69 free MCQs on Mechanical Properties of Solids with worked answers and explanations. Stress, strain, Hookes law, and the elastic moduli that describe how solids deform and recover under load.
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A typical stress-strain curve: the initial straight line (Hooke's Law region, slope = Young's modulus) ends at the elastic limit; beyond the yield point, deformation becomes permanent, peaking at the ultimate tensile strength before the wire finally fractures.
Q1.
What is the SI unit of stress?
Answer: B. Pascal (N/m²)
Why: Stress is force per unit area, so its SI unit is N/m², called the pascal (Pa).
Q2.
Strain is defined as the ratio of:
Answer: B. Change in dimension to original dimension
Why: Strain is the fractional change in dimension (length, shape, or volume) and is dimensionless.
Q3.
Which type of stress changes only the shape of a body, not its volume?
Answer: B. Shearing stress
Why: Shearing stress is tangential to the surface and deforms shape while keeping volume constant.
Q4.
Hookes law states that within the elastic limit:
Answer: B. Stress is directly proportional to strain
Why: Hookes law: stress is proportional to strain within the elastic limit, with the modulus as the constant of proportionality.
Q5.
Youngs modulus is defined for which type of deformation?
Answer: C. Longitudinal (tensile or compressive)
Why: Youngs modulus Y = longitudinal stress / longitudinal strain, applicable to stretching or compression along a length.
Q6.
The bulk modulus relates pressure change to:
Answer: C. Fractional change in volume
Why: Bulk modulus B = -ΔP/(ΔV/V), describing the resistance of a material to uniform compression.
Q7.
Which material would have the smallest bulk modulus?
Answer: D. Air (a gas)
Why: Gases are highly compressible, hence they have a very small bulk modulus compared to liquids and solids.
Q8.
Poissons ratio is the ratio of:
Answer: B. Lateral strain to longitudinal strain
Why: Poissons ratio σ = lateral strain / longitudinal strain, and is dimensionless.
Q9.
On a stress-strain curve, the point beyond which the material does not return to its original shape after removing load is called the:
Answer: B. Elastic limit
Why: The elastic limit is the maximum stress a material can withstand and still return to its original dimensions.
Q10.
The SI unit of the modulus of elasticity (Y, G, or B) is the same as that of:
Answer: B. Stress
Why: Since strain is dimensionless, modulus = stress/strain has the same unit as stress, i.e. N/m² or Pa.
Q11.
Which of these materials is the most ductile (shows large plastic deformation before fracture)?
Answer: B. Copper
Why: Copper is a ductile metal showing a long plastic region on the stress-strain curve before it fractures.
Q12.
Elastic potential energy stored per unit volume of a stretched wire is given by:
Answer: B. ½ × Y × (strain)²
Why: Elastic PE per unit volume = ½ × stress × strain = ½ × Y × (strain)² using Hookes law.
Q13.
Which of the following best describes a perfectly plastic material beyond its yield point?
Answer: B. It continues to deform without much increase in stress
Why: Beyond the yield point, plastic materials undergo significant deformation (strain) with little additional increase in stress, before eventually fracturing.
Q14.
The property by which a body regains its original shape after the deforming force is removed is:
Answer: A. elasticity
Why: Elasticity is the tendency to return to the original shape once the load is removed.
Q15.
Stress is defined as the deforming force acting per unit:
Answer: A. area
Why: Stress = force / area, so its SI unit is the pascal.
Q16.
Strain is the ratio of the change in a dimension to its:
Answer: A. original dimension
Why: Strain = change in dimension / original dimension, a dimensionless ratio.
Q17.
The SI unit of stress is the:
Answer: A. pascal
Why: Stress has units of force per area, i.e. N/m² = pascal.
Q18.
Strain is a quantity that has:
Answer: A. no units
Why: Being a ratio of like quantities, strain is dimensionless and unitless.
Q19.
A material that keeps its new shape after the deforming force is removed is described as:
Answer: A. plastic
Why: Plastic materials retain the deformation once the load is removed.
Q20.
Within the elastic limit, stress is directly proportional to strain. This statement is:
Answer: A. Hooke's law
Why: Hooke's law: within the elastic limit, stress ∝ strain.
Q21.
A wire of length 2 m and cross-sectional area 1 mm² is stretched by 1 mm under a load of 100 N. What is its Youngs modulus?
Answer: B. 2 × 10¹¹ Pa
Why: Y = (F/A)/(ΔL/L) = (100/1×10⁻⁶)/(0.001/2) = (1×10⁸)/(5×10⁻⁴) = 2×10¹¹ Pa.
Q22.
Two wires of the same material have radii in ratio 2:1 and lengths in ratio 1:2. If stretched by the same force, the ratio of their elongations (ΔL<sub>1</sub>:ΔL<sub>2</sub>) is:
Answer: A. 1:8
Why: ΔL = FL/(AY). A ∝ r². Ratio ΔL<sub>1</sub>/ΔL<sub>2</sub> = (L<sub>1</sub>/L<sub>2</sub>)×(A2/A1) = (1/2)×(1/4) = 1/8, so ratio is 1:8.
Q23.
A solid sphere of volume V is submerged in a fluid such that pressure increases by ΔP, causing a volume decrease ΔV. If bulk modulus is B, which expression is correct?
Answer: B. ΔV = -V·ΔP/B
Why: B = -ΔP/(ΔV/V), rearranging gives ΔV = -V·ΔP/B (volume decreases as pressure increases).
Q24.
A metal cube of side 10 cm is subjected to a shearing force on its top face, causing a lateral displacement of 0.05 cm. What is the shearing strain?
Answer: A. 0.005
Why: Shearing strain = lateral displacement / side length = 0.05 cm / 10 cm = 0.005 (dimensionless, since strain is a ratio).
Q25.
Two springs of force constants k1 and k2 connected in series behave like a single spring of effective Youngs-modulus analog with stiffness k. Which formula gives k?
Answer: B. 1/k = 1/k1 + 1/k2
Why: For springs (or wires) in series under the same force, compliances (inverse stiffness) add: 1/k = 1/k1 + 1/k2.
Q26.
A wire stretches by 1 mm under a certain load. Another wire of the same material, but with double the length and double the diameter, stretches under the same load by:
Answer: A. 0.5 mm
Why: ΔL = FL/(AY). Doubling L doubles ΔL; doubling diameter makes A four times larger, dividing ΔL by 4. Net effect: ΔL<sub>new</sub> = 1mm × 2/4 = 0.5 mm.
Q27.
If the Poissons ratio of a material is 0.5, the material is best described as:
Answer: B. Perfectly incompressible (volume does not change on stretching)
Why: A Poissons ratio of 0.5 (the theoretical upper limit) corresponds to an incompressible material, like rubber approximately.
Q28.
A rubber cord has a cross-sectional area 1 mm² and total unstretched length 10 cm. It is stretched to 12 cm. If Youngs modulus of rubber is 5 × 10⁸ Pa, the tension in the cord is:
Answer: B. 100 N
Why: Strain = 2/10 = 0.2. Stress = Y × strain = 5×10⁸ × 0.2 = 1×10⁸ Pa. Force = stress × area = 1×10⁸ Pa × 1×10⁻⁶ m² = 100 N.
Q29.
For a given material, which modulus generally has the largest numerical value?
Answer: C. Youngs modulus (for most solids, comparable to or greater than shear modulus)
Why: For most solids, Youngs modulus Y is typically larger than the shear modulus G, while gases have a very small bulk modulus by comparison.
Q30.
A beam supported at both ends sags under its own weight. To minimize the sag (depression) without significantly increasing material used, engineers use a beam with cross-section shaped like:
Answer: B. An I-shaped (I-beam) cross-section
Why: I-beams concentrate material away from the neutral axis, maximizing the moment of inertia of the cross-section for a given amount of material, reducing bending.
Q31.
A wire is stretched within its elastic limit and then the load is removed. The wire returns to its original length. This behaviour is called:
Answer: B. Elasticity
Why: Elasticity is the property of a material to regain its original shape and size after the removal of the deforming force, within the elastic limit.
Q32.
If a material has Youngs modulus Y and shear modulus G with G < Y always, this physically implies that compared to stretching, a given material:
Answer: A. Deforms more easily under shear (shape change) than under tension
Why: Since shear modulus is generally smaller than Youngs modulus for the same material, materials typically deform (change shape) more readily under shearing stress than they stretch under tensile stress of the same magnitude.
Q33.
Young’s modulus is the ratio of tensile (longitudinal) stress to:
Answer: A. longitudinal strain
Why: Young’s modulus Y = longitudinal stress / longitudinal strain.
Q34.
The bulk modulus of a material relates stress to a change in:
Answer: A. volume
Why: The bulk modulus describes resistance to a change in volume under pressure.
Q35.
The modulus of rigidity is associated with ___ strain:
Answer: A. shear
Why: The rigidity (shear) modulus relates shear stress to shear strain.
Q36.
The reciprocal of the bulk modulus of a material is called its:
Answer: A. compressibility
Why: Compressibility = 1 / bulk modulus.
Q37.
Steel is more elastic than rubber because steel has a ___ Young’s modulus:
Answer: A. higher
Why: A higher Young’s modulus means a larger stress is needed for the same strain, so steel is more elastic.
Q38.
The stress beyond which a material begins to deform permanently is called the:
Answer: A. yield point
Why: Past the yield point the material undergoes permanent (plastic) deformation.
Q39.
The dimensional formula of stress is the same as that of:
Answer: A. pressure
Why: Both stress and pressure are force per unit area, with dimensions [ML⁻¹T⁻²].
Q40.
For a stretched wire within the elastic limit, the ratio of stress to strain gives the:
Answer: A. Young's modulus
Why: For a wire under tension this ratio is Young’s modulus.
Q41.
A uniform wire of length L and cross-section A hangs vertically under its own weight (density ρ). The total elongation due to its own weight is:
Answer: A. ρgL²/(2Y)
Why: For a hanging wire under its own weight, integrate the varying tension along the length: total elongation = ρgL²/(2Y), half of what it would be if the full weight acted uniformly.
Q42.
Two wires A and B of the same material and same length, but radius of A is twice that of B, are stretched by the same force. The ratio of elastic potential energy stored in A to that in B is:
Answer: B. 1:4
Why: Elastic PE = F²L/(2AY). Since A ∝ r², energy ratio = A<sub>B</sub>/A<sub>A</sub> = (r<sub>B</sub>/r<sub>A</sub>)² = (1/2)² = 1/4, so ratio of A:B is 1:4.
Q43.
A composite rod is made of two equal-length segments, one of Youngs modulus Y1 and area A, and the other of Youngs modulus Y2 and the same area A, joined end to end and stretched by force F. The effective Youngs modulus of the composite rod is:
Answer: B. 2Y1Y2/(Y1 + Y2)
Why: Equal-length segments under the same force act like springs in series; the harmonic-mean-like combination gives effective Y = 2Y1Y2/(Y1+Y2).
Q44.
A spherical ball of volume V made of a material with bulk modulus B is dropped into a lake to depth h. The fractional decrease in volume (ΔV/V) at that depth (ignoring atmospheric pressure) is approximately:
Answer: A. ρ_water·g·h/B
Why: Pressure increase at depth h is ΔP = ρgh. Using B = ΔP/(ΔV/V), fractional volume decrease ΔV/V = ΔP/B = ρgh/B.
Q45.
A metal wire of Youngs modulus Y is stretched by a load such that the strain is x. If the load is increased so the strain doubles (still within elastic limit), how does the elastic energy stored per unit volume change?
Answer: C. Quadruples
Why: Energy per unit volume = ½Y(strain)². Doubling the strain increases energy by a factor of 2² = 4, so it quadruples.
Q46.
A rod fixed between two rigid walls is heated so that it would expand by ΔL if free, but the walls prevent any expansion. If Youngs modulus is Y, area A, and coefficient of linear expansion is α, the thermal stress (compressive force per area) developed is:
Answer: A. Y·α·ΔT
Why: Since the rod cannot expand, the strain equivalent is α·ΔT (the strain it would have had if free), so thermal stress = Y × (α·ΔT).
Q47.
In a tensile test, a metal sample shows necking (localized thinning) just before fracture. This occurs at a stress level called the:
Answer: C. Ultimate tensile strength
Why: Necking begins at the ultimate tensile strength, the maximum stress on the curve, after which the engineering stress appears to decrease until fracture even though true stress increases.
Q48.
A wire is replaced by another wire of the same material but with half the diameter, used to support the same load, suspended for the same length. Compared to the original, the new wire is:
Answer: B. Four times more likely to cross its elastic limit due to higher stress
Why: Halving diameter quarters the cross-sectional area (A ∝ d²), so for the same load, stress (F/A) becomes four times larger, making it much more likely to exceed the elastic limit.
Q49.
Steel is preferred over copper for the same length and load-bearing cable application primarily because, for steel:
Answer: B. The Youngs modulus is higher, so it stretches less and resists deformation better
Why: Steel has a higher Youngs modulus than copper, meaning it deforms (stretches) less under the same stress, making it more suitable for load-bearing applications like cables and bridges.
Q50.
A cube of side a is subjected to three mutually perpendicular equal stresses sigma on its faces (hydrostatic compression). If Youngs modulus is Y and Poissons ratio is sigma_p, the fractional decrease in volume in terms of bulk modulus B is given by ΔV/V = 3σ/Y multiplied by a factor. That factor, expressed using Poissons ratio, is:
Answer: A. (1 - 2σ_p)
Why: For hydrostatic stress on a cube, the volumetric strain works out to ΔV/V = (3σ/Y)(1-2σ_p), connecting bulk modulus B = Y/[3(1-2σ_p)] to Youngs modulus and Poissons ratio.
Q51.
From Y = F·L/(A·ΔL), doubling the length of a wire (other factors fixed) makes the extension:
Answer: A. double
Why: Since ΔL = F·L/(A·Y), the extension is directly proportional to the original length.
Q52.
Poisson’s ratio is the ratio of lateral strain to:
Answer: A. longitudinal strain
Why: Poisson’s ratio = lateral strain / longitudinal strain.
Q53.
The elastic potential energy stored per unit volume in a stretched wire equals:
Answer: A. ½ × stress × strain
Why: Energy density = ½ × stress × strain.
Q54.
A material that can be drawn out into thin wires is said to be:
Answer: A. ductile
Why: Ductile materials (e.g. copper) can be drawn into wires.
Q55.
The area under a stress–strain curve represents the ___ stored per unit volume:
Answer: A. energy
Why: The area under the curve gives the elastic energy stored per unit volume.
Q56.
Two wires of the same material, one twice as thick as the other, carry the same load. The thicker wire extends:
Answer: A. one quarter as much
Why: ΔL ∝ 1/A ∝ 1/r², so doubling the radius reduces the extension to one quarter.
Q57.
The breaking stress of a material depends on the:
Answer: A. material, not its dimensions
Why: Breaking stress is a property of the material itself, independent of the specimen’s size.
Q58.
Within the proportional limit, the graph of stress against strain is a:
Answer: A. a straight line
Why: Hooke’s law makes stress proportional to strain, giving a straight line through the origin.
Q59.
The three elastic moduli are Young’s modulus, bulk modulus and the:
Answer: A. shear (rigidity) modulus
Why: The shear (rigidity) modulus is the third of the three elastic moduli.
Q60.
A perfectly rigid body would have a Young’s modulus of:
Answer: A. infinity
Why: A perfectly rigid body cannot be deformed, so its Young’s modulus is infinite.
Q61.
Elastic hysteresis refers to the lag of strain behind stress during loading and unloading, seen most clearly in:
Answer: A. rubber
Why: Rubber shows a large hysteresis loop between its loading and unloading curves.
Q62.
A weight stretches a wire by 1 mm. A second wire of the same material but double the length and double the radius carries the same weight. Its elongation is:
Answer: B. 0.5 mm
Why: ΔL = FL/(AY), A ∝ r². New ΔL = old·(2L/L)·(A/4A) = old·(2/4) = 0.5 mm.
Q63.
A rod is rigidly clamped between two fixed walls and its temperature is raised by ΔT. The thermal stress developed (Y = Young modulus, α = expansion coefficient) is:
Answer: A. YαΔT
Why: Prevented strain = αΔT, so stress = Y·strain = YαΔT, independent of length and area.
Q64.
The elastic energy stored per unit volume of a stretched wire, in terms of stress σ and Young modulus Y, is:
Answer: A. σ²/2Y
Why: Energy density = (1/2)·stress·strain = (1/2)σ(σ/Y) = σ²/2Y.
Q65.
A wire is stretched so that its longitudinal strain is 0.1%. If Poisson ratio is 0.3, the magnitude of the lateral (radial) strain is:
Answer: A. 0.03%
Why: Lateral strain = Poisson ratio × longitudinal strain = 0.3 × 0.1% = 0.03%.
Q66.
A pressure of 10⁷ Pa is applied to a material of bulk modulus 2×10⁹ Pa. The fractional change in its volume is:
Answer: B. 5×10⁻³
Why: ΔV/V = P/B = 10⁷/(2×10⁹) = 5×10⁻³.
Q67.
The breaking force of a wire is proportional to its cross-sectional area. If the radius of a wire is doubled, its breaking force becomes:
Answer: B. 4×
Why: Breaking force ∝ area ∝ r², so doubling r multiplies the breaking force by 4.
Q68.
A force of 100 N stretches a wire by 2 mm. The work done in stretching it is:
Answer: B. 0.1 J
Why: W = (1/2)·F·ΔL = 0.5·100·0.002 = 0.1 J.
Q69.
A steel wire (Y = 2×10¹¹ Pa) carries a stress of 2×10⁸ Pa. The percentage longitudinal strain is:
Answer: B. 0.1%
Why: Strain = stress/Y = 2×10⁸/2×10¹¹ = 10⁻³ = 0.1%.