NEET Physics: Gravitation and Properties of Matter · Chapter 2

Mechanical Properties of Solids

What to remember

Within the elastic limit, stress is proportional to strain (Hooke's law). The ratio stress/strain is a constant called the modulus of elasticity of the material.
Young's modulus Y = (F/A)/(ΔL/L) = FL/(AΔL) applies to a change in length; shear modulus G = (F/A)/θ applies to a change in shape; bulk modulus B = −Δp/(ΔV/V) applies to a change in volume. Compressibility is 1/B.
Elastic potential energy stored in a stretched wire is ½ × stress × strain per unit volume, which equals ½ F ΔL for the whole wire.
For a wire, the extension is ΔL = FL/(AY). It is directly proportional to the force and the length, and inversely proportional to the area of cross-section and the Young's modulus.

Elastic behaviour, stress and strain

A body that regains its original size and shape when the deforming force is removed is called elastic. A body that does not regain its shape and stays deformed is called plastic. No real body is perfectly elastic or perfectly plastic. Steel is more elastic than rubber because it needs a much larger force for the same strain.

Stress is the restoring force developed per unit area of cross-section inside a deformed body. Stress = F/A. Its SI unit is N m⁻² or pascal (Pa), and its dimensions are [M L⁻¹ T⁻²].

Strain is the fractional change in size or shape. It has no unit and no dimensions.

There are three types of stress and strain.

Type	Force direction	Strain	Definition of strain
Longitudinal (tensile or compressive)	Normal to the area, along the length	Longitudinal strain	ΔL/L
Shear (tangential)	Parallel to a face	Shear strain	Δx/L = tan θ, about θ for small angles
Volumetric (hydraulic)	Equal pressure normal to the whole surface	Volume strain	ΔV/V

In a longitudinal stress, a stretching force gives tensile stress and a compressing force gives compressive stress. In a shear stress, the shape changes but the volume does not (for small deformation). In a hydraulic stress, such as a solid body deep in a liquid, the shape stays the same, but the volume changes. The hydraulic stress is the pressure p.

Hooke's law and the stress–strain curve

Hooke's law. For small deformations, stress is directly proportional to strain.

Stress = modulus of elasticity × strain

The constant is called the modulus of elasticity. It depends on the material but not on the size of the specimen.

Stress–strain curve for a metal wire (described in words). Stress is plotted on the vertical axis and strain on the horizontal axis.

Region	What happens
Straight line from the origin	Stress is proportional to strain. Hooke's law holds. The end of this line is the proportional limit.
Slightly curved part after the proportional limit	Stress is no longer proportional to strain, but the wire still returns to its original length when the load is removed.
Elastic limit (yield point)	The largest stress up to which the wire still recovers fully. Beyond it, the wire has a permanent set (a permanent change in length) after unloading. The stress here is called the yield strength.
Plastic region	Strain increases rapidly for a small increase in stress. The deformation is permanent.
Ultimate tensile stress (tensile strength)	The highest point of the curve, the maximum stress the wire can bear.
Fracture point	Beyond the highest point the wire thins at one place (necking) and breaks.

The slope of the straight part of the curve is the Young's modulus.

Types of material by the curve.

Ductile materials have a large plastic region between the elastic limit and the fracture point. They can be drawn into wires. Examples: copper, aluminium, steel (mild).
Brittle materials have a very small plastic region. They break soon after the elastic limit. Example: glass, cast iron.
Elastomers show a large strain for small stress and have no well-defined plastic region. Example: rubber, the tissue of the aorta. They do not obey Hooke's law over the whole range.

Moduli of elasticity

Young's modulus (Y). The ratio of longitudinal stress to longitudinal strain.

Y = (F/A) / (ΔL/L) = F L / (A ΔL)

Unit: N m⁻² (Pa). For a wire of diameter d, A = πd²/4, so Y = 4FL/(πd²ΔL). Young's modulus is a property of solids; liquids and gases do not have it. It is the same for tension and compression in the elastic range for most metals.

Material	Young's modulus (N m⁻²)
Steel	about 2.0 × 10¹¹
Copper	about 1.1 × 10¹¹
Aluminium	about 0.7 × 10¹¹

Steel has a larger Young's modulus than copper, so for the same load and the same size a steel wire extends less than a copper wire. A material with a larger Young's modulus is "more elastic" in the sense of resisting change in length.

Shear modulus (G, modulus of rigidity). The ratio of shear stress to shear strain.

G = (F/A) / (Δx/L) = F / (A θ)

Here F is the tangential force, A is the area of the face on which it acts, Δx is the displacement of the top face, L is the height, and θ is the angle of shear in radian. Unit: N m⁻² (Pa). Only solids have a shear modulus, because fluids cannot sustain shear stress at rest. For most metals G is smaller than Y (about one third of Y).

Bulk modulus (B). The ratio of hydraulic stress (pressure change) to volume strain.

B = − Δp / (ΔV/V) = − p V / ΔV

The negative sign shows that the volume decreases when the pressure increases, so B is positive. Unit: N m⁻² (Pa).

Compressibility (k) is the reciprocal of the bulk modulus.

k = 1/B = −(1/V)(ΔV/Δp). Unit: Pa⁻¹ or N⁻¹ m².

Solids have large bulk modulus (steel about 1.6 × 10¹¹ N m⁻²). Liquids have smaller values (water about 2.2 × 10⁹ N m⁻²). Gases have the smallest bulk modulus and are the most compressible. Solids and liquids are nearly incompressible compared with gases.

Modulus	Kind of strain	Applies to	Formula
Young's modulus	Change in length	Solids only	FL/(AΔL)
Shear modulus	Change in shape	Solids only	F/(Aθ)
Bulk modulus	Change in volume	Solids, liquids and gases	−Δp/(ΔV/V)

Poisson's ratio (in brief). When a wire is stretched, its length increases and its diameter decreases. The lateral strain is the fractional change in diameter. Poisson's ratio σ is

σ = lateral strain / longitudinal strain

It is a pure number with no unit. For common materials it lies between 0 and 0.5.

Elastic potential energy in a stretched wire

When a wire is stretched slowly, work is done against the restoring force. This work is stored as elastic potential energy. The force rises from zero to F as the extension rises from zero to ΔL, so the average force is F/2.

U = ½ F ΔL

Using F = YAΔL/L:

U = ½ (Y A / L) (ΔL)²

The wire behaves like a spring of force constant k = YA/L. The energy per unit volume (energy density) is

u = U / (A L) = ½ × (F/A) × (ΔL/L) = ½ × stress × strain = ½ Y (strain)² = (stress)² / 2Y

Unit of u: J m⁻³. The area under the stress–strain curve up to a given strain equals the energy stored per unit volume. The energy stored depends on the square of the strain, so doubling the extension makes the stored energy four times as large.

Applications of elastic behaviour

Cranes and ropes. The rope of a crane must not be loaded beyond its elastic limit, otherwise it will take a permanent set. The maximum stress allowed is the yield strength divided by a safety factor. The rope is made of many thin strands twisted together, and its total area of cross-section is chosen so that the load stays well below the yield strength. A heavier load needs a thicker rope (a larger area).

Beams. A beam of length l, breadth b and depth d, supported at its two ends and loaded with a weight W in the middle, bends (sags). The depression at the middle is

δ = W l³ / (4 b d³ Y)

So the sag is less if the beam is shorter, broader and deeper, and if Y is large. The depth d is cubed, so increasing the depth is more effective than increasing the breadth. For this reason, bridges and rail tracks use beams of an I-shape, which have a large depth and save material.

Mountain height. The base of a mountain is under a huge stress due to its own weight. If the stress is above the elastic limit of the rock, the rock flows. For a mountain of height h and density ρ, the pressure at the base is hρg. For rock this stress cannot exceed about 3 × 10⁸ N m⁻². Setting hρg equal to this limit gives a maximum height of about 10 km, which agrees with the height of Mount Everest (about 9 km).

Worked examples

Example 1 (extension of a wire). A steel wire of length 2 m and cross-sectional area 1 mm² (1 × 10⁻⁶ m²) carries a load of 100 N. Take Y = 2 × 10¹¹ N m⁻².

ΔL = F L / (A Y) = (100 × 2) / (1 × 10⁻⁶ × 2 × 10¹¹) = 200 / (2 × 10⁵) = 1 × 10⁻³ m = 1 mm.

Example 2 (energy stored). For the same wire, the stored energy is U = ½ F ΔL = ½ × 100 × 1 × 10⁻³ = 0.05 J.

Check using energy density: stress = F/A = 10⁸ N m⁻²; strain = ΔL/L = 5 × 10⁻⁴; u = ½ × 10⁸ × 5 × 10⁻⁴ = 2.5 × 10⁴ J m⁻³. Volume = A L = 2 × 10⁻⁶ m³. U = 2.5 × 10⁴ × 2 × 10⁻⁶ = 0.05 J.

Example 3 (bulk modulus). The pressure on a block of water is increased by 2.2 × 10⁷ N m⁻². Take B = 2.2 × 10⁹ N m⁻². The fractional decrease in volume is Δp/B = 2.2 × 10⁷ / 2.2 × 10⁹ = 0.01, that is 1%.

Example 4 (shear). A tangential force of 10⁴ N acts on the top face of a cube of side 0.1 m, with the bottom face fixed. Take G = 2 × 10¹⁰ N m⁻². Shear stress = 10⁴ / 0.01 = 10⁶ N m⁻². Shear strain = 10⁶ / (2 × 10¹⁰) = 5 × 10⁻⁵ rad. The top face moves by Δx = θ L = 5 × 10⁻⁵ × 0.1 = 5 × 10⁻⁶ m.

Example 5 (two wires). Two wires of the same material and the same load have lengths in the ratio 1 : 2 and diameters in the ratio 1 : 2. Since ΔL ∝ L/d², the ratio of their extensions is (1/1²) : (2/2²) = 1 : 0.5 = 2 : 1.

Formula summary

Quantity	Formula	Unit
Stress	F/A	Pa
Longitudinal strain	ΔL/L	none
Shear strain	Δx/L ≈ θ	none (radian)
Volume strain	ΔV/V	none
Young's modulus	FL/(AΔL) = 4FL/(πd²ΔL)	Pa
Shear modulus	F/(Aθ)	Pa
Bulk modulus	−Δp/(ΔV/V)	Pa
Compressibility	1/B	Pa⁻¹
Poisson's ratio	lateral strain / longitudinal strain	none
Extension of a wire	ΔL = FL/(AY)	m
Force constant of a wire	k = YA/L	N m⁻¹
Elastic energy	½ F ΔL = ½ (YA/L)(ΔL)²	J
Energy per unit volume	½ × stress × strain = ½ Y (strain)²	J m⁻³
Sag of a loaded beam	W l³/(4 b d³ Y)	m

NEET traps

1. Young's modulus measures the resistance to a change in length, not the strength; a material with a large Y is not necessarily the one with a large breaking stress.
2. Steel is more elastic than rubber (larger Y, larger restoring force for the same strain), even though rubber stretches more.
3. Stress is force per unit area of cross-section and has the same unit as pressure, but stress is not a vector and not a force.
4. Strain has no unit; the modulus has the same unit as stress.
5. Shear modulus and Young's modulus apply only to solids. Bulk modulus applies to solids, liquids and gases.
6. Compressibility is the reciprocal of the bulk modulus, so gases have the greatest compressibility.
7. The yield point and the ultimate tensile stress are different points: yield point comes first, the ultimate stress is the highest stress.
8. For a given load, a wire that is twice as long extends twice as much; a wire with twice the diameter extends one quarter as much, not one half.
9. In a beam, the sag depends on d³, so a deep beam is far better than a broad one for the same material.
10. Elastic energy is ½ F ΔL, not F ΔL; the force grows from zero to F, so the average force is F/2.

One-liners

1. Stress = restoring force per unit area, in N m⁻² (Pa).
2. Strain = change in dimension divided by the original dimension, with no unit.
3. Hooke's law: stress is proportional to strain within the proportional limit.
4. Y = FL/(AΔL), and for steel it is about 2 × 10¹¹ N m⁻².
5. Shear modulus G = F/(Aθ), and it exists only for solids.
6. Bulk modulus B = −Δp/(ΔV/V); compressibility = 1/B.
7. Water has a bulk modulus of about 2.2 × 10⁹ N m⁻².
8. Gases are much more compressible than solids and liquids.
9. Beyond the elastic limit, the wire shows a permanent set.
10. The highest stress on the stress–strain curve is the ultimate tensile stress.
11. A ductile material can be drawn into wires; a brittle material breaks soon after the elastic limit.
12. Elastomers such as rubber show large strain and no well-defined plastic region.
13. Energy stored per unit volume = ½ × stress × strain.
14. Poisson's ratio = lateral strain / longitudinal strain, and it has no unit.
15. The height of a mountain is limited to about 10 km by the elastic limit of rock.

Practice questions

The dimensional formula of stress is
1. [M L² T⁻²]
2. [M L⁻¹ T⁻²]
3. [M L T⁻²]
4. [M⁰ L⁰ T⁰]
Answer
B. [M L⁻¹ T⁻²]
Stress = force/area = [M L T⁻²]/[L²] = [M L⁻¹ T⁻²].
Poisson's ratio of a material is
1. measured in N m⁻¹
2. measured in N m⁻²
3. measured in Pa⁻¹
4. a pure number with no unit
Answer
D. a pure number with no unit
It is lateral strain divided by longitudinal strain, a ratio of two strains, so it has no unit.
Three wires of steel, copper and aluminium have the same length and area of cross-section and carry the same load. Taking Y(steel) = 2.0 × 10¹¹, Y(copper) = 1.1 × 10¹¹ and Y(aluminium) = 0.7 × 10¹¹ N m⁻², the wire with the least extension is
1. all three extend equally
2. aluminium
3. steel
4. copper
Answer
C. steel
ΔL = FL/(AY) is smallest for the largest Y, which is steel.
A steel wire of length 3 m and cross-sectional area 2 × 10⁻⁶ m² carries a load of 200 N. If Y = 2 × 10¹¹ N m⁻², the extension is
1. 1.5 mm
2. 15 mm
3. 0.75 mm
4. 3.0 mm
Answer
A. 1.5 mm
ΔL = FL/(AY) = (200 × 3)/(2 × 10⁻⁶ × 2 × 10¹¹) = 600/(4 × 10⁵) = 1.5 × 10⁻³ m. Check: strain = stress/Y = 10⁸/(2 × 10¹¹) = 5 × 10⁻⁴, and 5 × 10⁻⁴ × 3 m = 1.5 mm.
A wire of length L and radius r extends by x under a load. A wire of the same material of length 2L and radius 2r carries the same load. Its extension is
1. 2x
2. x/2
3. x/4
4. x
Answer
B. x/2
ΔL ∝ L/r². The ratio is 2/4 = 1/2, so the extension is x/2.
Which of the following moduli of elasticity is possessed by solids, liquids and gases?
1. Young's modulus
2. Bulk modulus
3. Shear modulus
4. Both Young's modulus and shear modulus
Answer
B. Bulk modulus
Fluids can resist a change in volume but not a change in length or shape at rest. Only the bulk modulus applies to all three states.
The bulk modulus of water is 2.2 × 10⁹ N m⁻². When the pressure on 500 mL of water is increased by 4.4 × 10⁷ N m⁻², the decrease in volume is
1. 10 mL
2. 20 mL
3. 1 mL
4. 5 mL
Answer
A. 10 mL
ΔV/V = Δp/B = 4.4 × 10⁷/2.2 × 10⁹ = 0.02. ΔV = 0.02 × 500 = 10 mL.
Compressibility of a material is
1. the product of bulk modulus and volume
2. equal to its bulk modulus
3. the reciprocal of its Young's modulus
4. the reciprocal of its bulk modulus
Answer
D. the reciprocal of its bulk modulus
k = 1/B, with unit Pa⁻¹. A gas has a very small B and hence a large compressibility.
A wire is under a stress of 2 × 10⁸ N m⁻² and a strain of 1 × 10⁻³. The elastic energy stored per unit volume is
1. 2 × 10¹¹ J m⁻³
2. 5 × 10⁴ J m⁻³
3. 1 × 10⁵ J m⁻³
4. 2 × 10⁵ J m⁻³
Answer
C. 1 × 10⁵ J m⁻³
u = ½ × stress × strain = ½ × 2 × 10⁸ × 10⁻³ = 10⁵ J m⁻³. Check: u = ½Y(strain)² with Y = 2 × 10¹¹ gives the same value.
A steel wire of length 1 m and cross-sectional area 1 × 10⁻⁶ m² (Y = 2 × 10¹¹ N m⁻²) is stretched by 1 mm. The elastic energy stored is
1. 1 × 10⁻⁴ J
2. 0.2 J
3. 0.1 J
4. 0.05 J
Answer
C. 0.1 J
U = ½(YA/L)(ΔL)² = ½ × 2 × 10⁵ × 10⁻⁶ = 0.1 J. Check: F = YAΔL/L = 200 N and ½FΔL = ½ × 200 × 10⁻³ = 0.1 J.
In the stress–strain curve of a metal wire, the highest point of the curve is called the
1. ultimate tensile stress
2. yield point
3. fracture point
4. proportional limit
Answer
A. ultimate tensile stress
The highest stress the wire can bear is the ultimate tensile stress; beyond it the wire necks and breaks.
The stress at the base of a rock mountain cannot exceed 3 × 10⁸ N m⁻². Taking the density of rock as 3 × 10³ kg m⁻³ and g = 10 m s⁻², the maximum height of the mountain is about
1. 100 km
2. 30 km
3. 1 km
4. 10 km
Answer
D. 10 km
hρg = 3 × 10⁸ gives h = 3 × 10⁸/(3 × 10³ × 10) = 10⁴ m = 10 km.
A beam supported at its two ends and loaded in the middle sags by δ. If only its depth is doubled, the sag becomes
1. δ/4
2. δ/8
3. δ/2
4. δ/16
Answer
B. δ/8
δ ∝ 1/d³, so doubling d gives δ/2³ = δ/8.
A steel wire of length 4 m and cross-sectional area 2 × 10⁻⁶ m² has Y = 2 × 10¹¹ N m⁻². Its force constant k = YA/L is
1. 2.5 × 10⁻⁶ N m⁻¹
2. 4 × 10⁵ N m⁻¹
3. 1 × 10⁵ N m⁻¹
4. 1 × 10¹⁷ N m⁻¹
Answer
C. 1 × 10⁵ N m⁻¹
k = YA/L = 2 × 10¹¹ × 2 × 10⁻⁶/4 = 1 × 10⁵ N m⁻¹. Check: a 100 N load gives 1 mm extension, and F/ΔL = 10⁵.
A wire of diameter 4 mm is stretched so that its length increases by 0.2%. If Poisson's ratio is 0.25, the decrease in its diameter is
1. 0.002 mm
2. 0.02 mm
3. 0.0005 mm
4. 0.008 mm
Answer
A. 0.002 mm
Lateral strain = 0.25 × 0.002 = 5 × 10⁻⁴. Decrease in diameter = 5 × 10⁻⁴ × 4 mm = 0.002 mm.
The slope of the straight-line part of the stress–strain graph of a wire gives its
1. shear modulus
2. bulk modulus
3. Poisson's ratio
4. Young's modulus
Answer
D. Young's modulus
In the proportional region stress/strain is constant and equals Young's modulus.
Given below are two statements. Statement I: Stress has the same unit as pressure. Statement II: Strain has no unit. Choose the correct option.
1. Both Statement I and Statement II are correct
2. Both Statement I and Statement II are incorrect
3. Statement I is correct but Statement II is incorrect
4. Statement I is incorrect but Statement II is correct
Answer
A. Both Statement I and Statement II are correct
Stress is in N m⁻² (Pa), the unit of pressure. Strain is a ratio, so it has no unit.
Given below are two statements. Statement I: The Young's modulus of steel is smaller than that of copper. Statement II: Under the same load and with the same dimensions, a steel wire extends more than a copper wire. Choose the correct option.
1. Both Statement I and Statement II are correct
2. Both Statement I and Statement II are incorrect
3. Statement I is correct but Statement II is incorrect
4. Statement I is incorrect but Statement II is correct
Answer
B. Both Statement I and Statement II are incorrect
Y(steel) is about 2 × 10¹¹, greater than Y(copper) about 1.1 × 10¹¹, so a steel wire extends less. Both are incorrect.
Given below are two statements. Statement I: A liquid has a shear modulus. Statement II: The bulk modulus is defined for solids, liquids and gases. Choose the correct option.
1. Both Statement I and Statement II are correct
2. Both Statement I and Statement II are incorrect
3. Statement I is correct but Statement II is incorrect
4. Statement I is incorrect but Statement II is correct
Answer
D. Statement I is incorrect but Statement II is correct
Fluids at rest cannot sustain shear stress, so I is incorrect. II is correct.
Given below are two statements. Statement I: Elastomers show a large strain for a small stress and have no well-defined plastic region. Statement II: Elastomers obey Hooke's law over their whole range of strain. Choose the correct option.
1. Both Statement I and Statement II are correct
2. Both Statement I and Statement II are incorrect
3. Statement I is correct but Statement II is incorrect
4. Statement I is incorrect but Statement II is correct
Answer
C. Statement I is correct but Statement II is incorrect
I describes rubber correctly. They do not obey Hooke's law over the whole range, so II is incorrect.
Which of the following statements are correct? 1. Ductile materials have a large plastic region. 2. Brittle materials break soon after the elastic limit. 3. Glass is a ductile material.
1. 1 and 2 only
2. 2 and 3 only
3. 1 and 3 only
4. 1, 2 and 3
Answer
A. 1 and 2 only
Statements 1 and 2 are correct. Glass is brittle, so 3 is wrong.
Which of the following statements are correct? 1. The elastic energy per unit volume of a stretched wire is ½ × stress × strain. 2. Doubling the extension of a wire doubles the energy stored. 3. The area under the stress–strain curve up to a given strain equals the energy stored per unit volume.
1. 1 and 2 only
2. 2 and 3 only
3. 1 and 3 only
4. 1, 2 and 3
Answer
C. 1 and 3 only
U ∝ (extension)², so doubling the extension gives four times the energy. Statement 2 is wrong.
Given below are two statements. Statement I: The elastic energy stored in a stretched wire is F ΔL, where F is the final force. Statement II: During slow stretching the average force is F/2. Choose the correct option.
1. Both Statement I and Statement II are correct
2. Both Statement I and Statement II are incorrect
3. Statement I is correct but Statement II is incorrect
4. Statement I is incorrect but Statement II is correct
Answer
D. Statement I is incorrect but Statement II is correct
The force grows from 0 to F, so the work done and energy stored is ½ F ΔL. I is incorrect and II is correct.
Given below are two statements. Statement I: Gases are more compressible than solids. Statement II: The bulk modulus of a gas is greater than that of a solid. Choose the correct option.
1. Both Statement I and Statement II are correct
2. Both Statement I and Statement II are incorrect
3. Statement I is correct but Statement II is incorrect
4. Statement I is incorrect but Statement II is correct
Answer
C. Statement I is correct but Statement II is incorrect
Gases have a very small bulk modulus, so a large compressibility. I is correct, II is incorrect.
Which of the following statements are correct? 1. The yield point is the limit beyond which a permanent set remains. 2. The ultimate tensile stress is the highest stress on the stress–strain curve. 3. Beyond the highest point the wire becomes thinner at one place and then breaks.
1. 1 and 2 only
2. 2 and 3 only
3. 1 and 3 only
4. 1, 2 and 3
Answer
D. 1, 2 and 3
All three statements describe the stress–strain curve of a metal wire correctly.
Given below are two statements. Statement I: The sag of a loaded beam is proportional to the square of its length. Statement II: The sag of a loaded beam is inversely proportional to its depth. Choose the correct option.
1. Both Statement I and Statement II are correct
2. Both Statement I and Statement II are incorrect
3. Statement I is correct but Statement II is incorrect
4. Statement I is incorrect but Statement II is correct
Answer
B. Both Statement I and Statement II are incorrect
δ = Wl³/(4bd³Y), so the sag is proportional to l³ and to 1/d³. Both statements are incorrect.
Assertion (A): Steel is more elastic than rubber. Reason (R): For the same strain, steel develops a much larger restoring stress than rubber.
1. Both A and R are true and R is the correct explanation of A
2. Both A and R are true but R is not the correct explanation of A
3. A is true but R is false
4. A is false but R is true
Answer
A. Both A and R are true and R is the correct explanation of A
Y(steel) is much larger, so a larger restoring stress for the same strain. R explains A.
Assertion (A): A liquid does not have a Young's modulus. Reason (R): Liquids are nearly incompressible.
1. Both A and R are true and R is the correct explanation of A
2. Both A and R are true but R is not the correct explanation of A
3. A is true but R is false
4. A is false but R is true
Answer
B. Both A and R are true but R is not the correct explanation of A
Both are true, but incompressibility is not the reason: a liquid at rest cannot hold a stretching or shearing stress.
Assertion (A): A wire stretched beyond its elastic limit does not regain its original length after unloading. Reason (R): Hooke's law holds beyond the elastic limit.
1. Both A and R are true and R is the correct explanation of A
2. Both A and R are true but R is not the correct explanation of A
3. A is true but R is false
4. A is false but R is true
Answer
C. A is true but R is false
A is true (permanent set). R is false because stress is no longer proportional to strain beyond the limit.
Assertion (A): Doubling the extension of a wire doubles the elastic energy stored in it. Reason (R): Elastic energy stored is proportional to the square of the strain.
1. Both A and R are true and R is the correct explanation of A
2. Both A and R are true but R is not the correct explanation of A
3. A is true but R is false
4. A is false but R is true
Answer
D. A is false but R is true
U ∝ (ΔL)², so doubling the extension gives four times the energy. A is false and R is true.
Assertion (A): Beams of I-shaped cross-section are used in bridges. Reason (R): The sag of a beam is inversely proportional to the cube of its depth.
1. Both A and R are true and R is the correct explanation of A
2. Both A and R are true but R is not the correct explanation of A
3. A is true but R is false
4. A is false but R is true
Answer
A. Both A and R are true and R is the correct explanation of A
A deep beam sags far less for the same material, which is why an I-shape is used. R explains A.
Assertion (A): Stress has the same unit as pressure. Reason (R): Strain has no unit.
1. Both A and R are true and R is the correct explanation of A
2. Both A and R are true but R is not the correct explanation of A
3. A is true but R is false
4. A is false but R is true
Answer
B. Both A and R are true but R is not the correct explanation of A
Both are true. Stress and pressure are both force per unit area; the unit of strain is not the reason.
Match List I with List II. List I: a. Young's modulus b. Shear modulus c. Bulk modulus d. Compressibility. List II: i. Change in shape ii. Change in volume iii. Change in length iv. Reciprocal of bulk modulus
1. a-iii, b-i, c-ii, d-iv
2. a-iii, b-i, c-iv, d-ii
3. a-iii, b-ii, c-i, d-iv
4. a-i, b-iii, c-ii, d-iv
Answer
A. a-iii, b-i, c-ii, d-iv
Young's modulus - length (iii); shear - shape (i); bulk - volume (ii); compressibility - 1/B (iv).
Match List I with List II. List I: a. Young's modulus b. Shear modulus c. Bulk modulus d. Poisson's ratio. List II: i. F/(Aθ) ii. FL/(AΔL) iii. −Δp/(ΔV/V) iv. Lateral strain/longitudinal strain
1. a-iii, b-i, c-ii, d-iv
2. a-ii, b-i, c-iii, d-iv
3. a-ii, b-i, c-iv, d-iii
4. a-i, b-ii, c-iii, d-iv
Answer
B. a-ii, b-i, c-iii, d-iv
Y = FL/(AΔL) (ii); G = F/(Aθ) (i); B = −Δp/(ΔV/V) (iii); σ = lateral/longitudinal strain (iv).
Match List I with List II. List I: a. Ductile material b. Brittle material c. Elastomer d. Largest Young's modulus among steel, copper and aluminium. List II: i. Rubber ii. Glass iii. Copper iv. Steel
1. a-ii, b-iii, c-i, d-iv
2. a-iii, b-i, c-ii, d-iv
3. a-iii, b-ii, c-iv, d-i
4. a-iii, b-ii, c-i, d-iv
Answer
D. a-iii, b-ii, c-i, d-iv
Copper is ductile, glass is brittle, rubber is an elastomer, and steel has the largest Y of the three metals.
Match List I with List II. List I: a. Proportional limit b. Yield point c. Ultimate tensile stress d. Fracture point. List II: i. Highest stress on the curve ii. End of the straight-line region iii. The wire breaks iv. Beyond this a permanent set remains
1. a-iv, b-ii, c-i, d-iii
2. a-ii, b-iv, c-iii, d-i
3. a-ii, b-iv, c-i, d-iii
4. a-ii, b-i, c-iv, d-iii
Answer
C. a-ii, b-iv, c-i, d-iii
Proportional limit: end of the straight line (ii); yield point: permanent set beyond (iv); ultimate stress: highest (i); fracture: breaks (iii).
Match List I with List II. List I: a. Energy stored in a stretched wire b. Energy per unit volume c. Force constant of a wire d. Sag of a loaded beam. List II: i. YA/L ii. ½ F ΔL iii. W l³/(4 b d³ Y) iv. ½ × stress × strain
1. a-i, b-iv, c-ii, d-iii
2. a-ii, b-iv, c-i, d-iii
3. a-iv, b-ii, c-i, d-iii
4. a-ii, b-iv, c-iii, d-i
Answer
B. a-ii, b-iv, c-i, d-iii
U = ½ FΔL (ii); u = ½ × stress × strain (iv); k = YA/L (i); δ = Wl³/(4bd³Y) (iii).
Match List I with List II. List I: a. Stress b. Compressibility c. Strain d. Force constant of a wire. List II: i. [M⁻¹ L T²] ii. [M⁰ L⁰ T⁰] iii. [M L⁻¹ T⁻²] iv. [M T⁻²]
1. a-iii, b-i, c-iv, d-ii
2. a-ii, b-i, c-iii, d-iv
3. a-iii, b-i, c-ii, d-iv
4. a-i, b-iii, c-ii, d-iv
Answer
C. a-iii, b-i, c-ii, d-iv
Stress [M L⁻¹ T⁻²]; compressibility = 1/stress = [M⁻¹ L T²]; strain is dimensionless; k = force/length = [M T⁻²].
Two wires of the same material carry the same load. The first has length 2L and radius r; the second has length L and radius 2r. The ratio of the extension of the first to that of the second is
1. 8 : 1
2. 4 : 1
3. 2 : 1
4. 1 : 8
Answer
A. 8 : 1
ΔL ∝ L/r². Ratio = (2/1)/(1/4) = 8.
A crane lifts a load that produces a force of 6 × 10⁴ N on its rope. The yield strength of the rope material is 3 × 10⁸ N m⁻², and a safety factor of 5 is to be used. The minimum area of cross-section of the rope is
1. 1 × 10⁻² m²
2. 5 × 10⁻³ m²
3. 2 × 10⁻⁴ m²
4. 1 × 10⁻³ m²
Answer
D. 1 × 10⁻³ m²
Allowed stress = 3 × 10⁸/5 = 6 × 10⁷ N m⁻². A = F/stress = 6 × 10⁴/6 × 10⁷ = 10⁻³ m². Ignoring the safety factor gives 2 × 10⁻⁴ m².
A steel wire and a copper wire have equal lengths and equal cross-sectional areas and carry the same load. Taking Y(steel) = 2.0 × 10¹¹ N m⁻² and Y(copper) = 1.1 × 10¹¹ N m⁻², the ratio of the extension of the steel wire to that of the copper wire is
1. 11 : 20
2. 1 : 1
3. 20 : 11
4. 11 : 40
Answer
A. 11 : 20
ΔL ∝ 1/Y, so ratio = Y(copper)/Y(steel) = 1.1/2.0 = 11/20.
A uniform wire has a force constant k. The wire is cut into two equal halves. The force constant of each half is
1. k/2
2. k
3. 2k
4. 4k
Answer
C. 2k
k = YA/L. Halving L doubles k. Check: for a given force each half extends half as much.
The elastic energy stored in a wire stretched by 2 mm is 4 J (within the elastic limit). The additional work needed to stretch it further from 2 mm to 3 mm is
1. 1 J
2. 5 J
3. 2 J
4. 9 J
Answer
B. 5 J
U ∝ x². U(3 mm) = 4 × (3/2)² = 9 J. Additional work = 9 − 4 = 5 J.
A steel sphere of volume 100 cm³ is taken deep into the sea where the pressure is greater by 8 × 10⁷ N m⁻². The bulk modulus of steel is 1.6 × 10¹¹ N m⁻². The decrease in its volume is
1. 5 cm³
2. 0.5 cm³
3. 0.005 cm³
4. 0.05 cm³
Answer
D. 0.05 cm³
ΔV = V Δp/B = 100 × 8 × 10⁷/(1.6 × 10¹¹) = 0.05 cm³.
A tangential force of 2 × 10⁴ N acts on a face of area 0.04 m² of a block whose opposite face is fixed. The shear modulus is 2 × 10¹⁰ N m⁻². The shear strain produced is
1. 2.5 × 10⁻⁵
2. 5 × 10⁻⁷
3. 1 × 10⁻⁵
4. 2.5 × 10⁻³
Answer
A. 2.5 × 10⁻⁵
Shear stress = 2 × 10⁴/0.04 = 5 × 10⁵ N m⁻². Strain = 5 × 10⁵/2 × 10¹⁰ = 2.5 × 10⁻⁵.

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