04) [PDF]

5.Elasticity Q.1 Define the following terms. 1) Deformation. 2) Deforming force 3) Internal Restoring force

2)

Ans:Deformation: Change in shape or size or both of the body when external force applied on it, is called deformation of the body. Deforming force: A force which produces deformation of in the body is called deforming force. Internal Restoring force: An internal force which tries to bring the body to the original dimension (shape or size) when the deforming force is removed is called internal restoring force.

3)

4)

Q.2 Define the terns:- Elasticity & Elastic Body. Give examples of elastic body. Ans: Elasticity:- Elasticity is the property of a body due to which a body opposes change in its shape or size or both when deforming force acts on it and it regains to the original dimension when the deforming force is removed. Elastic Body:- A material body which deforms under the action of external deforming force and it regains the original dimension when the deforming force is removed, is called as elastic body. e.g. 1) All thin metal wires, 2) Rubber ball. 3) Rubber balloon. 4) Foot ball bladder. Q.3.Define the terms plasticity & plastic Body. Give examples of plastic body. Ans: Plasticity:- Plasticity is the property of the body due to which a body easily deforms when the deforming force acts on it , but it does not regain to the original dimension when the deforming force is removed. Plastic Body:- A material body which easily deforms under the action of external deforming force, but does not regain to its original dimension. When the deforming force is removes, is called as plastic body. e.g. 1) Clay, 2) Wax, 3) Mud, 4) Plasticine Q.4 Distinguish between Elastic Body and plastic Body. Ans: Sr. Elastic Body Sr. Plastic Body No No. . 1) A body regains 1) The body does not to original regain to original shape and size shape and size

5)

6)

(Marks: 03/04)

when deforming force is removed. Internal restoring force develops inside the body Large force is required for deformation of the body Deformation of the body is not permanent. Body opposes deformation. Intermolecular force is large. e.g. rubber, gold, copper

when deforming force is removed . 2)

3)

4)

5)

6)

Internal restoring force does not develop inside the body. Small force is required for deformation of the body Deformation of the body is permanent. Body does not oppose deformation. Intermolecular force is small/less. e.g. Clay, wax, mud.

Q.5 Define the term stress. Give its S.I. unit & dimension. Ans: Stress:- The internal restoring force per unit area is called stress. 𝐹 ∴ Stress = 𝐴 Units and Dimensions: S. I. unit:- N/m2 CGS unit:- dyne/cm2 Dimension:- [M1 L-1 T-2] When the body is in equilibrium position, the magnitude of internal restoring force is same as that of the applied force. ∴ Stress = = Q.6 Ans:

Internal Restoring Force Area Externally Applied Force Area

State and Explain different types of stress. There are three types of stress:

1)Longitudinal stress/Tensile stress. 2)Volume stress/Bulk stress. 3)Tangential stress/ Shearing stress. 1)Longitudinal stress (Tensile stress):If the deforming force produces change in length of the body, the stress developed in the body is called longitudinal stress or tensile stress. OR 1

The ratio of normal applied force to the area of cross-section of the body is called as longitudinal stress. Explanation:-

Volume stress = Total

applied force surface area of body

𝐹

=𝐴 = Change in pressure = dp 3)Shearing stress (or Tangential Stress): If the tangentially applied force produces change in shape or shearing of the body, the stress produced is called as shearing stress. OR

The ratio of tangentially applied force to the surface area (parallel to force) is called as shearing stress. Tangentially applied force ∴ Shearing stress = Surface area

If the wire of length L is suspended from the rigid support and a load M is attached to the force end of the wire. Then the force ‘Mg’ acts on the wire. Let ‘l’ is the elongation in the wire and A is suppose area of cross-section of the wire. But, A = 𝜋r2 Where, r = radius of the wire. ∴ Longitudinal stress= area 𝐹

F

=A Q.7 Distinguish between Deforming force and stress. Ans: Sr. Deforming Sr. Stress. No force No. 1) It is an 1) It is an internal externally restoring force applied force. per unit area. 2) It tries to 2) It opposes the deform the deformation. body. 3) S. I. unit: 3) S. I. unit :N/m2 newton (N) 4) Dimension: 4) Dimension: 1 1 -2 [M L T ] [M1L-1T-2] 5) It is a vector 5) It is a tenser quantity. quantity.

applied force of cross −section of wire 𝑀𝑔

= 𝐴 = 𝜋𝑟 2 3)Volume stress (orBulk Stress):

If the deforming force produces change in volume of the body, then the stress developed in that body is called as volume stress. OR Ratio of normal applied force to the total surface area of the body is called as volume stress. Explanation:-

Q.8 Define the term strain. Give its unit and dimensions. Ans: Strain: The ratio of change in dimension to the original dimension of the body is called as strain. Change in dimension ∴ Strain = Original dimension .

As strain is the ratio of two similar quantities , hence it has no unit and no dimension. The change in volume is produced by increasing or decreasing pressure on the body. Let, ‘dp’ = Change in pressure.

Q.9.State and Explain different types of strain: Ans: There are three types of strain: 1)Longitudinal strain (Tensile strain) 2)Volume strain (Bulk strain). 3)Tangential strain (Shearing strain). 2

1)Longitudinal strain(Tensile strain):The ratio of change in length to the original length of the body is called as longitudinal strain.

∴ Longitudinal strain =

Change in length Original length

Explanation: Consider the rectangular block ABCD. The tangential force F is applied to the layer AB while the layer CD is kept fixed. Let AA is the lateral displacement (x) of the layer AB. ∴ Shearing strain = =

AA 1 AD

=

𝑥 𝐿

= tan 𝜃

∴ Shearing strain = 𝜃 Where, 𝜃 = angle through which upper layer is displaced = angle of shear.

2)Volume strain (Bulk Strain):The ratio of change in volume to the original volume of the body is called volume strain.

Q.10.What is elastic limit? Ans: Elastic Limit: The maximum stress to which a body can be subjected without permanent deformation is called elastic limit of that body. Explanation: When the stress is within the elastic limit, the body is perfectly elastic and it regain to the original dimension when the applied force is removed. When stress crosses the elastic limit, the body deforms permanently up to certain extent.

Change in volume Original volume

Explanation:-

Q.11.State the Hooke’s Law. Define the modulus of elasticity. State S.I. unit, CGS unit and dimension of modulus of elasticity. Ans:Hooke’s Law: Within the elastic limit, stress is directly proportional to the strain. i.e. stress ∝ strain stress = (constant of proportionality)× Strain

Let V be original volume of the spherical body, dV be change in volume (decrease in volume) of the body. dv ∴ Volume strain = V 3)Shearing strain (Tangential strain):The ratio of lateral displacement of any layer to its distance from the fixed layer is called shearing strain.

∴ Shearing strain =

Its distance from fixed layer

If 𝜃 is very small then, tan 𝜃 ≈ 𝜃

Let L be the original length of the wire, l be the change (increase) in length of the wire due to the force F. 𝑙 ∴ Longitudinal strain = 𝐿

Volume strain =

lateral displacement of a layer

This constant of proportionality is called modulus of elasticity. Modulus of elasticity is also called as elastic constant or coefficient of elasticity. Modulus of Elasticity:

lateral displacement of any layer Its distance from fixed layer

Explanation:3

Within the elastic limit, the ratio of stress to the strain is called modulus of elasticity. ∴ modulus of elasticity =

stress strain

Units and Dimensions: SI unit: N/m2 CGS unit : dyne/cm2 Dimension: [M1L-1T-2]

Q.12. State and Define different types of Modulus of Elasticity. Give their units and dimension. Ans: There are three types of modulus of elasticity:

Consider a uniform wire of length L and area of cross-section A suspended from a rigid support. Let M be the mass applied to the free end of the wire. Due to this, suppose the length of the wire increases by length ‘l’.

1)Young’s modulus (Y) 2)Bulk modulus (K) 3)Modulus of Rigidity (η) 1)Young’s modulus (Y): Young’s Modulus is defined as the ratio of longitudinal stress to the longitudinal strain. ∴ Young’s Modulus(Y) =

Longitudinal stress = =

longitudinal stress longitudinal strain

Units and Dimensions: SI unit: N/m2 CGS unit: dyne/cm2 Dimension: [M1L-1T-2]

Normal applied force area of cross section F A Mg

= πr 2 And change in length Longitudinal strain = original length

2)Bulk Modulus (K): Bulk modulus is defined as the ratio of volume stress to the volume strain. volume stress ∴ Bulk Modulus(K) = volume strain

𝑙

= L But, young’s modulus of the material of the wire is given by longitudinal stress Y = longitudinal strain

Units and Dimensions: SI unit: N/m2 CGS unit: dyne/cm2 1 -1 -2 Dimension: [M L T ]

=

3)Modulus of Rigidity (η) Modulus of rigidity is defined as the ratio of tangential stress to the tangential strain.

Mg /πr 2 𝑙/L 𝐌𝐠 𝐋

Y = 𝜋 𝐫𝟐𝒍

𝐭𝐚𝐧𝐠𝐞𝐧𝐭𝐢𝐚𝐥 𝐬𝐭𝐫𝐞𝐬𝐬

Q.14.Derive an expression for the Bulk Modulus for the material of a body. Ans:

∴ Modulus of rigidity (η) = 𝐭𝐚𝐧𝐠𝐞𝐧𝐭𝐢𝐚𝐥 𝐬𝐭𝐫𝐚𝐢𝐧

Units and Dimensions: SI unit: N/m2 CGS unit: dyne/cm2 Dimension: [M1L-1T-2] Q.13 Derive an expression for the young’s modulus of material of the wire. Ans:

Consider a spherical body of original volume V. when this body is subjected to the normal force F, there is change in its volume i.e. dV. 4

Let, AA1 is the lateral displacement of the upper layer through the small angle 𝜃.

Let A surface area of the body Volume stress =

normal applied force surface area

∴ Shearing strain =

𝐹

=𝐴

=

lateral displacenent of a layer Its distance from fixed layer 𝐴𝐴1

Generally 𝜃 is very small

= Applied pressure (change in pressure)

𝐴𝐷

= tan 𝜃

∴ tan 𝜃 ≈ 𝜃 = dp

∴ Shearing strain = 𝜃.

Now, Volume strain = =

change in volume

Therefore, the modulus of rigidity is given by

original volume

Tangential stress

dV

Modulus of rigidity(η) = Tangential

V

But, the bulk modulus is given by

𝑭/𝑨

∴ η = 𝒙/𝒉

volume stress

Bulk modulus(K) = volume

strain

strain

𝑭

𝑭𝒉

∴ η = 𝑨𝜽= 𝑨𝒙

dp

K = dv /v K=

Q.16.What is Poisson’s Ratio? Obtain its expression. Give its units & dimensions. Ans: Poisson’s Ratio: Within the elastic limit, the ratio of the lateral strain to the longitudinal strain is called as Poisson’s Ratio.

𝐕 𝐝𝐩 𝐝𝐕

Q.15.Derive an expression for Modulus of Rigidity of the material of the body. Ans:

Lateral strain

∴ Poisson’s ratio(𝜎) = Longitudinal

strain

Expression: Consider a wire of original length L suspended from a rigid support. Apply the force F to its free end. Its length increases by (l) in the direction of force. The strain produced in the direction of force is called as longitudinal strain. Therefore,

Consider a rectangular block ABCD. Let a tangential force F is applied to the upper layer AB while the lower layer CD is kept fixed. Due to the force F, the layer AB is displaced laterally in the direction of force. The new position of the block is A1B1CD. Let A is the area of the upper layer Tangentially applied ∴ Tangential stress = surface area

Longitudinal strain =

increases in length original length 𝒍

∴ longitudinal strain = 𝑳 When length of the wire increases, its diameter decreases in the perpendicular direction of the applied force. The strain produced in the direction perpendicular to that of force is called as lateral strain. Let, D = original diameters of the wire d = decrease in diameter of the wire

force

𝐹

=𝐴

5

∴ Lateral strain =

2) The diameter of the wire (D) is measured at different points along its length with the help of micrometer screw gauge. From these reading, the mean radius (r) of the wire is calculated. 3) When the weight is loaded in the pan attached to the experimental wire then it elongates. The elongation produced in the wire is measured by vernier V. The reference wire is used to eliminate the error due to the change in the room temperature. Because of temperature variation, length of both wires increases or decreased. 3) Both the wire are loaded in order to make wires straight and taut and vernier reading is measured. Now the load applied to the experimental wire increasing the weight. Every time the vernier scale is used to take readings. The difference between the two vernier readings gives the elongation produced in the experimental wire for the given load. Formula: Let , r= radius of the wire. L= length of the wire. A=πr 2 = area of cross section of the wire M= mass attached to the experimental wire. l = change in length of the wire. Therefore the young’s modulus is given by

decrease in diameter original diameter 𝑑

=𝐷 The Poisson’s Ratio is given by lateral strain

Poisson’s Ratio (𝜎) = longitudinal

strain

d/D

= 𝑙 /L dL

= D𝑙 It has no unit and no dimension. Q.17 Determination of the young’s modulus of the wire. Ans: Young’s modulus of the material of the wire:Following figure shows the typical arrangement for the measurement of the young’s modulus of the material of the wire.

Y=

Mg L πr 2 𝑙

In this way we obtain the young’s modulus of the material of the wire. Q.19 Explain the behavior of the wire under the increasing load using stress-strain graph. Ans: Behavior of the wire under the increasing load:-

Two long straight thin wires P and Q of same length , radius and material (usually steel) are suspended from fixed rigid support. Wire P is the reference wire is kept taut by the weight W attached to its lower end and carries a scale M graduated in milliammeter. The wire Q is the experimental wire to which the pan in which the weights can be placed. Vernier scale V is attached at the bottom of experimental wire , whereas main scale M is fixed to reference wire. Procedure (working):

E: Elastic limit OP: Permanent set Y: Yield point B: Breaking stress B1: Breaking point The behavior of the wire under increasing load can be strained by plotting a graph of stress against strain. 1)Elastic limit (Limit of Proportionality): Initially the graph is straight line upto the point E. It shows that up to the point E, stress is directly proportional to strain. Hence, Hooke’s law is obeyed. The

1) The original length L of the experimental wire is measured by the meter scale. 6

wire is perfectly elastic and it completely regains its original length when the load is removed. The point E is called elastic limit. 2)Permanent set (OP): Just beyond the elastic limit (E), up to the point E’ shows that for a small increase in stress, the strain increases more rapidly than indicated by Hook’s law and the graph bends towards the strain axis. If the stress is removed, then the new straight line graph (E’P) is obtained. This shows that the wire will still regain its elastic properties, but the strain has a value OP even when the stress becomes zero. This permanent strain of is called permanent set(or set). 3)Yield point (Y): If the load increased further, then the graph reaches to point Y at which a tangent drawn to the curve becomes parallel to the strain axis. This shows that strain goes on increasing without any increase in the stress. At this yield point, the wire is said to have plastic flow. The behavior of the wire between points E and Y is partly elastic and partly plastic. Thus, the yield point (Y) is defined as the point at which the strain in the wire begins to increase even without any increase in the stress.

The substance which breaks just after the elastic limit is called as brittle substance. If the material breaks soon after the elastic limit crossed the material is said to be brittle. e.g. glass, carbon etc. Elastomers:The substances such as tissue of aorta or rubber , etc. which can be subjected to cause large strains are called elastomers. Following graph shows the stress- strain curve for the elastomers such as tissue of aorta. It is see on from the curve that although the elastic region is large ,the elastomers do not obey the Hooke’s law. Relation between Young’s modulus , coeffiecient of linear expansion , thermal stress:Consider a metallic rod of length L0 at 0℃. If the temperature of rod is increased by ∆𝜃,then the final length of the rod will be L = L0 (1 + α∆θ) ---------------(1) Where ,𝛼 is the coefficient of linear expansion of material of the rod. We know that longitudinal strain is given by 𝐶ℎ𝑎𝑛𝑔𝑒 𝑖𝑛 𝑙𝑒𝑛𝑔𝑡 ℎ Longitudinal strain = 𝑂𝑟𝑖𝑔𝑖𝑛𝑎𝑙 𝑙𝑒𝑛𝑔𝑡 ℎ

4)Breaking stress (B): If the stress is further increased, the diameter of wire decreases uniformly. At point B, a neck or constriction begins to form and this is the maximum stress beared by the wire. Thus, the breaking stress is maximum stress which can be applied to a wire without breaking. 5)Breaking point (B’): Due to the formation of neck or constriction in the wire, the wire becomes too weak and the elongation of the wire continues even though the stress is decreasing due to which the wire breaks at a certain lower stress than that of breaking stress, denoted by point B’, is called breaking point. Thus, the point on stress strain curve at which the wire breaks is called breaking point.

𝐹𝑖𝑛𝑎𝑙 𝑙𝑒𝑛𝑔𝑡 ℎ−𝐼𝑛𝑖𝑡𝑖𝑎𝑙 𝑙𝑒𝑛𝑔𝑡 ℎ

=

𝐼𝑛𝑖𝑡𝑖𝑎𝑙 𝑙𝑒𝑛𝑔𝑡 ℎ 𝐿−𝐿

0 = ----------(2) 𝐿 But from the equation (1) , we have, 𝐿 = 𝐿0 + 𝛼∆𝜃𝐿0

𝐿 − 𝐿0 = 𝛼∆𝜃𝐿0 𝐿−𝐿0 𝐿0

= 𝛼∆𝜃

Therefore above equation (2) becomes, Longitudinal strain(Thermal strain)= 𝛼∆𝜃 We know that

Q. Define 1)Ductile substance and 2) Brittle substance. Give their examples. Ans: Ductile substance: The substance which sufficiently elongates on the application of the force before it breaks is known as ductile substance. If the plastic region of from point E’ to point B in the above graph , the material is said to be ductile. e.g. lead, gold, copper and iron etc. 2)Brittle substance:

𝐿𝑜𝑛𝑔𝑖𝑡𝑢𝑖𝑑𝑖𝑎𝑙 𝑠𝑡𝑟𝑒𝑠𝑠

Young’s modulus =𝑙𝑜𝑛𝑔𝑖𝑡𝑢𝑑𝑖𝑛𝑎𝑙 Here we can write 𝑇ℎ𝑒𝑟𝑚𝑎𝑙 Young’s modulus =𝑇ℎ𝑒𝑟𝑚𝑎𝑙 Y=

𝑠𝑡𝑟𝑎𝑖𝑛

𝑠𝑡𝑟𝑒𝑠𝑠 𝑠𝑡𝑟𝑎𝑖𝑛

𝑇ℎ𝑒𝑟𝑚𝑎𝑙 𝑠𝑡𝑟𝑒𝑠𝑠 𝛼∆𝜃

𝑇ℎ𝑒𝑟𝑚𝑎𝑙 𝑠𝑡𝑟𝑒𝑠𝑠 = Y𝛼∆𝜃 7

𝐹 𝐴

advantageous to increase its depth rather than the breadth of the beam. Consider a beam of length l and of rectangular cross-section having breadth b and depth d. When the beam is loaded in the middle with a load W as shown in the figure below. ,the it get depressed by an amount 𝛿 (also called sag)is given by

= Y𝛼∆𝜃

Therefore the force exerted by the rod on heating is F= 𝐘𝐀𝜶∆𝜽 Application of the Elastic Behavior of materials:-

𝑊𝑙 3

𝛿 = 4𝑌𝑏𝑑 3

1)Any metallic part of a machinery is never subjected to a stress beyond the elastic limit of the material . In case , the metallic part of the machinery is subjected to a stress beyond the elastic limit , it will get permanently deformed and hamper its working. 2) The thickness of metallic ropes used in cranes to lift and move heavy weights is decided on the basis of the elastic limit of the material of the rope an the factor of safety. In order to explain the above , let us consider that in a crane , the rope made of steel is used to lift load upto 10 metric ton i.e. 104 kg. We may not attach a load more than 104 kg but usually the rope is designed for a factor of safety of 10 i.e. it should not break even when a load of 104 × 10 i.e. 105 kg or a force of 105 × 9.8N is applied to it. If r is the radius of the rope , then 𝑀𝑔

From the above equation it is clear that 1)sag(𝛿) is inversely proportional to 𝑑 3 . 2)sag(𝛿) is inversely proportional to 𝑏. 3)sag(𝛿) is inversely proportional to 𝑌. 4)sag(𝛿) is directly proportional to 𝑙 3 . Since the sag (𝛿) is inversely proportional to Y, the young’s modulus of the material should be large and the depression can be reduced more effectively by increasing the depth of the beam than increasing its breadth. If a load is placed exactly at the centre of the beam and the depth of the bow is increased , then beam will bend as shown in the above figure. This is called buckling. To avoid the buckling , the cross sectional shape of the beam or bar is resemble letter I. The I beam minimizes both the weight and stress. An I beam can be much lighter , but almost as strong as for bending . I beam has a large load bearing surface , which prevents buckling and is enough to reduce too much buckling . I beam provides high bending moment and lot of material is saved. A hollow circular Pole(shaft) is stronger than a solid shaft made of same and equal material. Q. Define strain Energy(Elastic energy). Derive the expression strain energy. Obtain the expression for strain energy density (strain energy per unit volume). Ans: Strain Energy: The work done in stretching wire by applied force is stored in the form of potential energy. This potential energy is called strain energy. Expression: Consider a uniform wire of length L and area of cross-section A. One end of the wire is fixed to a rigid support and the other end is subjected to the force F. Elongation produced in the wire is suppose

10 5 ×9.8

Ultimate strength = 𝐴 = 𝜋𝑟 2 The ultimate strength should not increase the breaking stress or even the elastic limit for steel. The elastic limit for steel is 30 × 107 N/m2. Therefore 10 5 ×9.8 𝜋𝑟 2

= 30 × 107

10 5 ×9.8 𝜋×30 ×10 7

= 𝑟2

0.00104 = 𝑟 2 ∴ r =0.0332 m= 3.32 cm In practice , to impart flexibility to the rope , it is made of large number of thin wires braided together. 3) The bridges are declared unsafe after long use. During its long use , a bridge undergoes quick alternating strains continually. It results in the loss of elastic strength of the bridge. After a long period , such a bridge starts developing large strains corresponding to the same usual value of stress and ultimately it may lead to the collapse of the bridge. So that it does not happen , the bridges are declared unsafe after their long use. 4) In designing a beam for its use to support a load (in construction of roofs and bridges) , it is 8

1

‘l’. During elongation, let x be the extension produced in the wire by the application of the force f . f Therefore, longitudinal stress = A

W = 2 Fl This work done is called as the strain energy. 1 ∴ strain energy = 2 Fl Expression for strain energy per unit volume OR strain energy density OR work done per unit volume: Let, A = Area of cross-section of wire L= Length of wire A×L = Volume of the wire Now, Strain energy per unit volume/strain energy strain energy density/work done Per unit volume = volume

𝑥

And longitudinal strain

=L longitudinal stress

∴ Young’s modulus (Y) = longitudinal =

strain

f/A 𝑥/L

fL

∴ Y = A𝑥 YA 𝑥

∴ f= ------------ (1) L If the wire is stretched further through a distance ‘dx’, then the work done is given by

= 1

We know that, 𝐬𝐭𝐫𝐞𝐬𝐬 Y = 𝐬𝐭𝐫𝐚𝐢𝐧 ∴ stress = Y × strain

d𝑥

L

1

If the total extension is ‘l’ in the wire when the applied force is F, then the total work (W) in stretching the wire from 0 to l is given by W=

∴ strain energy density = 2 Y (strain) (strain) 1

= Y (strain)2 ----------- (3) 2

𝑙 𝑑𝑤 0

Also, Strain =

= =

=

=

𝑙 YA 𝑥 0 B

𝑑𝑥

L

𝑙 0

YA

𝑥2

𝑙

L

2

0

YA

𝑙2

L

2

YA

1 YA 𝑙

=2

L

𝐘

1

∴ strain energy per unit volume = 2 (stress). 1 (𝐒𝐭𝐫𝐞𝐬𝐬)𝟐

= 2.

Relation Between Elastic constants: 9 3 1 1) 𝑌 = 𝜂 + 𝐾 𝑌

2) 𝜎 = 2𝜂 − 1 𝑌

3) K = 3(1−26)

------------ (2)

But, for fully stretched wire, x = l and f = F ∴ equation (1) becomes, YA 𝑙 F= L ∴

𝐒𝐭𝐫𝐞𝐬𝐬

𝑥. 𝑑𝑤

𝑙

1 𝐹 𝑙

= 2 × stress × strain ----------- (2)

dW = F. d𝑥 YA 𝑥

𝐴𝐿

∴ Strain energy density = 2 . 𝐴 . 𝐿

Work done = Force × displacement

=

1 𝐹𝑙 2

Problems:-

Type I

Equation (2) becomes. 9

𝐘

𝐒𝐭𝐫𝐞𝐬𝐬 𝐘

1. A wire is one metre long and 1 mm in diameter. When it is stretched by a weight of 4 kg, its length increases by 0.25 mm. Find the values of stress and strain produced in the wire. (Ans. 4.99 x 107 N/m2 ; 2.5 x 10-4) 2. The area of the upper face of a rectangular block is 0.25 m2 and the lower face is fixed. The height of the block is 1 cm. A shearing force applied at the top face produces a displacement of 0.015 mm. Find the strain, stress and the shearing force. Given modulus of rigidity = 4.5 x 1010 N/m2. (Ans. 1.5 x10-3; 6.7 x 107 N/m2, 1.69 x 107 N) 3. What would be the greatest length of steel wire which when fixed at one end can hang free without breaking? (Density of steel = 7800 kg/m3, Breaking stress for steel = 7.8 x 108 N/m2) ( Ans. 1.02 x 104 m).

each of the same magnitude. Compare the extensions produced if the radius of the first wire is twice that of the second wire.(Ans. l1 : l2= 1:4) 10. There are two wires of the same material. Their radii and lengths are both in the ratio 1 :2 If the extensions produced are equal, what is the ratio of loads? (Ans. 1 : 2) 11. Two wires of the same material are subjected to the same tension. Compare the extensions produced if the length and radius of first wire are twice as large as those of the second wire.(Ans.1 : 2) 12. Two wires have lengths in the ratio 3 : 2, diameters in the ratio 2 : 1 and Young’s moduli in the ratio 1 : 2. Calculate the rato of the elongations produced in the wires when they are subjected to the same stretching force.(Ans.3 : 4) 13. Two wires P and Q are of the same diameter having lengths in the ratio 5 : 3 When the wires are subjected to the same load, the corresponding extensions are in the ratio 4:3 Compare Y of the material of P and Q. (Ans. 5 : 4) 14. Two wires have lengths in the ratio 4:3, diameters in the ratio 2:1 and Young’s modulus in the ratio 1:2. Calculate the ratio of elongations produced in the wires when they are subjected to the same stretching force. (Ans. 2 : 3) 15. When a load of 3 kg -wt is applied to the end of wire of length 1.5 m, the extension produced in the wire is 3mm.If the diameter of the wire is 0.3 mm, calculate the Young’s modulus of the material of the wire. (Ans. 2.07 x 1011 N/m2) 16. A volume of 5 x 10-3 m3 of water is compressed by a pressure of 20 atmosphere. If bulk modulus of water is 2 x 109 N/m2, find the change produced in the volume of water( 1 atm = 1.013 x 105 N/m2)(Ans. dV = 5.065 x 10-6 m2).

Type II 5. A wire of length 4m and area of cross-section 2mm2 is extended by 2mm when a stretching force of 80N is applied to it. Calculate the tensile stress, tensile strain and Young’s modulus of material of the wire. (Ans. 4 x 107N/m2, 5x10-4, 8x1010N/m2) 6. The volume of 2 litres of a liquid changes by 1.2 cm3 when it is subjected to a change of pressure of 106 N/m2. Determine the bulk modulus of the liquid.(Ans. 1.66 x109 N/m2). 8. A uniform brass wire and a uniform steel wire, each of length 3.14 m and diameter 2 x 10-3 are joined at one end to form a composite wire 6.28 m long. The composite wire is loaded until the extension length is 6 x 10-3 m. Calculate the extension of the brass and steel wires and the force applied to the wire (Young’s modulus for brass = 10 x 1010 N/m2 and for steel = 20 x 1010 N/m2) (Ans. 4 x 10-3m, 2 x 10-3m, 400 N). 9.Two wires of the same material and of the same length are stretched by longitudinal forces 10

17. What pressure should be applied to a lead block to reduce its volume by 10%.The bulk modulus of lead =6x109 N/m2 (Ans.6 x 108N/m2) 18. A wire extends by 1.5 mm when subjected to a certain load. Another wire of the same material is subjected to a load which is 50% greater than that applied to the first wire. If the length of the second than that first wire by 20% and its radius is greater than that of the first wire by 40%, find the extension produced in the second wire. (Ans. 1.378 mm) 19. A steel wire of diameter 0.4mm is heated to 350 0C and then rigidly clamped at its ends when hot. Find the pull exerted on the clamps when it cools to 300C. Coefficient of linear expansion os steel is 2 x 10-5per 0C and Young’s modulus of steel is 2 x 1011N/m2. (Ans. 160.8N) 20. An object of mass 0.2kg is fastened to one end of a rubber cord of radius 5mm and revolved with uniform speed in a horizontal circle around the other end of the cord, which is kept fixed. If the object performs 180 revolutions per minute and the radius of the circle is 20cm,determine the unstretched length of the cord.(Ans.19.93 cm) 21. Calculate the shearing strain produced in a block of copper when it is subjected to a shearing stress of 2.4 x 108 N/m2 Modulus of rigidity of copper is 4.8 x1010 N/m2 (Ans 0.005) 22. A 5cm cube of a substance has its upper surface displaced by 0.65cm by a tangential force of 0.25 N .Calculate the modulus of rigidity of the substance. (769.2 N/m2) 23. Prove that the stress needed to double the length of a wire is equal to its Young’s modulus. Assume that Hooke’s law is obeyed during the increase of length. 24. If a weight of 1kg stretches a wire, of diameter 1mm, by 0.5mm, how far will a wire of the same material and the same length but of

twice the diameter be stretched by a weight of 8 kg? (Ans. 1 mm) 25. A brass wire of length 5m and cross-section 1mm2 is hung from a rigid support with a brass weight of volume 1000c.c. hanging from the other end. Find the decrease in length of the wire when brass weight is completely immersed in water.( Y=1011 N/m2 and g = 9.8 m/s2) (Ans. 0.5 mm) 26. The elastic limit of copper is 1.5 x 104 N/m2. Find the minimum radius a copper wire must have, if its elastic limit is not to be exceeded under a load of 10 kg. (Ans. 0.04561 m) 27. A steel wire of length 7m and cross-section 1mm2 is hung from a rigid support with a steel weight of volume 1000 cc. hanging from the other end. Find the decrease in length of wire when steel weight is completely immersed in water.( Y=2 x1011 N/m2) (Ans. 3.43 x 10-4 m) 28. The length of a wire is 1.2m when stretched by a weight of 4kg and 1.3m when stretched by a weight of 6kg. What is the unstretched length of the wire? (Ans. 1 m) 29. What must be the elongation of a wire 5m long so that the strain is 1% of 0.1? If the wire has a cross-section of 1mm2 and is stretched by 10kgwt, what is the stress? (Ans. 5 mm; 9.8 x 107 N/m2) 30. Metal rod of coefficient of linear expansion 1.6 x 10-5 / °C has it’s temperature raised from 80°C to 100°C. Calculate compressive stress required to prevent the expansion of the rod. ( Y=2 x1011 N/m2) (Ans. 6.4 x 107 N/m2) 31. One cm3 of water is taken from surface to the bottom of lake 200 m deep. The bulk modulus of water is 2.2x 104 atmosphere and density of

11

38. For a given material, theYoung’s modulus is 2.4 times that of rigidity modulus. Calculate its Poisson’s ratio.Hint : Y = 2(1 + σ)(σ = 0.2) 39. Calculate the Poisson’s ratio for silver given its Young’s modulus is 7.25 x 1010 N/m2 and Bulk modulus is 11 x 1010 N/m2.(Ans. 0.39) 40. A wire of length 2m and diameter 2cm is suspended from a rigid support with a weight of 1000kg hanging from the free end. Find the lateral strain in the wire if Poisson’s ratio =0.2 Y=18x 1010N/m2. (3.47 x 10-3) 41. For steel, Y=20 x 1010N/m2 and bulk modulus is 13.3 x 1010N/m2. Calculate the Poisson’s ratio modulus or rigidity for steel.(0.25, 8x1010N/m2) 42. Calculate Poisson’s ratio for silver from the following data : Y=7.25 1010N/m2; k=11 x 1010N/m2. (Ans. 0.39)

water is 103 kg/m3. Calculate the change in volume (1 atm = 105 N/m2).(Ans. 8.9 x 10-4 cm3) 31. If young’s modulus of material of wire is 1.2 x 1011 N/m2. Calculate workdone in stretching the more of length 3m and cross sectional area 4 mm2.When it is suspended vertically and a load 8 kg is attached to it’s lower end.(Ans.1.92x10-2 J) 32. Steel wire has diameter 6 mm and Young’s modulus 1.2 x 1011 N/m2. What mass must be attached to free end of wire by keeping other end fixed to produced elongation 0.5% of it’s original length?(Ans. 17.3 km) 33. The extensions produced in the wires of length 4.7 m, having cross sectional area 0.3 cm2 and another wire of length 3.5 m having cross sectional area 0.4 cm2 under same load. compare their young’s moduli. (Ans. 1.79)

Type III

Type IV 43. The wire of length 3 m and having cross sectional area 4 mm2 is subjected to load of 8 kg wt. Calculate workdone during stretching wire. (Y = 2 x 1011 N/m2) (Ans. 4.61 x 10-2 J) 44. Calculate the strain energy of a steel wire of length 5 metre and radius 2 mm, when it is loaded by 10 kg-wt. (Given, Y for steel = 2.5 x 1011 N/m2) (Ans. 6.172 x 10-3 J) 45. A metal wire of length 2.5m and area of cross section 1.5 x 10-6 m2 is stretched through 2mm. Calculate the work done during the stretching. (Y = 1.25 x 1011 N/m2) (Ans. 0.15J) 46. A wire 2.5m long and of diameter 0.8mm is stretched by a load of 10N. Find the work done in stretching the wire if Young’s modulus of the material of the wire is 2 x 1011N/m2. (Ans. 1.244 x 10-3J) 47. A copper wire is stretched by 0.5 % of its length Calculate the energy stored per unit of its volume (Y = 12 x 1011 N/m2) (1.5 X 106 J/m3)

34. Find the longitudinal stress to be applied to the wire to decrease its diameter uniformly by 10%. (Poisson’s ratio = 0.25 and Young’s modulus = 2 x 1011N/m2.) (Ans. 8 x 1010N/m2) 35. A copper wire of length 4 m and diameter 2 mm is stretched by a weight of 8kg. Determine the longitudinal extension and lateral contraction produced, if Young’s modulus of copper is 1.25 x 1011N/m2 and Poisson’s ratio for copper is 0.25. (Ans. 8 x 10-4m, 5 x 10-8m) 36. A 3 cm long copper wire is stretched to increase its length by 0.3cm. Find the lateral strain produced in the wire, if the Poisson’s ratio for copper is 0.26. (Ans. 2.6 x 10-4) 37. The Young’s modulus of the material of a wire is 6 x 1012 N/m2 and there is no transverse strain in it, calculate its modulus of rigidity. Hint : Y = 2(1 + σ) (Ans. 3 x 1012 N/m2).

12

48. The strain energy per unit volume of a stretched wire is 2.5x 10-5 J/m3.Find (1) the stress (2) the strain in the wire if Y= 2x 1010 N/m2 (1)103N/m2 (2)5x10-8) 49. A wire of metal whose Young’s modulus is 1.2 x 1011 N/m2 is stretched by a force of 6kg wt. If the length of the wire is 2.4m and area of cross-section is 3mm2, find the elongation and the work done per unit volume in stretching the wire. (Ans. 0.392mm, 1601J/m3) 50. A wire of uniform cross-sectional area of 1mm2 and length 2m is stretched through 1mm by a weight of 2kg. Calculate the energy stored per unit volume of the wire.(Ans. 4900 J/m3) 51. An aluminum wire is stretched by 0.5% of its length. If Young’s modulus of aluminum is 7x 1010N/m2, calculate the work done per unit volume in stretching the wire.(Ans.8.75x 105J/m3) 52. A steel wire of length 5 m is pulled to have extension 1 mm.It’s young’s modulus is 1.9 x 104 N/m2. Determine strain energy per unit volume. (Ans. 3.8 x 10-4 J/m3)

values of stress and strain produced in the wire. (Ans. 4.99 x 107 N/m2 ; 2.5 x 10-4) 57. A wire of length 1.5 metre and radius 0.4 mm is stretched by 1.2 mm on loading. If the Young’s modulus of its material is 12.5 x 1010 N/m2, find the stretching force.(Ans. 50.24 N) 58. An aluminium wire and a steel wire of the same length and diameter are joined end to end. The composite wire is hung from a rigid support and a load is suspended from the free end. If the increase in length of the composite wire is 2.7 mm, find the increase in length of each wire.(Y for steel = 2 x 1011 N/m2 ,Y for aluminium = 7 x 1011 N/m2) (Ans. 2mm, 0.7 mm) 59. A uniform wire of length 4 m and diameter 2 x 10-3 m has a mass of 15.7 kg hung on it. If the extension in the length is 10-3m, calculate (a) stress (b) strain (c) Young’s modulus of elasticity. (Ans. (a) 49 x 106 N/m2, (b) 2.5 x 10-4, (c) 19.6 x 1010 N/m2) 60. An extension of 0.3 cm is produced when a 900 gm weight is hung on a rod of length 5 m and area of cross section 10.179 mm2. Find the stress, strain and Young’s modulus for the material of the rod.(Ans. 8.665 x 105 Nm-2, 6.0 x 10-4, 1.444 x 109 Nm-2) 61. A wire 5 m long and 10-3 in diameter, when stretched by a weight of 3 kg has its length increased by 0.9 mm. Find stress, strain and Young’s modulus of the material of the wire. (Ans. 3.743 x 107 N/m2,1.8 x 10-4,2.08 x 1011 N/m2) 62. A load of 2 kg-wt is applied to the end of a wire of length 2 m and diameter 2 x 10-4 m. If Y of the material of the wire is 2 x1011 N/m2, find the extension produced in the wire. (g = 9.8 m/s2) (Ans. 6.24 mm). 63. A volume of 10-3 m3 of water is subjected to a pressure of 10 atmosphere. The change in volume is 10-6 m3. Find the bulk modulus of

Problems for Practice:53. A wire is subjected to a tension of 400 gram wt. If its diameter is 0.5mm, find the stress exerted upon it. (Ans. 1.997 x 107N/m2) 54. Find the greatest length of a copper wire that can hang vertically without breaking. (Breaking stress for copper = 5 x 108N/m2, density of copper = 9 x 103kg/m3) (Ans. 5.669 x 103m) 55. Calculate the greatest length of a copper wire which can hang vertically without breaking. The breaking stress of copper is 2.2 x 108N/m2 and density of copper is 8900 kg/m3. (Ans. 2.522 x 103 m) 56. A wire is one metre long and 1 mm in diameter. When it is stretched by a weight of 4 kg, its length increases by 0.25 mm. Find the 13

water. Take atm. pressure = 105 N/m2.(Ans. 109 N/m2) 64. A volume of 5 x 10-3 m3 of water is compressed by a pressure of 20 atmosphere. If bulk modulus of water is 2 x 109 N/m2, find the change produced in the volume of water (Ans. 5.064 x 10-6 m2) 65. Find the change in volume of 2 litres of water when subjected to a pressure of 20 atmospheres. (bulk modulus of water = 2x109 N/m2 Atmospheric pressure = 1.013 x105N/m2)(Ans 2.026 x 10-6N/m3) 66. Determine the pressure required to reduce a given volume of water by 1% Bulk modulus of water is 2x 109 N/m2 (Ans 2x 107 N/m) 67. When a pressure of 1.33 x 106 N/m2 is exerted on 2.4 litres of liquid its volume decreases by 2cm3. Find bulk modulus of water (Ans 1.596x 109N/m2) 68. If Young’s modulus for a wire is 2 x 1011N/m2, what mass must be suspended from the lower end of the stretch it by 0.8mm, if its length is 4m and diameter is 1mm?(Ans. 3.204 kg) 69. A wire of radius 0.5 mm has an initial length of 120cm. What force must be applied to the wire to stretch it to length of 121.2cm? (Y=1.28 x 1011N/m2) (Ans. 1005N) 70. A metal wire is observed to strecth by one part in a million when subjected to a stress of 8x104 N/m2. Calculate the ‘Y’ of the metal.(Ans. 8 x 1010N) 71. The lower surface of a metal cube having each side of length 12 cm is kept fixed and a tangential force is applied to its upper surface to displace it through 0.1 mm If the modulus of rigidity of the metal is 8.4 x 1010 ,find the magnitude of the applied tangential force. (Ans 1.008 x 106 N)

72. A tangential force of 2100 N is applied on a substance of area 3x 10-6m2, which is 0.1 m away from a fixed face. The force produces a shift of 7mm of upper surface w.r.to the bottom. Calculate the modulus of rigidity of the substance.(Ans. 1010N/m2) 73. A metal plate has an area is fixed and a tangential force is of 1cm. One face of larger area is fixed and a tangential force is applied to the opposite face. The displacement of the edge produced is 0.005 cm. Find the shearing strain, stress and magnitude of tangential force applied. Modulus of rigidity of the metal is n = 8.4 x 1010N/m2. (Ans. 5 x 10-3, 6 x 107 N/m2 6 x 107N/) 74. Young’s modulus of the material of a wire is 7 x 1010N/m2. The wire is stretched by a load so that the longitudinal strain produced in it is 5 x 10-3. Find the corresponding stress.(Ans. 3.5 x 108N/m2) 75. When a load of 5kg wt is applied to the end of a wire of length 1.5m, suspended from a rigid support, the extension produced in it is 6mm. If teh diameter of the wire is 3 x 10-4m, find (i) the longitudinal stress (ii) the longitudinal strain (iii) Young’s modulus of the material of the wire. [Ans. (i) 6.936 x 108N/m2, (ii) 4 x 10-3 (iii) 1.734 x 1011N/m2] 76. A metal wire of diameter 1mm and length 2m is stretched by applying a force of 2 m is stretched by applying a force of 2kg wt. Calculate (i) the increase in the lengthof the wire (ii) the strain (iii) the stress.[Ans. (i) 2.497 x 104 m (ii) 1.2485 x 10-4 (iii) 2.497 x 107N/m2] 77. The radius of a copper wire is 0.4 mm. Find the force required to stretch the wire by 0.2% of its length, assuming that the elastic limit is not exceeded. (Ans. 1.206 x 102N) 14

(Ans. 1.02 x 1011 N/m2). 87. Two wires of same material of lengths in the ratio 2 :3 and diameters in the ratio 3 : 2 are stretched with equal forces. What is the ratio of their extensions produced? (Ans. 8 : 27) 88. What mass must be suspended from the force end of steel wire of length 2 m and diameter 1 mm to stretch it by 1 mm. (Y for steel = 2 x 1011 N/m2) (Ans. 8.010 kg). 89. The steel wire of length 3.14m , 1mm in diameter is stretched by 5 kg wt. If extension produced in it is 1 mm. Determine young’s modulus. (Ans.1.96 x 1011 N/m2) 90. An extension of 0.3 cm is produced when a 900 gm weight is hung on a rod of length 5 m and area of cross section 1.0179 mm2. Find the stress, strain and Young’s modulus for the material of the rod. (Ans. 8.665 x 105 Nm-2, 6.0 x 10-4, 1.444 x 109 Nm-2) 91. Calculate poisson’s ratio for steel. If young’s modulus for steel is 2 x 1011 N/m2 and modulus of rigidity 8 x 1010 N/m2. (Ans. 0.25) 92. A metal has poison’s ratio 0.2 If uniform rod elongates to develop is longitudinal strain 2 x 103 then find percentage change in volume(0.12 %) 93. A wire of length 3 m and diameter 1 mm is stretched so that its length increases by 2mm.Find the decrease in the diameter of the Wire if Poisson’s ratio for the material of the wire is 0.36. (Ans. 2.4 x10-7 m) 94. A ball falling in a lake of depth 200 m shows 0.1% decrease in its volume at the bottom. What is the bullk modulus of the material of the ball. (Ans. 19.6 x 108 N/m2). 95. When a brass rod of diameter 6mm is subjected to a tension in of 5x103N, the diameter changes by 3.6 x 104 cm. Calculate the longitudinal strain and Poisson’s ratio for brass given that Y for brass is 9x1010N/m2. (Ans. 1.97 x 10-3; 0.3)

78. The diameter of a steel rod is 8 x 10-3m. What force will stretch it by 0.3% of its length? (Ans. 3.014 x 104N) 79. A steel wire of length 1m and diameter 0.2 mm is elongated by 5mm due to a weight of 3.14kg. Determine Young’s modulus of steel. (Ans. 1.96 x 1011N/m2) 80. How much will a steel wire having length 2m and diameter 1mm stretch under a load of 5kg wt. (Y = 2 x 1011N/m2) (Ans. 6.243 x 10-4 m) 81. The radius of a wire is 0.2mm, Find the force required to produce an elongation of 0.2% of the initial length of the wire. (Y= 1.7 x 1011N/m2) (Ans. 42.7 N) 82. A metal wire of length 5m and of cross section 1mm2 is suspended from a rigid support with a weight (of the same metal) of volume 1000 cm3 hanging from the other end of the wire. Find the decrease in the length of the wire when the weight is completely immersed in water.(Y for the metal = 1 x 1011N/m2 and density of water = 1 gram/ cm3)(Ans. 0.49 mm) 83. Two wires have diameters in the ratio 1:2, lengths in the ratio 3:4 and Young’s moduli in the ratio 1:1.5. Calculate the ratio of elongations produced in the wires subjected to the same stretching force. (Ans. l1 : l2 = 8 : 1) 84. Two wires of the same material are subjected to the same tension. Compare the extensions produced if the length of the first wire is double that of the other, while its radius is half that of the other. (Ans. l1 : l2 = 8 : 1) 85. Under application of tangential at force of 5 kg. weight of cube of one side 5 cm, the upper face is displaced through 0.5cm find the modulus of rigidity. (Ans. 1.96 x 105 N/m2). 86. Calculate bulk modulus if spherical ball contracts by 0.0098% when subjected to pressure 102 atmosphere.1 atm= 105N/m2. 15

cross-sectional area 4mm2, when it is suspended vertically and a load of 8kg is attached to its lower end. (Ans. 1.921 x 10-2J) 106. Find the strain energy in a copper wire of length 2.5 m and diameter 1mm, when stretched by a weight of 5kg. (Ans. 0.0319 J) 107. When the load applied to suspended wire is increased gradually from 3kg wt to 5kg wt, the

96. A metal wire of length 1.5 m is loaded and an elongation of 2 mm is produced. If the diameter of the wire is 1mm, find the change in diameter of the wire when elongated. Poisson’s ratio = 0.24 (3.2 x 10-7m) 97.A metallic wire(Y=20 x1010N/m2& 𝜎 =0.26) of length 3m and diameter 0.1 cm is stretched by a load of 10kg. Calculate the decrease in diameter of the wire.(1.6x10-4 mm) 98. Find the longitudinal stress to be applied to a wire to decrease its diameter by 10% Poisson’s ratio = 0.25. Young’s modulus = 2 x 1011N/m2. (8 x 1010N/m2) 99. A wire 0.5 m long and 10-6 m2 in cross section is stretched through 10-3m. How much workdone ? Y=12.4x1010 N/m2(Ans.W =0.124 J) 100. Find the work done in stretching a wire of length 2m and of sectional area 1mm2 through 1mm if the Young’s modulus of the material of the wire is 2x 1011N/m2. (Ans. 0.05J) 101. Calculate the work done in stretching a wire of length 3m and cross-sectional area 4mm2 when it is suspended from a rigid support at one end and a load of 8kg is attached at the free end. Y=12x1010N/m2 and g= 9.8 m/s2(1.921x 10-2J) 102. When the load on a wire is increased slowly from 3 to 5 kg wt the elongation increases from 0.6 mm to 1mm How much work is done during the extension ? g=9.8 m/s2 . 103. A wire 4m long and 0.3 mm in diameter is stretched by a load of 0.8 kg If extension caused in the wire is 1.5 mm find the strain energy per unit volume of the wire g=9.8 m/s2 104. Find the energy stored in a stretched brass wire of cross-sectional area 3mm2 and initial length 1 m, when a load of 2kg wt is applied to it. (Ans. 6.533 x 10-4 J) 105. If Young’s modulus for the material of wire of wire is 1.2 x 1011N/m2, calculate the work done in stretchin gthe wire of length 3m and

elongation increases from 0.6mm to 1mm. How much work is done during this extension of the wire? (Ans. 1.568 x 10-2J) 108. A metal wire of length 2.5 m and area of crosssection 1.5 x 10-6 m2 is stretched through 2mm. Calculate the work done during the stretching. (Ans. 0.15 J) 109. A copper wire is stretched by 0.5% of its length. Calculate the energy stored per unit volume in the wire. (Y = 12 x 1011 N/m2) (Ans. 1.5 x 106J) 110. A copper rod 2 m long is stretched by 1 mm. Deter unit volume (Y = 1.2 x 1011 N/m2) (Ans. 15 x 103 J/m3). 111. Strain energy per unit volume of wire on stretching becomes 10 x 10-3 J/m3.If young’s modulus of wire 1.8 x 1011 N/m2.Find tensile stress and tensile strain. (Ans. 6 x 106 N/m2, 3.333 x 10-7) 112. The wire of length 3 mand having cros sectional area 4 mm2 is subjected to load of 8 kg wt. Calculate workdone eduring stretching wire. (Y = 2 x 1011 N/m2) (Ans. 4.61 x 10-2 J)

4. A steel wire of diameter 1 mm and length 2 mis stretched by applying of force of 2 kg wt. Calculate i)increase in length ii) Stress iii) Strain.(Ans. 2.484 x 10-4 m, 2.482 x 107 N/m2, 1.242x104) BOARD BOOK PROBLEMS 113. A mild steel wire of radius 0.5 mm and length 3 m is stretched by a force of 49 N. Calculate i) longitudinal stress ii) longitudinal strain 16

iii) elongation produced in the wire ( Ymild steel = 2.1 x 1011 N/m2 ) 114. Calculate the change in volume of a lead block of volume 1m3 subjected to pressure of 10 atmosphere. Also calculate compressibility of lead. (1 atmosphere = 1.013 x 105 N/m2 K = 8 x 105 N/m2 ) 115. A square steel plate has area 1m2 and thickness 5 cm. The lower surface is fixed. A tangential force applied to top surface displaces it through 0.005 cm. Find shearing stress, shearing strain, if the modulus of rigidity of steel is 4.2 x 106 N/m2 . 116. Copper and steel wires of same length and same area of cross section are joined end to end. The composite wire is suspended from a rigid support and the load is applied to its free end. If the increase in length of composite wire is 3.1 mm, find the increase in length of copper wire and steel wire. ( Ycu – 1.1 x 1011 N/m2 , Y steel = 2 x 1011 N/m2 ) 117. Calculate the strain energy per unit volume in a brass wire of length 2.0 m and cross sectional area 0.5 mm2 , when it is stretched by 2 mm and a force of 5 kg. wt is applied to its free end. 118. Compute the greatest length of steel wire that can hang vertically without breaking under its own weight. (Breaking stress of steel) wire 7.2 x 108 N/m2 , density of steel = 7800 kg/m3 ) 119. A steel wire having cross sectional are 1 mm2 is stretched by 10 N. Find the lateral strain produced in the wire. (Given : Ysteel = 2 x 1011 N/m2 , 𝜎 = 0.291) 120. A sphere of mass 1 kg is attached at one end of a steel wire having length 0.5 m and radius 1 mm. It is whirled in a vertical circle with an angular velocity of 120 r.p.m. What is the elongation of wire, when sphere is at lowest point in its path? ( Ysteel = 20 x 1010 N/m2 ) 121.Find the maximum load which may be placed on a tungsten wire of diameter 2 mm so that permitted strain 1 must not exceed 1000 ( Ytungsten = 5 x 1010 N/m2 ) 122. A mass of 2 kg is hung from a steel wire of radius 0.5 mm and length 3m. Compute the extension produced. What should be the minimum radius of wire so that elastic limit is not exceeded? ( Elastic limit of steel = 2.4 x 108 N/m2 Ysteel = 20 x 1010 N/m2 ) 123. A compressive force 4 x 104 N is exerted at the end of a bone of length 30 cm and 4 cm2 cross sectional area.

125.Find the increase in pressure required to decrease volume of mercury by 0.001 % . (Bulk modulus of mercury = 2.8 x 1010 N/m2 )( Ans : 2.8 x 105 N/m2 ) 126. A copper metal cube has each side of length 1 m. The bottom edge of cube is fixed and tangential force 4.2 x 108 N is applied t o top surface. Calculate lateral displacement of top surface, if modulus of rigidity of copper is 14 x 1010 N/m2 ) ( Ans : 3 mm ) 127. A copper wire 4 m long has diameter of 1 mm. if a load of 10 kg wt is attached at other end. What extension is produced, if poisson’s ratio is 0.26? How much lateral compression is produced in it? ( Ycu 1.25 x 1010 N/m2 ) ( Ans : 3.992 mm, 2.595 x 10-4 mm) 128. Calculate the work done in stretching steel wire of length 2m and of cross sectional area 0.0225 mm2, when a load of 100 N is applied slowly to its free end. (Young’s modulus of steel = 20 x 1010 N/m2) ( Ans : 2.222J) 129. A solid brass sphere of volume 0.305 m3 is dropped in ocean, where water pressure is 2 x 107 N/m2. The bulk modulus of water is 6.1 x 1010 N/m2. What is change in volume of sphere? ( Ans. 10-4 m3) 130. Two wires of equal cross section one made up of aluminium and other of brass are joined end to end. When the combination of wires is kept under tension the elongation in wires are found to be equal. Find the ratio of lengths of two wires. ( YAL = 7 x 1010 N/m2 and Ybrass = 9.1 x 1010 N/m2) 131. The length of wire increases by 9 mm when weight of 2.5 kg is hung from the free end of wire. If all conditions are kept the same and the radius of wire is made thrice the original radius, find the increase in length. ( Ans : 1 mm) 132. One end of steel wire is fixed to a ceiling and a load of 2.5 kg is attached to the free end of the wire. Another identical wire is attached to the bottom of load and another load of 2.0 kg; is attached to the lower end of this wire. Compute the longitudinal strain produced in both the wires, if the crosssectional area of wires is 10-4 m2. ( Ysteel = 20 x 1010 N/m2)

What will happen to bone? Calculate change in length of bone. ( Compressive strength of bone = 7.7 x 108 N*/m2 , Young’s modulus of bone = 1.5 x 1010 N/m2 ) Problems for practice 124. A wire of length 2 m and cross sectional area 104 m2 is stretched by a load 102 kg. The wire is stretched by 0.1 cm. Calculate longitudinal stress, longitudinal strain, Young’s modulus of material of wire.( Ans : 1 x 107 N/m2 , 5 x 10-4 , 20 x 109 N/m2 )

(Ans : Strain1 = 1.225 x 10-6 , strain2 = 2.205x10-6 )

17