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

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

Application Question

Medium difficulty • Concept in a practical situation

Medium

Question 1

Applied Concept

A structural steel rod of radius 10 mm and length 1.0 m is stretched by a 100 kN force along its length. With reference to the definition of stress and Young's modulus of steel (2.0 × 10¹¹ N m⁻²), calculate the stress and elongation of the rod.

Stress = F/A = F/(πr²) = (100×10³)/(π×(10⁻²)²) = 3.18×10⁸ N m⁻².
Elongation ΔL = (F/A × L)/Y = (3.18×10⁸ × 1)/(2×10¹¹) = 1.59×10⁻³ m = 1.59 mm.
Strain = ΔL/L = 1.59×10⁻³/1 = 1.59×10⁻³, equivalent to 0.16% — well within the elastic range of steel.
This confirms that steel can sustain very high stresses (over 300 MPa) with only minute elongation, validating its use in structural applications.

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Application Question

Hard difficulty • Concept in a practical situation

Hard

Question 2

Applied Concept

A copper wire (length 2.2 m) and a steel wire (length 1.6 m), both of diameter 3.0 mm, are connected end to end. When stretched by a load, the net elongation is 0.70 mm. With reference to Hooke's law and Young's moduli from the chapter, determine the load applied.

Since both wires have the same cross-section and carry the same tension W, using Y = stress/strain: W/A = Y_c(ΔL_c/L_c) = Y_s(ΔL_s/L_s).
The ratio ΔL_c/ΔL_s = (Y_s/Y_c)×(L_c/L_s) = (2.0/1.1)×(2.2/1.6) = 2.5; combined with ΔL_c + ΔL_s = 0.70 mm, we get ΔL_c = 0.50 mm and ΔL_s = 0.20 mm.
Load W = (A × Y_c × ΔL_c)/L_c = π(1.5×10⁻³)² × (5.0×10⁻⁴ × 1.1×10¹¹)/2.2 ≈ 1.8×10² N.
This application illustrates that wires with different elastic properties in series share the same tension but distribute elongation according to their Young's moduli and lengths.

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Application Question

Hard difficulty • Concept in a practical situation

Hard

Question 3

Applied Concept

In a human pyramid at a circus, a performer lying on his back supports a combined load of 220 kg on his two thighbones (each of length 0.5 m and radius 2.0 cm). With reference to Young's modulus of bone (9.4 × 10⁹ N m⁻²), find the compression in each thighbone.

Force on each thighbone = (220 × 9.8)/2 = 1078 N; cross-sectional area A = π(2×10⁻²)² = 1.26×10⁻³ m².
Compression ΔL = FL/(YA) = (1078 × 0.5)/(9.4×10⁹ × 1.26×10⁻³) = 4.55×10⁻⁵ m ≈ 0.046 mm.
Fractional compression ΔL/L = 0.000091 or 0.0091%, an extremely small change demonstrating the high stiffness of bone under biological loads.
This example shows that bone's elastic properties are optimised by nature: despite a seemingly large load, the compression is negligible, keeping the musculoskeletal system within the elastic (recoverable) regime.

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Application Question

Medium difficulty • Concept in a practical situation

Medium

Question 4

Applied Concept

A square lead slab (side 50 cm, thickness 10 cm) is subjected to a shearing force of 9.0 × 10⁴ N on its narrow face, with its lower edge riveted to the floor. Using shear modulus of lead (5.6 × 10⁹ N m⁻²), calculate the lateral displacement of the upper edge.

Area over which shear force is applied: A = 0.5 m × 0.1 m = 0.05 m²; shearing stress = F/A = 9.0×10⁴/0.05 = 1.80×10⁶ N m⁻².
Shearing strain = stress/G = 1.80×10⁶/(5.6×10⁹) = 3.21×10⁻⁴.
Displacement Δx = shearing strain × L = 3.21×10⁻⁴ × 0.5 m = 1.6×10⁻⁴ m = 0.16 mm.
Despite the large applied force, the displacement is small because even lead (the softest common metal, G = 5.6 GPa) resists shear significantly; this illustrates why structural elements must be checked for shear as well as bending.

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Application Question

Medium difficulty • Concept in a practical situation

Medium

Question 5

Applied Concept

The average depth of the Indian Ocean is about 3000 m. With reference to the bulk modulus of water (2.2 × 10⁹ N m⁻²), calculate the fractional compression ΔV/V of water at the ocean bottom.

Pressure at depth 3000 m: p = hρg = 3000 × 1000 × 10 = 3×10⁷ N m⁻².
Fractional compression ΔV/V = p/B = (3×10⁷)/(2.2×10⁹) = 1.36×10⁻² or 1.36%.
Even at the extreme depth of 3000 m with a pressure of 30 MPa, water is compressed by only 1.36%, demonstrating its high bulk modulus and relative incompressibility compared to gases.
This result is important in oceanography and deep-sea engineering: despite enormous pressures, water's nearly constant density simplifies calculations of buoyancy and pressure at depth.

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Application Question

Medium difficulty • Concept in a practical situation

Medium

Question 6

Applied Concept

With reference to the formula for beam sag, explain why railway tracks use an I-shaped cross-section. If you double the depth of a rectangular beam while keeping the breadth and material constant, by what factor does sag reduce?

Sag formula: δ = Wl³/(4bd³Y); since δ ∝ d⁻³, doubling d reduces sag by a factor of 2³ = 8.
If breadth were doubled instead, δ ∝ b⁻¹ gives only a factor of 2 reduction, making depth increase far more efficient.
However, a very deep rectangular beam buckles under off-centre loads; the I-section maintains large effective depth (via flanges) while the web prevents buckling — combining the advantages of both.
Railway tracks (and bridge beams) use I-sections for this reason: they are light, strong against bending, stable against buckling, and economical in material.

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Application Question

Medium difficulty • Concept in a practical situation

Medium

Question 7

Applied Concept

A crane is required to lift a load of 10 tonnes (metric). With reference to the yield strength of mild steel (300 × 10⁶ N m⁻²) from the chapter, determine the minimum cross-sectional area and radius of the steel rope. Why is a rope of radius about 3 cm actually used?

Minimum cross-sectional area: A ≥ W/σ_y = (10⁴ × 9.8)/(300×10⁶) ≈ 3.3×10⁻⁴ m², corresponding to radius r ≈ 1 cm.
A safety factor of approximately 10 is applied to account for dynamic loads (sudden jerks), material imperfections, fatigue, and long-term use.
Applying the safety factor: effective required area ≈ 10 × 3.3×10⁻⁴ = 3.3×10⁻³ m², giving radius ≈ 3 cm.
A single solid wire of 3 cm radius would be rigid and unmanageable; instead, many thin wires are braided together to achieve the necessary area while remaining flexible enough for use on drums and pulleys.

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Application Question

Medium difficulty • Concept in a practical situation

Medium

Question 8

Applied Concept

A wire is stretched within its elastic limit and then released. With reference to the stress-strain curve (region O to A), explain what happens to the elastic potential energy stored, and write its expression per unit volume.

Within the elastic limit (O to A), all the work done by the external force is stored as elastic potential energy in the wire, with no dissipation.
When the wire is released, this stored elastic potential energy is fully recovered and the wire returns to its original length — characteristic of the elastic (reversible) region.
Elastic potential energy per unit volume: u = ½σε = ½ × stress × strain, where σ = F/A is stress and ε = ΔL/L is longitudinal strain.
It can also be written as u = ½ × Y × ε², showing that stored energy increases with the square of strain and is proportional to Young's modulus, so stiffer materials store more energy for the same strain.

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Application Question

Hard difficulty • Concept in a practical situation

Hard

Question 9

Applied Concept

With reference to the chapter, explain why the maximum height of a mountain on Earth is approximately 10 km. What role does the elastic limit of rock play in this estimate?

At the base of a mountain of height h, the shear stress is approximately hρg, where ρ is rock density and g is gravitational acceleration.
For rocks to remain stable (not flow), this shear stress must not exceed the elastic limit of rock, which is approximately 30×10⁷ N m⁻².
Setting hρg = 30×10⁷ with ρ = 3×10³ kg m⁻³ and g = 10 m s⁻²: h = 30×10⁷/(3×10³×10) = 10 km.
This is consistent with the fact that Mt. Everest (~8.85 km) is the tallest mountain on Earth — any taller mountain would have a base stress exceeding the elastic limit of rock, causing the rock to flow and the mountain to collapse.

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Application Question

Medium difficulty • Concept in a practical situation

Medium

Question 10

Applied Concept

A student claims that rubber is more elastic than steel because it stretches more. Using the scientific definition of elasticity and data from the chapter, evaluate this claim.

The student's claim is a common misconception; scientifically, elasticity is measured by Young's modulus — the ratio of stress to strain — not by the amount of stretch.
Steel has Y ≈ 200×10⁹ N m⁻² and rubber has an extremely small effective modulus; steel requires a force of 2000 N to stretch a 0.1 cm² wire by 0.1%, while rubber stretches enormously under much smaller forces.
A material with higher Young's modulus (steel) is scientifically more elastic because it has a stronger restoring force for a given deformation.
The term 'elastic' in daily life refers to the ability to stretch and return, but in physics it refers to how strongly a material resists deformation — making steel far more elastic than rubber in the correct technical sense.

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Chapter 8: Mechanical Properties of Solids | Application Questions

A structural steel rod of radius 10 mm and length 1.0 m is stretched by a 100 kN force along its length. With reference to the definition of stress and Young's modulus of steel (2.0 × 10¹¹ N m⁻²), calculate the stress and elongation of the rod.

A copper wire (length 2.2 m) and a steel wire (length 1.6 m), both of diameter 3.0 mm, are connected end to end. When stretched by a load, the net elongation is 0.70 mm. With reference to Hooke's law and Young's moduli from the chapter, determine the load applied.

In a human pyramid at a circus, a performer lying on his back supports a combined load of 220 kg on his two thighbones (each of length 0.5 m and radius 2.0 cm). With reference to Young's modulus of bone (9.4 × 10⁹ N m⁻²), find the compression in each thighbone.

A square lead slab (side 50 cm, thickness 10 cm) is subjected to a shearing force of 9.0 × 10⁴ N on its narrow face, with its lower edge riveted to the floor. Using shear modulus of lead (5.6 × 10⁹ N m⁻²), calculate the lateral displacement of the upper edge.

The average depth of the Indian Ocean is about 3000 m. With reference to the bulk modulus of water (2.2 × 10⁹ N m⁻²), calculate the fractional compression ΔV/V of water at the ocean bottom.

With reference to the formula for beam sag, explain why railway tracks use an I-shaped cross-section. If you double the depth of a rectangular beam while keeping the breadth and material constant, by what factor does sag reduce?

A crane is required to lift a load of 10 tonnes (metric). With reference to the yield strength of mild steel (300 × 10⁶ N m⁻²) from the chapter, determine the minimum cross-sectional area and radius of the steel rope. Why is a rope of radius about 3 cm actually used?

A wire is stretched within its elastic limit and then released. With reference to the stress-strain curve (region O to A), explain what happens to the elastic potential energy stored, and write its expression per unit volume.

With reference to the chapter, explain why the maximum height of a mountain on Earth is approximately 10 km. What role does the elastic limit of rock play in this estimate?

A student claims that rubber is more elastic than steel because it stretches more. Using the scientific definition of elasticity and data from the chapter, evaluate this claim.

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