Chapter 8: Mechanical Properties of Solids Long Questions | physics Class 11 CBSE

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Chapter 8: Mechanical Properties of Solids Long Questions | physics Class 11 CBSE | ByteNotes.in

Long Answer

Medium difficulty • Structured explanation

Medium

Question 1

Long Form

Describe the stress-strain curve for a metallic wire under increasing tensile load, identifying all key regions and points. How does this curve help classify materials as ductile or brittle?

From O to A (proportional limit), the stress-strain curve is linear: Hooke's law is obeyed, and the material behaves as a perfectly elastic body, fully recovering its shape on unloading.
From A to B, stress and strain are no longer proportional, but the material still returns to its original dimensions when the load is removed; B is the yield point (elastic limit) with yield strength σ_y.
Beyond B, plastic deformation sets in; if unloaded at any point C between B and D, a permanent set remains — the strain is non-zero even when stress is zero.
Point D is the ultimate tensile strength (σ_u), the maximum stress the material can withstand; beyond D, the material necks and additional strain is produced even with reduced force.
Fracture occurs at point E; if D and E are close together, the material is brittle (e.g., glass), while materials where D and E are far apart are ductile (e.g., mild steel or copper).
Elastomers like rubber have a very large elastic region but do not obey Hooke's law and have no distinct plastic region, unlike metals whose curve shows clear proportional, elastic, plastic, and fracture zones.

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Long Answer

Medium difficulty • Structured explanation

Medium

Question 2

Long Form

Compare Young's modulus, shear modulus, and bulk modulus with respect to their definitions, formulas, applicable states of matter, and what each physically represents about a material.

Young's modulus (Y = F×L/A×ΔL) is the ratio of tensile/compressive stress to longitudinal strain; it represents resistance to change in length and is applicable only to solids, which alone possess fixed shapes.
Shear modulus (G = F×L/A×Δx = F/Aθ) is the ratio of shearing stress to shearing strain; it represents resistance to change in shape without change in volume, and is applicable only to solids since fluids cannot sustain shear.
Bulk modulus (B = −p/(ΔV/V)) is the ratio of hydraulic pressure to volume strain; it represents resistance to uniform compression and applies to solids, liquids, and gases.
Metals have large Y (steel: 200 GPa, aluminium: 70 GPa) and correspondingly large G (steel: 84 GPa); for most materials G ≈ Y/3, implying it is easier to shear than to stretch.
Gases have extremely small bulk moduli (air at STP: 1.0×10⁻⁴ GPa) compared to solids (steel: 160 GPa), making gases about a million times more compressible; solids are the least compressible due to tight inter-atomic coupling.
All three moduli share the same SI unit (N m⁻² or Pa) and are intrinsic material properties; a material with a larger modulus is stiffer and harder to deform in the corresponding mode.

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Long Answer

Medium difficulty • Structured explanation

Medium

Question 3

Long Form

Explain the three types of stress — tensile/compressive, shearing, and hydraulic — including their conditions of application, the corresponding strains, and the elastic moduli associated with each.

Tensile stress arises when equal and opposite forces are applied normal to the cross-sectional area, stretching a body; compressive stress arises when these forces compress it. Both are termed longitudinal stress, and the corresponding strain is longitudinal strain (ΔL/L).
Shearing stress arises when equal and opposite forces act parallel to the cross-sectional area of a body; it causes relative displacement (Δx) between opposite faces. The corresponding shearing strain is Δx/L = tan θ ≈ θ for small angles.
Hydraulic stress arises when a solid body is submerged in a pressurised fluid, with forces acting perpendicular to every point on its surface; this causes a uniform decrease in volume without change in shape. The corresponding strain is volume strain (ΔV/V).
The elastic modulus for longitudinal stress is Young's modulus Y = (F/A)/(ΔL/L); for shearing stress it is shear modulus G = F/(Aθ); for hydraulic stress it is bulk modulus B = −p/(ΔV/V).
Longitudinal and shearing stresses change shape but not volume; hydraulic stress changes volume but not shape. Shearing stress is unique to solids; bulk modulus applies to all states of matter.
In each case Hooke's law holds in the elastic region: the ratio of stress to its corresponding strain is constant (the respective modulus), enabling calculation of deformation for known applied forces.

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Long Answer

Hard difficulty • Structured explanation

Hard

Question 4

Long Form

Derive the expression for elastic potential energy stored in a stretched wire and show that the energy per unit volume equals half the product of stress and strain.

When a wire of original length L and cross-sectional area A is stretched by a force F through an elongation l, by Hooke's law: F = YA(l/L), where Y is Young's modulus.
For an infinitesimal additional elongation dl, the work done is dW = F dl = (YAl/L) dl.
Total work done in stretching from l = 0 to l = l is W = ∫₀ˡ (YAl/L) dl = YA/(2L) × l².
This can be rewritten as W = ½ × Y × (l/L)² × AL = ½ × Young's modulus × strain² × volume of wire.
Since stress σ = F/A = Y(l/L) and strain ε = l/L, we have W = ½ × σ × ε × volume.
Therefore, the elastic potential energy per unit volume of the wire is u = W/Volume = ½σε, confirming that energy density equals half the product of stress and strain.

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Long Answer

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Medium

Question 5

Long Form

Discuss the applications of elastic behaviour in the design of bridges and buildings, explaining why beams with I-shaped cross-sections and pillars with distributed ends are preferred.

A beam of length l, breadth b, and depth d, loaded at its centre by load W, sags by δ = Wl³/(4bd³Y); to minimise sag, one should maximise Young's modulus Y and the depth d of the beam.
Since δ ∝ d⁻³ but only ∝ b⁻¹, increasing depth is much more effective than increasing breadth in reducing deflection for a given volume of material.
However, increasing depth alone causes a thin deep bar to buckle (bend sideways) when the load is not precisely centred, as is typical in bridges with moving traffic.
The I-shaped cross-section resolves this conflict: the wide flanges provide a large load-bearing surface and the vertical web gives adequate depth to resist bending, while removing excess material from the web reduces weight and cost without sacrificing strength.
For columns and pillars, a distributed (flared) base rather than a rounded base provides a larger contact area at the ends, allowing the column to support significantly greater compressive loads.
The precise design of a bridge or building must account for traffic loads, wind forces, self-weight, long-term reliability, and cost of materials, all governed by the elastic properties — particularly Young's modulus and yield strength — of the materials used.

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Long Answer

Hard difficulty • Structured explanation

Hard

Question 6

Long Form

Explain the concept of Poisson's ratio. How does a deforming force in one direction produce strain in a perpendicular direction, and why cannot a single elastic constant describe this situation?

When a wire is stretched longitudinally, its length increases but its diameter simultaneously decreases; the strain perpendicular to the applied force is called lateral strain.
Simon Poisson showed that within the elastic limit, lateral strain is directly proportional to longitudinal strain; their ratio is Poisson's ratio: μ = (Δd/d)/(ΔL/L).
Poisson's ratio is a pure dimensionless number (ratio of two strains); its value depends only on the material — for steel it is 0.28–0.30 and for aluminium alloys it is about 0.33.
A single elastic constant (like Young's modulus) only relates stress and the strain in the direction of applied force; it cannot account for the lateral contraction that simultaneously occurs in perpendicular directions.
In general, a deforming force in one direction produces strains in other directions too, requiring at least two independent elastic constants (e.g., Y and μ, or Y and G) to fully describe the elastic behaviour.
This multi-directional strain response is important in precision engineering — for instance, the change in radius of cross-section of a loaded wire must be accounted for when designing components sensitive to dimensional tolerances.

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Long Answer

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Medium

Question 7

Long Form

Analyse the differences in compressibility among solids, liquids, and gases using the concept of bulk modulus, and explain the molecular basis for these differences.

Bulk modulus B = −p/(ΔV/V) measures a material's resistance to uniform compression; the larger the bulk modulus, the less compressible the material.
Solids have the largest bulk moduli (e.g., steel: 160 GPa, copper: 140 GPa) — they are the least compressible because neighbouring atoms are tightly coupled through strong inter-atomic bonds.
Liquids have intermediate bulk moduli (e.g., water: 2.2 GPa, glycerine: 4.76 GPa); their molecules are bound to neighbours but less strongly than in solids, allowing moderate compression.
Gases have extremely small bulk moduli (air at STP: 1.0×10⁻⁴ GPa), making them about a million times more compressible than solids; gas molecules are very poorly coupled — widely spaced and weakly interacting.
Gases are further distinctive in that their bulk moduli vary with pressure and temperature, unlike the essentially constant bulk moduli of solids and liquids.
Compressibility k = 1/B is highest for gases, intermediate for liquids, and lowest for solids — a direct reflection of the strength of inter-molecular interactions in each phase.

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Long Answer

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Hard

Question 8

Long Form

Using the concept of elastic behaviour of rocks, show how the maximum possible height of a mountain on Earth can be estimated. What assumptions are made in this analysis?

At the base of a mountain of height h and density ρ, the force per unit area due to the weight of rock above is hρg, acting vertically downward.
The sides of the mountain are free, so this is not a case of uniform hydraulic compression; instead, there is a significant shear component at the base, approximately equal to hρg itself.
For rocks to remain stable, this shear stress must be less than the elastic limit (critical shearing stress) of the rock, which is approximately 30×10⁷ N m⁻².
Setting hρg = 30×10⁷ N m⁻² and using ρ = 3×10³ kg m⁻³ and g = 10 m s⁻², the maximum height is h = 30×10⁷/(3×10³×10) = 10 km.
This value exceeds the height of Mt. Everest (~8.85 km), confirming that Everest is within the maximum theoretical limit for a mountain on Earth's surface.
The key assumptions are that rock density is uniform (3×10³ kg m⁻³), g = 10 m s⁻², the shear stress equals hρg, and the elastic limit of rock is taken as 30×10⁷ N m⁻²; the base of a real mountain experiences non-uniform stress, making the actual limit somewhat different.

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Long Answer

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Hard

Question 9

Long Form

Discuss the elastic behaviour of the aorta tissue as an elastomer. How does its stress-strain curve differ from that of a metal, and what physical significance does this have for its biological function?

The elastic tissue of the aorta (large blood vessel carrying blood from the heart) is an elastomer — it can be stretched to very large strains and returns to its original shape after the force is removed.
Its stress-strain curve shows a very large elastic region, but unlike metals, the curve is highly non-linear over most of this range, meaning Hooke's law is not obeyed.
There is no well-defined proportional limit, yield point, or plastic region in the aorta tissue's curve — the material transitions smoothly between elastic states without a sharp onset of permanent deformation.
In contrast, a metal's stress-strain curve has a distinct linear (proportional) region (O to A), a non-linear but still elastic region (A to B), a plastic region (B to D), and a fracture point (E).
Biologically, this non-linear elasticity is ideal for the aorta: it allows the vessel to expand widely under the high pressure of each heartbeat and recoil forcefully to maintain blood pressure between beats, acting as a 'pressure reservoir'.
The absence of a plastic region ensures the aorta does not suffer permanent deformation under repeated cyclic loading, which is critical for long-term cardiovascular health.

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Long Answer

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Medium

Question 10

Long Form

Explain the role of Young's modulus and yield strength in the design of a steel rope for a crane with a lifting capacity of 10 tonnes. Why is a safety factor applied, and how does it affect the final design?

A crane with a 10-tonne capacity must support a weight W = 10⁴ kg × 9.8 m s⁻² ≈ 10⁵ N; the rope must not be deformed beyond its elastic limit, so the cross-sectional area must satisfy A ≥ W/σ_y.
Using the yield strength of mild steel σ_y ≈ 300×10⁶ N m⁻², the minimum area is A = 10⁵/300×10⁶ = 3.3×10⁻⁴ m², corresponding to a rope radius of about 1 cm.
In practice, a safety factor of about 10 is applied to account for dynamic loads (jerks, swings), material imperfections, and prolonged use; this increases the effective design load to 10 times the actual load.
With the safety factor, the required radius becomes approximately 3 cm, which for a single solid wire would be nearly as rigid as a metal rod — impossible to wind on a drum or pass over a pulley.
Therefore, the rope is made of many thin wires braided together (like pigtails); this achieves the required cross-sectional area and strength while maintaining flexibility for practical operation.
Braiding also distributes stress among wires, so that failure of a single wire does not immediately cause catastrophic collapse, improving overall reliability and safety.

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Chapter 8: Mechanical Properties of Solids | Long Questions

Describe the stress-strain curve for a metallic wire under increasing tensile load, identifying all key regions and points. How does this curve help classify materials as ductile or brittle?

Compare Young's modulus, shear modulus, and bulk modulus with respect to their definitions, formulas, applicable states of matter, and what each physically represents about a material.

Explain the three types of stress — tensile/compressive, shearing, and hydraulic — including their conditions of application, the corresponding strains, and the elastic moduli associated with each.

Derive the expression for elastic potential energy stored in a stretched wire and show that the energy per unit volume equals half the product of stress and strain.

Discuss the applications of elastic behaviour in the design of bridges and buildings, explaining why beams with I-shaped cross-sections and pillars with distributed ends are preferred.

Explain the concept of Poisson's ratio. How does a deforming force in one direction produce strain in a perpendicular direction, and why cannot a single elastic constant describe this situation?

Analyse the differences in compressibility among solids, liquids, and gases using the concept of bulk modulus, and explain the molecular basis for these differences.

Using the concept of elastic behaviour of rocks, show how the maximum possible height of a mountain on Earth can be estimated. What assumptions are made in this analysis?

Discuss the elastic behaviour of the aorta tissue as an elastomer. How does its stress-strain curve differ from that of a metal, and what physical significance does this have for its biological function?

Explain the role of Young's modulus and yield strength in the design of a steel rope for a crane with a lifting capacity of 10 tonnes. Why is a safety factor applied, and how does it affect the final design?

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