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Class 11 · Tamil Nadu State Board

Properties of Matter Exercise Solutions | KalviGavani

Samacheer Kalvi 11th Physics Properties of Matter exercise solutions — elasticity, stress-strain, Young's modulus, viscosity, surface tension, Bernoulli's theorem.

Updated 15 June 202616 min read

How to Use This Page

1. Solve first, check later — Attempt every exercise question on your own before reading the solutions below. 2. Follow the sequence — Work through sections I to VI in order; later sections build on earlier ones. 3. Understand the method — Focus on the reasoning and formula selection, not just the final answer. 4. Practise numericals — Re-work the worked examples with different values until you can set them up without hints. 5. Review mistakes — Read the "Common Mistakes to Avoid" section before every exam.

Key Concepts (Quick Recap)

Elasticity and Moduli

QuantityFormulaSI UnitDimensional Formula
Stress (longitudinal)F/APa (N/m²)[ML⁻¹T⁻²]
Strain (longitudinal)ΔL/LNo unit (dimensionless)[M⁰L⁰T⁰]
Young's modulus (Y)(F/A)/(ΔL/L)Pa[ML⁻¹T⁻²]
Bulk modulus (K)−ΔP/(ΔV/V)Pa[ML⁻¹T⁻²]
Shear modulus or Rigidity modulus (η)(F/A)/θPa[ML⁻¹T⁻²]
Poisson's ratio (σ)Lateral strain / Longitudinal strainNo unit[M⁰L⁰T⁰]
Compressibility1/KPa⁻¹[M⁻¹LT²]

Key Relations

RelationExpression
Y = 3K(1 − 2σ)Relates Y, K, and σ
Y = 2η(1 + σ)Relates Y, η, and σ
9/Y = 3/K + 1/ηConnects all three moduli
Stress = Y × StrainHooke's law within elastic limit

Viscosity and Fluid Dynamics

QuantityFormulaSI Unit
Viscous force (Stokes' law)F = 6πηrvN
Coefficient of viscosity (η)F/A = η(dv/dx)Pa·s (N·s/m²)
Terminal velocityv_t = 2r²(ρ−σ)g / 9ηm/s
Poiseuille's formula (volume flow)Q = πPr⁴ / 8ηLm³/s
Reynold's numberN_R = ρvd / ηDimensionless

Surface Tension and Bernoulli

QuantityFormulaSI Unit
Surface tension (S)Force/Length or Work/AreaN/m or J/m²
Excess pressure inside a dropΔP = 2S/rPa
Excess pressure inside a bubbleΔP = 4S/rPa
Capillary riseh = 2S cosθ / (rρg)m
Bernoulli's equationP + ½ρv² + ρgh = const—
Pascal's lawP = F/A (transmitted equally)Pa

I. Choose the Best Answer (MCQs)

1. The SI unit of Young's modulus is:

  • (a) N·m
  • (b) N/m²
  • (c) N/m
  • (d) N·m²
Answer(b) N/m²
ExplanationYoung's modulus = stress/strain. Stress has SI unit N/m² (Pa), and strain is dimensionless, so Y also has the unit N/m².

2. The ratio of lateral strain to longitudinal strain is called:

  • (a) Young's modulus
  • (b) Bulk modulus
  • (c) Poisson's ratio
  • (d) Rigidity modulus
Answer(c) Poisson's ratio
ExplanationPoisson's ratio (σ) is defined as the ratio of lateral (transverse) strain to longitudinal (axial) strain. It is a dimensionless quantity and typically lies between 0 and 0.5 for most materials.

3. A wire is stretched to double its length. Its strain is:

  • (a) 0.5
  • (b) 1.0
  • (c) 2.0
  • (d) 0.25
Answer(b) 1.0
ExplanationStrain = ΔL/L. If the wire is stretched to double its length, ΔL = L (original length). So strain = L/L = 1.0.

4. The terminal velocity of a sphere falling through a viscous medium is proportional to:

  • (a) r
  • (b) r²
  • (c) r³
  • (d) 1/r
Answer(b) r²
ExplanationTerminal velocity v_t = 2r²(ρ − σ)g / 9η. Since all other quantities are constant for a given medium and sphere material, v_t ∝ r².

5. The excess pressure inside a soap bubble of radius r and surface tension S is:

  • (a) 2S/r
  • (b) 4S/r
  • (c) S/r
  • (d) S/2r
Answer(b) 4S/r
ExplanationA soap bubble has two surfaces (inner and outer), each contributing an excess pressure of 2S/r. So total excess pressure = 2 × 2S/r = 4S/r. For a liquid drop (one surface only), it would be 2S/r.

6. Bernoulli's theorem is applicable to:

  • (a) Viscous, compressible, turbulent flow
  • (b) Non-viscous, incompressible, streamline flow
  • (c) Viscous, incompressible, streamline flow
  • (d) Non-viscous, compressible, turbulent flow
Answer(b) Non-viscous, incompressible, streamline flow
ExplanationBernoulli's equation is derived assuming the fluid is ideal — non-viscous (no friction losses), incompressible (constant density), and flowing in steady streamlines (no turbulence).

7. According to Pascal's law, if a force F is applied on area A in a confined fluid, the pressure is transmitted:

  • (a) Only in the direction of the force
  • (b) Only perpendicular to the walls
  • (c) Equally in all directions throughout the fluid
  • (d) Only to the bottom of the container
Answer(c) Equally in all directions throughout the fluid
ExplanationPascal's law states that pressure applied to an enclosed fluid is transmitted undiminished to every portion of the fluid and the walls of the container. This principle is the basis of hydraulic presses and hydraulic brakes.

II. Fill in the Blanks

1. The modulus that relates volume strain to normal stress is called the Bulk modulus. 2. The value of Poisson's ratio for a perfectly incompressible material is 0.5. 3. The viscous force between liquid layers is proportional to the velocity gradient (dv/dx). 4. The SI unit of coefficient of viscosity is Pa·s (or N·s/m²). 5. If the radius of a capillary tube is doubled, the capillary rise becomes half (h ∝ 1/r). 6. The excess pressure inside a liquid drop is 2S/r. 7. Bernoulli's equation is based on the conservation of energy per unit volume. 8. The breaking stress of a wire depends on the material of the wire. 9. The restoring force in Hooke's law region is proportional to strain. 10. In a hydraulic press, force is multiplied but work remains the same. 11. The phenomenon of rise or fall of liquid in a capillary tube is called capillarity. 12. Stoke's law is valid for motion of a body through a viscous medium at low Reynolds number (laminar/creeping flow).

III. True or False (with corrections)

1.
AnswerFalse — Stress is a scalar quantity. Correction: Stress is a tensor quantity (it has both magnitude and direction depending on the plane considered).
2.
AnswerFalse — Young's modulus increases with temperature. Correction: Young's modulus decreases with increasing temperature because interatomic bonds weaken at higher temperatures, making the material less stiff.
3.
AnswerTrue — Strain has no unit and no dimensions.
4.
AnswerFalse — A soap bubble and a liquid drop of the same radius have the same excess pressure inside. Correction: A soap bubble has excess pressure 4S/r (two surfaces), while a liquid drop has excess pressure 2S/r (one surface).
5.
AnswerFalse — Bernoulli's equation applies to turbulent flow. Correction: Bernoulli's equation applies only to streamline (laminar), non-viscous, incompressible, and steady flow — not to turbulent flow.

IV. Match the Following

Column AColumn B
1. Young's modulus(d) Ratio of longitudinal stress to longitudinal strain
2. Bulk modulus(f) Ratio of volumetric stress to volume strain
3. Rigidity modulus(e) Ratio of tangential stress to shearing strain
4. Poisson's ratio(a) Ratio of lateral strain to longitudinal strain
5. Coefficient of viscosity(c) Ratio of shearing stress to velocity gradient
6. Surface tension(b) Force per unit length acting perpendicular to an imaginary line on the liquid surface

V. Answer Briefly (2-Mark Questions)

1. Distinguish between stress and strain.
Answer

Stress is the internal restoring force per unit area (SI unit: Pa), while strain is the ratio of change in dimension to the original dimension (dimensionless). Stress causes strain; strain is the effect of stress. Stress has dimensions [ML⁻¹T⁻²]; strain has no dimensions.

2. Define coefficient of viscosity and state its SI unit.
Answer

The coefficient of viscosity (η) is defined as the tangential force per unit area required to maintain a unit velocity gradient between two parallel layers of the liquid. η = F/(A × dv/dx). Its SI unit is Pa·s (or N·s/m²). In CGS, it is measured in poise (1 Pa·s = 10 poise).

3. State Pascal's law with one application.
Answer

Pascal's law states that the pressure applied to an enclosed, incompressible fluid is transmitted equally and undiminished in all directions throughout the fluid. Application: In a hydraulic press, a small force applied on a small piston creates pressure that is transmitted to a larger piston, producing a much larger force — F₂ = (A₂/A₁) × F₁.

4. What is capillarity? Give the formula for capillary rise.
Answer

Capillarity is the phenomenon of rise or depression of a liquid in a capillary tube due to surface tension. If the adhesive forces between liquid and tube exceed cohesive forces, the liquid rises; otherwise, it is depressed. The capillary rise formula is h = 2S cosθ / (rρg), where S is surface tension, θ is the contact angle, r is the tube radius, ρ is the liquid density, and g is acceleration due to gravity.

VI. Answer in Detail (5-Mark Questions)

1. Describe stress-strain curve with a neat diagram and explain the terms.

When a body is subjected to a gradually increasing load, the resulting stress and strain follow a characteristic curve with the following regions:

  • Proportional limit (A): Up to this point, stress is directly proportional to strain (Hooke's law holds). The curve is a straight line.
  • Elastic limit (B): Beyond the proportional limit but still within the elastic limit, the body returns to its original shape when the load is removed, though Hooke's law is no longer strictly valid.
  • Yield point (C): Beyond the elastic limit, the material begins to deform permanently. The yield point is where a small increase in stress causes a large increase in strain.
  • Ultimate tensile strength (D): The maximum stress the material can withstand. Beyond this, necking begins.
  • Fracture point (E): The material breaks.

The slope of the initial linear portion gives Young's modulus (Y = stress/strain). The area under the curve up to the elastic limit represents the elastic potential energy stored per unit volume (resilience).

Worked Example: A steel wire of length 2 m and cross-sectional area 1 mm² is stretched by 0.5 mm under a load. Find the stress, strain, and Young's modulus.

Given: L = 2 m, A = 1 mm² = 1 × 10⁻⁶ m², ΔL = 0.5 mm = 0.5 × 10⁻³ m Load F is not given directly — we use Y of steel ≈ 2 × 10¹¹ Pa Alternate problem with given force: F = 100 N, L = 2 m, A = 1 × 10⁻⁶ m², ΔL = 0.5 × 10⁻³ m Step 1: Calculate stress Stress = F / A = 100 / (1 × 10⁻⁶) = 1 × 10⁸ Pa = 100 MPa Step 2: Calculate strain Strain = ΔL / L = (0.5 × 10⁻³) / 2 = 2.5 × 10⁻⁴ Step 3: Calculate Young's modulus Y = Stress / Strain = (1 × 10⁸) / (2.5 × 10⁻⁴) = 4 × 10¹¹ Pa

2. Derive an expression for terminal velocity using Stokes' law and solve a numerical.

When a small sphere falls through a viscous fluid, three forces act on it:

  • Weight (W) acting downward: W = (4/3)πr³ρg
  • Upthrust or buoyant force (U) acting upward: U = (4/3)πr³σg
  • Viscous force (F) acting upward: F = 6πηrv (Stokes' law)

At terminal velocity, net force = 0:

W = U + F (4/3)πr³ρg = (4/3)πr³σg + 6πηrv_t (4/3)πr³g(ρ − σ) = 6πηrv_t Solving for v_t: v_t = 2r²(ρ − σ)g / 9η

Worked Example: A steel ball of radius 2 mm and density 7800 kg/m³ falls through an oil of density 900 kg/m³ and viscosity 0.5 Pa·s. Find the terminal velocity. (g = 9.8 m/s²)

Given: r = 2 mm = 2 × 10⁻³ m ρ = 7800 kg/m³ (steel) σ = 900 kg/m³ (oil) η = 0.5 Pa·s g = 9.8 m/s² Step 1: Write the formula v_t = 2r²(ρ − σ)g / 9η Step 2: Substitute values v_t = 2 × (2 × 10⁻³)² × (7800 − 900) × 9.8 / (9 × 0.5) Step 3: Simplify numerator r² = 4 × 10⁻⁶ m² (ρ − σ) = 6900 kg/m³ Numerator = 2 × 4 × 10⁻⁶ × 6900 × 9.8 = 2 × 4 × 10⁻⁶ × 67620 = 540960 × 10⁻⁶ = 0.54096 Step 4: Simplify denominator Denominator = 9 × 0.5 = 4.5 Step 5: Calculate terminal velocity v_t = 0.54096 / 4.5 ≈ 0.1202 m/s ≈ 0.12 m/s Answer: Terminal velocity ≈ 0.12 m/s

3. State and prove Bernoulli's theorem. Apply it to find the speed of efflux.

Bernoulli's theorem states that for an incompressible, non-viscous fluid in steady streamline flow, the total energy per unit volume (sum of pressure energy, kinetic energy, and potential energy) remains constant along a streamline.

Proof (by work-energy theorem):

Consider a fluid flowing through a pipe of varying cross-section and height. Let the fluid at position 1 have pressure P₁, velocity v₁, and height h₁, and at position 2 have P₂, v₂, and h₂.

Work done by pressure force: W = P₁V₁ − P₂V₂ (for equal volumes V₁ = V₂ = V) Change in kinetic energy: ΔKE = ½ρVv₂² − ½ρVv₁² Change in potential energy: ΔPE = ρVgh₂ − ρVgh₁

By work-energy theorem: W = ΔKE + ΔPE

P₁V − P₂V = ½ρV(v₂² − v₁²) + ρVg(h₂ − h₁) Dividing by V: P₁ − P₂ = ½ρ(v₂² − v₁²) + ρg(h₂ − h₁) Rearranging: P₁ + ½ρv₁² + ρgh₁ = P₂ + ½ρv₂² + ρgh₂ Therefore: P + ½ρv² + ρgh = constant

Application — Speed of efflux (Torricelli's theorem):

For a tank with a small hole at depth h below the free surface, applying Bernoulli's equation between the surface (point 1) and the hole (point 2):

At surface: P₁ = P_atm, v₁ ≈ 0, h₁ = h At hole: P₂ = P_atm, v₂ = v, h₂ = 0 P_atm + 0 + ρgh = P_atm + ½ρv² + 0 ρgh = ½ρv² v = √(2gh) Answer: Speed of efflux v = √(2gh)

Worked Example: Water flows through a horizontal pipe of cross-sectional area 0.02 m² at 2 m/s. The pipe narrows to 0.01 m². Find the speed in the narrow section and the pressure difference. (ρ = 1000 kg/m³)

Given: A₁ = 0.02 m², v₁ = 2 m/s, A₂ = 0.01 m², ρ = 1000 kg/m³ Step 1: Apply continuity equation A₁v₁ = A₂v₂ 0.02 × 2 = 0.01 × v₂ v₂ = 0.04 / 0.01 = 4 m/s Step 2: Apply Bernoulli's equation (horizontal, h₁ = h₂) P₁ + ½ρv₁² = P₂ + ½ρv₂² P₁ − P₂ = ½ρ(v₂² − v₁²) = ½ × 1000 × (16 − 4) = 500 × 12 = 6000 Pa Answer: v₂ = 4 m/s, Pressure drop = 6000 Pa = 6 kPa

Common Mistakes to Avoid

1. Confusing stress and pressure — Stress is internal restoring force per unit area (a tensor); pressure is external applied force per unit area (a scalar). They have the same SI unit but are conceptually different.

2. Forgetting that strain is dimensionless — Strain = ΔL/L is a ratio and has no unit. Do not assign units to strain values.

3. Using wrong excess pressure formula for bubble vs drop — A liquid drop has one surface (ΔP = 2S/r) while a soap bubble has two surfaces (ΔP = 4S/r). Always check which one the question refers to.

4. Mixing up the moduli — Young's modulus is for longitudinal stress-strain (stretching), Bulk modulus is for volumetric stress-strain (compression), and Rigidity modulus is for shearing stress-strain (twisting). Each applies to a different type of deformation.

5. Incorrect unit conversion — Always convert mm² to m² (multiply by 10⁻⁶), mm to m (multiply by 10⁻³), and cm² to m² (multiply by 10⁻⁴) before substituting into formulas. Forgetting these conversions is the most common source of wrong numerical answers.

6. Applying Bernoulli's equation to viscous or turbulent flow — Bernoulli's theorem requires the fluid to be non-viscous, incompressible, and in steady streamline flow. Do not apply it to turbulent or viscous flow situations.

7. Forgetting to subtract buoyant force in terminal velocity — The terminal velocity formula uses (ρ − σ), the difference between the body density and the fluid density. Using ρ alone gives a wrong answer.

8. Confusing Poisson's ratio range — Poisson's ratio σ lies between 0 and 0.5 for most materials. It cannot be negative (for real materials) and cannot exceed 0.5 (which corresponds to a perfectly incompressible material).

Exam Checklist

  • [ ] I can define and distinguish between stress, strain, and pressure
  • [ ] I can state Hooke's law and identify the proportional and elastic limits on a stress-strain curve
  • [ ] I can calculate Young's modulus, Bulk modulus, and Rigidity modulus from given data
  • [ ] I can relate Y, K, η, and σ using the three inter-moduli relations
  • [ ] I can derive terminal velocity from Stokes' law and solve numericals
  • [ ] I can state and apply Pascal's law to hydraulic press problems
  • [ ] I can calculate capillary rise and excess pressure inside drops and bubbles
  • [ ] I can state Bernoulli's theorem and apply it to Venturimeter and speed-of-efflux problems
  • [ ] I can correctly convert units (mm², cm², mm) before substituting into formulas
  • [ ] I can identify which modulus or formula applies to a given physical situation (stretching, compression, or shearing)

Frequently Asked Questions

Both have the same SI unit (N/m² or Pa), but stress is the internal restoring force per unit area within a deformed body, while pressure is the external force per unit area applied on a surface. Stress is a tensor; pressure is a scalar.
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