate excess pressure inside a soap bubble of radius 0.01 m with surface tension 0.07 N/m.
  • ΔP = 4 × 0.07/0.01 = 28 N/m².

Practice Set

• Q1: Excess pressure inside a liquid drop is:
  • A) 4S/R
  • B) 2S/R
  • C) S/R
  • D) 8S/R
• Q2: Excess pressure inside a soap bubble is:
  • A) 2S/R
  • B) 4S/R
  • C) S/R
  • D) 6S/R
• Q3: Excess pressure inside an air bubble in liquid is:
  • A) 4S/R
  • B) 2S/R
  • C) S/R
  • D) 8S/R

Answer Key

• Q1: B - Excess pressure inside liquid drop is 2S/R.
• Q2: B - Excess pressure inside soap bubble is 4S/R.
• Q3: B - Excess pressure inside air bubble in liquid is 2S/R.

Memory Aids

• Liquid drop: ΔP = 2S/R
• Soap bubble: ΔP = 4S/R
• Air bubble in liquid: same as liquid drop

Excess Pressure Inside Liquid Drop

• Consider a liquid drop of radius R and surface tension S.
• Pressure difference: ΔP = P - P₀.
• Excess pressure inside drop: ΔP = 2S/R.
• Work done in increasing radius ΔR: ΔW = ΔP × 4 πR²ΔR.
• Increase in potential energy due to surface tension: ΔU = 8 πSRΔR.
• At equilibrium, ΔU = ΔW leads to ΔP = 2S/R.

Excess Pressure Inside Soap Bubble

• Soap bubble has two surfaces in contact with air.
• Excess pressure inside bubble: ΔP = 4S/R.
• Work done: ΔW = ΔP × 4 πR²ΔR.
• Increase in potential energy: ΔU = 16 πSRΔR.
• At equilibrium, ΔU = ΔW leads to ΔP = 4S/R.

Excess Pressure Inside Air Bubble in Liquid

• Similar to liquid drop (one surface in contact with air).
• Excess pressure: ΔP = 2S/R.

Solved Examples

• Ex 1: Calculate excess pressure inside a drop of radius 0.01 m with surface tension 0.07 N/m.
  • ΔP = 2 × 0.07/0.01 = 14 N/m².
• Ex 2: Calculate excess pressure inside a soap bubble of radius 0.01 m with surface tension 0.07 N/m.
  • ΔP = 4 × 0.07/0.01 = 28 N/m².

Practice Set

• Q1: Excess pressure inside a liquid drop is:
  • A) 4S/R
  • B) 2S/R
  • C) S/R
  • D) 8S/R
• Q2: Excess pressure inside a soap bubble is:
  • A) 2S/R
  • B) 4S/R
  • C) S/R
  • D) 6S/R
• Q3: Excess pressure inside an air bubble in liquid is:
  • A) 4S/R
  • B) 2S/R
  • C) S/R
  • D) 8S/R

Answer Key

• Q1: B - Excess pressure inside liquid drop is 2S/R.
• Q2: B - Excess pressure inside soap bubble is 4S/R.
• Q3: B - Excess pressure inside air bubble in liquid is 2S/R.

Memory Aids

• Liquid drop: ΔP = 2S/R
• Soap bubble: ΔP = 4S/R
• Air bubble in liquid: same as liquid drop

Excess Pressure Inside Liquid Drop

• Consider a liquid drop of radius R and surface tension S.
• Pressure difference: ΔP = P - P₀.
• Excess pressure inside drop: ΔP = 2S/R.
• Work done in increasing radius ΔR: ΔW = ΔP × 4 πR²ΔR.
• Increase in potential energy due to surface tension: ΔU = 8 πSRΔR.
• At equilibrium, ΔU = ΔW leads to ΔP = 2S/R.

Excess Pressure Inside Soap Bubble

• Soap bubble has two surfaces in contact with air.
• Excess pressure inside bubble: ΔP = 4S/R.
• Work done: ΔW = ΔP × 4 πR²ΔR.
• Increase in potential energy: ΔU = 16 πSRΔR.
• At equilibrium, ΔU = ΔW leads to ΔP = 4S/R.

Excess Pressure Inside Air Bubble in Liquid

• Similar to liquid drop (one surface in contact with air).
• Excess pressure: ΔP = 2S/R.

Solved Examples

• Ex 1: Calculate excess pressure inside a drop of radius 0.01 m with surface tension 0.07 N/m.
  • ΔP = 2 × 0.07/0.01 = 14 N/m².
• Ex 2: Calculate excess pressure inside a soap bubble of radius 0.01 m with surface tension 0.07 N/m.
  • ΔP = 4 × 0.07/0.01 = 28 N/m².

Practice Set

• Q1: Excess pressure inside a liquid drop is:
  • A) 4S/R
  • B) 2S/R
  • C) S/R
  • D) 8S/R
• Q2: Excess pressure inside a soap bubble is:
  • A) 2S/R
  • B) 4S/R
  • C) S/R
  • D) 6S/R
• Q3: Excess pressure inside an air bubble in liquid is:
  • A) 4S/R
  • B) 2S/R
  • C) S/R
  • D) 8S/R

Answer Key

• Q1: B - Excess pressure inside liquid drop is 2S/R.
• Q2: B - Excess pressure inside soap bubble is 4S/R.
• Q3: B - Excess pressure inside air bubble in liquid is 2S/R.

Memory Aids

• Liquid drop: ΔP = 2S/R
• Soap bubble: ΔP = 4S/R
• Air bubble in liquid: same as liquid drop

Excess Pressure Inside Liquid Drop

• Consider a liquid drop of radius R and surface tension S.
• Pressure difference: ΔP = P - P₀.
• Excess pressure inside drop: ΔP = 2S/R.
• Work done in increasing radius ΔR: ΔW = ΔP × 4 πR²ΔR.
• Increase in potential energy due to surface tension: ΔU = 8 πSRΔR.
• At equilibrium, ΔU = ΔW leads to ΔP = 2S/R.

Excess Pressure Inside Soap Bubble

• Soap bubble has two surfaces in contact with air.
• Excess pressure inside bubble: ΔP = 4S/R.
• Work done: ΔW = ΔP × 4 πR²ΔR.
• Increase in potential energy: ΔU = 16 πSRΔR.
• At equilibrium, ΔU = ΔW leads to ΔP = 4S/R.

Excess Pressure Inside Air Bubble in Liquid

• Similar to liquid drop (one surface in contact with air).
• Excess pressure: ΔP = 2S/R.

Solved Examples

• Ex 1: Calculate excess pressure inside a drop of radius 0.01 m with surface tension 0.07 N/m.
  • ΔP = 2 × 0.07/0.01 = 14 N/m².
• Ex 2: Calculate excess pressure inside a soap bubble of radius 0.01 m with surface tension 0.07 N/m.
  • ΔP = 4 × 0.07/0.01 = 28 N/m².

Practice Set

• Q1: Excess pressure inside a liquid drop is:
  • A) 4S/R
  • B) 2S/R
  • C) S/R
  • D) 8S/R
• Q2: Excess pressure inside a soap bubble is:
  • A) 2S/R
  • B) 4S/R
  • C) S/R
  • D) 6S/R
• Q3: Excess pressure inside an air bubble in liquid is:
  • A) 4S/R
  • B) 2S/R
  • C) S/R
  • D) 8S/R

Answer Key

• Q1: B - Excess pressure inside liquid drop is 2S/R.
• Q2: B - Excess pressure inside soap bubble is 4S/R.
• Q3: B - Excess pressure inside air bubble in liquid is 2S/R.

Memory Aids

• Liquid drop: ΔP = 2S/R
• Soap bubble: ΔP = 4S/R
• Air bubble in liquid: same as liquid drop

Excess Pressure Inside Liquid Drop

• Consider a liquid drop of radius R and surface tension S.
• Pressure difference: ΔP = P - P₀.
• Excess pressure inside drop: ΔP = 2S/R.
• Work done in increasing radius ΔR: ΔW = ΔP × 4 πR²ΔR.
• Increase in potential energy due to surface tension: ΔU = 8 πSRΔR.
• At equilibrium, ΔU = ΔW leads to ΔP = 2S/R.

Excess Pressure Inside Soap Bubble

• Soap bubble has two surfaces in contact with air.
• Excess pressure inside bubble: ΔP = 4S/R.
• Work done: ΔW = ΔP × 4 πR²ΔR.
• Increase in potential energy: ΔU = 16 πSRΔR.
• At equilibrium, ΔU = ΔW leads to ΔP = 4S/R.

Excess Pressure Inside Air Bubble in Liquid

• Similar to liquid drop (one surface in contact with air).
• Excess pressure: ΔP = 2S/R.

Solved Examples

• Ex 1: Calculate excess pressure inside a drop of radius 0.01 m with surface tension 0.07 N/m.
  • ΔP = 2 × 0.07/0.01 = 14 N/m².
• Ex 2: Calculate excess pressure inside a soap bubble of radius 0.01 m with surface tension 0.07 N/m.
  • ΔP = 4 × 0.07/0.01 = 28 N/m².

Practice Set

• Q1: Excess pressure inside a liquid drop is:
  • A) 4S/R
  • B) 2S/R
  • C) S/R
  • D) 8S/R
• Q2: Excess pressure inside a soap bubble is:
  • A) 2S/R
  • B) 4S/R
  • C) S/R
  • D) 6S/R
• Q3: Excess pressure inside an air bubble in liquid is:
  • A) 4S/R
  • B) 2S/R
  • C) S/R
  • D) 8S/R

Answer Key

• Q1: B - Excess pressure inside liquid drop is 2S/R.
• Q2: B - Excess pressure inside soap bubble is 4S/R.
• Q3: B - Excess pressure inside air bubble in liquid is 2S/R.

Memory Aids

• Liquid drop: ΔP = 2S/R
• Soap bubble: ΔP = 4S/R
• Air bubble in liquid: same as liquid drop

Excess Pressure Inside Liquid Drop

• Consider a liquid drop of radius R and surface tension S.
• Pressure difference: ΔP = P - P₀.
• Excess pressure inside drop: ΔP = 2S/R.
• Work done in increasing radius ΔR: ΔW = ΔP × 4 πR²ΔR.
• Increase in potential energy due to surface tension: ΔU = 8 πSRΔR.
• At equilibrium, ΔU = ΔW leads to ΔP = 2S/R.

Excess Pressure Inside Soap Bubble

• Soap bubble has two surfaces in contact with air.
• Excess pressure inside bubble: ΔP = 4S/R.
• Work done: ΔW = ΔP × 4 πR²ΔR.
• Increase in potential energy: ΔU = 16 πSRΔR.
• At equilibrium, ΔU = ΔW leads to ΔP = 4S/R.

Excess Pressure Inside Air Bubble in Liquid

• Similar to liquid drop (one surface in contact with air).
• Excess pressure: ΔP = 2S/R.

Solved Examples

• Ex 1: Calculate excess pressure inside a drop of radius 0.01 m with surface tension 0.07 N/m.
  • ΔP = 2 × 0.07/0.01 = 14 N/m².
• Ex 2: Calculate excess pressure inside a soap bubble of radius 0.01 m with surface tension 0.07 N/m.
  • ΔP = 4 × 0.07/0.01 = 28 N/m².

Practice Set

• Q1: Excess pressure inside a liquid drop is:
  • A) 4S/R
  • B) 2S/R
  • C) S/R
  • D) 8S/R
• Q2: Excess pressure inside a soap bubble is:
  • A) 2S/R
  • B) 4S/R
  • C) S/R
  • D) 6S/R
• Q3: Excess pressure inside an air bubble in liquid is:
  • A) 4S/R
  • B) 2S/R
  • C) S/R
  • D) 8S/R

Answer Key

• Q1: B - Excess pressure inside liquid drop is 2S/R.
• Q2: B - Excess pressure inside soap bubble is 4S/R.
• Q3: B - Excess pressure inside air bubble in liquid is 2S/R.

Memory Aids

• Liquid drop: ΔP = 2S/R
• Soap bubble: ΔP = 4S/R
• Air bubble in liquid: same as liquid drop

Excess Pressure Inside Liquid Drop

• Consider a liquid drop of radius R and surface tension S.
• Pressure difference: ΔP = P - P₀.
• Excess pressure inside drop: ΔP = 2S/R.
• Work done in increasing radius ΔR: ΔW = ΔP × 4 πR²ΔR.
• Increase in potential energy due to surface tension: ΔU = 8 πSRΔR.
• At equilibrium, ΔU = ΔW leads to ΔP = 2S/R.

Excess Pressure Inside Soap Bubble

• Soap bubble has two surfaces in contact with air.
• Excess pressure inside bubble: ΔP = 4S/R.
• Work done: ΔW = ΔP × 4 πR²ΔR.
• Increase in potential energy: ΔU = 16 πSRΔR.
• At equilibrium, ΔU = ΔW leads to ΔP = 4S/R.

Excess Pressure Inside Air Bubble in Liquid

• Similar to liquid drop (one surface in contact with air).
• Excess pressure: ΔP = 2S/R.

Solved Examples

• Ex 1: Calculate excess pressure inside a drop of radius 0.01 m with surface tension 0.07 N/m.
  • ΔP = 2 × 0.07/0.01 = 14 N/m².
• Ex 2: Calculate excess pressure inside a soap bubble of radius 0.01 m with surface tension 0.07 N/m.
  • ΔP = 4 × 0.07/0.01 = 28 N/m².

Practice Set

• Q1: Excess pressure inside a liquid drop is:
  • A) 4S/R
  • B) 2S/R
  • C) S/R
  • D) 8S/R
• Q2: Excess pressure inside a soap bubble is:
  • A) 2S/R
  • B) 4S/R
  • C) S/R
  • D) 6S/R
• Q3: Excess pressure inside an air bubble in liquid is:
  • A) 4S/R
  • B) 2S/R
  • C) S/R
  • D) 8S/R

Answer Key

• Q1: B - Excess pressure inside liquid drop is 2S/R.
• Q2: B - Excess pressure inside soap bubble is 4S/R.
• Q3: B - Excess pressure inside air bubble in liquid is 2S/R.

Memory Aids

• Liquid drop: ΔP = 2S/R
• Soap bubble: ΔP = 4S/R
• Air bubble in liquid: same as liquid drop

Excess Pressure Inside Liquid Drop

• Consider a liquid drop of radius R and surface tension S.
• Pressure difference: ΔP = P - P₀.
• Excess pressure inside drop: ΔP = 2S/R.
• Work done in increasing radius ΔR: ΔW = ΔP × 4 πR²ΔR.
• Increase in potential energy due to surface tension: ΔU = 8 πSRΔR.
• At equilibrium, ΔU = ΔW leads to ΔP = 2S/R.

Excess Pressure Inside Soap Bubble

• Soap bubble has two surfaces in contact with air.
• Excess pressure inside bubble: ΔP = 4S/R.
• Work done: ΔW = ΔP × 4 πR²ΔR.
• Increase in potential energy: ΔU = 16 πSRΔR.
• At equilibrium, ΔU = ΔW leads to ΔP = 4S/R.

Excess Pressure Inside Air Bubble in Liquid

• Similar to liquid drop (one surface in contact with air).
• Excess pressure: ΔP = 2S/R.

Solved Examples

• Ex 1: Calculate excess pressure inside a drop of radius 0.01 m with surface tension 0.07 N/m.
  • ΔP = 2 × 0.07/0.01 = 14 N/m².
• Ex 2: Calculate excess pressure inside a soap bubble of radius 0.01 m with surface tension 0.07 N/m.
  • ΔP = 4 × 0.07/0.01 = 28 N/m².

Practice Set

• Q1: Excess pressure inside a liquid drop is:
  • A) 4S/R
  • B) 2S/R
  • C) S/R
  • D) 8S/R
• Q2: Excess pressure inside a soap bubble is:
  • A) 2S/R
  • B) 4S/R
  • C) S/R
  • D) 6S/R
• Q3: Excess pressure inside an air bubble in liquid is:
  • A) 4S/R
  • B) 2S/R
  • C) S/R
  • D) 8S/R

Answer Key

• Q1: B - Excess pressure inside liquid drop is 2S/R.
• Q2: B - Excess pressure inside soap bubble is 4S/R.
• Q3: B - Excess pressure inside air bubble in liquid is 2S/R.

Memory Aids

• Liquid drop: ΔP = 2S/R
• Soap bubble: ΔP = 4S/R
• Air bubble in liquid: same as liquid drop

Excess Pressure Inside Liquid Drop

• Consider a liquid drop of radius R and surface tension S.
• Pressure difference: ΔP = P - P₀.
• Excess pressure inside drop: ΔP = 2S/R.
• Work done in increasing radius ΔR: ΔW = ΔP × 4 πR²ΔR.
• Increase in potential energy due to surface tension: ΔU = 8 πSRΔR.
• At equilibrium, ΔU = ΔW leads to ΔP = 2S/R.

Excess Pressure Inside Soap Bubble

• Soap bubble has two surfaces in contact with air.
• Excess pressure inside bubble: ΔP = 4S/R.
• Work done: ΔW = ΔP × 4 πR²ΔR.
• Increase in potential energy: ΔU = 16 πSRΔR.
• At equilibrium, ΔU = ΔW leads to ΔP = 4S/R.

Excess Pressure Inside Air Bubble in Liquid

• Similar to liquid drop (one surface in contact with air).
• Excess pressure: ΔP = 2S/R.

Solved Examples

• Ex 1: Calculate excess pressure inside a drop of radius 0.01 m with surface tension 0.07 N/m.
  • ΔP = 2 × 0.07/0.01 = 14 N/m².
• Ex 2: Calculate excess pressure inside a soap bubble of radius 0.01 m with surface tension 0.07 N/m.
  • ΔP = 4 × 0.07/0.01 = 28 N/m².

Practice Set

• Q1: Excess pressure inside a liquid drop is:
  • A) 4S/R
  • B) 2S/R
  • C) S/R
  • D) 8S/R
• Q2: Excess pressure inside a soap bubble is:
  • A) 2S/R
  • B) 4S/R
  • C) S/R
  • D) 6S/R
• Q3: Excess pressure inside an air bubble in liquid is:
  • A) 4S/R
  • B) 2S/R
  • C) S/R
  • D) 8S/R

Answer Key

• Q1: B - Excess pressure inside liquid drop is 2S/R.
• Q2: B - Excess pressure inside soap bubble is 4S/R.
• Q3: B - Excess pressure inside air bubble in liquid is 2S/R.

Memory Aids

• Liquid drop: ΔP = 2S/R
• Soap bubble: ΔP = 4S/R
• Air bubble in liquid: same as liquid drop

Excess Pressure Inside Liquid Drop

• Consider a liquid drop of radius R and surface tension S.
• Pressure difference: ΔP = P - P₀.
• Excess pressure inside drop: ΔP = 2S/R.
• Work done in increasing radius ΔR: ΔW = ΔP × 4 πR²ΔR.
• Increase in potential energy due to surface tension: ΔU = 8 πSRΔR.
• At equilibrium, ΔU = ΔW leads to ΔP = 2S/R.

Excess Pressure Inside Soap Bubble

• Soap bubble has two surfaces in contact with air.
• Excess pressure inside bubble: ΔP = 4S/R.
• Work done: ΔW = ΔP × 4 πR²ΔR.
• Increase in potential energy: ΔU = 16 πSRΔR.
• At equilibrium, ΔU = ΔW leads to ΔP = 4S/R.

Excess Pressure Inside Air Bubble in Liquid

• Similar to liquid drop (one surface in contact with air).
• Excess pressure: ΔP = 2S/R.

Solved Examples

• Ex 1: Calculate excess pressure inside a drop of radius 0.01 m with surface tension 0.07 N/m.
  • ΔP = 2 × 0.07/0.01 = 14 N/m².
• Ex 2: Calculate excess pressure inside a soap bubble of radius 0.01 m with surface tension 0.07 N/m.
  • ΔP = 4 × 0.07/0.01 = 28 N/m².

Practice Set

• Q1: Excess pressure inside a liquid drop is:
  • A) 4S/R
  • B) 2S/R
  • C) S/R
  • D) 8S/R
• Q2: Excess pressure inside a soap bubble is:
  • A) 2S/R
  • B) 4S/R
  • C) S/R
  • D) 6S/R
• Q3: Excess pressure inside an air bubble in liquid is:
  • A) 4S/R
  • B) 2S/R
  • C) S/R
  • D) 8S/R

Answer Key

• Q1: B - Excess pressure inside liquid drop is 2S/R.
• Q2: B - Excess pressure inside soap bubble is 4S/R.
• Q3: B - Excess pressure inside air bubble in liquid is 2S/R.

Memory Aids

• Liquid drop: ΔP = 2S/R
• Soap bubble: ΔP = 4S/R
• Air bubble in liquid: same as liquid drop

Excess Pressure Inside Liquid Drop

• Consider a liquid drop of radius R and surface tension S.
• Pressure difference: ΔP = P - P₀.
• Excess pressure inside drop: ΔP = 2S/R.
• Work done in increasing radius ΔR: ΔW = ΔP × 4 πR²ΔR.
• Increase in potential energy due to surface tension: ΔU = 8 πSRΔR.
• At equilibrium, ΔU = ΔW leads to ΔP = 2S/R.

Excess Pressure Inside Soap Bubble

• Soap bubble has two surfaces in contact with air.
• Excess pressure inside bubble: ΔP = 4S/R.
• Work done: ΔW = ΔP × 4 πR²ΔR.
• Increase in potential energy: ΔU = 16 πSRΔR.
• At equilibrium, ΔU = ΔW leads to ΔP = 4S/R.

Excess Pressure Inside Air Bubble in Liquid

• Similar to liquid drop (one surface in contact with air).
• Excess pressure: ΔP = 2S/R.

Solved Examples

• Ex 1: Calculate excess pressure inside a drop of radius 0.01 m with surface tension 0.07 N/m.
  • ΔP = 2 × 0.07/0.01 = 14 N/m².
• Ex 2: Calculate excess pressure inside a soap bubble of radius 0.01 m with surface tension 0.07 N/m.
  • ΔP = 4 × 0.07/0.01 = 28 N/m².

Practice Set

• Q1: Excess pressure inside a liquid drop is:
  • A) 4S/R
  • B) 2S/R
  • C) S/R
  • D) 8S/R
• Q2: Excess pressure inside a soap bubble is:
  • A) 2S/R
  • B) 4S/R
  • C) S/R
  • D) 6S/R
• Q3: Excess pressure inside an air bubble in liquid is:
  • A) 4S/R
  • B) 2S/R
  • C) S/R
  • D) 8S/R

Answer Key

• Q1: B - Excess pressure inside liquid drop is 2S/R.
• Q2: B - Excess pressure inside soap bubble is 4S/R.
• Q3: B - Excess pressure inside air bubble in liquid is 2S/R.

Memory Aids

• Liquid drop: ΔP = 2S/R
• Soap bubble: ΔP = 4S/R
• Air bubble in liquid: same as liquid drop

Excess Pressure Inside Liquid Drop

• Consider a liquid drop of radius R and surface tension S.
• Pressure difference: ΔP = P - P₀.
• Excess pressure inside drop: ΔP = 2S/R.
• Work done in increasing radius ΔR: ΔW = ΔP × 4 πR²ΔR.
• Increase in potential energy due to surface tension: ΔU = 8 πSRΔR.
• At equilibrium, ΔU = ΔW leads to ΔP = 2S/R.

Excess Pressure Inside Soap Bubble

• Soap bubble has two surfaces in contact with air.
• Excess pressure inside bubble: ΔP = 4S/R.
• Work done: ΔW = ΔP × 4 πR²ΔR.
• Increase in potential energy: ΔU = 16 πSRΔR.
• At equilibrium, ΔU = ΔW leads to ΔP = 4S/R.

Excess Pressure Inside Air Bubble in Liquid

• Similar to liquid drop (one surface in contact with air).
• Excess pressure: ΔP = 2S/R.

Solved Examples

• Ex 1: Calculate excess pressure inside a drop of radius 0.01 m with surface tension 0.07 N/m.
  • ΔP = 2 × 0.07/0.01 = 14 N/m².
• Ex 2: Calculate excess pressure inside a soap bubble of radius 0.01 m with surface tension 0.07 N/m.
  • ΔP = 4 × 0.07/0.01 = 28 N/m².

Practice Set

• Q1: Excess pressure inside a liquid drop is:
  • A) 4S/R
  • B) 2S/R
  • C) S/R
  • D) 8S/R
• Q2: Excess pressure inside a soap bubble is:
  • A) 2S/R
  • B) 4S/R
  • C) S/R
  • D) 6S/R
• Q3: Excess pressure inside an air bubble in liquid is:
  • A) 4S/R
  • B) 2S/R
  • C) S/R
  • D) 8S/R

Answer Key

• Q1: B - Excess pressure inside liquid drop is 2S/R.
• Q2: B - Excess pressure inside soap bubble is 4S/R.
• Q3: B - Excess pressure inside air bubble in liquid is 2S/R.

Memory Aids

• Liquid drop: ΔP = 2S/R
• Soap bubble: ΔP = 4S/R
• Air bubble in liquid: same as liquid drop

Excess Pressure Inside Liquid Drop

• Consider a liquid drop of radius R and surface tension S.
• Pressure difference: ΔP = P - P₀.
• Excess pressure inside drop: ΔP = 2S/R.
• Work done in increasing radius ΔR: ΔW = ΔP × 4 πR²ΔR.
• Increase in potential energy due to surface tension: ΔU = 8 πSRΔR.
• At equilibrium, ΔU = ΔW leads to ΔP = 2S/R.

Excess Pressure Inside Soap Bubble

• Soap bubble has two surfaces in contact with air.
• Excess pressure inside bubble: ΔP = 4S/R.
• Work done: ΔW = ΔP × 4 πR²ΔR.
• Increase in potential energy: ΔU = 16 πSRΔR.
• At equilibrium, ΔU = ΔW leads to ΔP = 4S/R.

Excess Pressure Inside Air Bubble in Liquid

• Similar to liquid drop (one surface in contact with air).
• Excess pressure: ΔP = 2S/R.

Solved Examples

• Ex 1: Calculate excess pressure inside a drop of radius 0.01 m with surface tension 0.07 N/m.
  • ΔP = 2 × 0.07/0.01 = 14 N/m².
• Ex 2: Calculate excess pressure inside a soap bubble of radius 0.01 m with surface tension 0.07 N/m.
  • ΔP = 4 × 0.07/0.01 = 28 N/m².

Practice Set

• Q1: Excess pressure inside a liquid drop is:
  • A) 4S/R
  • B) 2S/R
  • C) S/R
  • D) 8S/R
• Q2: Excess pressure inside a soap bubble is:
  • A) 2S/R
  • B) 4S/R
  • C) S/R
  • D) 6S/R
• Q3: Excess pressure inside an air bubble in liquid is:
  • A) 4S/R
  • B) 2S/R
  • C) S/R
  • D) 8S/R

Answer Key

• Q1: B - Excess pressure inside liquid drop is 2S/R.
• Q2: B - Excess pressure inside soap bubble is 4S/R.
• Q3: B - Excess pressure inside air bubble in liquid is 2S/R.

Memory Aids

• Liquid drop: ΔP = 2S/R
• Soap bubble: ΔP = 4S/R
• Air bubble in liquid: same as liquid drop

Excess Pressure Inside Liquid Drop

• Consider a liquid drop of radius R and surface tension S.
• Pressure difference: ΔP = P - P₀.
• Excess pressure inside drop: ΔP = 2S/R.
• Work done in increasing radius ΔR: ΔW = ΔP × 4 πR²ΔR.
• Increase in potential energy due to surface tension: ΔU = 8 πSRΔR.
• At equilibrium, ΔU = ΔW leads to ΔP = 2S/R.

Excess Pressure Inside Soap Bubble

• Soap bubble has two surfaces in contact with air.
• Excess pressure inside bubble: ΔP = 4S/R.
• Work done: ΔW = ΔP × 4 πR²ΔR.
• Increase in potential energy: ΔU = 16 πSRΔR.
• At equilibrium, ΔU = ΔW leads to ΔP = 4S/R.

Excess Pressure Inside Air Bubble in Liquid

• Similar to liquid drop (one surface in contact with air).
• Excess pressure: ΔP = 2S/R.

Solved Examples

• Ex 1: Calculate excess pressure inside a drop of radius 0.01 m with surface tension 0.07 N/m.
  • ΔP = 2 × 0.07/0.01 = 14 N/m².
• Ex 2: Calculate excess pressure inside a soap bubble of radius 0.01 m with surface tension 0.07 N/m.
  • ΔP = 4 × 0.07/0.01 = 28 N/m².

Practice Set

• Q1: Excess pressure inside a liquid drop is:
  • A) 4S/R
  • B) 2S/R
  • C) S/R
  • D) 8S/R
• Q2: Excess pressure inside a soap bubble is:
  • A) 2S/R
  • B) 4S/R
  • C) S/R
  • D) 6S/R
• Q3: Excess pressure inside an air bubble in liquid is:
  • A) 4S/R
  • B) 2S/R
  • C) S/R
  • D) 8S/R

Answer Key

• Q1: B - Excess pressure inside liquid drop is 2S/R.
• Q2: B - Excess pressure inside soap bubble is 4S/R.
• Q3: B - Excess pressure inside air bubble in liquid is 2S/R.

Memory Aids

• Liquid drop: ΔP = 2S/R
• Soap bubble: ΔP = 4S/R
• Air bubble in liquid: same as liquid drop

Excess Pressure Inside Liquid Drop

• Consider a liquid drop of radius R and surface tension S.
• Pressure difference: ΔP = P - P₀.
• Excess pressure inside drop: ΔP = 2S/R.
• Work done in increasing radius ΔR: ΔW = ΔP × 4 πR²ΔR.
• Increase in potential energy due to surface tension: ΔU = 8 πSRΔR.
• At equilibrium, ΔU = ΔW leads to ΔP = 2S/R.

Excess Pressure Inside Soap Bubble

• Soap bubble has two surfaces in contact with air.
• Excess pressure inside bubble: ΔP = 4S/R.
• Work done: ΔW = ΔP × 4 πR²ΔR.
• Increase in potential energy: ΔU = 16 πSRΔR.
• At equilibrium, ΔU = ΔW leads to ΔP = 4S/R.

Excess Pressure Inside Air Bubble in Liquid

• Similar to liquid drop (one surface in contact with air).
• Excess pressure: ΔP = 2S/R.

Solved Examples

• Ex 1: Calculate excess pressure inside a drop of radius 0.01 m with surface tension 0.07 N/m.
  • ΔP = 2 × 0.07/0.01 = 14 N/m².
• Ex 2: Calculate excess pressure inside a soap bubble of radius 0.01 m with surface tension 0.07 N/m.
  • ΔP = 4 × 0.07/0.01 = 28 N/m².

Practice Set

• Q1: Excess pressure inside a liquid drop is:
  • A) 4S/R
  • B) 2S/R
  • C) S/R
  • D) 8S/R
• Q2: Excess pressure inside a soap bubble is:
  • A) 2S/R
  • B) 4S/R
  • C) S/R
  • D) 6S/R
• Q3: Excess pressure inside an air bubble in liquid is:
  • A) 4S/R
  • B) 2S/R
  • C) S/R
  • D) 8S/R

Answer Key

• Q1: B - Excess pressure inside liquid drop is 2S/R.
• Q2: B - Excess pressure inside soap bubble is 4S/R.
• Q3: B - Excess pressure inside air bubble in liquid is 2S/R.

Memory Aids

• Liquid drop: ΔP = 2S/R
• Soap bubble: ΔP = 4S/R
• Air bubble in liquid: same as liquid drop

Excess Pressure Inside Liquid Drop

• Consider a liquid drop of radius R and surface tension S.
• Pressure difference: ΔP = P - P₀.
• Excess pressure inside drop: ΔP = 2S/R.
• Work done in increasing radius ΔR: ΔW = ΔP × 4 πR²ΔR.
• Increase in potential energy due to surface tension: ΔU = 8 πSRΔR.
• At equilibrium, ΔU = ΔW leads to ΔP = 2S/R.

Excess Pressure Inside Soap Bubble

• Soap bubble has two surfaces in contact with air.
• Excess pressure inside bubble: ΔP = 4S/R.
• Work done: ΔW = ΔP × 4 πR²ΔR.
• Increase in potential energy: ΔU = 16 πSRΔR.
• At equilibrium, ΔU = ΔW leads to ΔP = 4S/R.

Excess Pressure Inside Air Bubble in Liquid

• Similar to liquid drop (one surface in contact with air).
• Excess pressure: ΔP = 2S/R.

Solved Examples

• Ex 1: Calculate excess pressure inside a drop of radius 0.01 m with surface tension 0.07 N/m.
  • ΔP = 2 × 0.07/0.01 = 14 N/m².
• Ex 2: Calculate excess pressure inside a soap bubble of radius 0.01 m with surface tension 0.07 N/m.
  • ΔP = 4 × 0.07/0.01 = 28 N/m².

Practice Set

• Q1: Excess pressure inside a liquid drop is:
  • A) 4S/R
  • B) 2S/R
  • C) S/R
  • D) 8S/R
• Q2: Excess pressure inside a soap bubble is:
  • A) 2S/R
  • B) 4S/R
  • C) S/R
  • D) 6S/R
• Q3: Excess pressure inside an air bubble in liquid is:
  • A) 4S/R
  • B) 2S/R
  • C) S/R
  • D) 8S/R

Answer Key

• Q1: B - Excess pressure inside liquid drop is 2S/R.
• Q2: B - Excess pressure inside soap bubble is 4S/R.
• Q3: B - Excess pressure inside air bubble in liquid is 2S/R.

Memory Aids

• Liquid drop: ΔP = 2S/R
• Soap bubble: ΔP = 4S/R
• Air bubble in liquid: same as liquid drop

Excess Pressure Inside Liquid Drop

• Consider a liquid drop of radius R and surface tension S.
• Pressure difference: ΔP = P - P₀.
• Excess pressure inside drop: ΔP = 2S/R.
• Work done in increasing radius ΔR: ΔW = ΔP × 4 πR²ΔR.
• Increase in potential energy due to surface tension: ΔU = 8 πSRΔR.
• At equilibrium, ΔU = ΔW leads to ΔP = 2S/R.

Excess Pressure Inside Soap Bubble

• Soap bubble has two surfaces in contact with air.
• Excess pressure inside bubble: ΔP = 4S/R.
• Work done: ΔW = ΔP × 4 πR²ΔR.
• Increase in potential energy: ΔU = 16 πSRΔR.
• At equilibrium, ΔU = ΔW leads to ΔP = 4S/R.

Excess Pressure Inside Air Bubble in Liquid

• Similar to liquid drop (one surface in contact with air).
• Excess pressure: ΔP = 2S/R.

Solved Examples

• Ex 1: Calculate excess pressure inside a drop of radius 0.01 m with surface tension 0.07 N/m.
  • ΔP = 2 × 0.07/0.01 = 14 N/m².
• Ex 2: Calculate excess pressure inside a soap bubble of radius 0.01 m with surface tension 0.07 N/m.
  • ΔP = 4 × 0.07/0.01 = 28 N/m².

Practice Set

• Q1: Excess pressure inside a liquid drop is:
  • A) 4S/R
  • B) 2S/R
  • C) S/R
  • D) 8S/R
• Q2: Excess pressure inside a soap bubble is:
  • A) 2S/R
  • B) 4S/R
  • C) S/R
  • D) 6S/R
• Q3: Excess pressure inside an air bubble in liquid is:
  • A) 4S/R
  • B) 2S/R
  • C) S/R
  • D) 8S/R

Answer Key

• Q1: B - Excess pressure inside liquid drop is 2S/R.
• Q2: B - Excess pressure inside soap bubble is 4S/R.
• Q3: B - Excess pressure inside air bubble in liquid is 2S/R.

Memory Aids

• Liquid drop: ΔP = 2S/R
• Soap bubble: ΔP = 4S/R
• Air bubble in liquid: same as liquid drop

Excess Pressure Inside Liquid Drop

• Consider a liquid drop of radius R and surface tension S.
• Pressure difference: ΔP = P - P₀.
• Excess pressure inside drop: ΔP = 2S/R.
• Work done in increasing radius ΔR: ΔW = ΔP × 4 πR²ΔR.
• Increase in potential energy due to surface tension: ΔU = 8 πSRΔR.
• At equilibrium, ΔU = ΔW leads to ΔP = 2S/R.

Excess Pressure Inside Soap Bubble

• Soap bubble has two surfaces in contact with air.
• Excess pressure inside bubble: ΔP = 4S/R.
• Work done: ΔW = ΔP × 4 πR²ΔR.
• Increase in potential energy: ΔU = 16 πSRΔR.
• At equilibrium, ΔU = ΔW leads to ΔP = 4S/R.

Excess Pressure Inside Air Bubble in Liquid

• Similar to liquid drop (one surface in contact with air).
• Excess pressure: ΔP = 2S/R.

Solved Examples

• Ex 1: Calculate excess pressure inside a drop of radius 0.01 m with surface tension 0.07 N/m.
  • ΔP = 2 × 0.07/0.01 = 14 N/m².
• Ex 2: Calculate excess pressure inside a soap bubble of radius 0.01 m with surface tension 0.07 N/m.
  • ΔP = 4 × 0.07/0.01 = 28 N/m².

Practice Set

• Q1: Excess pressure inside a liquid drop is:
  • A) 4S/R
  • B) 2S/R
  • C) S/R
  • D) 8S/R
• Q2: Excess pressure inside a soap bubble is:
  • A) 2S/R
  • B) 4S/R
  • C) S/R
  • D) 6S/R
• Q3: Excess pressure inside an air bubble in liquid is:
  • A) 4S/R
  • B) 2S/R
  • C) S/R
  • D) 8S/R

Answer Key

• Q1: B - Excess pressure inside liquid drop is 2S/R.
• Q2: B - Excess pressure inside soap bubble is 4S/R.
• Q3: B - Excess pressure inside air bubble in liquid is 2S/R.

Memory Aids

• Liquid drop: ΔP = 2S/R
• Soap bubble: ΔP = 4S/R
• Air bubble in liquid: same as liquid drop

Excess Pressure Inside Liquid Drop

• Consider a liquid drop of radius R and surface tension S.
• Pressure difference: ΔP = P - P₀.
• Excess pressure inside drop: ΔP = 2S/R.
• Work done in increasing radius ΔR: ΔW = ΔP × 4 πR²ΔR.
• Increase in potential energy due to surface tension: ΔU = 8 πSRΔR.
• At equilibrium, ΔU = ΔW leads to ΔP = 2S/R.

Excess Pressure Inside Soap Bubble

• Soap bubble has two surfaces in contact with air.
• Excess pressure inside bubble: ΔP = 4S/R.
• Work done: ΔW = ΔP × 4 πR²ΔR.
• Increase in potential energy: ΔU = 16 πSRΔR.
• At equilibrium, ΔU = ΔW leads to ΔP = 4S/R.

Excess Pressure Inside Air Bubble in Liquid

• Similar to liquid drop (one surface in contact with air).
• Excess pressure: ΔP = 2S/R.

Solved Examples

• Ex 1: Calculate excess pressure inside a drop of radius 0.01 m with surface tension 0.07 N/m.
  • ΔP = 2 × 0.07/0.01 = 14 N/m².
• Ex 2: Calculate excess pressure inside a soap bubble of radius 0.01 m with surface tension 0.07 N/m.
  • ΔP = 4 × 0.07/0.01 = 28 N/m².

Practice Set

• Q1: Excess pressure inside a liquid drop is:
  • A) 4S/R
  • B) 2S/R
  • C) S/R
  • D) 8S/R
• Q2: Excess pressure inside a soap bubble is:
  • A) 2S/R
  • B) 4S/R
  • C) S/R
  • D) 6S/R
• Q3: Excess pressure inside an air bubble in liquid is:
  • A) 4S/R
  • B) 2S/R
  • C) S/R
  • D) 8S/R

Answer Key

• Q1: B - Excess pressure inside liquid drop is 2S/R.
• Q2: B - Excess pressure inside soap bubble is 4S/R.
• Q3: B - Excess pressure inside air bubble in liquid is 2S/R.

Memory Aids

• Liquid drop: ΔP = 2S/R
• Soap bubble: ΔP = 4S/R
• Air bubble in liquid: same as liquid drop

Excess Pressure Inside Liquid Drop

• Consider a liquid drop of radius \(R\) and surface tension \(S\).
• Pressure difference: \( \Delta P = P - P_0 \).
• Excess pressure inside drop: \( \Delta P = \frac{2S}{R} \).
• Work done in increasing radius \( \Delta R \): \( \Delta W = \Delta P \times 4 \pi R^2 \Delta R \).
• Increase in potential energy due to surface tension: \( \Delta U = 8 \pi S R \Delta R \).
• At equilibrium, \( \Delta U = \Delta W \) leads to \( \Delta P = \frac{2S}{R} \).

Excess Pressure Inside Soap Bubble

• Soap bubble has two surfaces in contact with air.
• Excess pressure inside bubble: \( \Delta P = \frac{4S}{R} \).
• Work done: \( \Delta W = \Delta P \times 4 \pi R^2 \Delta R \).
• Increase in potential energy: \( \Delta U = 16 \pi S R \Delta R \).
• At equilibrium, \( \Delta U = \Delta W \) leads to \( \Delta P = \frac{4S}{R} \).

Excess Pressure Inside Air Bubble in Liquid

• Similar to liquid drop (one surface in contact with air).
• Excess pressure: \( \Delta P = \frac{2S}{R} \).

Solved Examples

• Ex 1: Calculate excess pressure inside a drop of radius 0.01 m with surface tension 0.07 N/m.
  • \( \Delta P = \frac{2 \times 0.07}{0.01} = 14 \text{ N/m}^2 \).
• Ex 2: Calculate excess pressure inside a soap bubble of radius 0.01 m with surface tension 0.07 N/m.
  • \( \Delta P = \frac{4 \times 0.07}{0.01} = 28 \text{ N/m}^2 \).

Practice Set

• Q1: Excess pressure inside a liquid drop is:
  • A) \( \frac{4S}{R} \)
  • B) \( \frac{2S}{R} \)
  • C) \( \frac{S}{R} \)
  • D) \( \frac{8S}{R} \)
• Q2: Excess pressure inside a soap bubble is:
  • A) \( \frac{2S}{R} \)
  • B) \( \frac{4S}{R} \)
  • C) \( \frac{S}{R} \)
  • D) \( \frac{6S}{R} \)
• Q3: Excess pressure inside an air bubble in liquid is:
  • A) \( \frac{4S}{R} \)
  • B) \( \frac{2S}{R} \)
  • C) \( \frac{S}{R} \)
  • D) \( \frac{8S}{R} \)

Answer Key

• Q1: B - Excess pressure inside liquid drop is \( \frac{2S}{R} \).
• Q2: B - Excess pressure inside soap bubble is \( \frac{4S}{R} \).
• Q3: B - Excess pressure inside air bubble in liquid is \( \frac{2S}{R} \).

Memory Aids

• Liquid drop: \( \Delta P = \frac{2S}{R} \)
• Soap bubble: \( \Delta P = \frac{4S}{R} \)
• Air bubble in liquid: same as liquid drop

Capillary Rise

Explanation and Formula

• Occurs when liquid rises or falls in a narrow tube due to surface tension and angle of contact.
• Pressure difference at meniscus: P_i - P₀ = 2S/R = 2S/(a sec θ) = 2S/a cos θ.
• Height of rise/fall: h = 2S/(ρga) cos θ, where a is tube radius, ρ is liquid density, g is acceleration due to gravity.
• If θ < 90° (concave meniscus), liquid rises (h > 0).
• If θ > 90° (convex meniscus), liquid falls (h < 0).

Solved Examples

• Ex 1: Calculate capillary rise for water (S=0.072,ρ=1000 kg/m³) in a tube of radius 0.001 m with contact angle 0°.
  • h = 2 × 0.072/(1000 × 9.8 × 0.001) cos 0° = 14.7 m.

Practice Set

• Q1: Capillary rise occurs when contact angle is:
  • A) < 90°
  • B) = 90°
  • C) > 90°
  • D) 180°
• Q2: Formula for capillary rise height is:
  • A) h = 2S/(ρga) cos θ
  • B) h = S/(ρga) sin θ
  • C) h = 2S/(ρga) sin θ
  • D) h = S/(ρga) cos θ
• Q3: If contact angle is 120°, capillary height is:
  • A) Positive (rise)
  • B) Zero
  • C) Negative (fall)
  • D) Infinite

Answer Key

• Q1: A - Capillary rise occurs for contact angle less than 90°.
• Q2: A - Formula is h = 2S/(ρga) cos θ.
• Q3: C - Contact angle greater than 90° causes liquid to fall (negative height).

Memory Aids

• Capillary rise formula: h = 2S/(ρga) cos θ
• θ < 90°: rise, θ > 90°: fall

Explanation and Formula

• Occurs when liquid rises or falls in a narrow tube due to surface tension and angle of contact.
• Pressure difference at meniscus: P_i - P₀ = 2S/R = 2S/(a sec θ) = 2S/a cos θ.
• Height of rise/fall: h = 2S/(ρga) cos θ, where a is tube radius, ρ is liquid density, g is acceleration due to gravity.
• If θ < 90° (concave meniscus), liquid rises (h > 0).
• If θ > 90° (convex meniscus), liquid falls (h < 0).

Solved Examples

• Ex 1: Calculate capillary rise for water (S=0.072,ρ=1000 kg/m³) in a tube of radius 0.001 m with contact angle 0°.
  • h = 2 × 0.072/(1000 × 9.8 × 0.001) cos 0° = 14.7 m.

Practice Set

• Q1: Capillary rise occurs when contact angle is:
  • A) < 90°
  • B) = 90°
  • C) > 90°
  • D) 180°
• Q2: Formula for capillary rise height is:
  • A) h = 2S/(ρga) cos θ
  • B) h = S/(ρga) sin θ
  • C) h = 2S/(ρga) sin θ
  • D) h = S/(ρga) cos θ
• Q3: If contact angle is 120°, capillary height is:
  • A) Positive (rise)
  • B) Zero
  • C) Negative (fall)
  • D) Infinite

Answer Key

• Q1: A - Capillary rise occurs for contact angle less than 90°.
• Q2: A - Formula is h = 2S/(ρga) cos θ.
• Q3: C - Contact angle greater than 90° causes liquid to fall (negative height).

Memory Aids

• Capillary rise formula: h = 2S/(ρga) cos θ
• θ < 90°: rise, θ > 90°: fall

Explanation and Formula

• Occurs when liquid rises or falls in a narrow tube due to surface tension and angle of contact.
• Pressure difference at meniscus: P_i - P₀ = 2S/R = 2S/(a sec θ) = 2S/a cos θ.
• Height of rise/fall: h = 2S/(ρga) cos θ, where a is tube radius, ρ is liquid density, g is acceleration due to gravity.
• If θ < 90° (concave meniscus), liquid rises (h > 0).
• If θ > 90° (convex meniscus), liquid falls (h < 0).

Solved Examples

• Ex 1: Calculate capillary rise for water (S=0.072,ρ=1000 kg/m³) in a tube of radius 0.001 m with contact angle 0°.
  • h = 2 × 0.072/(1000 × 9.8 × 0.001) cos 0° = 14.7 m.

Practice Set

• Q1: Capillary rise occurs when contact angle is:
  • A) < 90°
  • B) = 90°
  • C) > 90°
  • D) 180°
• Q2: Formula for capillary rise height is:
  • A) h = 2S/(ρga) cos θ
  • B) h = S/(ρga) sin θ
  • C) h = 2S/(ρga) sin θ
  • D) h = S/(ρga) cos θ
• Q3: If contact angle is 120°, capillary height is:
  • A) Positive (rise)
  • B) Zero
  • C) Negative (fall)
  • D) Infinite

Answer Key

• Q1: A - Capillary rise occurs for contact angle less than 90°.
• Q2: A - Formula is h = 2S/(ρga) cos θ.
• Q3: C - Contact angle greater than 90° causes liquid to fall (negative height).

Memory Aids

• Capillary rise formula: h = 2S/(ρga) cos θ
• θ < 90°: rise, θ > 90°: fall

Explanation and Formula

• Occurs when liquid rises or falls in a narrow tube due to surface tension and angle of contact.
• Pressure difference at meniscus: P_i - P₀ = 2S/R = 2S/(a sec θ) = 2S/a cos θ.
• Height of rise/fall: h = 2S/(ρga) cos θ, where a is tube radius, ρ is liquid density, g is acceleration due to gravity.
• If θ < 90° (concave meniscus), liquid rises (h > 0).
• If θ > 90° (convex meniscus), liquid falls (h < 0).

Solved Examples

• Ex 1: Calculate capillary rise for water (S=0.072,ρ=1000 kg/m³) in a tube of radius 0.001 m with contact angle 0°.
  • h = 2 × 0.072/(1000 × 9.8 × 0.001) cos 0° = 14.7 m.

Practice Set

• Q1: Capillary rise occurs when contact angle is:
  • A) < 90°
  • B) = 90°
  • C) > 90°
  • D) 180°
• Q2: Formula for capillary rise height is:
  • A) h = 2S/(ρga) cos θ
  • B) h = S/(ρga) sin θ
  • C) h = 2S/(ρga) sin θ
  • D) h = S/(ρga) cos θ
• Q3: If contact angle is 120°, capillary height is:
  • A) Positive (rise)
  • B) Zero
  • C) Negative (fall)
  • D) Infinite

Answer Key

• Q1: A - Capillary rise occurs for contact angle less than 90°.
• Q2: A - Formula is h = 2S/(ρga) cos θ.
• Q3: C - Contact angle greater than 90° causes liquid to fall (negative height).

Memory Aids

• Capillary rise formula: h = 2S/(ρga) cos θ
• θ < 90°: rise, θ > 90°: fall

Explanation and Formula

• Occurs when liquid rises or falls in a narrow tube due to surface tension and angle of contact.
• Pressure difference at meniscus: P_i - P₀ = 2S/R = 2S/(a sec θ) = 2S/a cos θ.
• Height of rise/fall: h = 2S/(ρga) cos θ, where a is tube radius, ρ is liquid density, g is acceleration due to gravity.
• If θ < 90° (concave meniscus), liquid rises (h > 0).
• If θ > 90° (convex meniscus), liquid falls (h < 0).

Solved Examples

• Ex 1: Calculate capillary rise for water (S=0.072,ρ=1000 kg/m³) in a tube of radius 0.001 m with contact angle 0°.
  • h = 2 × 0.072/(1000 × 9.8 × 0.001) cos 0° = 14.7 m.

Practice Set

• Q1: Capillary rise occurs when contact angle is:
  • A) < 90°
  • B) = 90°
  • C) > 90°
  • D) 180°
• Q2: Formula for capillary rise height is:
  • A) h = 2S/(ρga) cos θ
  • B) h = S/(ρga) sin θ
  • C) h = 2S/(ρga) sin θ
  • D) h = S/(ρga) cos θ
• Q3: If contact angle is 120°, capillary height is:
  • A) Positive (rise)
  • B) Zero
  • C) Negative (fall)
  • D) Infinite

Answer Key

• Q1: A - Capillary rise occurs for contact angle less than 90°.
• Q2: A - Formula is h = 2S/(ρga) cos θ.
• Q3: C - Contact angle greater than 90° causes liquid to fall (negative height).

Memory Aids

• Capillary rise formula: h = 2S/(ρga) cos θ
• θ < 90°: rise, θ > 90°: fall

Explanation and Formula

• Occurs when liquid rises or falls in a narrow tube due to surface tension and angle of contact.
• Pressure difference at meniscus: P_i - P₀ = 2S/R = 2S/(a sec θ) = 2S/a cos θ.
• Height of rise/fall: h = 2S/(ρga) cos θ, where a is tube radius, ρ is liquid density, g is acceleration due to gravity.
• If θ < 90° (concave meniscus), liquid rises (h > 0).
• If θ > 90° (convex meniscus), liquid falls (h < 0).

Solved Examples

• Ex 1: Calculate capillary rise for water (S=0.072,ρ=1000 kg/m³) in a tube of radius 0.001 m with contact angle 0°.
  • h = 2 × 0.072/(1000 × 9.8 × 0.001) cos 0° = 14.7 m.

Practice Set

• Q1: Capillary rise occurs when contact angle is:
  • A) < 90°
  • B) = 90°
  • C) > 90°
  • D) 180°
• Q2: Formula for capillary rise height is:
  • A) h = 2S/(ρga) cos θ
  • B) h = S/(ρga) sin θ
  • C) h = 2S/(ρga) sin θ
  • D) h = S/(ρga) cos θ
• Q3: If contact angle is 120°, capillary height is:
  • A) Positive (rise)
  • B) Zero
  • C) Negative (fall)
  • D) Infinite

Answer Key

• Q1: A - Capillary rise occurs for contact angle less than 90°.
• Q2: A - Formula is h = 2S/(ρga) cos θ.
• Q3: C - Contact angle greater than 90° causes liquid to fall (negative height).

Memory Aids

• Capillary rise formula: h = 2S/(ρga) cos θ
• θ < 90°: rise, θ > 90°: fall

Explanation and Formula

• Occurs when liquid rises or falls in a narrow tube due to surface tension and angle of contact.
• Pressure difference at meniscus: P_i - P₀ = 2S/R = 2S/(a sec θ) = 2S/a cos θ.
• Height of rise/fall: h = 2S/(ρga) cos θ, where a is tube radius, ρ is liquid density, g is acceleration due to gravity.
• If θ < 90° (concave meniscus), liquid rises (h > 0).
• If θ > 90° (convex meniscus), liquid falls (h < 0).

Solved Examples

• Ex 1: Calculate capillary rise for water (S=0.072,ρ=1000 kg/m³) in a tube of radius 0.001 m with contact angle 0°.
  • h = 2 × 0.072/(1000 × 9.8 × 0.001) cos 0° = 14.7 m.

Practice Set

• Q1: Capillary rise occurs when contact angle is:
  • A) < 90°
  • B) = 90°
  • C) > 90°
  • D) 180°
• Q2: Formula for capillary rise height is:
  • A) h = 2S/(ρga) cos θ
  • B) h = S/(ρga) sin θ
  • C) h = 2S/(ρga) sin θ
  • D) h = S/(ρga) cos θ
• Q3: If contact angle is 120°, capillary height is:
  • A) Positive (rise)
  • B) Zero
  • C) Negative (fall)
  • D) Infinite

Answer Key

• Q1: A - Capillary rise occurs for contact angle less than 90°.
• Q2: A - Formula is h = 2S/(ρga) cos θ.
• Q3: C - Contact angle greater than 90° causes liquid to fall (negative height).

Memory Aids

• Capillary rise formula: h = 2S/(ρga) cos θ
• θ < 90°: rise, θ > 90°: fall

Explanation and Formula

• Occurs when liquid rises or falls in a narrow tube due to surface tension and angle of contact.
• Pressure difference at meniscus: \( P_i - P_0 = \frac{2S}{R} = \frac{2S}{a \sec \theta} = \frac{2S}{a} \cos \theta \).
• Height of rise/fall: \( h = \frac{2S}{\rho g a} \cos \theta \), where \(a\) is tube radius, \(\rho\) is liquid density, \(g\) is acceleration due to gravity.
• If \( \theta < 90^\circ \) (concave meniscus), liquid rises (\(h > 0\)).
• If \( \theta > 90^\circ \) (convex meniscus), liquid falls (\(h < 0\)).

Solved Examples

• Ex 1: Calculate capillary rise for water (\(S=0.072\,\text{N/m}\), \(\rho=1000\,\text{kg/m}^3\)) in a tube of radius 0.001 m with contact angle 0°.
  • \( h = \frac{2 \times 0.072}{1000 \times 9.8 \times 0.001} \cos 0^\circ = 14.7 \text{ m} \).

Practice Set

• Q1: Capillary rise occurs when contact angle is:
  • A) < 90°
  • B) = 90°
  • C) > 90°
  • D) 180°
• Q2: Formula for capillary rise height is:
  • A) \( h = \frac{2S}{\rho g a} \cos \theta \)
  • B) \( h = \frac{S}{\rho g a} \sin \theta \)
  • C) \( h = \frac{2S}{\rho g a} \sin \theta \)
  • D) \( h = \frac{S}{\rho g a} \cos \theta \)
• Q3: If contact angle is 120°, capillary height is:
  • A) Positive (rise)
  • B) Zero
  • C) Negative (fall)
  • D) Infinite

Answer Key

• Q1: A - Capillary rise occurs for contact angle less than 90°.
• Q2: A - Formula is \( h = \frac{2S}{\rho g a} \cos \theta \).
• Q3: C - Contact angle greater than 90° causes liquid to fall (negative height).

Memory Aids

• Capillary rise formula: \( h = \frac{2S}{\rho g a} \cos \theta \)
• \( \theta < 90^\circ \): rise, \( \theta > 90^\circ \): fall