Mechanical Properties of Solids
1. Elasticity & Hooke's Law
Property of body to regain original shape after removal of deforming force.
Stress (σ): Restoring force per unit area (F/A).
Strain (ε): Ratio of change in dimension to original dimension (ΔL/L).
Hooke's Law: Within elastic limit, Stress ∝ Strain.
Stress = E × Strain (E is Modulus of Elasticity).
Strain (ε): Ratio of change in dimension to original dimension (ΔL/L).
Hooke's Law: Within elastic limit, Stress ∝ Strain.
Stress = E × Strain (E is Modulus of Elasticity).
Types of Modulus
Young's Modulus (Y): Longitudinal Stress / Longitudinal Strain.
Bulk Modulus (B): Volumetric Stress / Volumetric Strain.
Shear Modulus (G): Tangential Stress / Shearing Strain.
Bulk Modulus (B): Volumetric Stress / Volumetric Strain.
Shear Modulus (G): Tangential Stress / Shearing Strain.
(Asked in NEET 2017, 2019)
2. Stress-Strain Curve
Graphical representation of behavior of wire under load.
Proportional Limit (A): Stress ∝ Strain (Hooke's Law valid).
Yield Point (B): Elastic limit. Futher stress causes permanent set.
Fracture Point (E): Wire breaks.
Ductile Material: Large plastic region (e.g., Copper).
Brittle Material: Small plastic region (e.g., Glass).
Yield Point (B): Elastic limit. Futher stress causes permanent set.
Fracture Point (E): Wire breaks.
Ductile Material: Large plastic region (e.g., Copper).
Brittle Material: Small plastic region (e.g., Glass).
(Asked in NEET 2016, 2020, 2022)
3. Elastic Energy
Work done in stretching a wire is stored as Potential Energy.
U = ½ × Stress × Strain × Volume.
Energy Density (u): Energy per unit volume.
u = ½ × Stress × Strain = ½ Y (ε)2 = (Stress)2 / 2Y.
Energy Density (u): Energy per unit volume.
u = ½ × Stress × Strain = ½ Y (ε)2 = (Stress)2 / 2Y.
4. Other Elastic Properties
Poisson's Ratio (σ): Lateral Strain / Longitudinal Strain.
Value lies between -1 and 0.5 (Practical 0 to 0.5).
Thermal Stress: Stress developed due to temperature change if expansion is prevented.
Stress = Y α ΔT. force = YA α ΔT.
Value lies between -1 and 0.5 (Practical 0 to 0.5).
Thermal Stress: Stress developed due to temperature change if expansion is prevented.
Stress = Y α ΔT. force = YA α ΔT.
(Asked in NEET 2018, 2021, 2023)
Numericals: Solids
Q1. A steel wire of length 4.7 m and cross-section 3.0 × 10-5 m2 stretches by
same amount as a copper wire of length 3.5 m and cross-section 4.0 × 10-5 m2
under same load. Find ratio of Young's modulus (Steel : Copper).
Solution:
ΔL = FL / AY. Given ΔL and F are same.
Ls / (As Ys) = Lc / (Ac Yc)
Ys / Yc = (Ls Ac) / (Lc As)
Ratio = (4.7 × 4 × 10-5) / (3.5 × 3 × 10-5) = 18.8 / 10.5 ≈ 1.79.
ΔL = FL / AY. Given ΔL and F are same.
Ls / (As Ys) = Lc / (Ac Yc)
Ys / Yc = (Ls Ac) / (Lc As)
Ratio = (4.7 × 4 × 10-5) / (3.5 × 3 × 10-5) = 18.8 / 10.5 ≈ 1.79.
Q2. A load of 4 kg is suspended from a ceiling wire of radius 2 mm. Find the tensile stress. (g=3.1π)
Solution:
Stress = F / A = mg / πr2.
F = 4 × 3.1π = 12.4π N.
A = π (2 × 10-3)2 = 4π × 10-6 m2.
Stress = 12.4π / 4π × 10-6 = 3.1 × 106 N/m2.
Stress = F / A = mg / πr2.
F = 4 × 3.1π = 12.4π N.
A = π (2 × 10-3)2 = 4π × 10-6 m2.
Stress = 12.4π / 4π × 10-6 = 3.1 × 106 N/m2.
Q3. A wire stretches by 1 mm under load. If Y = 2 × 1011 N/m2, find energy
density if Strain is 10-3?
Solution:
Energy Density u = ½ Y (Strain)2.
u = 0.5 × 2 × 1011 × (10-3)2
u = 1011 × 10-6 = 105 J/m3.
Energy Density u = ½ Y (Strain)2.
u = 0.5 × 2 × 1011 × (10-3)2
u = 1011 × 10-6 = 105 J/m3.
Q4. A rope has breaking force of 104 N. What should be the radius of a rope of same material to
withstand 4 × 104 N? (Original radius r)
Solution:
Breaking Stress is constant for material.
F1/A1 = F2/A2 → F ∝ A ∝ r2.
F2 / F1 = (r2 / r1)2
4 = (r2 / r)2 → r2 = 2r.
Breaking Stress is constant for material.
F1/A1 = F2/A2 → F ∝ A ∝ r2.
F2 / F1 = (r2 / r1)2
4 = (r2 / r)2 → r2 = 2r.
Q5. A steel rod is fixed at ends and cooled by 20°C. Find thermal stress. (Y = 2×1011,
α = 10-5)
Solution:
Stress = Y α ΔT.
Stress = 2 × 1011 × 10-5 × 20
Stress = 40 × 106 = 4 × 107 N/m2.
Stress = Y α ΔT.
Stress = 2 × 1011 × 10-5 × 20
Stress = 40 × 106 = 4 × 107 N/m2.
Q6. Find depth in ocean where density of water increases by 0.1%. (B = 2×109 Pa)
Solution:
Δρ/ρ = ΔV/V = 0.1/100 = 10-3.
Pressure P = B (ΔV/V) = 2×109 × 10-3 = 2×106 Pa.
P = hρg → 2×106 = h × 1000 × 10
h = 2000000 / 10000 = 200 m.
Δρ/ρ = ΔV/V = 0.1/100 = 10-3.
Pressure P = B (ΔV/V) = 2×109 × 10-3 = 2×106 Pa.
P = hρg → 2×106 = h × 1000 × 10
h = 2000000 / 10000 = 200 m.
Q7. Longitudinal strain is 2 × 10-3. If Poisson's ratio is 0.4, find lateral strain.
Solution:
σ = Lateral Strain / Longitudinal Strain.
Lateral = σ × Longitudinal = 0.4 × 2 × 10-3 = 8 × 10-4.
σ = Lateral Strain / Longitudinal Strain.
Lateral = σ × Longitudinal = 0.4 × 2 × 10-3 = 8 × 10-4.
Q8. A uniform wire of length L and mass M extends by l under its own weight. What is strain at the midpoint?
Solution:
Stress at mid point is due to weight of lower half (Mg/2).
Stress = (Mg/2) / A.
Strain = Stress / Y = Mg / 2AY.
(We know total l = MgL/2AY), this relates to that.
Stress at mid point is due to weight of lower half (Mg/2).
Stress = (Mg/2) / A.
Strain = Stress / Y = Mg / 2AY.
(We know total l = MgL/2AY), this relates to that.
Q9. A spring stretches by x when loaded with M. Stored energy is U. If stretched by M + M, what is new
energy?
Solution:
U = ½ kx2. Load M → x. Load 2M → Extension 2x.
U' = ½ k(2x)2 = 4 (½ kx2) = 4U.
U = ½ kx2. Load M → x. Load 2M → Extension 2x.
U' = ½ k(2x)2 = 4 (½ kx2) = 4U.
Q10. Two wires A and B are in series. Radius rA = 2rB. Find Ratio of Stress.
Solution:
In series, Force F is same.
Stress S = F/A = F/πr2.
SA / SB = (rB / rA)2
Ratio = (1/2)2 = 1:4.
In series, Force F is same.
Stress S = F/A = F/πr2.
SA / SB = (rB / rA)2
Ratio = (1/2)2 = 1:4.
Important Formulae
20 NEET Golden Facts
- 1. Elasticity: Is not constant. Decreases with increase in temperature (Intermolecular forces weaken).
- 2. Steel vs Rubber: Steel is more elastic than rubber because for same strain, steel requires more stress.
- 3. Solids/Liquids/Gases: Solids have Young's & Shear Modulus. Liquids/Gases have only Bulk Modulus.
- 4. Poisson's Ratio: Theoretical range -1 to 0.5. Practical range 0 to 0.5. Can never be greater than 0.5.
- 5. Incompressible: For perfectly incompressible body, Poisson's ratio is 0.5.
- 6. Shear: Shape changes, Volume remains constant.
- 7. Quartz: Nearest approach to a perfectly elastic body. Putty is nearest to plastic body.
- 8. Breaking Stress: Depends on material, independent of dimension (Length or Area). Breaking Force depends on Area.
- 9. Compressibility: Reciprocal of Bulk Modulus. Gases are most compressible.
- 10. Slopes: Slope of Stress-Strain graph gives Modulus of Elasticity. Steep slope = More elastic.
- 11. Area under Curve: Area under Stress-Strain curve gives Potential Energy Density (Energy per unit volume).
- 12. Elastomers: Materials like rubber that can be stretched to large strains. Does not obey Hooke's Law well.
- 13. Hysteresis: Loss of energy as heat during stretching and unstretching cycle (Area of loop). Tyre rubber has low hysteresis.
- 14. Hammering: Hammering or rolling increases elasticity. Annealing decreases elasticity.
- 15. Hollow Shaft: Hollow shaft is stronger than solid shaft of same mass.
- 16. Spring Constant (k): F=kx. k = YA/L. Cutting spring in half doubles k.
- 17. Sagging: Beam sags under load. Depth δ = WL3 / 4bd3Y.
- 18. Isothermal vs Adiabatic: Adiabatic Elasticity (Eφ) > Isothermal Elasticity (Eθ). Ratio = γ = Cp/Cv.
- 19. Yield Point: Point where permanent deformation begins.
- 20. Pressure of Gas: Bulk modulus of ideal gas at constant temperature is equal to its Pressure P.
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