Properties of Solids

Q: What are the elastic properties of solids?

They describe how a solid resists and recovers from deformation: elasticity, the elastic limit, Hooke's law, the three moduli (Young's, bulk and shear), Poisson's ratio, the stress-strain curve and the elastic energy stored in a stretched body.

Q: What is the difference between stress and pressure?

Both are force per unit area with the unit pascal. Pressure is the external normal force of a fluid on a surface. Stress is the internal restoring force per area inside a deformed solid, and it can be tensile, compressive or shearing.

Q: Is steel more elastic than rubber?

Yes. In physics a more elastic material has a larger modulus: it develops a larger restoring stress for the same strain. Steel's Young's modulus is about 2 x 10^11 Pa, thousands of times that of rubber.

Q: What is the difference between the proportional limit and the elastic limit?

Up to the proportional limit stress is exactly proportional to strain, so Hooke's law holds. Between it and the elastic limit the wire still returns to its length, but not in proportion. Beyond the elastic limit it is permanently stretched.

Q: Why is the bulk modulus defined with a negative sign?

An increase in pressure decreases the volume, so the volume strain is negative. The minus sign makes the bulk modulus a positive number, as every modulus of elasticity should be.

Q: What is Poisson's ratio?

It is the ratio of lateral strain to longitudinal strain when a body is stretched: the fractional decrease in thickness divided by the fractional increase in length. It has no unit and lies between 0 and 0.5 in practice.

Q: What properties of solids questions come in JEE Main and JEE Advanced?

JEE Main asks Young's modulus numericals, ratios of extensions of wires, energy stored and bulk modulus. JEE Advanced adds rods stretching under their own weight, thermal stress in clamped rods, wires in series or parallel, Poisson's ratio and the relations between Y, K and G.

Q: Which properties of solids topics are important for NEET?

NEET regularly asks the stress-strain curve and its points, Young's modulus numericals on wires, the ratio of extensions when length or radius changes, bulk modulus and compressibility, and elastic potential energy. Most are short numericals using delta L = FL / (A Y).

PhysicsProperties Of Solids And FluidsFor JEE aspirants

The properties of solids in this chapter are their elastic properties: how a solid resists being stretched, twisted or squeezed and springs back. Hooke's law ( $stress = E \times strain$ ), the stress-strain curve, Young's, bulk and shear moduli, Poisson's ratio and stored elastic energy are the core properties of solids asked in NEET and JEE Main, usually as quick numericals on wires.

On this page1Elasticity and plasticity2Stress and strain3Hooke's law4Stress-strain curve5Three moduli6Poisson's ratio7Elastic energy8Searle's method9Applications

Key Formulas - Quick Reference

★ Must learnStress $= \frac{F}{A}$ ( $N m^{- 2}$ = Pa); strain $= \frac{Δ L}{L}$ , $\frac{Δ V}{V}$ or $θ$ (no unit)
★ Must learnHooke's law: stress $= E \times$ strain (within the proportional limit)
★ Must learnYoung's modulus $Y = \frac{F / A}{Δ L / L} = \frac{F L}{A Δ L}$ ; for a wire $Δ L = \frac{F L}{π r ^{2} Y}$
Bulk modulus $K = - \frac{p}{Δ V / V}$ ; compressibility $= \frac{1}{K}$
Modulus of rigidity (shear modulus) $G = \frac{F / A}{θ} = \frac{F L}{A Δ x}$
Poisson's ratio $σ = \frac{Δ d / d}{Δ L / L}$ (between $0$ and $0.5$ )
★ Must learnElastic energy $U = \frac{1}{2} F Δ L$ ; energy density $u = \frac{1}{2} stress \times strain = \frac{1}{2} Y (strain)^{2}$
Interatomic force constant $k = Y r_{0}$
Thermal stress in a clamped rod $= Y α Δ T$

1. Elasticity, Plasticity and Their Cause

A deforming force changes the relative positions of the molecules of a body, and so changes its length, volume or shape.

Term	Meaning
Elasticity	Property by which a body regains its original length, volume or shape when the deforming force is removed
Plasticity	Property by which a body keeps the deformation after the force is removed (putty, mud)
Perfectly elastic body	Regains its original shape immediately and completely. Quartz fibre and phosphor bronze come closest
Perfectly plastic body	Does not regain its shape at all, however small the deforming force

Cause of elasticity. Atoms in a solid sit at an equilibrium spacing $r_{0}$ where the interatomic force is zero. Pull them apart and they attract; push them closer and they repel. For small displacements the restoring force is proportional to the displacement, so each bond acts like a tiny spring. The interatomic force constant is

k = \frac{interatomic force}{change in interatomic distance} = Y r_{0}

Figure 1: Near the equilibrium spacing

r_{0}

the interatomic force is linear in the displacement, like a spring: this is the origin of elasticity and of Hooke's law (curve plotted from a Lennard-Jones model).

Key idea

Elasticity comes from the spring-like bonds between atoms; Hooke's law is the straight part of the interatomic force curve near

r_{0}

2. Stress and Strain

★ Must learn

Stress is the internal restoring force per unit area set up in a deformed body; in equilibrium it equals the applied deforming force per unit area: $stress = \frac{F}{A}$ . SI unit $N m^{- 2}$ (pascal), dimensions $[ML^{- 1} T^{- 2}]$ .

Strain is the fractional change in size or shape, $\frac{change in dimension}{original dimension}$ . It is a ratio of like quantities, so it has no unit and no dimensions.

Type	Stress	Strain	Effect
Tensile / compressive (longitudinal)	Normal force per area, $\frac{F}{A}$	Longitudinal strain $\frac{Δ L}{L}$	Change in length
Shearing (tangential)	Tangential force per area of the face, $\frac{F}{A}$	Shear strain $θ \approx tan θ = \frac{Δ x}{L}$	Change in shape
Hydraulic (volume)	Uniform pressure $p$ on all faces	Volume strain $\frac{Δ V}{V}$	Change in volume

Figure 2: Stress is force per area; the matching strains are

\frac{Δ L}{L}

(tensile),

θ \approx \frac{Δ x}{L}

(shearing) and

\frac{Δ V}{V}

(volume).

The elastic limit is the largest stress up to which a body regains its original form completely when the force is removed. Beyond it the body is permanently deformed. The elastic limit belongs to a body; elasticity is a property of its material.

3. Hooke's Law

★ Must learn

Hooke's law: within the elastic (strictly, proportional) limit, stress is directly proportional to strain:

stress = E \times strain

$E$ is the modulus of elasticity. It depends on the material and on the kind of deformation, and has the unit of stress (Pa).

Hooke first stated it for a stretched wire (extension $\propto$ load); it was later found to hold for compression, bending and twisting too, which is why it is written in the general stress-strain form.

4. The Stress-Strain Curve

Load a metal wire in steps and plot stress against strain. The curve shows how the material behaves from small stretching to breaking.

Figure 3: OA: Hooke's law; A: proportional limit; B: elastic limit (yield point); BC: plastic deformation; C: ultimate tensile strength; CD: plastic flow; D: fracture (breaking stress). Unloading beyond B leaves a permanent set OE.

Region or point	What happens
OA	Straight line: Hooke's law holds; the wire returns to its original length
A (proportional limit)	Highest stress up to which stress is proportional to strain
AB	Still elastic, but Hooke's law no longer holds
B (elastic limit / yield point)	Beyond B the deformation is permanent; the strain here is small (about 1%). Stress at B is the yield strength
BC	Plastic deformation: strain grows quickly; unloading follows a line parallel to OA and leaves a permanent set OE
C (ultimate tensile strength)	Greatest stress the wire can bear
CD	Plastic flow: the wire necks and stretches even as the load is reduced
D (fracture point)	The wire breaks; the stress here is the breaking stress

A material with a long plastic region (C far from D) is ductile and can be drawn into wires (copper, aluminium). A material that fractures soon after the elastic limit is brittle (glass, cast iron).

4.1 Elastomers

Substances that can be stretched to very large strains are called elastomers, for example rubber and the elastic tissue of the aorta, the largest artery carrying blood from the heart. Their stress-strain curve has no straight region, and although they stretch a lot they still return to their original length.

Figure 4: Stress (vertical) against strain (horizontal). Ductile metals have a wide plastic range, brittle materials fracture near the elastic limit, and elastomers (rubber, aorta tissue) reach large strains without a linear region (shapes schematic).

Quick Recall: tap to check

Which point on the stress-strain curve gives the yield strength?

B, the elastic limit (yield point).

What is the permanent set?

The strain OE left when a wire loaded beyond the elastic limit is unloaded.

Name two elastomers.

Rubber and the elastic tissue of the aorta.

5. The Three Moduli of Elasticity

Each kind of strain has its own modulus, $modulus = \frac{stress}{strain}$ within the elastic limit.

5.1 Young's modulus $Y$

For a wire of length $L$ and radius $r$ stretched by $Δ L$ under a force $F$ :

Y = \frac{normal stress}{longitudinal strain} = \frac{F / π r ^{2}}{Δ L / L} = \frac{F L}{π r ^{2} Δ L}

5.2 Bulk modulus $K$ and compressibility

A body of volume $V$ under an extra uniform pressure $p$ (normal stress $F / A = p$ ) shrinks by $Δ V$ :

K = - \frac{p}{Δ V / V} = - \frac{p V}{Δ V}

The minus sign makes $K$ positive, because the volume decreases ( $Δ V < 0$ ) when pressure increases. The reciprocal of the bulk modulus is the compressibility, $\frac{1}{K}$ . Solids are the least compressible and gases the most; bulk modulus applies to solids, liquids and gases alike.

5.3 Modulus of rigidity (shear modulus) $G$

A tangential force $F$ on a face of area $A$ shifts that face by $Δ x$ relative to the opposite face a distance $L$ away, through the angle $θ$ :

G = \frac{tangential stress}{shear strain} = \frac{F / A}{θ} = \frac{F L}{A Δ x}

Only solids have a modulus of rigidity: fluids cannot resist a steady shear. For most materials $G \approx \frac{Y}{3}$ .

Material	$Y$ ( $1 0^{9}$ Pa)	$G$ ( $1 0^{9}$ Pa)	$K$ ( $1 0^{9}$ Pa)
Steel	200	84	160
Copper	120	42	140
Aluminium	70	25	72
Glass	65	23	37
Water	-	-	2.2

Solids

Have all three moduli, $Y$ , $G$ and $K$ . They resist changes in length, shape and volume.

Liquids and gases

Have only a bulk modulus $K$ . They cannot sustain a tensile or shearing stress, so $Y$ and $G$ are not defined.

Exam Trick

Same material, same load: $Δ L \propto \frac{L}{r ^{2}}$ . Doubling the diameter cuts the extension to one quarter; doubling the length doubles it. The maximum load within the elastic limit is $\propto r^{2}$ , so a rope twice as thick holds four times as much.

Figure 5: Same material, same load:

Δ L = \frac{F L}{π r ^{2} Y} \propto \frac{L}{r ^{2}}

. For steel,

F = 100 N

L = 1 m

r = 0.5 mm

Δ L = 0.64 mm

; doubling

L

gives

1.27 mm

and doubling

r

gives

0.16 mm

(extensions exaggerated, ratios to scale).

Figure 6: Pick the modulus from what changes: length (

Y

), shape (

G

) or volume (

K

). Liquids and gases have only

K

6. Poisson's Ratio

A stretched wire gets longer and thinner. The ratio of lateral strain to longitudinal strain is Poisson's ratio:

σ = \frac{lateral strain}{longitudinal strain} = \frac{Δ d / d}{Δ L / L}

It has no unit. For most metals it lies between $0.25$ and $0.35$ (steel about $0.28$ to $0.30$ ); theoretically $- 1 < σ < 0.5$ , and in practice $0 < σ < 0.5$ .

Figure 7: Stretching lengthens the wire and narrows it. Poisson's ratio

σ = \frac{Δ d / d}{Δ L / L}

(dashed = before).

JEE Advanced

The moduli of an isotropic material are linked through Poisson's ratio: $Y = 3 K (1 - 2 σ)$ and $Y = 2 G (1 + σ)$ . For $σ = 0.5$ , $K \to \infty$ : the material is incompressible (rubber is close to this). Rod hanging under its own weight (length $L$ , density $ρ$ ): stress grows linearly from the bottom, and the extension is $Δ L = \frac{ρ g L ^{2}}{2 Y}$ , half of what a load equal to the rod's weight at the end would produce. Thermal stress: a rod clamped between rigid walls and heated by $Δ T$ cannot expand, so it carries a compressive stress $Y α Δ T$ .

Figure 8: At height

x

above the free end, the rod holds up only the part below: stress

= ρ g x

. Adding up the stretch of every element gives

Δ L = \frac{ρ g L ^{2}}{2 Y} = \frac{M g L}{2 A Y}

, half the stretch of a load

M g

hung at the end. A

10 m

steel rod stretches only

19.1 μ m

Figure 9: A clamped rod is squeezed back by exactly its free expansion

α L Δ T

, a strain

α Δ T

, so the stress is

Y α Δ T

and the force on each wall

F = Y A α Δ T

(independent of length). Steel,

Δ T = 50 K

1.2 \times 1 0^{8} Pa

7. Factors Affecting Elasticity; Elastic Fatigue

Temperature: elasticity generally decreases as temperature rises. Invar steel (short for "invariable") is an exception: its elasticity hardly changes with temperature.
Impurities: a more elastic impurity increases the elasticity of the material; a more plastic impurity decreases it.
Hammering or rolling increases elasticity; annealing (alternate heating and slow cooling) decreases it.
Elastic fatigue: a body subjected to repeated stress and strain temporarily loses some of its elastic strength (it recovers more slowly). If left undisturbed for some time it regains its properties, so elastic fatigue is a temporary effect. This is why bridges are declared unsafe after long use.

8. Elastic Potential Energy in a Stretched Wire

Work done against the internal restoring forces while stretching a wire is stored as elastic potential energy.

The restoring force grows from $0$ to $F$ as the extension grows to $Δ L$ , so the average force is $\frac{F}{2}$ .
Work done $= \frac{F}{2} \times Δ L$ , stored as
$U = \frac{1}{2} F Δ L$
Write it with stress, strain and volume $A L$ : $U = \frac{1}{2} \cdot \frac{F}{A} \cdot \frac{Δ L}{L} \cdot A L = \frac{1}{2} stress \times strain \times volume$ .
Energy per unit volume (energy density):
$u = \frac{1}{2} stress \times strain = \frac{1}{2} Y (strain)^{2} = \frac{( stress ) ^{2}}{2 Y}$

Figure 10: Within the elastic limit

F \propto Δ l

, so the stored energy is the triangle area

U = \frac{1}{2} F Δ l

Exam Trick

Pick the form of $U$ that fits the data. Given the load: $U = \frac{F ^{2} L}{2 A Y}$ . Given the extension: $U = \frac{Y A ( Δ L ) ^{2}}{2 L}$ . Given only stress or strain: $u = \frac{( stress ) ^{2}}{2 Y}$ per unit volume. For the same load a thinner wire stores more energy.

Key idea

Stored energy is the area under the force-extension line:

U = \frac{1}{2} F Δ L

, and per unit volume

\frac{1}{2} stress \times strain

9. Searle's Apparatus: Measuring Young's Modulus

Two identical long wires hang from the same rigid support. The reference wire carries a fixed dead weight to keep it taut. The experimental wire (length $L$ , radius $r$ ) carries slotted weights. A spirit level rests on frames attached to the two wires; a micrometer screw on the experimental side is turned until the bubble is centred again after each load is added, and the change in reading gives the extension $l$ .

Y = \frac{M g L}{π r ^{2} l}

Using a reference wire cancels any sag of the support and the effect of temperature changes, because both wires are affected equally. To stay safely inside the elastic limit, the maximum load used is about one-third of the breaking load: breaking load $= breaking stress \times π r^{2}$ , so maximum load $\approx \frac{1}{3} breaking stress \times π r^{2}$ .

Figure 11: Searle's method. The reference wire cancels temperature and support sag; the micrometer reads the extension

l

, and

Y = \frac{M g L}{π r ^{2} l}

10. Applications of Elastic Behaviour

Crane ropes: the rope must keep the stress below the yield strength with a safety factor (often 10). For a $1 0^{4} kg$ load and yield strength $3 \times 1 0^{8} Pa$ the rope needs a radius of about $3 cm$ ; in practice many thin wires are braided for flexibility.
Beams and bridges: a beam of length $l$ , breadth $b$ and depth $d$ loaded by $W$ at the middle sags by $δ = \frac{W l ^{3}}{4 b d ^{3} Y}$ . Increasing the depth $d$ reduces sagging most, which is why beams have an I-section: deep, with material placed where the stress is largest, but light.
Height of mountains: the pressure at the base of a mountain of height $h$ is $h ρ g$ . It must stay below the elastic limit of rock (about $3 \times 1 0^{8} Pa$ ), so $h < \frac{3 \times 1 0 ^{8}}{( 3 \times 1 0 ^{3} ) ( 9.8 )} \approx 10 km$ . Mount Everest (about $8.8 km$ ) is close to this limit.

Figure 12: A beam loaded at the middle sags by

δ = \frac{W l ^{3}}{4 b d ^{3} Y}

(NCERT; sag exaggerated, shape from the elastic-beam formula). Doubling the depth

d

cuts the sag to

\frac{1}{8}

but doubling the breadth

b

only to

\frac{1}{2}

, so girders use a deep, light I-section.

Quick Recall: tap to check

Why do beams have an I-shaped cross-section?

Sag

δ \propto \frac{1}{b d ^{3}}

: depth matters most, so a deep, light section resists bending best.

Why is a reference wire used in Searle's apparatus?

To cancel sagging of the support and temperature effects.

Which modulus is defined for liquids?

Only the bulk modulus.

Figure 13: Revision map of the properties of solids: stress

=

modulus

\times

strain within the elastic limit, with the right modulus for the right change.

11. Solved Examples

Solved Example 1

A cable is replaced by another of the same length and material but twice the diameter. How does this affect the elongation under a given load, and the maximum load it can support without exceeding the elastic limit?

Solution:

$Y = \frac{M g l}{π ( D /2 ) ^{2} Δ l} = \frac{4 M g l}{π D ^{2} Δ l}$ , so $Δ l = \frac{4 M g l}{π D ^{2} Y}$ , i.e. $Δ l \propto \frac{1}{D ^{2}}$ .

The maximum load is set by the maximum stress: $M g = stress \times \frac{π D ^{2}}{4}$ , so $M g \propto D^{2}$ .

Answer: doubling the diameter makes the elongation one-fourth and the maximum load four times.

Solved Example 2

A steel wire

4.0 m

long is stretched by

2.0 mm

. Its cross-sectional area is

2.0 mm^{2}

and

Y = 2.0 \times 1 0^{11} N m^{- 2}

. Find (a) the energy density of the wire and (b) the elastic potential energy stored.

Solution:

Strain $= \frac{2 \times 1 0 ^{- 3}}{4} = 5 \times 1 0^{- 4}$ ; $A = 2.0 \times 1 0^{- 6} m^{2}$ .

(a) $u = \frac{1}{2} Y (strain)^{2} = \frac{1}{2} (2.0 \times 1 0^{11}) (5 \times 1 0^{- 4})^{2} = 2.5 \times 1 0^{4} J m^{- 3}$ .

(b) $U = u \times A L = (2.5 \times 1 0^{4}) (2.0 \times 1 0^{- 6}) (4.0)$ .

Answer: (a) $2.5 \times 1 0^{4} J m^{- 3}$ ; (b) $0.20 J$ .

Solved Example 3

A steel wire of length

2 m

and radius

1 mm

carries a

10 kg

load. Find its extension. (

Y = 2 \times 1 0^{11} Pa

g = 9.8 m s^{- 2}

)

Solution:

$Δ L = \frac{F L}{π r ^{2} Y} = \frac{( 98 ) ( 2 )}{π ( 1 0 ^{- 3} ) ^{2} ( 2 \times 1 0 ^{11} )} = \frac{196}{6.28 \times 1 0 ^{5}}$ .

Answer: $Δ L \approx 3.1 \times 1 0^{- 4} m = 0.31 mm$ .

Solved Example 4

By what fraction does the volume of water shrink at a depth where the extra pressure is

100 atm

? (

K_{water} = 2.2 \times 1 0^{9} Pa

1 atm = 1.013 \times 1 0^{5} Pa

)

Solution:

$\frac{Δ V}{V} = \frac{p}{K} = \frac{100 ( 1.013 \times 1 0 ^{5} )}{2.2 \times 1 0 ^{9}}$ .

Answer: $\approx 4.6 \times 1 0^{- 3}$ , i.e. about $0.46%$ . Water is nearly incompressible.

Solved Example 5

A brass cube of side

10 cm

has its bottom face fixed. A tangential force of

4000 N

acts on its top face. Find the shear strain and the displacement of the top face. (

G = 3.6 \times 1 0^{10} Pa

)

Solution:

Shear stress $= \frac{4000}{( 0.1 ) ^{2}} = 4 \times 1 0^{5} Pa$ .

Shear strain $θ = \frac{stress}{G} = \frac{4 \times 1 0 ^{5}}{3.6 \times 1 0 ^{10}} = 1.1 \times 1 0^{- 5} rad$ .

$Δ x = θ L = (1.1 \times 1 0^{- 5}) (0.1)$ .

Answer: $θ \approx 1.1 \times 1 0^{- 5}$ ; $Δ x \approx 1.1 \times 1 0^{- 6} m$ .

Solved Example 6

Two wires of the same material carry the same load. Wire A has length

L

and radius

2 r

; wire B has length

2 L

and radius

r

. The ratio of their extensions

Δ L_{A} : Δ L_{B}

is
(A)

1 : 2

(B)

1 : 4

(C)

1 : 8

(D)

8 : 1

Solution:

Answer: (C). $Δ L \propto \frac{L}{r ^{2}}$ : $\frac{Δ L _{A}}{Δ L _{B}} = \frac{L / ( 2 r ) ^{2}}{2 L / r ^{2}} = \frac{1}{8}$ .

Solved Example 7

Young's modulus of steel is

2 \times 1 0^{11} Pa

and the interatomic spacing is

2.8 \times 1 0^{- 10} m

. Estimate the interatomic force constant.

Solution:

$k = Y r_{0} = (2 \times 1 0^{11}) (2.8 \times 1 0^{- 10})$ .

Answer: $k \approx 56 N m^{- 1}$ .

Solved Example 8

A wire of length

2 m

and diameter

1 mm

is stretched by

1 mm

. If Poisson's ratio is

0.3

, find the decrease in diameter.

Solution:

Longitudinal strain $= \frac{1 0 ^{- 3}}{2} = 5 \times 1 0^{- 4}$ . Lateral strain $= σ \times 5 \times 1 0^{- 4} = 1.5 \times 1 0^{- 4}$ .

$Δ d = (1.5 \times 1 0^{- 4}) (1 0^{- 3} m)$ .

Answer: $Δ d = 1.5 \times 1 0^{- 7} m$ .

Solved Example 9

In Searle's experiment the wire has radius

0.5 mm

and breaking stress

4 \times 1 0^{8} Pa

. What is the largest load that should be used? (

g = 9.8 m s^{- 2}

)

Solution:

Breaking load $= (4 \times 1 0^{8}) π (0.5 \times 1 0^{- 3})^{2} = 314 N$ . Maximum load $\approx \frac{1}{3} (314) = 105 N$ .

Answer: about $105 N$ , i.e. about $10.7 kg$ .

Solved Example 10

A crane must lift

1 0^{4} kg

. The steel rope has yield strength

3 \times 1 0^{8} Pa

and a safety factor of

10

is required. Find the minimum radius of the rope. (

g = 9.8 m s^{- 2}

)

Solution:

Allowed stress $= \frac{3 \times 1 0 ^{8}}{10}$ , so $A = \frac{( 1 0 ^{4} ) ( 9.8 ) ( 10 )}{3 \times 1 0 ^{8}} = 3.27 \times 1 0^{- 3} m^{2}$ .

$r = \frac{A}{π}$ .

Answer: $r \approx 3.2 cm$ .

Solved Example 11

Estimate the greatest possible height of a mountain on Earth. (Elastic limit of rock

3 \times 1 0^{8} Pa

, density

3 \times 1 0^{3} kg m^{- 3}

g = 9.8 m s^{- 2}

)

Solution:

Stress at the base $h ρ g$ must not exceed the elastic limit: $h = \frac{3 \times 1 0 ^{8}}{( 3 \times 1 0 ^{3} ) ( 9.8 )}$ .

Answer: $h \approx 1.0 \times 1 0^{4} m \approx 10 km$ .

Solved Example 12

A steel rod is clamped between two rigid walls at room temperature and then heated by

5 0^{\circ} C

. Find the thermal stress. (

Y = 2 \times 1 0^{11} Pa

α = 1.2 \times 1 0^{- 5} K^{- 1}

)

Solution:

The walls stop the free expansion $α L Δ T$ , so the rod is compressed by a strain $α Δ T$ : stress $= Y α Δ T = (2 \times 1 0^{11}) (1.2 \times 1 0^{- 5}) (50)$ .

Answer: $1.2 \times 1 0^{8} Pa$ (compressive).

Solved Example 13

A uniform rod of mass

M

hangs vertically from a rigid support. Its extension due to its own weight, compared with the extension of the same rod (treated as massless) carrying a load

M g

at its lower end, is
(A) equal
(B) half
(C) double
(D) one-quarter

Solution:

Answer: (B). The tension grows from $0$ at the free end to $M g$ at the top, so on average only half the weight stretches the rod: $Δ L = \frac{M g L}{2 A Y}$ , half of $\frac{M g L}{A Y}$ .

Solved Example 14

A rectangular beam rests on two supports and carries a load at its centre. If its depth is doubled (length, breadth, material and load unchanged), the sag at the centre becomes
(A)

\frac{1}{2}

(B)

\frac{1}{4}

(C)

\frac{1}{8}

(D)

\frac{1}{16}

of its old value

Solution:

Answer: (C). $δ = \frac{W l ^{3}}{4 b d ^{3} Y} \propto \frac{1}{d ^{3}}$ , so doubling $d$ divides the sag by $2^{3} = 8$ .

Solved Example 15

A steel rod

10 m

long hangs from a ceiling. Find its extension due to its own weight. (

ρ = 7800 kg m^{- 3}

Y = 2 \times 1 0^{11} Pa

g = 9.8 m s^{- 2}

)

Solution:

$Δ L = \frac{ρ g L ^{2}}{2 Y} = \frac{( 7800 ) ( 9.8 ) ( 10 ) ^{2}}{2 ( 2 \times 1 0 ^{11} )} = \frac{7.644 \times 1 0 ^{6}}{4 \times 1 0 ^{11}}$ .

Answer: $Δ L \approx 1.9 \times 1 0^{- 5} m = 19 μ m$ . The answer does not depend on the cross-section.

Practice Questions

A load of $4 kg$ stretches a wire of length $2 m$ and area $2 \times 1 0^{- 6} m^{2}$ by $0.2 mm$ . Find $Y$ . ( $g = 10 m s^{- 2}$ )Answer: $Y = 2 \times 1 0^{11} Pa$
What pressure reduces the volume of a rubber ball ( $K = 9.1 \times 1 0^{8} Pa$ ) by $0.1%$ ?Answer: $9.1 \times 1 0^{5} Pa$
Find the energy stored in a wire stretched by $1 mm$ under a force of $100 N$ .Answer: $0.05 J$
Why is steel more elastic than rubber?Answer: For the same stress steel has a much smaller strain, i.e. a much larger modulus
The length of a wire is doubled by stretching (volume constant). By what factor does the extension under the same load change?Answer: $4$ times ( $L$ doubles, area halves)
Which modulus is involved when a rod is twisted, and which when a solid is squeezed uniformly?Answer: Shear modulus; bulk modulus

Common Mistakes to Avoid

Watch out

Using the diameter in place of the radius in $A = π r^{2}$ , or forgetting to convert $mm^{2}$ to $m^{2}$ ( $1 mm^{2} = 1 0^{- 6} m^{2}$ ).
Thinking rubber is more elastic than steel. Elasticity means a large modulus; steel has a far larger $Y$ .
Dropping the minus sign in $K = - \frac{p}{Δ V / V}$ , which gives a negative bulk modulus.
Using $U = F Δ L$ instead of $\frac{1}{2} F Δ L$ : the force grows from zero.
Calling the proportional limit and the elastic limit the same point; Hooke's law stops at A, elasticity at B.
Assigning $Y$ or $G$ to liquids; fluids have only a bulk modulus.
Writing the dimensions of stress as $[MLT^{- 2}]$ (force). Stress is force per area: $[ML^{- 1} T^{- 2}]$ .

Frequently Asked Questions

What are the elastic properties of solids?

They describe how a solid resists and recovers from deformation: elasticity, the elastic limit, Hooke's law, the three moduli (Young's, bulk and shear), Poisson's ratio, the stress-strain curve and the elastic energy stored in a stretched body.

What is the difference between stress and pressure?

Both are force per unit area with the unit pascal. Pressure is the external normal force of a fluid on a surface. Stress is the internal restoring force per area inside a deformed solid, and it can be tensile, compressive or shearing.

Is steel more elastic than rubber?

Yes. In physics a more elastic material has a larger modulus: it develops a larger restoring stress for the same strain. Steel's Young's modulus is about 2 x 10^11 Pa, thousands of times that of rubber.

What is the difference between the proportional limit and the elastic limit?

Up to the proportional limit stress is exactly proportional to strain, so Hooke's law holds. Between it and the elastic limit the wire still returns to its length, but not in proportion. Beyond the elastic limit it is permanently stretched.

Why is the bulk modulus defined with a negative sign?

An increase in pressure decreases the volume, so the volume strain is negative. The minus sign makes the bulk modulus a positive number, as every modulus of elasticity should be.

What is Poisson's ratio?

It is the ratio of lateral strain to longitudinal strain when a body is stretched: the fractional decrease in thickness divided by the fractional increase in length. It has no unit and lies between 0 and 0.5 in practice.

What properties of solids questions come in JEE Main and JEE Advanced?

JEE Main asks Young's modulus numericals, ratios of extensions of wires, energy stored and bulk modulus. JEE Advanced adds rods stretching under their own weight, thermal stress in clamped rods, wires in series or parallel, Poisson's ratio and the relations between Y, K and G.

Which properties of solids topics are important for NEET?

NEET regularly asks the stress-strain curve and its points, Young's modulus numericals on wires, the ratio of extensions when length or radius changes, bulk modulus and compressibility, and elastic potential energy. Most are short numericals using delta L = FL / (A Y).

Previous year questions on Properties of Solids

24 questions from past papers, each with a step-by-step solution.

Show all 24 questions

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1. Elasticity, Plasticity and Their Cause

2. Stress and Strain

3. Hooke's Law

4. The Stress-Strain Curve

4.1 Elastomers

5. The Three Moduli of Elasticity

5.1 Young's modulus Y

5.2 Bulk modulus K and compressibility

5.3 Modulus of rigidity (shear modulus) G

6. Poisson's Ratio

7. Factors Affecting Elasticity; Elastic Fatigue

8. Elastic Potential Energy in a Stretched Wire

9. Searle's Apparatus: Measuring Young's Modulus

10. Applications of Elastic Behaviour

11. Solved Examples

Common Mistakes to Avoid

Frequently Asked Questions

What are the elastic properties of solids?

What is the difference between stress and pressure?

Is steel more elastic than rubber?

What is the difference between the proportional limit and the elastic limit?

Why is the bulk modulus defined with a negative sign?

What is Poisson's ratio?

What properties of solids questions come in JEE Main and JEE Advanced?

Which properties of solids topics are important for NEET?

Previous year questions on Properties of Solids

Ready to master Properties Of Solids And Fluids?

5.1 Young's modulus $Y$

5.2 Bulk modulus $K$ and compressibility

5.3 Modulus of rigidity (shear modulus) $G$