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UNIT – 01

Properties of Matter andHydrodynamics

Properties of Matter

01.1 INTRODUCTION

There are some bodies, which do not change in shape or size on the application of an external force,however large. Such bodies are called rigid bodies. Of course, this is an idealization since there is nomaterial in nature which would not deform at all under the influence of very large forces. However, forthe range of forces encountered in practice, many bodies deform under the influence of an externalforce, but regain their original shape and size when the external force is withdrawn. The maximumexternal force up to which the body regains its original shape and size after the external force isremoved is a measure of the elastic limit of the body. If the force exceeds this limit, the body does notreturn to its original shape and size after the deforming force has ceased to act. The body then getspermanently deformed and is said to be in a permanent set. Materials which regain their original shapeand size after the removal of external force are called elastic materials and those acquiring permanentset are called plastic materials. No material in nature is perfectly elastic or perfectly plastic. Somematerials are more elastic while others are more plastic. Most metals are good examples of elasticmaterials, while rubber has small degree of elasticity. Clay and polymers are good examples of plasticmaterials.

01.2 STRESS AND STRAIN

Stress: When deforming forces act on an elastic body, forces of reaction come into play internally dueto relative displacement of the molecules. The restoring forces per unit area set up inside the body iscalled stress. The restoring force is equal in magnitude but opposite in direction to the externallyapplied force so long as there is no permanent change produced in the body. Hence stress is measuredby the applied force per unit area. If a force F is applied to a surface of area A, then

Stress = AF

2 A Textbook of Engineering Physics

Strain: Strain is the fractional deformation resulting from a stress. It is measured by the ratio ofthe change in some dimension of a body to the original dimension in which change took place. Since itis the ratio of two similar quantities, it has no unit or dimension. Three types of strain are possible foran elastic body and hence three kinds of elastic modulii are possible. They are discussed below:

(i) Longitudinal strain and Young’s modulus

When the length of a body is changed by the application of an external force, it is said to be strainedlongitudinally. The change in length per unit length is called longitudinal strain. If l is the originallength of the wire and x, the change in length caused by the deforming force,

Linear or longitudinal strain = l

x=

lengthOriginal

length in Change

Hence strain has no dimension.If F is the magnitude of the force to produce the strain in a wire and A is the cross-sectional area,

then AF

is the stress. Young’s modulus is the ratio of the longitudinal stress to the longitudinal strain

and is usually represented by Y. Thus

Y = Ax

FL

LxA

F==

)/(Strain

Stress

Its unit is newton/m2 and dimension is .212

2−−

−

= TMLL

MLT

(ii) Shearing strain and rigidity modulus

Fig. 01.1 Shearing strain

When a body undergoes a change in shape without any change in volume, it is said to suffer a shearingstrain. A section of rectangular block with the bottom face fixed is shown in the figure.

It the upper surface is urged forward by a force tangential to the surface, the block is sheared.The face ABCD is now distorted to A B1 C1 D. We thus observe that BC changes to B1 C1 with an angleθ.

B B1 F C C1

θ

A D

θ

01 Properties of Matter and Hydrodynamics 3

The ratio AB

BB1 or

CD

CC1 is the shearing strain. The angle θ in radian is called angle of shear with

the assumption BB1 and CC1 are small. Thus shearing stress = AF

, where A is the cross-sectional area.

Angle of shear

θ = CD

CC

AD

BB 11 =

Thus modulus of rigidity

η = θ

=θ A

FAF

(iii) Volume strain and bulk modulus

Fig. 01.2 Volume strain

When a body is subjected to normal external forces from all sides, the body experiences a change in itsvolume. Let us consider a cube of side l and volume V (= L3) with a pressure acting on all its faces.

If the pressure on all its faces is increased to P + ∆P by applying compressive force F normallyon all its faces as shown in the figure, the volume of the cube would decrease by certain amount, say∆V. Stress and strain experienced by the body in this case are called volume stress and volume strainrespectively. The volume stress on the body is given by

Stress = PL

F∆=

2

F

F

F

F

F

F


The volume strain produced in the body due to this stress is

Strain = VV∆

The ratio of the volume stress to volume strain is called the bulk modulus of elasticity and isusually denoted by the symbol B, i.e.,

B = strain Volumestress Volume

=

∆∆

−=

∆

∆V

PV

VV

P

The negative sign signifies the decrease in the volume as the pressure is increased. The aboveequation in terms of the derivative, can be written as

B = dVdP

V−

01.3 HOOKE'S LAW

Based on experimental observations on the elastic properties of materials, Prof. Hooke pronounced, in1678, an empirical relation between stress and strain. According to Hooke's law, the ratio of stress andstrain is a constant.

StrainStress

= E

where E is the modulus of elasticity of the material. The above relation holds good within theelastic limit of the material.

Let a wire be fixed at one end and loaded at the other end. Let the load be gradually increased andthe corresponding extension be noted for every load. A graph is drawn between the stress and strain.The graph represents the behaviour of the wire. The graph is found to be a straight line sloping upto Aindicating that stress is proportional to the strain and Hooke's law is obeyed. A is the limit of elasticityi.e., within the range OA, the wire will regain its original length on removal of the stretching force(load).

Beyond A, the curve bends a little showing that the strain (elongation) is much more than what ithas been in the range OA. A part of this elongation is permanent and if the load is removed at B, thedotted path BC will be traced. The wire will not regain its original length but acquires a permanentextension of length which remains as residual strain OC which is what we recall permanent set.

On loading the wire further, even a small increase of load will produce a large but erratic extensionup to D and the metal wire seems to flow under the action of the load. The stage B at which this beginsto take place is called the yield point and the stress corresponding to it is known as the yielding stress.

At the yield point and beyond it the wire becomes thinner and thinner, thereby the force per unitarea (stress) becomes greater and greater for the same stretching force. At this stage, say at F, aconstriction develops at some weak point in the wire. The wire breaks ultimately beyond constriction.This point F is called Breaking point.


Fig. 01.3 Shearing strain

The area under the stress-strain diagram (area OABDEFG) represents the work done per unitvolume during the process. It is a measure of toughness of the material.

01.4 TWISTING COUPLE ON A WIRE AND TORSION PENDULUM

Let a cylindrical wire be clamped rigidly at the top. It is said to be under torsion. An elastic couplecalled torsional couple comes into play. The wire will be in equilibrium in the deformed state.

Let l be the length of the wire, a its radius and η the rigidity modulus of the material of the wire.A couple of moment C twists the wire at the bottom since the wire is fixed at the top. Let us nowconsider an annular tube of radius r and thickness dr. When the wire is twisted through an angle θ, theradius OD turns through an angle to OD'. The longitudinal line along its length rotates by an angleDAD' = φ which is the angle of shear.

Now

sin φ = φ = lDD ′

; (or) DD' = l φ

Similarly

sin θ = θ = ;rDD ′

(or) DD' = r θ

i.e., l φ = r θ

or φ = l

r θ

Thus rigidity modulus,

η = φ

=Stress

shear of Angle

Stress

Y

XG

E

FBA

D

Elastic Limit

Elastic Range

PlasticRange

Residual Strain

Strain O C

Stre

ss


Thus the stress produced η φ = lr θη

The corresponding force = stress × area

=

θη

l

r 2 π r dr

Moment of the force about the axis is rl

r

θπη 22 .

Total moment of all the force on the entire system is

∫πηθ

a

o

drrl

32=

a

o

r

l

πηθ4

2 4

l

a

2

4θπη=

Fig. 01.4 Twisting couple

In the equilibrium condition, this is equal to the elastic couple on the wire. If C is the couplerequired to produce unit twist, then

C θ = la

2

4θπη

aO r

φ

l

θD

O

D'


Now the couple for unit twist

C = la

2

4πη

Torsional Pendulum

Fig. 01.5 Torsional pendulum

(i) Determination of rigidity modulus of the given wire with uniform cross-section

Suspend the disc horizontally by means of the experimental wire as shown in the figure. The other endof the wire is clamped to a rigid support. The length of the wire from the bottom of the support to thedisc is adjusted for a convenient value, say l.

Two equal masses m each are placed symmetrically on the disc at a distance d1 from the centre ofthe system. The wire is twisted by rotating the disc. On releasing the disc it will perform torsionaloscillations. The period of torsional oscillations T1 is found out. The experiment is repeated for a largedistance d2 between the centre of the mass and that of the disc. The corresponding period is alsoworked out.

Let Io be the moment of inertia of the suspended disc alone about its axis a suspension and io the

moment of inertia of the cylinder about its own axis and m 21d is the moment of inertia of the cylinder

with the wire as the axis. Thus the total moment inertia is

I = Io + 2io + 2m 21d

T1 = C

dmiI oo2122

2++

π ...(1)

and

T2 = C

dmiI oo2222

2++

π ...(2)

m m

d1 d1

m m

d2 d2


Thus

21

22 TT − = ( )2

122

2

24

ddmC

−π...(3)

2

21

22

4π

− TT=

Cddm )(2 2

122 −

i.e., with C = l

r

2

4πη

2

21

22

4π

− TT=

4

21

22 2)(2

r

lddm

πη×−

η = )(

)(162

12

24

21

22

TTr

ddml

−

−π...(4)

Measuring l and r (using a screw gauge) and knowing m, d1 and d2 , η the rigidity modulus of thewire may be determined. The main precaution is T1 and T2 must be the average values of the periodsby finding T1 and T2 for 10, 15 and 20 oscillations.

(ii) To find out the moment inertia of the disc

Equation (3) is

2

21

22

4π

− TT=

C

ddm )(2 21

22 −

...(A)

Without the cylinders, the period of torsional pendulum is determined. Let it be T. In this case

T = C

Iπ2

2

2

4π

T=

CI

...(B)

Divide Equation (A) by (B)

2

21

22

T

TT −=

Iddm )(2 2

122 −

I = )(

)(22

12

2

221

22

TT

Tddm

−

−

All quantities on the R.H.S. are known and hence I can be calculated.

0.1.5 BENDING OF BEAMS

A rod of uniform cross-section whose length is large compared to its other dimensions is called a beam.When such a beam is deformed by an external couple, forces of reactions are called into play within the


beam, tending to resist the deformation. When there is equilibrium, all the elastic reactions (forces)constitute a couple and the moment of this couple due to the elastic reactions balances the externalcouple and it is called bending moment.

Let the longitudinal section of the portion of a bent beam be represented by ABCD. Here PQ isrepresenting a filament at the neutral surface (length remains constant), P'Q' be another filament at adistance (y) from PQ. In a bent beam PQ and P'Q' will be arcs of circles, having their centres at O andradius R and (R + y) respectively. If θ be the angle POQ, then

PQ = R θ and P'Q' = (R + y ) θ

Fig. 01.6 Bending of beams

Extension produced in the filament P'Q' due to bendingP'Q' – PQ = (R + y) θ – R θ = y θ

The strain = R

y

R

y=

θθ

=length original

extension

If q is the Young’s modulus of the meterial of the beam then longitudinal modulus

q = StrainStress

Stress = q × Strain = R

yq×

If a is the area of the cross-section of the filament, the longitudinal force on this filament =stress × a

i.e.,R

ayq ××

Moment of this force = Rq

y × a × y = Rq

ay2

If we take all the filaments into account

T.M.C. = ∑ 2ayRq

B

P

D C

Q

O

θ

A

P'

P

D

BQ'

Q

C

yP'Q '

A


C = ∑ = 22 AKRq

ayRq

Here Σ ay2 is the moment of inertia of the rectangular bar about the central axis and is calledgeometrical moment of inertia and it is represented as AK2 where A is the area of the cross-section and

K is radius of gyration. If b and d are the respective dimensions, then A = bd and .12

22 d

K =

Therefore

∑ 2ay = AK2 = bd 12

2d12

3bd=

For circular cross-section A = πr2, 4

2rK =

∑ 2ay = 44

422 rrr π=×π

Depression for the free end of a cantilever

A light beam clamped horizontally at one end and loaded with a weight W = Mg at the free end is calleda cantilever. We can derive an expression for the depression of the free end of the cantilever in terms ofthe weight W, Young's modulus q, and the dimensions of the beam. The expression for the depression

in such a case is δ = 2

3

3qAK

Mgl with AK2 = bd (d2/12) for rectangular cross-section. This equation is a

basic one which is used directly or with some modifications in many experiments. Two of them are

discussed below.

(i) Beam supported on two knife-edges and load in the middle (non-uniform bending)

Consider a beam supported symmetrically on two knife-edges at a distance l apart with mass at

the middle. The reaction on the knife edges are each equal to 2

Mg. Now the depression δ is got by

substituting Mg = 2

Mg and

2l

l =

i.e.,

δ = 2

3

3

1

82 qAK

lMg

δ = 3

3 12

48 qbd

Mgl


δ = 3

3

4qbd

Mgl for rectangular cross-section.

and

δ = )4/(

1

48 22

3

rrq

Mgl

π

δ = )4/(3

1

82 22

3

rrq

lMg

π

for circular cross-section.

i.e.,

δ = 4

3

12 rq

Mgl

π

Now by conducting an experiment, δ can be calculated from which q can be computed.

Fig. 01.7

(ii) Beam supported symmetrically on two knife edges and loaded equally at the free ends (Uniform bending)

Fig. 01.8

Mg2

—— Mg2——

l—2

l—2

δ

a

C

A

h

l

D

a

B


Consider the beam CD supported symmetrically on two knife-edges at A and B and loaded withequal weights Mg at each end. The reactions at the knife edges are each equal to Mg. The forces at Cand A constitute a couple = Mga.

Bending moment = R

qAK2

RqAK2

= Mga

If l is the distance between the knife edges and h is the elevation of the mid point,then

h(2R – h) = 4

2l

If h is small,

2 Rh = 4

2l

or R = h

l8

2

ButR

qAK2

= Mga

h = 2

2

8qAK

aMgl with AK2

12

3bd= for rectangular cross-section

h = 3

2

2

3

qbd

aMgl

h = 4

2

22

2

2)4/(8 rg

aMgl

rrq

aMgl

π=

π for circular cross-section

(iii) Experimental determination of Young's modulus of a beam (non-uniform bending)

Fig. 01.9


The beam may be supported on two knife-edges in the same horizontal level, so that equal lengthsare projecting beyond the supports. A pin or small needle is formed at the centre of the beam verticallyby means of some wax. A loop is suspended exactly at the centre of the beam and a hanger is attached.A travelling microscope is focused on the tip of the needle and the tip of the pin is made to coincidewith the horizontal cross-wire. First a load of 50 gm is added to the weight hanger. The total mass iscalled dead load. The microscope is vertically moved down so that the pin is again coinciding with thehorizontal cross-wire. The reading is taken. Now another 50 gm added, the microscope is adjusted sothat the well defined image of the tip of the pin coincides with the horizontal cross-wire. The newreading is taken. The experiment is repeated by adding the load in steps by 50 gm. Once seven or eightreadings are taken, the weights are decreased by 50 gm steps and the readings are taken. The average ormean reading is also obtained. A tabulation is drawn as shown below.

Reading of microscope Depression

Load Load Load Mean for

gm increasing decreasing (metre) 200 gm

x α1 β1 x1 δ1

x + 50 α2 β2 x2 δ2

x + 100 α3 β3 x3 δ3

x + 150 α4 β4 x4 δ4

x + 200 α5 β5 x5

x + 250 α6 β6 x6

x + 300 α7 β7 x7

x + 350 α8 β8 x8

Mean, δ =

Measure the distance between the knife edges, say l. Also measure the thickness and breadth ofthe beam (say a metre scale)

We know

δ = 3

3

2

3

48

121

48 bdq

Mgl

qAk

Mgl

××

×=

q = 3

3

4 bd

Mgl

δwith M = 200 × 10–3 kg all other parameters in SI units.

The only unknown quantity is q which can be easily calculated using the above relation.

(iv) Young's modulus by uniform bending

In this case the given beam is placed on two knife edges in the same horizontal level, equal lengthprojecting on either side of the knife edges.


Fig. 01.10

Two weight hangers are arranged at two points, having equal distances from the knife edges. Aneedle or pin is fixed at the mid point. Equal weights must be added to both hangers. The experimentis conducted as in the previous case. The elevation for a particular load is given by

h = 2

2

8qAK

aMgl

q = 12/8 3

2

hbd

aMgl

q =

3

2

3

4

hbd

aMgl

q – can be calculated using the above formula.

01.6 HYDRODYNAMICS

The study of the mechanics of flow of a fluid requires the knowledge of the motion of each of itsparticles during the flow. Obtaining such a detailed knowledge is, however, a difficult task. Though itmay not be difficult to assign position coordinates to a particle at any instant of time and to follow itsmotion for certain interval of time, but to focuss attention on all the particles at a time and to studytheir mechanics together during the flow is mathematically a stupendous job. Usually a simpler methodadopted to understand the behaviour of flow is to concentrate one's attention at certain points withinthe liquid and to study the behaviour of all the particles passing through these points.

01.7 STREAMLINE FLOW

Streamline flow of a liquid is a steady flow in which each layer of liquid follows the same path and hasthe same velocity as that of its predecessor. The path followed by an element of a moving fluid iscalled line of flow. In general, the velocity of the element changes in both magnitude and directionalong the line of flow. In a streamline flow every element passing through a given point follows thesame line of flow as that of the preceding elements. In such a flow the velocity at each point remainsa constant, although the velocity of a particular particle of the fluid may change as it moves from onepoint to another.

a a

l


In such a flow if we consider a line along which a particle of the liquid moves, the direction ofthe line at any point is the direction of the velocity of the liquid at that point. Such a line is called astreamline. A streamline may be defined as a curve, the tangent to which at any point gives thedirection of the flow of liquid at that point. The streamline may be straight or curved, according to thelateral thrust on it which may be the same throughout or different.

The steady or streamline flow is possible only when the liquid velocity is less than a limitingvalue known as critical velocity.

01.8 TURBULENT FLOW

When the velocity of flow of a liquid is greater than the critical value, the flow of the liquid becomesdisorderly and zig-zag and is called turbulent flow.

01.9 REYNOLD'S NUMBER-CRITICAL VELOCITY

The flow of a liquid is steady or streamline only when the velocity of the liquid is less than the criticalvalue. It was Osborne Reynolds who first showed that the critical velocity v of a liquid flowingthrough a narrow tube in given by the relation,

v = r

Re

ρη

where Re is a constant called Reynold number, η is the coefficient of viscosity of the liquid (fluid), r isthe radius of the tube and ρρ is the density of the liquid. For narrow tubes, the value of Re is of the orderof 1000.

01.10 VISCOSITY

Most of the fluids are not ideal ones and offer some resistance to motion. This resistance to fluidmotion is like an internal friction analogous to friction when a solid moves on a surface. This is usuallycalled by the single name viscosity. This force exists when there is a relative motion between layers ofthe liquid. Suppose we consider a fluid like oil enclosed between two glass plates. The bottom plate isfixed while the top plate is moved with a constant velocity v relative to the fixed plate. If oil is replacedby honey, a greater force in required to move the plate with the same velocity. Hence we say that honeyis more viscous than oil. The fluid in contact with the surface has the same velocity as that of thesurface. Hence the layer of the liquid in contact with top surface moves with a velocity v and the layerof the liquid in contact with the fixed surface in stationary. Thus the velocities of the layers increaseuniformly from bottom (zero velocity) to the top layer (velocity v). For any layer of liquid, its upperlayer pulls it forward while lower layer pulls it backward. This results in a force between the layers.When a fluid is flowing in a pipe or a tube, then velocity of the liquid layer along the axis of the tubeis maximum and decreases gradually as we move towards the walls where it becomes zero. The conclusionis that in each and every layer, there is a backward dragging force acting tangentially on the layers.Therefore for maintaining the relative velocity between the layers, an external force must be applied toovercome this backward dragging force. In the absence of any such outside force, the relative motionbetween the layers is destroyed and the flow of the liquid ceases. This property of the liquid by virtueof which it opposes the relative motion between its different layers is called viscosity or internalfriction of the liquid.


01.11 COEFFICIENT OF VISCOSITY

The backward dragging or viscous force acting tangentially on any liquid layer is directly proportionalto its surface area A and velocity v and inversely proportional to its distance x from the stationary layer.

If F is the viscous force, F ∝ A and F ∝ v and F ∝x1

i.e.,

F ∝ x

Av

F = x

Avη−

The negative sign indicates that the direction of the force is opposite to that of velocity. η iscalled coefficient of viscosity depending on the nature of the liquid.

xv

may be replaced by dxdv

which gives the rate of change of velocity with distance and called

velocity gradient.

F =

η−

dx

dA

v

If A = unity, ,1v =

dxd

then F = η

or the coefficient of viscosity of a liquid may be defined as the tangential force required per unitarea to maintain a unit velocity gradient.

01.12 RATE OF FLOW OF LIQUID IN A CAPILLARY TUBE (POISEUILLE'S FORMULA)

We now try to derive an expression for the volume of the liquid flowing/sec through a capillary tube interms of the coefficient of viscosity and the difference of pressure between the two ends of a capillarytube. The expression holds good only for streamline motion.

Fig. 01.11

x+x=2x

l

2 a


Consider a cylindrical layer of the liquid at a distance x from the axial line. Let a be the radius ofthe capillary tube.

Now the viscous force acting on that layer is

F = –

πη

dx

dvlx2

If P is the pressure difference between the two ends, then the hydrostatic pressure due to the areaπx2 is πx2 P. For steady flow of the liquid, the two forces must be equal.

–

ηπ

dx

dxl

v2 = Px2π

η

dx

dl

v2 = – Px

or

dxdv

= –ηl

Px

2or

dv = xdxl

P

η−

2Integrating

∫ vd = ∫η− xdx

l

P

2

v = Cx

l

P+

η−

22

2

...(1)

When x = a, v = 0i.e.,

0 = Ca

l

P+

η−

22

2

C = ηl

Pa

4

2

Substituting this value of C in Eq. (1)

v = η

+η

−l

Pax

l

P

422

22

or

v = η−l

xaP

4

)( 22

...(2)

This is the velocity of the flow of the liquid at a distance x from the axis of the capillary tube.


01.13 VOLUME OF THE LIQUID (V) FLOWING PER SEC

Fig. 01.12

Consider a co-axial layer of the liquid at a distance (x + dx) from the centre. Volume of the liquidflowing per sec through the annular space between the layers x and x + dx = area of cross-section ×velocity

= 2π x dx × v

= 2π x dx )(4

22 xal

P−

η

Total volume of the liquid flowing / sec through the entire cross-section of the capillary tube.

V = ∫ −η

πa

o

dxxxal

P)(

232

=

−

ηπ

422

422 aa

al

P

or = η

π=

−

ηπ

l

Paaa

l

P

8422

444

η = lV

Pa

××π

8

4

Thus by measuring the volume of the liquid flowing per sec, η may be determined.

01.14 DETERMINATION OF THE COEFFICIENT OF VISCOSITY OF A LIQUID

The arrangement is shown in the figure. AB is a capillary tube of uniform cross-section attachedhorizontally at the bottom of the vessel of large size. When the conditions become steady, the liquid isallowed to flow and collected for a time of t sec. Now find the mass of the liquid collected. The densityof the liquid is found with Hare's apparatus.

Mass of the liquid collected = mTime of collection = tDensity of liquid = ρ

x

dx

a


Fig. 01.13

Distance between the surface of the liquid and the axis of the tube = h.Radius of the capillary tube = a and length of the capillary tube = l

The mass collected for one sec is

t

m

Now

P = hρg ;η

π=

l

PaV

8

4

V = η

π=

ρ=

ρ l

Pa

t

mM

8

4

η = lm

tgahlm

tPa88

44 ρρπ=ρπ

η = lm

tgah8

42ρπ

From this η can be determined.

Solved Problems

1. A uniform disc is suspended by steel wire, allowed to oscillate torsionally with a period of4 sec. Find the period if(a) Length of wire is doubled

h

A

B


(b) Two chucks each of mass equal to 41

th the mass of the disc are placed at diametricallyy

opposite points on the circumference of the disc.Solution :

(a) General Formula

Period, T = C

Iπ2

Case (i) 4 = C

Iπ2

4

22

nr

lI

π×

π=

Case (ii) When length is doubled

T1 = 4

)2(22

nr

lI

π

×π

41T

= 2

T1 = 24×

(b) When the two cylinders of mass each 41

the mass of the disc are placed, then moment

of inertia of the system

I = 2222

442MR

MRMRMR =++

T2 = C

I12π

42T

= 2)2/( 2

21 ==

MR

MR

I

I

T2 = 24× Ans.

2. A gold wire 0.32 × 10–3 m in diameter elongates by 1 × 10–3 m when stretched by force of0.33 kg. wt and twists through one radian when equal and opposite torques of 145 × 10–7

Nm are applied at its ends. Find the value of Poisson's ratio of gold.Solution :Young's modulus

q = 323 10)1016.0(

8.933.0−− ×××π

××=

L

l

L

A

F


The couple for unit twist is

C = lr

2

4πη

η = 43

7

4 )1016.0(

2101452−

−

×π×××=

π× L

r

LC

Poisson's ratio, σ = 12

−ηq

= 122101452

)1016.0(

)1016.0(

108.933.07

43

23−

×××××π

×××π

×××−

−

−

L

429.0=σ Ans.

3. In an experiment to determine Young's modulus of a rod of diameter 1.26 × 10–2 m it wassupported on knife edges placed 0.7 m apart. On applying a load of 0.9 kg exactly mid waybetween the knife edges, the depression of midpoint was observed to be 0.25 cm. Calculateq for material of the rod.Solution :The depression at the midpoint

δ = 4

3

12 rq

Wl

π

q = 422

3

4

3

)1063.0(1025.012

7.08.99.0

12 −− ××××π×××

=δπr

Mgl

q = 842

3

1063.01025.012

7.08.99.0−− ××××π×

××

26 N/m1009.8 ×=q Ans.

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