Chapter 18 Mechanical Properties of Solids – Physics free study material by TEACHING CARE online tuition and coaching classes
Chapter 18 Mechanical Properties of Solids – Physics free study material by TEACHING CARE online tuition and coaching classes
Interatomic Forces.
The forces between the atoms due to electrostatic interaction between the charges of the atoms are called interatomic forces. These forces are electrical in nature and these are active if the distance between the two atoms is of the order of atomic size i.e. 10^{–10} metre.
 Every atom is electrically neutral, the number of electrons (negative charge) orbiting around the nucleus is equal to the number of proton (positive charge) in the So if two atoms are placed at a very large distance from each other then there will be a very small (negligible) interatomic force working between them.
 When these two atoms are brought close to each other to a distance of the order of 10^{–10} m, the distances between their positive nuclei and negative electron clouds get disturbed, and due to this, attractive interatomic force is produced between two
 This attractive force increases continuously with decrease in r and becomes maximum for one value of r
called critical distance, represented by x (as shown in the figure). Beyond this the attractive force starts decreasing rapidly with further decrease in the value of r.
 When the distance between the two atoms becomes r_{0}, the interatomic force will be This distance r_{0} is called normal or equilibrium distance.
(r_{0} = 0.74 Å for hydrogen).
 When the distance between the two atoms further decreased, the interatomic force becomes repulsive in nature and increases very rapidly with decrease in distance between two
 The potential energy U is related with the interatomic force F by the following
F = – dU
dr
 When two atoms are at very large distance, the potential energy is negative and becomes more negative as r is
 When the distance between the two atoms becomes r_{0}, the potential energy of the system of two atoms becomes minimum (e. attains maximum negative value). As the state of minimum potential energy is the state of equilibrium, hence the two atoms at separation r_{0} will be in a state of equilibrium.
( U0 = 7.2 ´ 10 ^{19} Joule
for hydrogen).
 When the distance between the two atoms is further decreased (e. r < r_{0}) the negative value of potential energy of the system starts decreasing. It becomes zero and then attains positive value with further decrease in r (as shown in the figure).
Intermolecular Forces.
The forces between the molecules due to electrostatic interaction between the charges of the molecules are called intermolecular forces. These forces are also called Vander Waal forces and are quite weak as compared to interatomic forces. These forces are also electrical in nature and these are active if the separation between two molecules is of the order of molecular size i.e. » 10^{–9} m.
 It is found that the force of attraction between molecules varies inversely as seventh power of the distance between them e.
F µ 1
or F
= – a
att r 7 att r7
The negative sign indicates that the force is attractive in nature.
 When the distance between molecules becomes less than r_{0}, the forces becomes repulsive in nature and is found to vary inversely as ninth power of the distance between them e.
F µ 1
or F = b .
rep r 9 rep r 9
Therefore force between two molecules is given by
F = Fatt
 Frep
= – a + b r 7 r 9
The value of constants a and b depend upon the structure and nature of molecules.
 Intermolecular forces between two molecules has the same general nature as shown in the figure for interatomic
 Potential Energy : Potential energy can be approximately expressed by the formula U = A– B
rn rm
where the term
 represents repulsive contribution and term
r n
 represents the attractive
rm
Constants A, B and numbers m and n are different for different molecules.
For majority of solids n = 12 and m = 6.
So potential energy can be expressed as U =
A – B r 12 r 6
Comparison Between Interatomic and Intermolecular Forces.
 Similarities
 Both the forces are electrical in
 Both the forces are active over short
 General shape of forcedistance graph is similar for both the
 Both the forces are attractive up to certain distance between atoms/molecules and become repulsive when the distance between them become less than that
 Dissimilarities
 Interatomic force depends upon the distance between the two atoms, whereas the intermolecular force depends upon the distance between the two molecules as well as their relative
 Interatomic forces are about 50 to100 times stronger than intermolecular
 The value of r_{0} for two atoms is smaller than the corresponding value for the Therefore one molecule is not restricted to attract only one molecule, but can attract many molecule. It is not so incase of atoms, since the atoms of one molecule cannot bind the atoms of other molecules.
States of Matter.
The three states of matter differ from each other due to the following two factors.
 The different magnitudes of the interatomic and intermolecular
 The extent of random thermal motion of atoms and molecules of a substance (which depends upon temperature).
Comparison Chart of Solid, Liquid and Gaseous States  
Property Shape Volume Density
Compressibility
Crystallinity Interatomic or intermolecular distance Relation between kinetic energy K and potential energy (U) Intermolecular force
Freedom of motion
Effect of temperature  Solid Definite Definite Maximum
Incompressible
Crystalline Constant
K < U
Strongest
Molecules vibrate about their mean position but cannot move freely. Matter remains in solid form below a certain temperature.  Liquid Not definite Definite Less than solids but more than gases. Less than gases but more than solids. Noncrystalline Not constant
K> U
Less than solids but more than gases. Molecules have limited free motion.
Liquids are found at temperatures more than that of solid.  Gas Not definite Not definite Minimum
Compressible
Not constant
K >> U
Weakest Molecules are free to move.
These are found at temperatures greater than that of solids and liquids. 
Note : @ The fourth state of matter in which the medium is in the form of positive and negative ions, is known as plasma. Plasma occurs in the atmosphere of stars (including the sun) and in discharge tubes.
Types of Solids.
A solid is that state of matter in which its constituent atoms or molecules are held strongly at the position of minimum potential energy and it has a definite shape and volume. The solids can be classified into two categories, crystalline and glassy or amorphous solids.
Comparison chart of Crystalline and Amorphous Solids  
Crystalline solids  Amorphous or glassy solids 
The constituent atoms, ions or molecules are arranged in a  The constituent atoms, ions or molecules are not arranged in a 
regular repeated three dimensional pattern, within the solid.  regular repeated three dimensional pattern, within the solid. 
Definite external geometric shape.  No regularity in external shape. 
All the bonds in ions, or atoms or molecules are equally strong.  All the bonds are not equally strong. 
They are anisotropic.  They are isotropic. 
They have sharp melting point.  They don’t have no sharp melting point. 
They have a longrange order of atoms or ions or  They don’t have a longrange order. 
molecules in them.  
They are considered true and stable solids.  They are not regarded as true and stable solids. 
Elastic Property of Matter.
 Elasticity : The property of matter by virtue of which a body tends to regain its original shape and size after the removal of deforming force is called
 Plasticity : The property of matter by virtue of which it does not regain its original shape and size after the removal of deforming force is called
 Perfectly elastic body : If on the removal of deforming forces the body regain its original configuration completely it is said to be perfectly
A quartz fibre and phosphor bronze (an alloy of copper containing 4% to 10% tin, 0.05% to 1% phosphorus) is the nearest approach to the perfectly elastic body.
 Perfectly plastic body : If the body does not have any tendency to recover its original configuration, on the removal of deforming force, it is said to be perfectly
Paraffin wax, wet clay are the nearest approach to the perfectly plastic body.
Practically there is no material which is either perfectly elastic or perfectly plastic and the behaviour of actual bodies lies between the two extremes.
 Reason of elasticity : In a solids, atoms and molecules are arranged in such a way that each molecule is acted upon by the forces due to neighbouring These forces are known
as intermolecular forces.
For simplicity, the two molecules in their equilibrium positions (at inter molecular distance r = r_{0}) (see graph in article 9.1) are shown by connecting them with a spring.
In fact, the spring connecting the two molecules represents the intermolecular
force between them. On applying the deforming forces, the molecules either come closer or go far apart from each other and restoring forces are developed. When the deforming force is removed, these restoring forces bring the molecules of the solid to their respective equilibrium position (r = r_{0}) and hence the body regains its original form.
 Elastic limit : Elastic bodies show their property of elasticity upto a certain value of deforming force. If we go on increasing the deforming force then a stage is reached when on removing the force, the body will not return to its original state. The maximum deforming force upto which a body retains its property of elasticity is called elastic limit of the material of
Elastic limit is the property of a body whereas elasticity is the property of material of the body.
 Elastic fatigue : The temporary loss of elastic properties because of the action of repeated alternating deforming force is called elastic
It is due to this reason
 Bridges are declared unsafe after a long time of their
 Spring balances show wrong readings after they have been used for a long
 We are able to break the wire by repeated
 Elastic after effect : The time delay in which the substance regains its original condition after the removal of deforming force is called elastic after effect. It is the time for which restoring forces are present after the removal of the deforming force it is negligible for perfectly elastic substance, like quartz, phosphor bronze and large for glass
Stress.
When a force is applied on a body there will be relative displacement of the particles and due to property of elasticity an internal restoring force is developed which tends to restore the body to its original state.
The internal restoring force acting per unit area of cross section of the deformed body is called stress.
At equilibrium, restoring force is equal in magnitude to external force, stress can therefore also be defined as external force per unit area on a body that tends to cause it to deform.
If external force F is applied on the area A of a body then,
Stress = Force = F
Area A
Unit : N / m^{2} (S.I.) , dyne / cm^{2} (C.G.S.)
Dimension : [ML^{1}T ^{2}]
Stress developed in a body depends upon how the external forces are applied over it. On this basis there are two types of stresses : Normal and Shear or tangential stress
 Normal stress : Here the force is applied normal to the
It is again of two types : Longitudinal and Bulk or volume stress
 Longitudinal stress
 It occurs only in solids and comes in picture when one of the three dimensions viz. length, breadth, height is much greater than other
 Deforming force is applied parallel to the length and causes increase in
 Area taken for calculation of stress is area of cross
 Longitudinal stress produced due to increase in length of a body under a deforming force is called tensile
 Longitudinal stress produced due to decrease in length of a body under a deforming force is called compressional
 Bulk or Volume stress
 It occurs in solids, liquids or
 In case of fluids only bulk stress can be
 It produces change in volume and density, shape remaining same.
 Deforming force is applied normal to surface at all
 Area for calculation of stress is the complete surface area perpendicular to the applied
 It is equal to change in pressure because change in pressure is responsible for change in
 Shear or tangential stress : It comes in picture when successive layers of solid move on each other e.
when there is a relative displacement between various layers of solid.
 Here deforming force is applied tangential to one of the
 Area for calculation is the area of the face on which force is
 It produces change in shape, volume remaining the same.
Problem 1. A and B are two wires. The radius of A is twice that of B. they are stretched by the same load. Then the stress on B is [MP PMT 1993]
 Equal to that on A (b) Four times that on A
(c) Two times that on A (d) Half that on A
Solution : (b) Stress = Force =
Area
F
pr ^{2}
1 (Stress) æ r ö 2
\ Stress µ Þ ^{B} = ç ^{A} ÷ = (2)^{2} Þ (Stress)_{B} = 4 × (stress)_{A} [As F = constant]
r ^{2} (Stress) _{A}
è rB ø
Problem 2. One end of a uniform wire of length L and of weight W is attached rigidly to a point in the roof and a weight W_{1} is suspended from its lower end. If S is the area of crosssection of the wire, the stress in the wire at a height 3L/4 from its lower end is [IITJEE 1992]
(a) W1 S
(b)
W_{1} + (W / 4)
S
(c)
W_{1} + (3W / 4)
S
(d)
W1 + W S
Solution : (c) As the wire is uniform so the weight of wire below point P is 3W
4
\ Total force at point
P = W1 + 3W
4
and area of crosssection = S
\ Stress at point P = Force =
Area
W1 + 3W
4
S
Problem 3. On suspending a weight Mg, the length l of elastic wire and area of crosssection A its length becomes double the initial length. The instantaneous stress action on the wire is
(a) Mg/A (b) Mg/2A (c) 2Mg/A (d) 4Mg/A
Solution : (c) When the length of wire becomes double, its area of cross section will become half because volume of wire is constant
(V = AL) .
So the instantaneous stress =
Force = Mg = 2Mg .
Area A / 2 A
Problem 4. A bar is subjected to equal and opposite forces as shown in the figure. PQRS is a plane making angle q with the crosssection of the bar. If the area of crosssection be ‘A’, then what is the tensile stress on PQRS
 F / A
 F cosq / A
 F cos^{2}q / A
 F / A cosq
Solution : (c) As tensile stress =
Normal force = F_{N}
Area AN
and here
AN = (A / cosq ) , FN
= Normal force = F cosq
So, Tensile stress =
F cosq
A / cosq
= F cos ^{2} q
A
Problem 5. In the above question, what is the shearing stress on PQ
(a) F / A cos q (b) F sin 2q / 2A (c) F / 2A sin 2q (d) F cosq / A
Solution : (b) Shear stress = Tangential force =
F sinq
= F sin q cosq
= F sin 2q
Area ( A /θcos ) A 2A
Problem 6. In the above question, when is the tensile stress maximum
(a)
q = 0^{o}
(b)
q = 30^{o}
(c)
q = 45^{o}
(d)
q = 90^{o}
Solution : (a) Tensile stress = F cos 2 q . It will be maximum when cos ^{2} q = max . i.e. cosq = 1 Þ q = 0 ^{o} .
A
Problem 7. In the above question, when is the shearing stress maximum
(a)
q = 0^{o}
(b)
q = 30^{o}
(c)
q = 45^{o}
(d)
q = 90^{o}
Solution : (c) Shearing stress = F sin 2q . It will be maximum when sin 2q = max
2A
i.e. sin 2q = 1 Þ 2q = 90 ^{o}
Þ q = 45 ^{o} .
Strain.
The ratio of change in configuration to the original configuration is called strain. Being the ratio of two like quantities, it has no dimensions and units.
Strain are of three types :
 Linear strain : If the deforming force produces a change in length alone, the strain produced in the body is called linear strain or tensile
Linear strain = Change in length(Dl)
Original length(l)
Linear strain in the direction of deforming force is called longitudinal strain and in a direction perpendicular to force is called lateral strain.
 Volumetric strain : If the deforming force produces a change in volume alone the strain produced in the body is called volumetric
Volumetric strain = Change in volume(DV)
Original volume(V )
 Shearing strain : If the deforming force produces a change in the shape of the body without changing its volume, strain produced is called shearing
It is defined as angle in radians through which a plane perpendicular to the fixed surface of the cubical body gets turned under the effect of tangential force.
f = x
L
Note : @ When a beam is bent both compression strain as well as an extension strain is produced.
Problem 8. A cube of aluminium of sides 0.1 m is subjected to a shearing force of 100 N. The top face of the cube is displaced through 0.02 cm with respect to the bottom face. The shearing strain would be [MP PAT 1990]
(a) 0.02 (b) 0.1 (c) 0.005 (d) 0.002
Solution : (d) Shearing strain f = x = 0.02cm = 0.002
L 0.1m
Problem 9. A wire is stretched to double its length. The strain is
(a) 2 (b) 1 (c) Zero (d) 0.5
Solution : (b) Strain = Change in length
Original length
= 2L – L = 1
L
Problem 10. The length of a wire increases by 1% by a load of 2 kgwt. The linear strain produced in the wire will be (a) 0.02 (b) 0.001 (c) 0.01 (d) 0.002
Solution : (c) Strain = Change in length = 1% of L = L / 100 = 0.01
Original length L L
Stressstrain Curve.
If by gradually increasing the load on a vertically suspended metal wire, a graph is plotted between stress (or load) and longitudinal strain (or elongation) we get the curve as shown in figure. From this curve it is clear that :
 When the strain is small (< 2%) (e., in region OP) stress is proportional to strain. This is the region where the so called Hooke’s law is obeyed. The point P is called limit of proportionality and slope of line OP gives the Young’s modulus Y of the material of the wire. If q is the angle of OP from strain
axis then Y = tanq .
 If the strain is increased a little bit, e., in the region PE, the stress is not proportional to strain. However, the wire still regains its original length after the removal of stretching force. This behaviour is shown up to point E known as elastic limit or yieldpoint. The region OPE represents the elastic behaviour of the material of wire.
 If the wire is stretched beyond the elastic limit E, e., between EA, the strain increases much more rapidly and if the stretching force is removed the wire does not come back to its natural length. Some permanent increase in length takes place.
 If the stress is increased further, by a very small increase in it a very large increase in strain is produced (region AB) and after reaching point B, the strain increases even if the wire is unloaded and ruptures at C. In the region BC the wire literally The maximum stress corresponding to B after which the wire begins to flow and breaks is called breaking or tensile strength. The region EABC represents the plastic behaviour of the material of wire.
 Stressstrain curve for different
C
E C E C
P P
O Strain
O Strain
O Strain
The plastic region between E and C is small for brittle material and it will break soon after the elastic limit is crossed.
The material of the wire have a good plastic range and such materials can be easily changed into different shapes and can be drawn into thin wires
Stress strain curve is not a straight line within the elastic limit for elastomers and strain produced is much larger than the stress applied. Such materials have no plastic range and the breaking point lies very close to elastic limit. Example rubber
Problem 11. The stressstrain curves for brass, steel and rubber are shown in the figure. The lines A, B and C are for
 Rubber, brass and steel respectively
 Brass, steel and rubber
 Steel, brass and rubber respectively
 Steel, rubber and brass
Solution : (c) From the graph tanq _{C} < tanq _{B} < tanq _{A}
Þ YC < YB < YA
\YRubber
< YBrass < YSteel
Problem 12. The strain stress curves of three wires of different materials are shown in the figure. P, Q and R are the elastic limits of the wires. The figure shows that
 Elasticity of wire P is maximum
 Elasticity of wire Q is maximum
 Tensile strength of R is maximum
 None of the above is true
Solution : (d) On the graph stress is represented on X– axis and strain Yaxis
So from the graph Y = cot q =
1 µ 1
tanq q
[where q is the angle from stress axis]
\YP
< YQ < YR
[As
q _{P} > q _{Q} > q _{R} ]
We can say that elasticity of wire P is minimum and R is maximum.
Hooke’s law and Modulus of Elasticity.
According to this law, within the elastic limit, stress is proportional to the strain.
i.e. stress µ strain or
stress = constant = E
strain
The constant E is called modulus of elasticity.
 It’s value depends upon the nature of material of the body and the manner in which the body is
 It’s value depends upon the temperature of the
 It’s value is independent of the dimensions (length, volume ) of the body.
There are three modulii of elasticity namely Young’s modulus (Y), Bulk modulus (K) and modulus of rigidity (h) corresponding to three types of the strain.
Young’s Modulus (Y).
It is defined as the ratio of normal stress to longitudinal strain within limit of proportionality.
Y = Normal stress = F / A = FL
longitudinal strain l / L Al
If force is applied on a wire of radius r by hanging a weight of mass M, then
Y = MgL
pr ^{2}l
 If the length of a wire is doubled,
Then longitudinal strain =
change in length(l) = final length – initial length = 2L – L = 1
initial length(L) Initial length L
\ Young’s modulus =
stress Þ Y = stress [As strain = 1]
strain
So young’s modulus is numerically equal to the stress which will double the length of a wire.
 Increment in the length of wire
l = FL
éAs Y = FL ù
pr ^{2}Y
ëê Al úû
So if same stretching force is applied to different wires of same material, l µ L
r 2
[As F and Y are constant]
i.e., greater the ratio
L , greater will be the elongation in the wire.
r 2
 Elongation in a wire by its own weight : The weight of the wire Mg act at the centre of gravity of the wire so that length of wire which is stretched will be L/2.
\ Elongation l = FL
= Mg(L / 2) =
MgL = L^{2}dg
[As mass (M) = volume (AL) × density (d)]
AY AY 2AY 2Y
 Thermal stress : If a rod is fixed between two rigid supports, due to change in temperature its length will change and so it will exert a normal stress (compressive if temperature increases and tensile if temperature decreases) on the This stress is called thermal stress.
As by definition, coefficient of linear expansion a =
l LDq
Þ thermal strain
l = aDq
L
So thermal stress = Ya Dq [As Y = stress/strain] And tensile or compressive force produced in the body = YAa Dq
Note : @ In case of volume expansion Thermal stress = KgDq
Where K = Bulk modulus, g = coefficient of cubical expansion
 Force between the two rods : Two rods of different metals, having the same area of cross section A, are placed end to end between two massive walls as shown in The first rod has a length L_{1}, coefficient of linear expansion a_{1} and young’s modulus Y_{1}. The corresponding quantities for second rod are L_{2}, a_{2} and Y_{2}. If the temperature of both the rods is now raised by T degrees.
Increase in length of the composite rod (due to heating) will be equal to
l1 + l2 = [L1a_{1} + L2a _{2} ]T
[As l = L a Dq]
and due to compressive force F from the walls due to elasticity,
decrease in length of the composite rod will be equal to é L1 + L2 ù F
éAs l = FL ù


ê Y1
Y2 ú A êë
AY úû
as the length of the composite rod remains unchanged the increase in length due to heating must be equal to
decrease in length due to compression i.e. F é L1 + L2 ù = [L a + L a ]T
A ê Y Y ú
1 1 2 2
ë 1 2 û
or F =
A[L1a_{1} + L2a _{2} ]T


é L1 + L2 ù
ê Y1 Y2 ú
 Force constant of wire : Force required to produce unit elongation in a wire is called force constant of material of It is denoted by k.
\ k = F
l
…..(i)
but from the definition of young’s modulus
Y = F / A Þ F = YA
…..(ii)
l / L l L
from (i) and (ii) k = YA
L
It is clear that the value of force constant depends upon the dimension (length and area of cross section) and material of a substance.
 Actual length of the wire : If the actual length of the wire is L, then under the tension T_{1}, its length becomes L_{1} and under the tension T_{2}, its length becomes L_{2}.
L1 = L + l1 Þ
L = L + T1
1 k
……(i) and
L2 = L + l2
Þ L2
= L + T2
k
..…(ii)
From (i) and (ii) we get L = L1T2 – L2T1
T2 – T1
Problem 13. The diameter of a brass rod is 4 mm and Young’s modulus of brass is 9 ´ 10^{10} N / m^{2} . The force required to stretch by 0.1% of its length is [MP PET 1991; BVP 2003]
(a) 360 pN (b) 36 N (c)
144p ´ 10^{3} N
(d)
36p ´ 10^{5} N
Solution : (a)
r = 2 ´ 10 ^{–}^{3} m,
Y = 9 ´ 10^{10} N / m^{2} ,
l = 0.1% L
Þ l = 0.001
L
As Y = F L \ F = YA l = 9 ´ 10^{10} ´ p (2 ´ 10 ^{–}^{3} )^{2} ´ 0.001 = 360p N
A l L
Problem 14. A wire of length 2m is made from 10 cm^{3} of copper. A force F is applied so that its length increases by 2 mm. Another wire of length 8 m is made from the same volume of copper. If the force F is applied to it, its length will increase by [MP PET 2003]
(a) 0.8 cm (b) 1.6 cm (c) 2.4 cm (d) 3.2 cm
Solution : (d)
l = FL
AY
= FL^{2}
VY
\l µ L^{2}
[As V, Y and F are constant]
l é L ù 2 æ 8 ö ^{2}
2 = ê 2 ú = ç
÷ = 16
Þ l_{2} = 16 l_{1} = 16 ´ 2 mm = 32 mm = 3.2 cm
l1 ë L1 û
è 2 ø
Problem 15. A wire of length L and radius r is rigidly fixed at one end. On stretching the other end of the wire with a force F, the increase in its length is l. If another wire of same material but of length 2L and radius 2r is stretched with a force of 2F, the increase in its length will be
[AIIMS 1980; MP PAT 1990; MP PET 1989, 92; MP PET/PMT 1988; MP PMT 1996, 2002; UPSEAT 2002]
 l (b) 2l (c) l
2
(d) l
4
FL l F L æ r ö 2
æ 1 ö ^{2}
Solution : (a)
l = Þ 2 = 2 2 ç 1 ÷
= 2 ´ 2 ´ ç ÷ = 1
\l 2 = l1
i.e. the increment in length will be same.
p r ^{2}Y
l1 F1
L1 è r2 ø
è 2 ø
Problem 16. Two wires A and B are of same materials. Their lengths are in the ratio 1 : 2 and diameters are in the ratio 2 : 1 when stretched by force F_{A} and F_{B} respectively they get equal increase in their lengths. Then the ratio F_{A}/F_{B} should be [Orissa JEE 2002]
(a) 1 : 2 (b) 1 : 1 (c) 2 : 1 (d) 8 : 1
Solution : (d)
Y = FL
pr ^{2}l
\ F = Ypr ^{2} l
L
F Y æ r
ö ^{2} æ l
ö æ L ö
æ 2 ö ^{2}
æ 2 ö
A = A ç A ÷
ç A ÷ ç B ÷
= 1 ´ ç ÷ ´ (1)´ ç
÷ = 8
FB YB è rB ø
è l B ø è LA ø
è 1 ø
è 1 ø
Problem 17. A uniform plank of Young’s modulus Y is moved over a smooth horizontal surface by a constant horizontal force F. The area of crosssection of the plank is A. the compressive strain on the plank in the direction of the force is [Kerala (Engg.) 2002]
 F
 2F
1 æ F ö
 3F
AY AY
2 ç AY ÷ AY
Solution : (a) Compressive strain =
Stress Young’ s modulus
è ø
= F / A = F
Y AY
Problem 18. A wire is stretched by 0.01 m by a certain force F. Another wire of same material whose diameter and length are double to the original wire is stretched by the same force. Then its elongation will be
[EAMCET (Engg.) 1995; CPMT 2001]
(a) 0.005 m (b) 0.01 m (c) 0.02 m (d) 0.002 m
Solution : (a)
l = FL
pr ^{2}Y
\ l µ L
r 2
[As F and Y are constants]
l æ L ö æ r ö ^{2}
æ 1 ö ^{2} 1
l 0.01
2 = ç 2 ÷ ç 1 ÷
= (2) ´ ç ÷ = Þ l _{2} = ^{1} =
= 0.005m .
l1 è L1 ø è r2 ø
è 2 ø 2 2 2
Problem 19. The length of an elastic string is a metres when the longitudinal tension is 4 N and b metres when the longitudinal tension is 5 N. The length of the string in metres when the longitudinal tension is 9 N is
[EAMCET 2001]
 a – b (b) 5b – 4a (c)
2b – 1 a
4
 4a – 3b
Solution : (b) Let the original length of elastic string is L and its force constant is k.
When longitudinal tension 4N is applied on it
and when longitudinal tension 5N is applied on it
L + 4 = a k
L + 5 = b k
…..(i)
…..(ii)
By solving (i) and (ii) we get k = 1 and L = 5a – 4b
b – a
Now when longitudinal tension 9N is applied on elastic string then its length =
L + 9 = 5a – 4b + 9(b – a) = 5b – 4a
k
Problem 20. The load versus elongation graph for four wires of the same material is shown in the figure. The thickest wire is represented by the line [KCET (Engg./Med.) 2001]
 OD
 OC
 OB
 OA
Solution : (a) Young’s modulus Y = FL
Al
\ l µ 1
A
(As Y, L and F are constant)
From the graph it is clear that for same load elongation is minimum for graph OD. As elongation (l) is minimum therefore area of crosssection (A) is maximum.
So thickest wire is represented by OD.
Problem 21. A 5 m long aluminum wire (Y = 7 ´ 10^{10} N / m^{2} )
of diameter 3 mm supports a 40 kg mass. In order to have
the same elongation in a copper wire (Y = 12 ´ 10^{10} N / m^{2})
of the same length under the same weight, the
diameter should now be, in mm [AMU 2000]
(a) 1.75 (b) 2.0 (c) 2.3 (d) 5.0
Solution : (c)
l = FL
pr ^{2}Y
= 4 FL
pd ^{2}Y
[As r = d / 2 ]
If the elongation in both wires (of same length) are same under the same weight then d ^{2} Y = constant
æ d ö ^{2} Y
ç Cu ÷
è d Al ø
= Al Þ dCu = d Al ´
YCu
= 3 ´
= 2.29 mm
Problem 22. On applying a stress of
20 ´ 10^{8} N / m^{2}
the length of a perfectly elastic wire is doubled. Its Young’s modulus
will be [MP PET 2000]
(a)
40 ´ 10^{8} N / m^{2}
(b)
20 ´ 10^{8} N / m^{2}
(c)
10 ´ 10^{8} N / m^{2}
(d)
5 ´ 10^{8} N / m^{2}
Solution : (b) When strain is unity then Young’s modulus is equal to stress.
Problem 23. The dimensions of four wires of the same material are given below. In which wire the increase in length will be maximum when the same tension is applied
[IIT–JEE 1981; NCERT 1976; CPMT 1983, 90; MP PMT 1992, 94, 97; MP PET/PMT 1998; MP PET 1989, 90, 99]
(a) Length 100 cm, diameter 1 mm (b) Length 200 cm, diameter 2 mm
(c) Length 300 cm, diameter 3 mm (d) Length 50 cm, diameter 0.5 mm
Solution : (d) If same force is applied on four wires of same material then elongation in each wire depends on the length
and diameter of the wire and given by l µ L
d 2
and the ratio of L is maximum for (d) option.
d 2
Problem 24. The Young’s modulus of a wire of length L and radius r is Y N/m^{2}. If the length and radius are reduced to L/2 and r/2, then its Young’s modulus will be [MP PMT 1985; MP PET 1997; KCET (Engg./Med.) 1999]
(a) Y/2 (b) Y (c) 2Y (d) 4Y
Solution : (b) Young’s modulus do not depend upon the dimensions of wire. It is constant for a given material of wire.
Problem 25. A fixed volume of iron is drawn into a wire of length L. The extension x produced in this wire by a constant force F is proportional to [MP PMT 1999]
(a) 1
L2
(b)
1 (c) L^{2}
L
(d) L
Solution : (c)
l = FL = FL^{2} = FL^{2}
for a fixed volume l µ L^{2}
AY ALY VY
Problem 26. A rod is fixed between two points at 20^{o} C . The coefficient of linear expansion of material of rod is
1.1´ 10^{–}^{5} / ^{o}C and Young’s modulus is 1.2 ´ 10^{11} N / m^{2} . Find the stress developed in the rod if temperature
of rod becomes 10^{o} C
[RPET 1997]
(a)
1.32 ´ 10^{7} N / m^{2}
(b)
1.10 ´ 10^{15} N / m^{2} (c)
1.32 ´ 10^{8} N / m^{2}
(d)
1.10 ´ 10^{6} N / m^{2}
Solution : (a) Thermal stress
F = Ya Dq
A
= 1.2 ´ 10^{11} ´ 1.1 ´ 10 ^{–}^{5} ´ (20 – 10) = 1.32 ´ 10^{7} N / m^{2}
Problem 27. The coefficient of linear expansion of brass and steel are a_{1} and a_{2}. If we take a brass rod of length L_{1} and steel rod of length L_{2} at 0^{o} C , their difference in length (L_{2} – L_{1} ) will remain the same at any temperature if
[EAMCET (Med.) 1995]
(a)
a_{1} L2 = a _{2} L1
(b)
a L^{2} = a L^{2} (c) a ^{2} L = a ^{2} L
(d)
a_{1} L1 = a _{2} L2
1 2 2 1 1 1 2 2
Solution : (d) Difference in lengths of rods will remain same if expansion is same in both the rods.
If expansion in first rod is l1 = L1a_{1}Dq and expansion in second rod is l 2 = L2a _{2}Dq
then L1a_{1}Dq = L2a _{2}Dq \L1a_{1} = L2a _{2}
Problem 28. The force required to stretch a steel wire of 1 cm^{2} crosssection to 1.1 times its length would be
(Y = 2 ´ 10^{11} Nm^{–}^{2})
[MP PET 1992]
(a)
2 ´ 10^{6} N
(b)
2 ´ 10^{3} N
(c)
2 ´ 10^{–}^{6} N
(d)
2 ´ 10^{–}^{7} N
Solution : (a)
L = 1.1 L
\Strain = l = L2 – L1
= 1.1L1 – L1 = 0.1.
^{2} ^{1} L1 L1 L1
F = YA l = 2 ´ 10^{11} ´ 1 ´ 10 ^{–}^{4} ´ 0.1
L
= 2 ´ 10^{6} N .
Problem 29. A two metre long rod is suspended with the help of two wires of equal length. One wire is of steel and its cross sectional area is 0.1 cm^{2} and another wire is of brass and its crosssectional area is 0.2 cm^{2}. If a load W is suspended from the rod and stress produced in both the wires is same then the ratio of tensions in them will be
(a) Will depend on the position of W
(b) T_{1} / T_{2} = 2
(c) T_{1} / T_{2} = 1
(d) T_{1} / T_{2} = 0.5
Solution : (d) Stress =
Tension
Area of crosssection
= constant
\ T1 = T2 Þ T1
= A1
= 0.1 = 1 = 0.5 .
A1 A2
T2 A2
0.2 2
Problem 30. Three blocks, each of same mass m, are connected with wires W_{1} and W_{2} of same crosssectional area a and Young’s modulus Y. Neglecting friction the strain developed in wire W_{2} is
(a)
(b)
(c)
(d)
 mg
 aY
3 mg
2 aY
 mg
3 aY
3mg aY
Solution : (a) If the system moves with acceleration a and T is the tension in the string W_{2}
then by comparing this condition
from standard case T =
m1m2 g m1 + m2
In the given problem m1 = (m + m) = 2m and m2 = m
\ Tension = m. 2m.g
m + 2m
= 2 mg
3
\ Stress = T =
a
 mg
3a
and Strain =
Stress Young’ s modulus
= 2 mg
 aY
Problem 31. A wire elongates by 1.0 mm when a load W is hanged from it. If this wire goes over a pulley and two weights
W each are hung at the two ends, the elongation of the wire will be
(a) 0.5 m (b) 1.0 mm (c) 2.0 mm (d) 4.0 mm Solution : (b) Elongation in the wire µ Tension in the wire

In first case T1 = W and in second case T2 = 2W ´ W = W
W W
As T1 = 1
T2
\ l1 = 1 Þ l2
l2
= l1
= 1.0mm
Problem 32. The Young’s modulus of three materials are in the ratio 2 : 2 : 1. Three wires made of these materials have their crosssectional areas in the ratio 1 : 2 : 3. For a given stretching force the elongation’s in the three wires are in the ratio
(a) 1 : 2 : 3 (b) 3 : 2 : 1 (c) 5 : 4 : 3 (d) 6 : 3 : 4
Solution : (d)
l = FL
AY
and for a given stretching force l µ 1
AY
Let three wires have young’s modulus 2Y, 2Y and Y and their cross sectional areas are A, 2A and 3 A respectively.
l : l : l =
1 : 1 : 1 = 1 : 1
: 1 = 1 : 1 : 1
1 2 3
A1Y1 A2 Y2 A3 Y3 A ´ 2Y 2A ´ 2Y
3 A ´ Y 2 4 3
= 6 : 3 : 4 .
Problem 33. A light rod with uniform crosssection of
104 m2
is shown in the adjoining figure. The rod consists of three
different materials whose lengths are 0.1 m, 0.2 m and 0.15 m respectively and whose Young’s modulii are
2.5 ´ 10^{10} N / m^{2} , 4 ´ 10^{10} N / m^{2} and 1´ 10^{10} N / m^{2} respectively. The displacement of point B will be
(a) 24 ´ 10^{–}^{6} m (b) 9 ´ 10^{–}^{6} m (c) 4 ´ 10^{–}^{6} m (d) 1´ 10^{–}^{6} m
Solution : (c) Increment in the length AB =
MgL =
AY
10 ´ 10 ´ 0.1
10 ^{–}^{4} ´ 2.5 ´ 10^{10}
= 4 ´ 10 ^{–}^{6} m
\ Displacement of point B = 4 ´ 10^{–}^{6} m
Problem 34. In the above problem, displacement of point C will be
(a)
24 ´ 10^{–}^{6} m
(b)
9 ´ 10^{–}^{6} m
(c)
4 ´ 10^{–}^{6} m
(d)
1´ 10^{–}^{6} m
Solution : (b) Increment in the length BC =
MgL = 10 ´ 10 ´ 0.2 = 5 ´ 106 m
AY 10^{–}^{4} ´ 4 ´ 10^{10}
\ Displacement of point C = 4 ´ 10^{–}^{6} + 5 ´ 10^{–}^{6} = 9 ´ 10^{–}^{6}m
Problem 35. In the above problem, the displacement of point D will be
(a)
24 ´ 10^{–}^{6} m
(b)
9 ´ 10^{–}^{6} m
(c)
4 ´ 10^{–}^{6} m
(d)
1´ 10^{–}^{6} m
Solution : (a) Increment in the length CD = MgL = 10 ´ 10 ´ 0.15
= 15 ´ 10^{–}^{6} m
AY 10^{–}^{4} ´ 1 ´ 10^{10}
\ Displacement of point D = 4 ´ 10 ^{–}^{6} + 5 ´ 10 ^{–}^{6} m + 15 ´ 10 ^{–}^{6} = 24 ´ 10 ^{–}^{6} m .
Problem 36. Two blocks of masses m_{1} and m_{2} are joined by a wire of Young’s modulus Y via a massless pulley. The area of cross section of the wire is S and its length is L. When the system is released, increase in length of the wire is
(a)
(b)
(c)
(d)
m1m2 gL YS (m1 + m2 )
2m1m2 gL YS (m1 + m2 )
(m1 – m2 )gL YS (m1 + m2 )
4m1m2 gL YS (m1 + m2 )
Solution : (b) Tension in the wire T =
2m1m2
m1 + m2
g \ stress in the wire = T =
S
2m1m2g S(m1 + m2)
\ Strain
l = Stress =
L Y
2m1m2 g YS(m1 + m2 )
Þ l =
2m1m2gL YS(m1 + m2)
Problem 37. A steel wire of diameter d, area of crosssection A and length 2L is clamped firmly at two points A and B which are 2L metre apart and in the same plane. A body of mass m is hung from the middle point of wire such that the middle point sags by x lower from original position. If Young’s modulus is Y then m is given by
(a)
1 YAx ^{2}
2 gL^{2}
(b)
1 YAL^{2}
2 gx ^{2}
(c)
(d)
YAx ^{3}
gL^{3}
YAL^{3}
gx ^{2}
Solution : (c) Let the tension in the string is T and for the equilibrium of mass m
2T sin q = mg
Þ T =
mg
2 sin q
= mgL
2x
[As q is small then sinq = x ]
L
Increment in the length l = AC – AB =
– L = (L^{2} + x^{2})^{1} ^{/} ^{2} – L
éæ 2 ö1 / 2 ù é
2 ù 2
= Lêç1 + x ÷ – 1ú = Lê1 + 1 x
– 1ú = x




ç ÷
ëè û
ëê 2 L^{2}
ûú 2L
As Young’s modulus Y = T L
A l
\ T = YAl
L
Substituting the value of T and l in the above equation we get
mgL = YA . x^{2} \ m = YAx^{3}
2x L 2L
gL^{3}
Problem 38. Two wires of equal length and crosssection are suspended as shown. Their Young’s modulii are Y_{1}
respectively. The equivalent Young’s modulus will be
and Y2
(a)
(b)
(c)
(d)
Y1 + Y2
Y1 + Y2
2
Y1Y2 Y1 + Y2
Solution : (b) Let the equivalent young’s modulus of given combination is Y and the area of cross section is 2A. For parallel combination k1 + k2 = keq.
Y A Y A Y 2A
1 + 2 =
L L L
Y1 + Y2 = 2Y , \Y = Y1 + Y2
2
Problem 39. If a load of 9kg is suspended on a wire, the increase in length is 4.5 mm. The force constant of the wire is
(a)
0.49 ´ 10^{4} N / m
(b)
1.96 ´ 10^{4} N / m
(c)
4.9 ´ 10^{4} N / m
(d)
0.196 ´ 10^{4} N / m
Solution : (b) Force constant k = F = mg =
9 ´ 9.8
Þ k = 1.96 ´ 10^{4} N / m
l l 4.5 ´ 10^{–}^{3}
Problem 40. One end of a long metallic wire of length L, area of crosssection A and Young’s modulus Y is tied to the ceiling. The other end is tied to a massless spring of force constant k. A mass m hangs freely from the free end of the spring. It is slightly pulled down and released. Its time period is given by
 2p
 2p
 2p
 2p
Solution : (d) Force constant of wire k1 = F = YA
and force constant of spring k_{2} = k
(given)
l L
Equivalent force constant for given combination 1 = 1 + 1 = L + 1
Þ keq = kYA
\ Time period of combination T = 2p
keq
= 2p
k1 k2
YA k
kL + YA
Problem 41. Two wires A and B have the same length and area of cross section. But Young’s modulus of A is two times the Young’s modulus of B. Then the ratio of force constant of A to that of B is
(a) 1 (b) 2 (c) 1
2
(d)
Solution : (b) Force constant of wire k = YA
Þ kA = YA = 2
[As L and A are same]
L
Work Done in Stretching a Wire.
kB YB
In stretching a wire work is done against internal restoring forces. This work is stored in the wire as elastic potential energy or strain energy.
If a force F acts along the length L of the wire of crosssection A and stretches it by x then
Y = stress = F / A = FL
Þ F = YA .x
strain x / L Ax L
So the work done for an additional small increase dx in length, dw = Fdx = YA x . dx
L
Hence the total work done in increasing the length by l,
W = òdW = ò Fdx = ò YA .xdx = 1 YA l ^{2}

l l
This work done is stored in the wire.
\ Energy stored in wire U = 1 YAl 2 = 1 Fl
0 0 0 L 2 L
éAs F = YAl ù
2 L 2
ëê L úû
Dividing both sides by volume of the wire we get energy stored in per unit volume of wire.
1 F l 1 1
U = ´ ´ = ´ stress ´ strain = ´ Y ´ (strain)^{2} =
1 (stress)^{2}
[As AL = volume of wire]
^{V} 2 A L 2
2 2Y
Note : @ If the force on the wire is increased from F_{1} to F_{2} and the elongation in wire is l then energy stored in the wire U = 1 (F1 + F2 ) l
2 2
@ Thermal energy density = Thermal energy per unit volume = 1 ´ Thermal stress ´ strain
2
= 1 F l =
2 A L
1 (YaDq )(aDq ) =
2
1 Ya ^{2}(Dq )^{2}
2
Problem 42. A wire suspended vertically from one of its ends is stretched by attaching a weight of 200 N to the lower end. The weight stretches the wire by 1 mm, then the elastic energy stored in the wire is [AIEEE 2003]
(a) 0.1 J (b) 0.2 J (c) 10 J (d) 20 J
Solution : (a) Elastic energy stored in wire = U = 1 Fl = 1 ´ 200 ´ 1 ´ 10 ^{–}^{3} = 0.1J
2 2
Problem 43. The graph shows the behaviour of a length of wire in the region for which the substance obeys Hooke’s law. P
and Q represent [AMU 2001]
 P = applied force, Q = extension
 P = extension, Q = applied force
 P = extension, Q = stored elastic energy
 P = stored elastic energy, Q = extension
Solution : (c) The graph between applied force and extension will be straight line because in elastic range applied force µ
extension, but the graph between extension and stored elastic energy will be parabolic in nature.
As U = 1 kx ^{2}
2
or U µ x ^{2}
Problem 44. When a 4 kg mass is hung vertically on a light spring that obeys Hooke’s law, the spring stretches by 2 cms. The work required to be done by an external agent in the stretching this spring by 5 cms will be (g = 9.8 m/s^{2})
[MP PMT 1995]
(a) 4.900 J (b) 2.450 J (c) 0.495 J (d) 0.245 J
Solution : (b) When a 4 kg mass is hung vertically on a spring, it stretches by 2 cm \k = F =
x
4 ´ 9.8
2 ´ 10 ^{–}^{2}
= 1960N / m
Now work done in stretching this spring by 5 cms
U = 1 kx ^{2} = 1 ´ 1960(5 ´ 10 ^{–}^{2} )^{2} = 2.45 J.
2 2
Problem 45. A rod of iron of Young’s modulus Y = 2.0 ´ 10^{11} N / m^{2} just fits the gap between two rigid supports 1m apart.
If the rod is heated through
100^{o} C
the strain energy of the rod is ( a = 18 ´ 10^{–}^{6} ^{o}C^{–}^{1}
and area of cross
section
A = 1 cm^{2} )
(a) 32.4 J (b) 32.4 mJ (c) 26.4 J (d) 26.4 mJ
Solution : (a)
U = 1 ´ Y ´ (strain)^{2} ´ volume = 1 ´ Y(a Dq )^{2} ´ A ´ L
æ Thermal strain l
= a Dq ö
ç ÷
2 2 è L ø
= 1 ´ (2 ´ 10^{11}) ´ (18 ´ 10 ^{–}^{6} ´ 100)^{2} ´ 1 ´ 10 ^{–}^{4} ´ 1 = 324 ´ 10 ^{–}^{1} = 32.4 J.
2
Problem 46. Which of the following cases will have the greatest strain energy (F is the stretching force, A is the area of cross section and s is the strain)
(a) F = 10 N, A = 1 cm^{2}, s = 10^{–3} (b) F = 15 N, A = 2 cm^{2}, s = 10^{–3}
(c) F = 10 N, A =
1 cm^{2}, s = 10^{–4} (d) F = 5 N, A = 3 cm^{2}, s = 10^{–3}
2
Solution: (b) Strain energy = 1 ´ stress × strain × volume = 1 ´ F × strain × AL = 1 ´ F ´ strain × L
2 2 A 2
For wire (a)
For wire (c)
U = 1 ´ 10 ´ 10 ^{–}^{3} ´ L = 5 ´ 10 ^{–}^{3} L ; For wire (b)
2
U = 1 ´ 10 ´ 10 ^{–}^{4} ´ L = 0.5 ´ 10 ^{–}^{3} L ; For wire (d)
2
U = 1 ´ 15 ´ 10 ^{–}^{3} ´ L = 7.5 ´ 10 ^{–}^{3} L
2
U = 1 ´ 5 ´ 10 ^{–}^{3} = 2.5 ´ 10 ^{–}^{3} L
2
For a given length wire (b) will have greatest strain energy.
Breaking of Wire.
When the wire is loaded beyond the elastic limit, then strain increases much more rapidly. The maximum stress corresponding to B (see stressstrain curve) after which the wire begin to flow and breaks, is called breaking stress or tensile strength and the force by application of which the wire breaks is called the breaking force.
 Breaking force depends upon the area of crosssection of the wire i.e., Breaking force µ A
\ Breaking force = P × A
Here P is a constant of proportionality and known as breaking stress.
 Breaking stress is a constant for a given material and it does not depends upon the dimension (length or thickness) of
 If a wire of length L is cut into two or more parts, then again it’s each
part can hold the same weight. Since breaking force is independent of the length of wire.
 If a wire can bear maximum force F, then wire of same material but double thickness can bear maximum force 4F because Breaking force µ pr^{2}.
 The working stress is always kept lower than that of a breaking
So that safety factor =
breaking stress working stress
may have large value.
 Breaking of wire under its own
Breaking force = Breaking stress ´ Area of cross section
Weight of wire = Mg = ALdg = PA [As mass = volume ´ density = ALd]
Þ Ldg = P \
L = P dg
This is the length of wire if it breaks by its own weight.
Problem 47. A wire of diameter 1 mm breaks under a tension of 1000 N. Another wire of same material as that of the first one, but of diameter 2 mm breaks under a tension of [Orissa JEE 2003]
(a) 500 N (b) 1000 N (c) 10000 N (d) 4000 N
Solution : (d) Breaking force µ area of crosssection (pr ^{2} ) µ d ^{2}
F æ d ö ^{2}
F æ 2mm ö ^{2}
2 = ç 2 ÷
Þ ^{ } ^{2} = ç ÷ Þ F_{2} = 1000 ´ 4 = 4000N.
F1 è d1 ø
1000 è 1mm ø
Problem 48. In steel, the Young’s modulus and the strain at the breaking point are
2 ´ 10^{11} Nm^{–}^{2}
and 0.15 respectively.
The stress at the breaking point for steel is therefore [MP PET 1990; MP PMT 1992; DPMT 2001]
(a)
1.33 ´ 10^{11} Nm^{–}^{2}
(b)
1.33 ´ 10^{12} Nm^{–}^{2}
(c)
7.5 ´ 10^{–}^{13} Nm^{–}^{2}
(d)
3 ´ 10^{10} Nm^{–}^{2}
Solution : (d)
Y = Stress
Strain
\ Stress = Y × Strain = 2 ´ 10^{11} ´ 0.15 = 0.3 ´ 10^{11} = 3 ´ 10^{10} N / m^{2}
Problem 49. To break a wire, a force of 10^{6} N / m^{2}
is required. If the density of the material is
 ´ 10^{3} kg / m^{3} , then the
length of the wire which will break by its own weight will be [Roorkee 1979]
(a) 34 m (b) 30 m (c) 300 m (d) 3 m
Solution : (a) Length of the wire which will break by its own weight
L = P =
10^{6} = 100
= 33.3 m ≃ 34 m.
dg 3 ´ 10^{3} ´ 10 3
Problem 50. The wires A and B shown in the figure are made of the same material and have radii r_{A} and r_{B} respectively.
The block between them has a mass m. When the force F is mg/3, one of the wires break
 A will break before B if r_{A} = r_{B}
 A will break before B if r_{A} < 2r_{B}
 Either A or B may break if r_{A} = 2r_{B}
 The lengths of A and B must be known to predict which wire will break
Solution :(a,b,c) When force F = mg
3
is applied at the lower end then

mg + mg
Stress in wire B = F
= mg
and stress in wire A = F
mg = 3 = 4 mg
pr ^{2} 3pr ^{2}
pr ^{2}
p r ^{2} 3 p r ^{2}
B B A A A
 if rA= rB = r (Let) then stress in wire B = mg and stress in wire A =
 .mg
3pr ^{2}
i.e. stress in wire A > stress in wire B so the A will break before B
 if rB = r , (let) then rA = 2r
3 pr ^{2}
Stress in wire B = mg
and Stress in wire A =
4mg
= mg
3pr ^{2} 3p (2r)^{2} 3pr ^{2}
i.e. stress in wire A = stress in wire B. It means either A or B may break.
 If rA< 2rB then stress in A will be more than B. e. A will break before B.
Problem 51. A body of mass 10 kg is attached to a wire 0.3 m long. Its breaking stress is
 ´ 10^{7} N / m^{2} . The area of
crosssection of the wire is 10^{–}^{6} m^{2} . What is the maximum angular velocity with which it can be rotated in the horizontal circle
 1 rad/sec (b) 2 rad/sec (c) 4 rad/sec (d) 8 rad/sec Solution : (c) Breaking force = centrifugal force
Breaking stress × area of crosssection = mw ^{2}l
4.8 ´ 10^{7} ´ 10 ^{–}^{6} = 10 ´ w ^{2} ´ 0.3
Þ w ^{2} = 16
Þ w = 4rad / sec
Problem 52. Two block of masses 1 kg and 4 kg are connected by a metal wire going over a smooth pulley as shown in the
figure. The breaking stress of the metal is break is
3.18 ´ 10^{10} N / m^{2} . The minimum radius of the wire so it will not
(a) 1´ 10^{–}^{5} m (b) 2 ´ 10^{–}^{5} m (c) 3 ´ 10^{–}^{5} m (d) 4 ´ 10^{–}^{5} m
Solution : (d) Tension in the wire T =
2m1m2 g m1 + m2
Þ T = 2 ´ 1 ´ 4 ´ 10 Þ T = 16N
1 + 4
Breaking force = Breaking stress × Area of crosssection Tension in the wire = 3.18 ´ 10^{10} ´ pr ^{2}
16 = 3.18 ´ 10^{10} ´ pr ^{2}
Þ r =
= 4 ´ 10 ^{–}^{5} m.
Bulk Modulus.
When a solid or fluid (liquid or gas) is subjected to a uniform pressure all over the surface, such that the shape remains the same, then there is a change in volume.
Then the ratio of normal stress to the volumetric strain within the elastic limits is called as Bulk modulus. This is denoted by K.
K = Normal stress volumetric strain
K = F / A = – pV
 DV / V DV
where p = increase in pressure; V = original volume; DV = change in volume
The negative sign shows that with increase in pressure p, the volume decreases by DV i.e. if p is positive, DV is negative. The reciprocal of bulk modulus is called compressibility.
C = compressibility = 1
K
= DV
pV
S.I. unit of compressibility is N^{–1}m^{2} and C.G.S. unit is dyne^{–1} cm^{2}.
Gases have two bulk moduli, namely isothermal elasticity E_{q} and adiabatic elasticity E_{f} .
 Isothermal elasticity (E_{q}) : Elasticity possess by a gas in isothermal condition is defined as isothermal For isothermal process, PV = constant (Boyle’s law)
Differentiating both sides PdV + VdP = 0 Þ PdV = – VdP
P = dP = stress
= Eq
(dV / V) strain
\ E_{q} = P
i.e., Isothermal elasticity is equal to pressure.
 Adiabatic elasticity (E_{f}) : Elasticity possess by a gas in adiabatic condition is defined as adiabatic
For adiabatic process,
PV ^{g} = constant (Poisson’s law)
Differentiating both sides, P g V ^{g} ^{1}dV + V ^{g} dP = 0 Þ g PdV + VdP = 0
g P = dP = stress
= Ef
æ – dV ö strain
ç ÷
è V ø
\ E_{f} = g P
i.e., adiabatic elasticity is equal to g times pressure. [where g
= Cp ]
Cv
Note : @ Ratio of adiabatic to isothermal elasticity Ef
Eq
= g P = g > 1
P
\ E_{f}
> E_{q}
i.e., adiabatic elasticity is always more than isothermal elasticity.
Density of Compressed Liquid.
If a liquid of density r , volume V and bulk modulus K is compressed, then its density increases.
As density
r = m so
V
Dr = – DV
r V
…..(i)
But by definition of bulk modulus
K = – VDP Þ
DV
– DV V
= DP
K
…..(ii)
From (i) and (ii)
Dr =
r
r ¢ – r
r
= DP
K
[As Dr = r ¢ – r ]
or r ¢ = r é1 + DP ù = r[1 + CDP]
éAs 1
= Cù
ëê K úû ëê K úû
Fractional Change in the Radius of Sphere.
A solid sphere of radius R made of a material of bulk modulus K is surrounded by a liquid in a cylindrical container.
A massless piston of area A floats on the surface of the liquid.
Volume of the spherical body V = 4 pR^{3}
3
DV = 3 DR V R
\ DR = 1 DV
…..(i)
R 3 V
Bulk modulus
K = – V DP
DV
\ DV = DP = mg
…..(ii)
éAs DP = mg ù
V K AK
ëê A úû
Substituting the value of DV
from equation (ii) in equation (i) we get
DR = 1 mg
V R 3 AK
Problem 53. When a pressure of 100 atmosphere is applied on a spherical ball of rubber, then its volume reduces to 0.01%. The bulk modulus of the material of the rubber in dyne/cm^{2} is
(a)
10 ´ 10^{12}
100 ´ 10^{12}
1 ´ 10^{12}
20 ´ 10^{12}
Solution : (c) 1 atm = 10^{5} N / m^{2}
\100 atm = 10^{7} N / m^{2} and DV = 0.01%V \
DV = 0.0001
V
K = P =
DV / V
10^{7}
0.0001
= 1 ´ 10^{11} N / m^{2} = 1 ´ 10^{12} Dyne .
cm^{2}
Problem 54. Coefficient of isothermal elasticity E_{q} and coefficient of adiabatic elasticity E_{f} are related by (g
= Cp / Cv )
[MP PET 2000]
(a)
Eq = g Ef
(b)
Ef = g Eq
(c)
Eq = g / Ef
(d)
Eq = g ^{2}Ef
Solution : (b) Adiabatic elasticity = g ´ isothermal elasticity Þ E_{f} = g E_{q} .
Problem 55. A uniform cube is subjected to volume compression. If each side is decreased by 1%,then bulk strain is
[EAMCET (Engg.) 1995; DPMT 2000]
(a) 0.01 (b) 0.06 (c) 0.02 (d) 0.03
Solution : (d) Volume of cube V = L^{3} \ Percentage change in V = 3 ´ (percentage change in L)= 3 (1%) = 3%
\DV = 3%
of V Þ Volumetric strain = DV =
V
3 = 0.03
100
Problem 56. A ball falling in a lake of depth 200 m shows 0.1% decrease in its volume at the bottom. What is the bulk modulus of the material of the ball [AFMC 1997]
(a)
19.6 ´ 10^{8} N / m^{2}
(b)
19.6 ´ 10^{–}^{10} N / m^{2}
(c)
19.6 ´ 10^{10} N / m^{2}
(d)
19.6 ´ 10^{–}^{8} N / m^{2}
Solution : (a)
K = P = hdg = 200 ´ 10^{3} ´ 9.8
= 19.6 ´ 10^{8} N / m^{2}
DV / V
DV / V
0.001
Problem 57. The ratio of the adiabatic to isothermal elasticities of a triatomic gas is [MP PET 1991]
 34
4 (c) 1 (d) 5
3 3
Solution : (b) For triatomic gas g
= 4 / 3 \ Ratio of adiabatic to isothermal elasticity g
= 4 .
3
Problem 58. A gas undergoes a change according to the law P = P_{0} e^{a}^{V} . The bulk modulus of the gas is
(a) P (b) aPV (c) aP (d) PV
a
Solution : (b)
P = Poe^{a}^{V} Þ
dP = Poe^{a}^{V}a = Pa
dV
[As P = Poe^{a}^{V} ]

dP V = Pa V
Þ ( dP
) = Pa V
\K = Pa V

Problem 59. The ratio of two specific heats of gas
Cp / Cv
for argon is1.6 and for hydrogen is 1.4. Adiabatic elasticity of
argon at pressure P is E. Adiabatic elasticity of hydrogen will also be equal to E at the pressure
(a) P (b) 8 P
7
Solution : (b) Adiabatic elasticity = g (pressure)
 7 P
8
 4 P

For Argon (Ef ) Ar = 1.6P and for Hydrogen (Ef )H = 1.4 P ¢
According to problem (Ef )H
= (E_{f} ) _{Ar} Þ 1.4 P ¢ = 1.6P Þ P ¢ = 16 P = 8 P .
2 14 7
Problem 60. The pressure applied from all directions on a cube is P. How much its temperature should be raised to maintain the original volume ? The volume elasticity of the cube is b and the coefficient of volume expansion is a
 P
ab
 Pa
b
 Pb
a
 ab
P
Solution : (a) Change in volume due to rise in temperature DV = Va Dq
\ volumetric strain = DV
V
= a Dq
But bulk modulus Þ b = stress =
strain
P
a Dq
\Dq = P
ab
Modulus of Rigidity.
Within limits of proportionality, the ratio of tangential stress to the shearing strain is called modulus of rigidity of the material of the body and is denoted by h, i.e. h = Shearing stress
Shearing strain
In this case the shape of a body changes but its volume remains unchanged.
Consider a cube of material fixed at its lower face and acted upon by a tangential force F at its upper surface having area A. The shearing stress, then, will be
Shearing stress = F = F
A A
This shearing force causes the consecutive horizontal layers of the cube to be slightly displaced or sheared relative to one another, each line such as PQ or RS in the cube is rotated through an angle f by this shear. The shearing strain is defined as the angle f in radians through which a line normal to a fixed surface has turned. For
small values of angle,
Shearing strain = f = QQ‘ = x
PQ L
So h = shear stress = F / A = F
shear strain f Af
Only solids can exhibit a shearing as these have definite shape.
Poisson’s Ratio.
When a long bar is stretched by a force along its length then its length increases and the radius decreases as shown in the figure.
Lateral strain : The ratio of change in radius to the original radius is called lateral strain.
Longitudinal strain : The ratio of change in length to the original length is
called longitudinal strain.
The ratio of lateral strain to longitudinal strain is called Poisson’s ratio (s).
i.e.
s = Lateral strain Longitudinal strain
s = – dr / r
dL / L
Negative sign indicates that the radius of the bar decreases when it is stretched. Poisson’s ratio is a dimensionless and a unitless quantity.
Relation Between Volumetric Strain, Lateral Strain and Poisson’s Ratio.
If a long bar have a length L and radius r then volume V = pr ^{2} L
Differentiating both the sides dV = pr ^{2}dL + p 2rL dr
Dividing both the sides by volume of bar
dV = pr ^{2}dL + p 2rL dr = dL + 2 dr
V pr ^{2} L pr ^{2} L L r
Þ Volumetric strain = longitudinal strain + 2(lateral strain)
Þ dV = dL + 2s dL = (1 + 2s ) dL
éAs s = dr / r Þ dr = s dLù
V L L L
ëê dL / L r
L úû
or s = 1 é1 – dV ù
[where A = crosssection of bar]
2 êë AdL úû
 If a material having s = – 5 then
dV = [1 + 2s ] dL = 0
V L
\ Volume = constant or K = ¥ i.e., the material is incompressible.
 If a material having s = 0, then lateral strain is zero e. when a substance is stretched its length increases without any decrease in diameter e.g. cork. In this case change in volume is maximum.
 Theoretical value of Poisson’s ratio – 1 < s < 5 .
 Practical value of Poisson’s ratio 0 < s < 5
Relation between Y, k, h and s.
Moduli of elasticity are three, viz. Y, K and h while elastic constants are four, viz, Y, K, h and s. Poisson’s ratio
s is not modulus of elasticity as it is the ratio of two strains and not of stress to strain. Elastic constants are found to depend on each other through the relations : Y = 3K(1 – 2s ) and Y = 2h(1 + s )
Eliminating s or Y between these, we get Y = 9Kh
3K + h
and s
= 3K – 2h
6K + 2h
Problem 61. Minimum and maximum values of Poisson’s ratio for a metal lies between [Orissa JEE 2003]
(a) – ¥ to + ¥ (b) 0 to 1 (c) – ¥ to 1 (d) 0 to 0.5
Solution : (d)
Problem 62. For a given material, the Young’s modulus is 2.4 times that of rigidity modulus. Its Poisson’s ratio is
[EAMCET 1990; RPET 2001]
(a) 2.4 (b) 1.2 (c) 0.4 (d) 0.2
Solution : (d)
Y = 2h(1 + s )
Þ 2.4h = 2h(1 + s ) Þ
1.2 = 1 + s
Þ s = 0.2
Problem 63. There is no change in the volume of a wire due to change in its length on stretching. The Poisson’s ratio of the material of the wire is
(a) + 0.50 (b) – 0.50 (c) + 0.25 (d) – 0.25
Solution : (b)
dV = dL + 2s dL = (1 + 2s ) dL = 0 [As there is no change in the volume of the wire]
V L L L
\1 + 2s = 0
Þ s = – 1
2
Problem 64. The values of Young’s and bulk modulus of elasticity of a material are respectively. The value of Poisson’s ratio for the material will be
8 ´ 10^{10} N / m^{2}
and 10 ´ 10^{10} N / m^{2}
(a) 0.25 (b) – 0.25 (c) 0.37 (d) – 0.37
Solution : (c)
Y = 3K(1 – 2s ) Þ
8 ´ 10^{10} = 3 ´ 10 ´ 10^{10} (1 – 2s ) Þ
s = 0.37
Problem 65. The Poisson’s ratio for a metal is 0.25. If lateral strain is 0.0125, the longitudinal strain will be (a) 0.125 (b) 0.05 (c) 0.215 (d) 0.0125
Solution : (b)
s = Lateral strain Longitudinal strain
\Longitudinal strain = Lateral strain
s
= 0.0125 = 0.05
0.25
Problem 66. The ‘s’ of a material is 0.20. If a longitudinal strain of volume change
4.0 ´ 10^{–}^{3}
is caused, by what percentage will the
(a) 0.48% (b) 0.32% (c) 0.24% (d) 0.50%
Solution : (c) Longitudinal strain = 4 ´ 10^{–}^{3} or 0.4%
Lateral strain = s ´ 0.4% = 0.2 ´ 0.4% = 0.08%
\ Volumetric strain = longitudinal strain – 2 lateral strain = 0.4 – 2 ´ (0.08) = 0.24%
\ Volume will change by 0.24%.
Torsion of Cylinder.
If the upper end of a cylinder is clamped and a torque is applied at the lower end the cylinder gets twisted by angle q. Simultaneously shearing strain f is produced in the cylinder.
 The angle of twist q is directly proportional to the distance from the fixed end of the
At fixed end q = 0^{o} and at free end q = maximum.
 The value of angle of shear f is directly proportional to the radius of the cylindrical shell. At the axis of cylinder f = 0 and at the outermost shell f =
 Relation between angle of twist (q) and angle of shear (f)
AB = rq = fl \ f = rq l
 Twisting couple per unit twist or torsional rigidity or torque required to produce unit
C = phr ^{4}
2l
\ C µ r ^{4} µ A^{2}
 Work done in twisting the cylinder through an angle q is W = 1Cq ^{2} = phr 4q 2
2 4l
Problem 67. Mark the wrong statement [MP PMT 2003]
 Sliding of molecular layer is much easier than compression or expansion
 Reciprocal of bulk modulus of elasticity is called compressibility
 It is difficult to twist a long rod as compared to small rod
 Hollow shaft is much stronger than a solid rod of same length and same mass
Solution : (c)
Problem 68. A rod of length l and radius r is joined to a rod of length l / 2 and radius r / 2 of same material. The free end of small rod is fixed to a rigid base and the free end of larger rod is given a twist of q, the twist angle at the joint will be [RPET 1997]
(a) q / 4 (b) q / 2 (c) 5q / 6 (d) 8q / 9
Solution : (d) If torque t is applied at the free end of larger rod and twist q is given to it then twist at joint is q _{1} and twist at the upper end (fixed base) q _{2}
æ r ö^{4} ( )
phr ^{4}(q – q )
phç ÷
2
q_{1} – q _{2}
t = ^{1} =
2l
è ø
2(l / 2)
Þ (q – q_{1}
) = (q_{1} – 0)
8
[As q _{2}
= 0]
Þ 8q – 8q_{1} = q_{1}
Þ 9q_{1} = 8q
Þ q_{1} = 8q .
9
Problem 69. The upper end of a wire of radius 4 mm and length 100 cm is clamped and its other end is twisted through an angle of 30^{o} . Then angle of shear is [NCERT 1990; MP PMT 1996]
(a)
12^{o}
(b)
0.12^{o}
(c)
1.2^{o}
(d)
0.012^{o}
Solution : (b)
Lf = rq
\f = rq
L
= 4 ´ 10^{–}^{3} ´ 30^{o}
1
= 0.12^{o}
Problem 70. Two wires A and B of same length and of the same material have the respective radii r_{1} and r_{2}. Their one end is fixed with a rigid support, and at the other end equal twisting couple is applied. Then the ratio of the angle of twist at the end of A and the angle of twist at the end of B will be [AIIMS 1980]

2


 ^{1} 2
phr ^{4}q
phr ^{4}q
2



 ^{2} 1
q
æ r ö4
4



 ^{2} 1
4



 ^{1} 2
Solution : (c)
t_{1} = t _{2}
Þ 1 1 = 2 2
Þ 1 = ç 2 ÷
2l1
2l2
q 2 ç r1 ÷
è ø
Problem 71. The work done in twisting a steel wire of length 25 cm and radius 2mm through 45^{o} will be (h = 8 ´ 10^{10} N / m^{2})
(a) 2.48 J (b) 3.1 J (c) 15.47 J (d) 18.79 J
Solution : (a)
W = 1 Cq ^{2} = phr 4q 2
= 3.14 ´ 8 ´ 10^{10} ´ (2 ´ 10 ^{–}^{3} )^{4} ´ (p / 4)^{2}
= 2.48 J
2 4l
4 ´ 25 ´ 10 ^{–}^{2}
Interatomic Force Constant.
Behaviour of solids with respect to external forces is such that if their atoms are connected to springs. When an external force is applied on a solid, this distance between its atoms changes and interatomic force works to restore the original dimension.
The ratio of interatomic force to that of change in interatomic distance is defined as the interatomic force
constant.
K = F
Dr
It is also given by K = Y ´ r_{0}
[Where Y = Young’s modulus, r_{0}
= Normal distance between the atoms of wire]
Unit of interatomic force constant is N/m and Dimension MT^{–2}
Note : @ The number of atoms having interatomic distance r_{0} in length l of a wire, N = l/r_{0}.

@ The number of atoms in area A of wire having interatomic separation r_{0} is N = A / r ^{2} .
Problem 72. The mean distance between the atoms of iron is 3 ´ 10^{–}^{10} m and interatomic force constant for iron is 7 N/m. The Young’s modulus of elasticity for iron is [JIPMER 2002]
(a)
2.33 ´ 10^{5} N / m^{2}
(b)
23.3 ´ 10^{10} N / m^{2} (c)
233 ´ 10^{10} N / m^{2}
(d)
2.33 ´ 10^{10} N / m^{2}
Solution : (d)
Y = k ro
= 7
3 ´ 10 ^{–}^{10}
= 2.33 ´ 10^{10} N / m ^{2} .
Problem 73. The Young’s modulus for steel is
force constant in N/ Å will be
Y = 2 ´ 10^{11} N / m^{2} . If the interatomic distance is 3.2Å, the inter atomic
(a)
6.4 ´ 10^{9}
(b)
6.4 ´ 10^{–}^{9}
(c)
3.2 ´ 10^{9}
(d)
3.2 ´ 10^{–}^{9}
Solution : (b)
k = Y ´ r_{0} = 2 ´ 10^{11} ´ 3.2 ´ 10 ^{–}^{10} = 6.4 ´ 10^{1} N / m = 6.4 ´ 10 ^{–}^{9} N / Å .
Elastic Hysteresis.
When a deforming force is applied on a body then the strain does not change simultaneously with stress rather it lags behind the stress. The lagging of strain behind the stress is defined as elastic hysteresis. This is the reason why the values of strain for same stress are different while increasing the load and while decreasing the load.
Hysteresis loop : The area of the stressstrain curve is called the hysteresis loop and it is numerically equal to the work done in loading the material and then unloading it.
If we have two tyres of rubber having different hysteresis loop then rubber B should be used for making the car tyres. It is because of the reason that area under the curve i.e. work done in case of rubber B is lesser and hence the car tyre will not get excessively heated and rubber A should be used to absorb vibration of the machinery because of the large area of the curve, a large amount of vibrational energy can be dissipated.
Factors Affecting Elasticity.
 Hammering and rolling : Crystal grains break up into smaller units by hammering and This result in increase in the elasticity of material.
 Annealing : The metals are annealed by heating and then cooling them Annealing results in decrease in the elasticity of material.
 Temperature : Intermolecular forces decreases with rise in temperature. Hence the elasticity decreases with rise in temperature but the elasticity of invar steel (alloy) does not change with change of
 Impurities : Due to impurities in a material elasticity can increase or The type of effect depends upon the nature of impurities present in the material.
Important Facts About Elasticity.
 The body which requires greater deforming force to produce a certain change in dimension is more elastic. Example : Ivory and steel balls are more elastic than
 When equal deforming force is applied on different bodies then the body which shows less deformation is more
Example : (i) For same load, more elongation is produced in rubber wire than in steel wire hence steel is more elastic than rubber.
 Water is more elastic than air as volume change in water is less for same applied
 Four identical balls of different materials are dropped from the same height then after collision balls rises upto different
The order of their height can be given by h_{ivory} > h_{steel} > h_{rubber} > h_{clay} because Y_{ivory} > Y_{steel} > Y_{rubber} > Y_{clay}.
 The value of moduli of elasticity is independent of the magnitude of the stress and It depends only on the nature of material of the body.
 For a given material there can be different moduli of elasticity depending on the type of stress applied and resulting
Name of substance  Young’s modulus (Y) 10^{10}N/m^{2}  Bulk modulus (K) 10^{10}N/m^{2}  Modulus of rigidity (h) 10^{10}N/m^{2} 
Aluminium  6.9  7.0  2.6 
Brass  9.0  6.7  3.4 
Copper  11.0  13.0  4.5 
Iron  19.0  14.0  4.6 
Steel  20.0  16.0  8.4 
Tungsten  36.0  20.0  15.0 
Diamond  83.0  55.0  34.0 
Water  –  0.22  – 
Glycerin  –  0.45  – 
Air  –  1.01  – 
 The moduli of elasticity has same dimensional formula and units as that of stress since strain is
dimensionless. \ Dimensional formula
ML1T 2
while units dyne/cm^{2} or Newton/m^{2}.
 Greater the value of moduli of elasticity more elastic is the But as Y µ (1/l), K µ (1/DV) and
h µ (1/f) for a constant stress, so smaller change in shape or size for a given stress corresponds to greater elasticity.
 The moduli of elasticity Y and h exist only for solids as liquids and gases cannot be deformed along one dimension only and also cannot sustain shear However K exist for all states of matter viz. solid, liquid or gas.
 Gases being most compressible are least elastic while solids are most e. the bulk modulus of gas is very low while that for liquids and solids is very high. K_{solid} > K_{liquid} > K_{gas}
 For a rigid body l, DV or f = 0 so Y, K or h will be ¥, e. elasticity of a rigid body is infinite. Diamond and carborundum are nearest approach to rigid bodies.
 In a suspension bridge there is a stretch in the ropes by the load of the Due to which length of rope changes. Hence Young’s modulus of elasticity is involved.
 In an automobile tyre as the air is compressed, volume of the air in tyre changes, hence the bulk modulus of elasticity is
 In transmitting power, an automobile shaft is sheared as it rotates, so shearing strain is set up, hence modulus of rigidity is
 The shape of rubber heels changes under stress, so modulus of rigidity is
Practical Applications of Elasticity.
 The metallic parts of machinery are never subjected to a stress beyond elastic limit, otherwise they will get permanently
 The thickness of the metallic rope used in the crane in order to lift a given load is decided from the knowledge of elastic limit of the material of the rope and the factor of
 The bridges are declared unsafe after long use because during its long use, a bridge under goes quick alternating strains It results in the loss of elastic strength.
 Maximum height of a mountain on earth can be estimated from the elastic behaviour of earth.
At the base of the mountain, the pressure is given by P = hrg and it must be less than elastic limit (K) of earth’s supporting material.
K > P > hrg \ h < K or
rg
hmax
= K
rg
 In designing a beam for its use to support a load (in construction of roofs and bridges), it is advantageous to increase its depth rather than the breadth of the beam because the depression in rectangular
d = Wl ^{3}
4Ybd ^{3}
To minimize the depression in the beam, it is designed as Ishaped girder.
 For a beam with circular crosssection depression is given by d =
WL^{3}
12p r ^{4} Y
 A hollow shaft is stronger than a solid shaft made of same mass, length and
Torque required to produce a unit twist in a solid shaft t = phr 4
.…..(i)
solid
2l
ph(r ^{4} – r ^{4} )
and torque required to produce a unit twist in a hollow shaft t _{hollow} = ^{2} ^{1}
2l
…..(ii)
From (i) and (ii), t hollow
t solid
r ^{4} – r ^{4}
= 2 1
r 4
(r ^{2} + r ^{2} )(r ^{2} – r ^{2} )
= 2 1 2 1
r 4
…..(iii)
Since two shafts are made from equal volume \ pr ^{2}l = p (r ^{2} – r ^{2} )l Þ r ^{2} = r ^{2} – r ^{2}
2 1 2 1
Substituting this value in equation (iii) we get, t hollow = r 2 + r 2 > \ t t
t solid
2 1 1
r 2
hollow >
solid
i.e., the torque required to twist a hollow shaft is greater than the torque necessary to twist a solid shaft of the same mass, length and material through the same angle. Hence, a hollow shaft is stronger than a solid shaft.