Best of luck to all the aspirants! - Target · PDF fileNo part of this book may be reproduced or transmitted in any form or ... (Reference of answer to each question is ... Pascal’s

No part of this book may be reproduced or transmitted in any form or by any means, C.D. ROM/Audio Video Cassettes or electronic, mechanicalincluding photocopying; recording or by any information storage and retrieval system without permission in writing from the Publisher.

PHYSICS First Year Diploma Semester - I

Printed at: Repro India Ltd., Mumbai

TEID : 917

Written as per the revised ‘G’ Scheme syllabus prescribed by the Maharashtra State Board of Technical Education (MSBTE) w.e.f. academic year 2012-2013

First Edition: June 2015

Basic Science

Salient Features

• Concise content with complete coverage of revised G-scheme syllabus. • Simple & lucid language.

• Neat, Labelled & authentic diagrams.

• Illustrative examples showing detailed solution of Numericals.

• MSBTE Theory Questions and Numericals from Winter-2006 to Summer-2015.

• MSBTE Question Papers of Summer-2014, Winter-2014, Summer-2015. • Three Model Question Papers for practice.

• Important Inclusions: Additional Theory Questions, Practice Problems, KnowledgeBank and ‘Physics behind…’

PREFACE Target’s “Basic Science Physics” is compiled with an aim of shaping engineering minds of students while catering to their needs. It is a complete & thorough book designed as per the new revised G-scheme of MSBTE curriculum effective from June 2012. Each unit from the syllabus is divided into chapters bearing ‘specific objectives’ in mind. The sub-topic wise classification of this book helps the students in easy comprehension. Each chapter includes the following features: Theory is provided in the form of pointers. Neat labelled diagrams have been provided wherever

required. Italicized definitions are hard to miss and help students map answers easily. Illustrative Examples are provided in order to understand the application of different concepts and

formulae. By introducing them after formulae, these examples enable students to gain command over formulae. An array of problems from simple to complex are included. (Examples here are similar to problems asked in previous years’ MSBTE Question Papers and also problems important from examination point of view)

Formulae are provided for quick recap and last minute revision. MSBTE Theory Questions covered in separate section to give a clear idea of the type of questions

asked. (Reference of answer to each question is provided.) MSBTE Numericals till latest year are included. Additional Theory Questions to help the student gain insight on the various levels of theory-based

questions. Problems for Practice (With final answers) covers a variety of questions from simple to complex. For the enrichment of students and igniting their interest in Physics, we have ensured inclusion of: Knowledge Bank which is designed to bridge the gap between the knowledge required to

understand the concept covered in syllabus but does not fall in the scope of syllabus. It acts as a base to understand the concept.

“Physics behind….” is an effort to make students aware of real life engineering situations where physics plays prominent role or day-to-day experiences ruled by physics.

MSBTE Question Papers of year 2014 and Summer-2015 are added at the end to make students familiar with the MSBTE examination pattern. A set of three Model Question Papers are designed as per MSBTE Pattern for thorough revision and to prepare the students for the final examination.

Best of luck to all the aspirants! From, Publisher

SYLLABUS

Topic and Contents Hours Marks Topic 1] Properties of solids Specific Objectives: [8 Marks]

05 08

Calculate the Young’s Modulus of material of wire.

Elasticity: Definitions of deforming force, restoring force, elasticity, plasticity, Factors affecting elasticity.

Stress: Tensile, Compressive, Volumetric and Shear stress. Strain: Tensile, Volumetric and Shear strain. Elastic limit, Hooke’s law. Elastic co-efficient-Young’s modulus, bulk modulus, modulus of rigidity and

relation between them

Stress-strain diagram, behavior of wire under continuously increasing load, yield point, ultimate stress, breaking stress, factor of safety, compressibility, Poisson’s ratio.

Topic 2] Properties of liquids Specific objectives:

09 12

Determine the surface tension of the given liquid Determine the coefficient of viscosity by Stoke’s method.

2.1 Fluid friction: [8 Marks]

Pressure , pressure-depth relation (P = ρ h g), atmospheric pressure, Pascal’s law, Archimedes’ principle.

Viscous force, definition of viscosity, velocity gradient, Newton’s law of viscosity, coefficient of viscosity and its SI unit.

Streamline and turbulent flow with examples, critical velocity, Reynold’s number and its significance.

Up thrust force, terminal velocity, Stoke’s law, and derivation of coefficient of viscosity by Stoke’s method, effect of temperature and adulteration on viscosity of liquid.

2.2 Surface tension: [4 Marks]

Cohesive and adhesive force, Laplace’s molecular theory of surface tension, surface Tension: definition and unit, effect of temperature on surface tension.

Angle of contact, capillarity and examples of capillary action, derivation of expression for surface tension by capillary rise method, applications of surface tension.

Topic 3] Thermal properties of matter Specific objectives:

08 12

Distinguish between isothermal and adiabatic process. Determine the relation between specific heats.

3.1 Modes of transformation of heat: [6 Marks]

Difference between heat and temperature, definition of calorie, Absolute zero, units of temperature: °C, °F, K, with their conversion.

Conduction, law of thermal conductivity, coefficient of thermal conductivity, good conductors of heat and insulators with suitable examples, applications of conduction. Convection, applications of convection. Radiation, applications of radiation.

3.2 Gas laws: [6 Marks]

Gas Laws: Boyle’s law, Charles’ law, Gay Lussac’s law (Statement and mathematical equation only) Perfect gas equation (PV = RT) (No derivation), specific heat of a substance, SI unit, specific heat of gas at constant volume (CV) specific heat of gas at constant pressure (CP), ratio of specific heats, Mayer’s relation between CP and CV, isothermal process, adiabatic process, difference between isothermal process and adiabatic process.

Topic 4] Optics Specific objectives:

04 06

Calculate refractive index of prism. Determine the numerical aperture of optical fiber

Refraction of light: [6 Marks]

Refraction of monochromatic light, Snell’s law, derivation of prism formula, total internal reflection, critical angle.

Optical fibre: principle, structure of optical fiber, propagation of light wave through optical fibre, derivation of numerical aperture and acceptance angle.

Topic 5] Wave motion Specific objectives:

06 12

Differentiate between transverse waves and longitudinal waves Derive expression for displacement, velocity and acceleration of a body executing S.H.M.

5.1 Wave motion: [6 Marks]

Definition of a wave, wave motion, wave velocity, wave period, wave frequency, wavelength, vibratory motion, periodic motion, amplitude of a vibrating particle, derivation of v = n λ

Simple harmonic motion (S.H.M.), examples of S.H.M., equation of S.H.M., expression of velocity and acceleration of a body executing S.H.M.

Types of progressive waves: transverse and longitudinal waves with examples.

5.2 Resonance: [6 Marks]

Stationary wave, formation of stationary wave, examples of stationary wave, characteristics of stationary waves, free and forced vibrations with examples.

Resonance: definition of resonance, examples of resonance, formula to calculate velocity of sound by resonance tube method.

TOTAL 32 50

1

Basic Physics (F.Y.Dip.Sem.-1) MSBTEUnit I: Properties of Solids

Publications Pvt. Ltd. Target Chapter-1 Properties of Solids 1.1 Introduction 1.2. Elasticity 1.2.(a) Deforming force 1.2.(b) Restoring force 1.2.(c) Elasticity 1.2.(d) Plasticity 1.2.(e) Comparison between elasticity and plasticity 1.2.(f) Factors affecting elasticity 1.3 Stress 1.3.(a) Longitudinal stress 1.3.(b) Volumetric stress 1.3.(c) Shear stress 1.4. Strain 1.4.(a) Tensile strain 1.4.(b) Volumetric strain 1.4.(c) Shear strain 1.5. Elastic Limit, Hooke’s law 1.5.(a) Elastic limit 1.5.(b) Hooke’s law 1.6. Elastic coefficient 1.6.(a) Young’s modulus 1.6.(b) Bulk modulus 1.6.(c) Compressibility 1.6.(d) Modulus of rigidity 1.6.(e) Relation between moduli of elasticity 1.6.(f) Comparision between moduli of elasticity 1.7. Poisson’s ratio 1.8 Stress-strain diagram 1.8.(a) Behaviour of wire under continuously increasing load 1.8.(b) Ultimate stress 1.8.(c) Factor of safety

Chapter - 1 Properties of Solids

Properties of SolidsUNIT I

2

Basic Science Physics (F.Y.Dip.Sem.-1) MSBTEPublications Pvt. Ltd. Target 1.1 Introduction i. When a body is subjected to a set of balanced forces, there is no motion of the body but the

force may produce change in its size, shape or both. ii. Due to applied force, some bodies undergo change in dimension (length, breadth, height, etc.)

easily, while some bodies oppose any change in dimension. iii. Some bodies show permanent change in their dimension after removal of applied force,

whereas some bodies preserve their original dimension, i.e., size, shape or both after removal of applied force.

1.2 Elasticity A solid body is an array of atoms or molecules. These atoms and molecules are bounded together by interatomic or intermolecular forces and remain in equilibrium positions. 1.2.(a) Deforming force: i. When an external force is applied on a body, it changes the dimensions of the body. The

change in size, shape or both of a body due to applied external force is called deformation.

Properties of Solids 01

Spring-ball model for the illustration of elastic behaviour of solids

Lattice point

Crystal structure

A hard, solid body with definite size and shape is said to be rigid if its dimensions do not change when a force is applied to it. In reality, no body is rigid. But for practical purposes, stone, iron, etc. are taken to be rigid bodies.

Knowledge Bank

Basic Physics (F.Y.Dip.Sem.-1) MSBTEChapter 01: Properties of Solids

3

Publications Pvt. Ltd. Target ii. Since the applied force changes the configuration of the body, it is termed as deforming

force. iii. The force required to produce deformation in a body is called deforming force. iv. When a body is deformed by applying an external force, the molecules are displaced

from their original positions of stable equilibrium. 1.2.(b) Restoring force: i. Under deformed condition, every shifted molecule tries to come back to its original position. ii. Due to elastic property of the body, internal molecular forces are set up within the body,

which tend to oppose the changes in size and shape of the body. iii. Restoring force is defined as internal force developed in a body, in order to regain its

original size and shape after application of deforming force. iv. Formula: Applied internal force = Restoring force 1.2.(c) Elasticity: i. Under deformed condition, the molecules of a body are displaced from their original

position. ii. The intermolecular distances change and restoring forces act on the molecules which

bring them back to their original position. iii. This leads to the phenomenon of elasticity in material body. iv. The property by virtue of which material bodies regain their original dimensions (size,

shape or both) after removal of deforming force is called elasticity. v. Elastic property is more in solid, less in liquid and least in gas. Elastic body: A body which possesses the property of elasticity i.e., the body which changes its size,

shape or both when a deforming force is applied and comes back to its original position as soon as deforming force is removed is called elastic body.

e.g. Steel wire, rubber band, sponge ball, etc. Perfectly elastic body: i. A body which regains its original shape, size or both completely after removal of deforming

force is called perfectly elastic body. ii. Perfectly elastic bodies do not exist in nature, however some bodies are considered to be

perfectly elastic. e.g. Quartz, phosphor bronze, etc. 1.2.(d) Plasticity: i. The property by virtue of which material bodies undergo permanent deformation even

after the removal of external deforming forces is called plasticity.

The force with which batsman hits the ball, acts as a deforming force. After certain overs, ball loses its elastic property and original shape. Hence, the players in a cricket match ask for change of ball after certain overs.

Physics behind changing the ball after certain overs in a cricket match…

4

Basic Science Physics (F.Y.Dip.Sem.-1) MSBTEPublications Pvt. Ltd. Target ii. Plastic bodies do not regain their original dimensions (size, shape or both) after removal

of deforming forces. iii. These bodies can be deformed to a large extent by a small deforming force. Perfectly plastic body: i. A body which does not regain its original configuration at all on the removal of

deforming force, how-so-ever small the deforming force may be, is called perfectly plastic body.

ii. Ideal perfectly plastic bodies do not exist in nature, however some bodies are considered to be perfectly plastic.

e.g. Wax, mud, putty, clay, plasticine, polythene plastic, etc. Note: Elastic properties of materials always lie between those of perfectly elastic and perfectly

plastic bodies. 1.2.(e) Comparison between elasticity and plasticity:

No. Elasticity Plasticity i. Body regains its original shape or size

after removal of external force. Body does not regain its original shape or size after removal of external force.

ii. External force changes the dimensions of the body temporarily.

External force changes the dimensions of the body permanently.

iii. Internal restoring force is set up inside the body.

Internal restoring force is not set up inside the body.

iv. Ratio of stress and strain remains constant.

Ratio of stress and strain does not remain constant.

1.2.(f) Factors affecting elasticity: Elasticity depends upon the material of the body, hence the structure of material greatly

affects its elastic properties. Though for a given material, elasticity is constant, following factors can affect its elasticity.

i. Change in Temperature: a. As the temperature of material increases, intermolecular distance increases. b. This decreases intermolecular force and in turn, internal restoring force. c. Thus, if temperature is increased, elasticity of material decreases and vice-versa.

e.g. When lead is cooled in liquid air, it becomes elastic. A carbon filament which is elastic at room temperature turns plastic when heated by passing current through it.

d. Exception: Invar steel, an alloy of nickel (36 %) and iron (64 %) shows elasticity independent of change in temperature.

ii. Impurities: a. The impurity affects elastic properties of material to which they are added. b. Sometimes suitable impurities are deliberately added to increase binding between

the lattice points (individual points that come together to constitute lattice structure) without disturbing their orientation.

c. In metals, impurities are added to increase their elastic properties. e.g. Addition of potassium in gold, addition of carbon in molten steel, etc. d. On the other hand, if impurity added is less elastic than the material itself, then

addition of impurity results in decrease in elastic properties of material.


5

Publications Pvt. Ltd. Target iii. Hammering and rolling: Operations like hammering and rolling break up crystal lattice into smaller units which

increase the elastic properties of material. iv. Annealing: a. Annealing is a process of bringing desired consistency, texture or hardness in a

material by gradually heating and cooling. b. Due to annealing, smaller crystals present inside the material get aligned to form

large crystal structure and softening of material is observed. c. Thus, annealing decreases elasticity of material. v. Recurring stress: If recurring (repeated) stress is applied on the same body, softening of material takes

place decreasing the elasticity. 1.3 Stress i. The internal elastic restoring force per unit cross-sectional area of a body is called stress.

OR Stress is defined as applied force per unit cross-sectional area of a body.

Stress = Applied force

Area of cross-section =

Elastic restoring force

Area of cross-section

ii. Formula: Stress = F

A

iii. Unit: N/m2 or Pa in SI system and dyne/cm2 in CGS system. iv. Dimensions: [M1L1T2] v. When an external force acts on a solid, there are three ways in which a solid may change its

dimension. Depending upon the type of change, we get three types of stress. a. Longitudinal stress b. Volumetric stress c. Shearing stress 1.3.(a) Longitudinal stress: i. If deforming force produces change in length of a body (wire, beam etc.) the stress

associated is called longitudinal stress.

ii. Formula: Longitudinal stress = Applied force on wire

Area of cross section=

F

A=

2

Mg

r

where, M = mass of the wire r = radius of cross-section of wire g = acceleration due to gravity. iii. It occurs only in solids like wire, rod, beam, etc and comes into picture when one of the

three dimensions viz. length, breadth or height is much greater than other two. iv. Longitudinal stress is either tensile or compressive. a. Tensile stress: Longitudinal stress produced due to increase in length of a body

under a deforming force is called tensile stress.

A bridge during its use undergoes recurring stress depending upon the movement ofvehicles on it. When bridge is used for long time, it loses its elastic strength and ultimately may collapse. Hence, the bridges are declared unsafe after long use.

Physics behind declaring bridges unsafe after long use…

6

Basic Science Physics (F.Y.Dip.Sem.-1) MSBTEPublications Pvt. Ltd. Target b. Compressive stress: Longitudinal stress produced due to decrease in length of a body

under a deforming force is called compressive stress. 1.3.(b) Volumetric or volume stress or bulk stress: i. If a deforming force produces change in volume of a body, the stress associated with it is

called volume stress. ii. This stress can be produced in solid, liquids and gases. iii. Let a gas balloon with original volume V be compressed by

additional pressure P on all sides. Due to applied additional pressure P, volume of gas balloon

decreases by V. Thus, new volume of gas in balloon = V V iv. If original pressure is P and new pressure applied is P + P then increase in pressure P

tries to restore the body back to its original dimensions. v. P generates an internal restoring force which acts on the walls of the balloon. vi. Applied force = Internal restoring force = P A

vii. Formula: Volume stress = Applied force

Area =

A P

A

= P

1.3.(c) Shearing (shear) or tangential stress: i. If deforming force produces change in shape of a body, the stress associated with it is

called shearing stress. OR The restoring force per unit area developed due to applied tangential force is called

shearing stress or tangential stress. ii. When a tangential force ‘F’ is applied on the

surface QRSV of a cube as shown in the figure then position of surface changes at an angle without change in volume of the cube.

In the figure, PQRSTUVW = original shape of cube

PQRSTUVW = new shape of cube in deformed state.

V V P

P

P

(a) A cylindrical body under tensile stress

elongates by L

F

F

L+LL

(b) Shearing stress on a

cylinder deforming it by an angle

x

F

F

L

(c) A body subjected to shearing stress

FF

Physics

(d) A solid body under a stress normal to the

surface at every point

V

P Q

R

ST

UV

V

U

ST

x

W

F

h

x


7

Publications Pvt. Ltd. Target F = tangential force TT = SS = UU = VV = x = lateral displacement of edge

iii. Formula: Shearing stress = Tangentialforce

Area=

F

A

Illustrative Examples: Example 1

A wire of length 3 m and a uniform area of cross section 1.5 106 m2 is stretched by a force of 50 N. Calculate the stress on the wire.

Solution: Given: L = 3 m, A = 1.5 106 m2, F = 50 N To find: Stress

Formula: Stress = A

F

Calculation: From formula,

Stress = 6

50

1.5 10 = 3.33 107 N/m2

Ans: The stress on the wire is 3.33 107 N/m2. Example 2

Find the maximum weight that can be attached to a copper wire of area of cross-section 2.4 mm2 if breaking stress for copper is 5 108 N/m2.

Solution: Given: A = 2.4 mm2 = 2.4 106 m2, Breaking stress = 5 108 N/m2

To find: Weight (F)

Formula: Stress = F

A

Calculation: In this case, weight attached is the force acting on the wire. The wire can bear the weight attached to it upto specific limit depending on its breaking stress. If the weight attached exceeds this limit, the wire will break.

From formula, F = Stress A = 5 108 2.4 106 F = 1200 N Ans: The maximum weight that can be attached to the wire is 1200 N. 1.4 Strain i. The change in dimension per unit original dimension of a body is called strain. OR Strain is the ratio of change in dimension to the original dimensions of the material.

ii. Formula: Strain = Change in dimension

Original dimension

iii. Strain is the ratio of two similar physical quantities, hence it has no unit and dimension. iv. Strain is classified into three types: a. Longitudinal strain b. Volume strain (Bulk strain) c. Shearing strain (Tangential strain)

8

Basic Science Physics (F.Y.Dip.Sem.-1) MSBTEPublications Pvt. Ltd. Target 1.4.(a) Longitudinal strain: i. The ratio of the change in length (l) to the

original length (L) is called longitudinal strain.

ii. In the figure, L = original length of wire

L = L + l = deformed length due to load Mg

Longitudinal strain = L L

L

=

L L

L

l

= L

l

iii. Formula: Longitudinal strain = L

l

iv. Longitudinal stress is either tensile or compressive. a. Tensile strain: Longitudinal strain produced due to increase in length of a body under a

deforming force is called tensile strain. b. Compressive strain: Longitudinal strain produced due to decrease in length of a body under a

deforming force is called compressive strain. v. It is observed in solid metal wires. vi. It has no units and dimensions. Illustrative Examples:

Example 1 A cylindrical bar of length 0.2 m deforms to 2 mm. What will be the strain developed in the bar?

Solution: Given: l = 2 mm = 0.002 m, L = 0.2 m To find: Strain

Formula: Strain = L

l


Strain = 0.002

0.2= 0.01

Ans: The strain developed in the bar will be 0.01. Example 2 A material wire elongates by 4% of its original length when loaded. Calculate tensile strain for the wire.

Solution:

Given: l = 4 % of L = 4

100L

To find: Tensile strain

L

B

l

L=

O

Mg

Deformed length

L

A

O

Initial length


9

Publications Pvt. Ltd. Target

Formula: Tensile strain = L

l


Tensile strain = 4L /100

L=

4

100= 0.04

Ans: Tensile strain for the wire is 0.04. 1.4.(b) Volumetric or volume strain or bulk strain: i. The ratio of change in volume per unit original volume of a body is called volume strain.

ii. Formula: Volume strain = V

V

where, V = original volume of gas, V = change in volume iii. It has no units and dimensions. Note: In this case, change in volume is without change in shape of body. Illustrative Example:

Calculate the magnitude of volume strain if volume changes by 2% of original volume. Solution:

Given: V = 2% V = 2

100V

To find: Magnitude of volume strain

Formula: Volume strain = V

V


Volume strain =

2V

100V

= 0.02

Ans: Magnitude of volume strain is 0.02. 1.4.(c) Shearing (shear) or tangential strain: i. The ratio of lateral displacement of any layer to its perpendicular distance from fixed

layer is called shearing (shear) strain.

Shearing strain =Relative displacement

Distance from fixed layer

ii. In the figure, base of the block AB is fixed. A tangential force ‘F’ is applied to the upper surface so that lateral displacement of the face is ‘x’.

In right angled ADQ,

Shearing strain = x

h = tan ,

where is called shear angle. If is very small, then tan

Shearing strain = = x

h

h

D Q C P

A B

x F

x

h

10

Basic Science Physics (F.Y.Dip.Sem.-1) MSBTEPublications Pvt. Ltd. Target

iii. Formula: Shearing strain = x

h

iv. It has no dimensions but expressed in radian. v. Shear strain is possible only for solid bodies. If the deforming force is applied parallel to

the liquid surface, it will begin to flow in the direction of applied force. e.g. On twisting a wire or rod, shearing strain is produced in it. Illustrative Example:

For a cube each of side 8 cm, the upper face is displaced by 1.2 cm due to a force of 1 N. Calculate shearing stress and shearing strain.

Solution: Given: F = 1 N, h = 8 cm = 8 10–2 m, A = h2 = 64 10–4 m2, x = 1.2 cm = 12 10–3 m To find: Shearing stress, shearing strain

Formulae: i. Shearing stress = F

A ii. Shearing strain =

x

h

Calculation: From formula (i),

Shearing stress = 4

1

64 10= 156.25 N/m

From formula (ii),

Shearing strain = 3

2

12 10

8 10

= 0.15

Ans: Shearing stress is 156.25 N/m and shearing strain is 0.15. 1.5 Elastic limit, Hooke’s law 1.5.(a) Elastic limit: i. The maximum stress to which an elastic body can be subjected without causing

permanent deformation is called as elastic limit. ii. The deformation produced in an elastic body depends upon the magnitude of deforming force.

If the deforming force is small or large, then deformation produced will also be small or large. iii. If the deforming force is removed, body regains its original shape and size. iv. However, if we go on increasing the deforming force, a stage is reached, where the body

does not regain its original dimensions, even after the removal of deforming force. The body is said to acquire permanent deformation called “set”.

v. When such a stage is reached, we say that body is stretched beyond the elastic limit. 1.5.(b) Hooke’s law:

i. Statement: Within the elastic limit, the stress developed in a body is directly proportional to the strain produced in it.

ii. Mathematically, Stress Strain iii. Formula: Stress = E (strain) where, E is constant of proportionality called modulus of

elasticity. iv. Value of E depends upon the nature of material of the body

and the manner in which the body is deformed.

Str

ess

StrainO


11

Publications Pvt. Ltd. Target 1.6 Elastic coefficient i. Elastic coefficient is defined as the ratio of stress to the corresponding strain. ii. It is also called as coefficient of elasticity or modulus of elasticity.

iii. Formula: Coefficient of elasticity = Stress

Strain

iv. Unit: N/m2 in SI system and dyne/cm2 in CGS system. v. Dimensions: [M1L1T2] vi. There are three moduli of elasticity namely Young’s modulus (Y), bulk modulus (K) and

modulus of rigidity () corresponding to three types of strain. 1.6.(a) Young’s modulus: i. Within elastic limit, the ratio of longitudinal stress to longitudinal strain is called as

Young’s modulus of elasticity. ii. It is denoted by Y.

iii. Formula: Y = 2

MgL

r l

iv. Expression for Young’s modulus: a. Let, L = original length of wire Mg = weight suspended to wire l = increase in length after stretching r = radius of the cross-section of wire

b. Longitudinal stress = Applied force

Area of cross-section=

F

A=

2

Mg

r

c. Longitudinal strain = Increase in length

Original length=

L

l

d. From definition,

Young’s modulus (Y) = longitudinal stress

longitudinal strain =

2Mg / r

/ L

l

Y = 2r

Mg

.L

l

v. Unit: N/m2 in SI system and dyne/cm2 in CGS system. vi. Dimensions: [M1L1T2] vii. Young’s modulus is property of solids only. viii. Greater the value of Young’s modulus of a material, larger is its elasticity.

r2

Mg

l

L

Steel is more elastic than rubber For elastic bodies, force of opposition starts building as deforming force acts on it. Between steel and rubber, steel offers more opposition to produce change in shape and size. Also, Young’s modulus of steel is 2 1011 N/m2 and of rubber is variable in range of 1 109 N/m2 to 2 109 N/m2. Due to greater value of Young’s modulus of steel, steel is more elastic than rubber.

Knowledge Bank

12

Basic Science Physics (F.Y.Dip.Sem.-1) MSBTEPublications Pvt. Ltd. Target Illustrative Examples: Example 1 When a load of 5 kg is attached to the free end of a suspended wire of length 4 m and

diameter 2 mm, the elongation produced is 0.5 mm. Calculate the longitudinal stress, longitudinal strain and Young’s modulus of the material of wire.

Solution: Given: M = 5 kg, L = 4 m, D = 2 mm, r = 1 mm = 1 103 m, l = 0.5 mm = 0.5 103 m To find: i. Longitudinal stress ii. Longitudinal strain iii. Young’s modulus (Y)

Formulae: i. Longitudinal stress = 2

Mg

r ii. Longitudinal strain =

L

l

iii. Y = longitudinalstress

longitudinalstrain


Longitudinal stress = 23 )10(14.3

8.95

= 1.56 107 N/m2

From formula (ii),

Longitudinal strain = 30.5 10

4

= 1.25 104

From formula (iii),

Y = 7

4

1.56 10

1.25 10

= 1.248 1011 N/m2

Ans: i. The longitudinal stress of the material of wire is 1.56 107 N/m2. ii. The longitudinal strain of the material of wire is 1.25 104. iii. Young’s modulus of the material of wire is 1.248 1011 N/m2. Example 2 A steel wire of length 4.7 m and cross-section 3.0 105 m2 is stretched by the same

amount as a copper wire of length 3.5 m and cross-section 4.0 105 m2 under a given load. What is the ratio of the Young’s modulus of steel to that of copper?

Solution: Given: LS = 4.7 m, AS = 3.0 105 m2, LC = 3.5 m, AC = 4.0 105 m2, lS = lC, FS = FC (same load applied)

To find: Ratio of Young’s moduli S

C

Y

Y

Formula: Y = FL

Al


YS = S S

S S

F L

A l = S

5S

F 4.7

3.0 10

l

YC = C C

C C

F L

A l= C

5C

F 3.5

4.0 10

l


13


S

C

Y

Y=

5S C

5S C

F 4.7 4.0 10

3.0 10 F 3.5

l

l

S

C

Y

Y= 1.79

Ans: The ratio of the Young’s modulus of steel to that of copper is 1.79. Example 3

A steel wire of length 4 m has a mass of 25 g. It is elongated by 1.25 mm when stretched by a weight of 5 kg. Calculate the Young’s modulus of steel. [Density of steel is 7.8 103 kg/m3]

Solution: Given: L = 4 m, l = 1.25 mm = 1.25 103 m, m = 25 g = 25 103 kg, M = 5 kg, = 7.8 103 kg/m3 To find: Young’s modulus (Y)

Formula: Y = MgL

A l

Calculation: Volume V = m

=

3

3

25 10

7.8 10

Also V = AL

A = L

V =

3

3

25 10

7.8 10 4

= 0.8 106 m2

From formula,

Y = 36 1025.1108.0

48.95

= 19.6 1010 N/m2

Ans: Young’s modulus of steel is 19.6 1010 N/m2. Example 4 Two wires of diameter 0.25 cm, one made of steel and the other made of brass are loaded

as shown in the figure. The unloaded length of steel wire is 1.5 m and that of brass wire is 1.0 m. Compute the elongations of the steel and the brass wires.

[YS = 2.0 1011 Nm2 , YB = 0.91 1011 Nm2] Solution: Given: D = 0.25 cm = 0.25 102 m, LS = 1.5 m, LB = 1 m, YS = 2.0 1011 N m2, YB = 0.91 1011 N m2 To find: i. Elongation of brass wire (lB) ii. Elongation of steel wire (lS)

Formula: Y = FL

Al

Calculation: Since, A = 2D

4

A = 2 2(0.25 10 )

4

A = 4.9 106 m2

14

Basic Science Physics (F.Y.Dip.Sem.-1) MSBTEPublications Pvt. Ltd. Target From formula,

l = FL

AY

For brass wire, load = 6 kg

lB = MgL

AY=

6 11

6 9.8 1

4.9 10 0.91 10

lB = 1.32 104 m For steel wire, load = 4 + 6 = 10 kg

lS = S

S

F L

AY =

MgL

AY

= 6 11

10 9.8 1.5

4.9 10 2 10

lS = 1.5 104 m Ans: The elongation of the steel wire is 1.5 104 m and that of brass wire is 1.32 104 m. Example 5 Find the weight attached to the lower end of wire having length 200 cm, radius 0.3 mm

and extension produced is 0.6 mm, if Young’s modulus of wire is 2.4 × 1011 N/m2. Solution: Given: L = 200 cm = 2 m, r = 0.3 mm = 3 104 m, l = 0.6 mm = 6 104 m, Y = 2.4 1011 N/m2 To find: Weight (Mg)

Formula: Y = (Mg)L

Al= 2

(Mg)L

r l


Mg = 2Y r

L

l=

11 4 2 42.4 10 3.14 (3 10 ) 6 10

2

Mg = 20.35 N Ans: The weight attached is 20.35 N. Example 6 A wire of length 2 m extends by 1 mm when stretched by a load of 4 kg. Find the area of

cross-section of the wire. (Given: Y = 2 1011 N/m2 and g = 9.81 m/s2). Solution: Given: L = 2 m, l = 1 mm = 103 m, M = 4 kg, Y = 2 1011 N/m2, g = 9.81 m/s2 To find: Area of cross-section (A)

Formula: Y = Mg L

Al


A = Mg L

Yl =

11 3

4 9.81 2

2 10 10

= 3.924 107 m2

Ans: The area of cross-section of wire is 3.924 107 m2.

1.5 mSteel

4.0 kg

1.0 m Brass

6.0 kg


15

Publications Pvt. Ltd. Target Example 7 A weight exerts force of 200 N on steel wire of cross-sectional area of A m2. If the

extension produced is 4 mm, for 5 m length of wire then find the diameter of wire. (Y = 2 1011 N/m2) Solution: Given: F = 200 N, A = A m2, Y = 2 1011 N/m2, l = 4 mm = 4 103 m, L = 5 m, To find: Diameter (D)

Formulae: i. Y = FL

Al ii. A =

2D

2


A = FL

Yl =

11 3

200 5

2 10 4 10

= 1.25 106 m2

From formula (ii),

D = 4A

=

64 1.25 10

3.14

= 1.262 103 m = 1.262 mm

Ans: The diameter of the wire is 1.262 mm. 1.6.(b) Bulk modulus: i. The ratio of volume stress to the volume strain is called bulk modulus. ii. From definition,

Bulk modulus, K = Volume stress

Volume strain =

PV

V

iii. Formula: K = P

VV

iv. Unit: N/m2 in SI system and dyne/cm2 in CGS system. v. Dimension: [M1L1T2] vi. Bulk modulus is the property of solids, liquids and gases. Illustrative Examples: Example 1 If the change in volume of a gas having original volume 500 cc is 0.008 cc at NTP, find

Bulk’s modulus of elasticity. Solution: Given: V = 0.008 cc, V = 500 cc, At NTP, P = 76 cm of Hg = hg = 76 cm 13.6 gm/c.c 980 cm/s2 = 1.013 106 dyne/cm2 To find: Bulk modulus (K)

Formula: K = P V

V


K = 61.013 10 500

0.008

= 6.33 1010 dyne/cm2

Ans: Bulk modulus of elasticity is 6.33 1010 dyne/cm2.

16

Basic Science Physics (F.Y.Dip.Sem.-1) MSBTEPublications Pvt. Ltd. Target Example 2 Calculate the decrease in the volume of 4 litre of water, when the change in pressure is

10 atmospheric pressure. [1 atm Pressure = 1.013 106 dyne/cm2 and bulk modulus of water = 2 1010 dyne/cm2]

Solution: Given: V = 4 litre = 4000 cm3, P = 10 atm = 10 1.013 106 dyne/cm2, K = 2 1010 dyne/cm2 To find: Decrease in volume (V)

Formula: K = VP

V


V = V P

K

= 6

10

10 1.013 10 4000

2 10

= 1.013 4

2

V = 2.026 cm3 Ans: The decrease in the volume of water is 2.026 cm3. Example 3 Find the increase in pressure required to decrease volume of mercury by 0.001 %. [Bulk modulus of mercury = 2.8 1010 N/m2] Solution: Given: V = 0.001 % of V

V

V

=

0.001

100 = 105, K = 2.8 1010 N/m2

To find: Increase in pressure (P)

Formula: K = P

VV


P = K V

V

P = 2.8 1010 105 P = 2.8 105 N/m2 Ans: The increase in pressure required to decrease volume of mercury by 0.001 % is

2.8 105 N/m2. Example 4 Determine the volume contraction of solid copper cube, 10 cm on an edge, when subjected

to a hydraulic pressure of 7.0 106 Pa. [Bulk modulus of copper = 140 109 Pa] Solution: Given: L = 10 cm = 0.1 m, V = L3 = (0.1)3 m3, P = 7 106 Pa, K = 140 109 Pa To find: Volume contraction (V)

Formula: K = P

VV


17

Publications Pvt. Ltd. Target Calculation: From formula,

V = 3 PL

K

= 6

3

9

7 100.1

140 10

V = 5 108 m3 Ans: The volume contraction of solid copper cube is 5 108 m3. 1.6.(c) Compressibility: i. The ratio of volume strain to volume stress is called compressibility. ii. Compressibilty is the reciprocal of bulk modulus of elasticity.

iii. Formula: Compressibility = 1

Bulk modulus(K)

iv. Higher the value of K, lower will be its compressibility. v. Unit: m2/N in SI system and cm2/dyne in CGS system. vi. Dimensions: [M1L1T2] vii. Solids are least compressible whereas gases are most compressible. Illustrative Example: The bulk modulus of water is 2.3 109 N/m2. Find its compressibility per atmospheric

pressure. Solution: Given: K = 2.3 109 N/m2, P0 = 1.013 105 N/m2

To find: Compressibility per atmospheric pressure 0

1

K / P

Formula: Compressibility = 1

K

Calculation: Bulk modulus per atmospheric pressure,

0

K

P=

9

5

2.3 10

1.013 10

2.27 104

Compressibility per atmospheric pressure,

0

1

K / P =

4

1

2.27 10 4.4 105

Ans: Compressibility per atmospheric pressure of water is 4.4 105. 1.6.(d) Modulus of rigidity: i. The ratio of shearing stress to the shearing strain is called modulus of rigidity or

rigidity modulus. ii. Now, modulus of rigidity,

= Shearing stress

Shearing strain =

A/F

18


iii. Formula: = A

F =

Fh F

Ax A tan

iv. Unit: N/m2 in SI system and dyne/cm2 in CGS system.

v. Dimensions: [M1L1T2] vi. Modulus of rigidity is property of solids only. Illustrative Examples: Example 1 A copper metal cube has each side of length 1 m. The bottom edge of cube is fixed and

tangential force 4.2 108 N is applied to top surface. Calculate the lateral displacement of the top surface, if modulus of rigidity of copper is 14 1010 N/m2.

Solution: Given: L = h = 1 m, A = L2 = 1 m2, F = 4.2 108 N, = 14 1010 = 1.4 1011 N/m2 To find : Lateral displacement (x)

Formula: = Fh

Ax


x = Fh

A

x = 8

11

4.2 10 1

1 1.4 10

= 3 103 m

x = 3 mm Ans: The lateral displacement of top is 3 mm. Example 2 A piece of copper having a rectangular cross-section of 15.2 mm 19.14 mm is pulled in

tension with 44500 N force, producing only elastic deformation. Calculate the resulting strain. [Rigidity modulus of copper = 42 109 N m2]

Solution: Given: A = 15.2 mm 19.14 mm = 15.2 19.14 106 m2, F = 44500 N, η = 42 109 N m2 To find: Strain (θ)

Formula: η = F

A


θ = F

A=

6 9

44500

(15.2 19.14 10 ) 42 10

θ = 3.64 103 Ans: The strain produced in the piece of copper is 3.64 103. 1.6.(e) Relation between moduli of elasticity: Relation between Young’s modulus (Y), bulk modulus (K) and rigidity modulus () of

elasticity is given by,

9 3 1

Y K

or Y = 9K

3K


19

Publications Pvt. Ltd. Target Illustrative Example: If the shear modulus and bulk modulus of a material are 0.42 1011 N/m2 and 0.21 1011 N/m2

respectively, find Young’s modulus of that material. Solution: Given: = 0.42 1011 N/m2, K = 0.21 1011 N/m2

Formula: Y = 9 K

3K

To find: Young’s modulus (Y) Calculation: From formula,

Y = 11 11

11 11

9 0.21 10 0.42 10

3 0.21 10 0.42 10

= 22

11

0.7938 10

1.05 10

= 0.756 1011

Y = 7.56 1010 N/m2 Ans: Young’s modulus of that material is 7.56 1010 N/m2. 1.6.(f) Comparision between moduli of elasticity:

Young’s modulus Bulk modulus Modulus of rigidity i. It is the ratio of

longitudinal stress to longitudinal strain.

It is the ratio of volume stress to volume strain.

It is the ratio of shearing stress to shearing strain.

ii. It is given by, Y =

2

MgL

πr l It is given by, K =

V P

V

It is given by, = F

Aθ

iii. It exists in solids. It exists in solids, liquids and gases. It exists in solids. iv. It relates to change in

length of a body. It relates to change in volume of a body.

It relates to change in shape of a body.

1.7 Poisson’s ratio i. When a deforming force is applied at the free end of a suspended wire of length L and diameter

D, then its length increases by l but its diameter decreases by D. ii. Now, two types of strain are produced by a single force. They are:

a. Longitudinal strain = L

l

b. Lateral strain = D

D

(Negative sign shows that diameter of wire decreases with

increase in length.) iii. The ratio of lateral strain to the longitudinal strain is called Poisson’s ratio (). iv. From definition,

=

= D / D

/ L

l

= D

D

L

l ….(1)

20

Basic Science Physics (F.Y.Dip.Sem.-1) MSBTEPublications Pvt. Ltd. Target Equation (1) can also be written as,

=

DL rL2

D r2

ll

,

where, r = radius of the cross-section of the wire.

v. Formula: = L D

D

l

= L r

r

l

(in magnitude)

vi. Theoretical value of lies between 1 and 0.5. But practically, there are no such substances which have negative Poisson’s ratio. Therefore, practical value of lies between 0 and 0.5.

vii. It is unitless and dimensionless quantity. Illustrative Examples: Example 1 A uniform steel wire of length 3 m and radius 1 mm is stretched by 0.18 mm and the

diameter decreases by 3.84 108 m. Find the Poisson’s ratio for the material of the wire. Solution: Given: L = 3 m, r = 1 mm = 1 103 m, D = 2 103 m, l = 0.18 mm = 0.18 103 m = 18 105 m, D = 3.84 108 m To find: Poisson’s ratio ()

Formula: = DL

D

l


= 8

3 5

3.84 10 3

2 10 18 10

= 11.52

36= 0.32

Ans: Poisson’s ratio for the material of the wire is 0.32. Example 2 A load of 1 kg produces a certain extension in the wire of length 3 m and radius

5 104 m. How much will be the lateral strain produced in the wire? [Y = 7.48 1010 N/m2 and = 0.291] Solution: Given: M = 1 kg, g = 9.8 m/s2, r = 5 104 m, Y = 7.48 1010 N/m2, = 0.291 To find: Lateral strain Formula: Lateral strain = longitudinal strain

Calculation: Longitudinal strain = F

AY =

2

Mg

r Y

From formula,

Lateral strain = 2

Mg

r Y=

4 2 10

0.291 1 9.8

3.14 (5 10 ) 7.48 10

= 8 10

0.291 9.8

3.14 25 10 7.48 10

Lateral strain = 4.856 105

Ans: The lateral strain produced in the wire is 4.856 105.


21

Publications Pvt. Ltd. Target 1.8 Stress-strain diagram i. Consider a body subjected to a uniform stress. As the stress increases, it results in a change in

dimension of the body. This means, stress produces strain on the body. ii. The relationship between applied stress and strain can be studied by drawing a curve between

strain and stress. The curve is called as stress-strain curve. Terms related to stress-strain curve: i. Point of proportionality: The point on stress-strain curve, upto which stress is directly

proportional to strain, obeying Hooke’s law completely is called point of proportionality. ii. Yield point: The point on stress-strain curve at which the strain begins to increase without any

increase in the stress (plastic flow begins) is called yield point. A metal after yield point becomes ductile.

iii. Breaking stress (ultimate strength): The maximum stress which a wire can bear is called breaking stress. OR

The maximum stress at which a wire breaks is called breaking stress. iv. Breaking weight: Product of breaking stress and area of cross-section is called breaking

weight. Formula: Breaking weight = Breaking stress Area of cross-section v. Breaking point: The point on stress-strain curve at which the wire breaks, is known as the

breaking point. vi. Permanent deformation (set): If the elastic limit is exceeded, the body does not preserve its

original dimension after removal of the deforming force. This state is called permanent deformation (set).

1.8.(a) Behaviour of wire under continuously increasing load: i. Consider a wire of uniform area of cross-section. Let the wire be suspended vertically

through one end and load be attached to other end of the wire. ii. Wire is loaded gradually by adding weights and corresponding extensions produced in

wire are measured. iii. From various observations, stress and strain is calculated and plotted. The resulting plot

resembles the curve given below: The plot consists of following sections: i. Proportional limit: a. The initial portion OP of the graph is a straight line indicating stress is directly

proportional to strain.

P : Proportional limit E : Elastic limit Y : Yield point B : Breaking stress C : Breaking point OO : Permanent set

B

O O

E Y C

Strain

Str

ess

P

E

22

Basic Science Physics (F.Y.Dip.Sem.-1) MSBTEPublications Pvt. Ltd. Target b. Thus Hooke’s law is completely obeyed in the region OP. c. Point P is called point of proportional limit. ii. Elastic limit: a. Beyond the point P, the stress-strain variation is not a straight line as indicated by

the part PE of the graph. b. If the wire is unloaded at point E, the graph between stress and strain follows the

reverse path EPO, and the point E is called elastic limit. c. The portion between O and E is called elastic region. iii. Permanent set: a. If the load is increased so that stress becomes greater than that corresponding to the

point E, the graph is no longer a straight line and the wire does not obey Hooke’s law. b. If the wire is strained upto E beyond point E and then the load is removed, the wire

does not regain its original length and there is a permanent increase in length. c. A small strain corresponding to OO is set up permanently in the wire, called

permanent set. d. However, the wire is still elastic and if loaded again, gives a linear relation shown

by the dotted line OE. iv. Yield point: a. As the stress is increased beyond the elastic limit, the graph is a curve and reaches

a point Y where the tangent to the curve is parallel to the strain axis. b. This shows that for the stress corresponding to point Y, the strain increases even

without any increase in the stress. This is known as plastic flow. c. Point Y on the curve is called yield point. d. The value of stress corresponding to yield point is called yield stress. v. Breaking stress: a. When the wire begins to flow, its cross-section decreases uniformly and hence, the

stress increases steadily. b. Later a neck or constriction begins to form at a weak point. c. The maximum stress corresponding to the point B is breaking stress. vi. Breaking point: Once the neck is formed, the wire goes on stretching even if the load is

reduced, until the breaking point C is reached when the wire breaks. Classification of material on the basis of stress-strain curve: From the study of stress-strain variation, different materials can be classified as below: i. Ductile materials: a. The materials which show large plastic range beyond elastic limit are called as

ductile materials. b. For such materials, the breaking point is widely separated from the point of elastic

limit on the stress-strain graph. c. Such materials can be used in making springs and sheets. d. e.g. copper, silver, iron, aluminium, etc. ii. Brittle materials: a. The materials which show very small plastic range beyond elastic limit are called

as brittle materials. b. For such materials, the breaking point lies close to the elastic limit. c. Such materials cannot be used in making springs and sheets. d. e.g. cast iron, glass, etc.


23

Publications Pvt. Ltd. Target iii. Elastomers: a. Within elastic limit, the materials for which stress and strain variation is not a

straight line are called as elastomers. b. These materials do not obey Hooke’s law over most of the regions. c. Such materials have no plastic range. The breaking point lies just close to the

elastic limit. d. e.g. rubber, the elastic tissue of aorta, etc. Elastic after-effect: i. On removing the deforming forces from the elastic bodies, they regain their original configuration. ii. Some elastic bodies regain their original configuration immediately, but a few others

show delay to recover their original configurations. iii. This temporary delay in regaining the original configuration by an elastic body after the

removal of a deforming force is called elastic after-effect. iv. The elastic after-effect is negligibly small for quartz and phosphor bronze. The elastic

after effect is very large for glass fibre. Hence, the suspensions made from quartz or phosphor bronze are used in moving coil galvanometers.

Elastic fatigue: i. Elastic fatigue is the property of an elastic body by virtue of which its behaviour

becomes less elastic under the action of repeated alternating deforming forces. ii. On stretching and releasing the rubber band again and again, its elasticity is lost due to

elastic fatigue. 1.8.(b) Ultimate stress: i. Ultimate stress is defined as the maximum stress that the specimen (system) can

withstand per initial cross-sectional area of specimen.

ii. Formula: Ultimate stress = Maximumstress thespecimen can withstand

Initialcross-sectionalarea

Working stress: i. Working stress is defined as the maximum stress to which specimen is actually subjected

per initial cross-sectional area.

ii. Formula: Working stress = Maximumstress to which specimen isactuallysubjected

Initialcross-sectionalarea

iii. Safe working stress is always less than the elastic limit for a given material. 1.8.(c) Factor of safety/safety factor: i. It is a measure of structural capacity of a system beyond the expected or actual loads. ii. It describes how much stronger the system is than it usually needs to be for an intended

load. e.g. Message on lift may be for 300 kgs only but its actual capacity may be of 500 kgs. iii. Factor of safety is defined as the ratio of maximum load that the structure can bear

(ultimate stress ) to the actual load on the structure(working stress).

iv. Formula: Factor of safety = Ultimatestress

Workingstress

v. Factor of safety depends on wear and tear of material, maximum load subjected to object, characteristics of load such as fixed or variable.

vi. Factor of safety must be greater than one.

24


MSBTE Theory QuestionsFormulae 1. Stress:

i. Longitudinal stress = F

A =

2

Mg

r

ii. Volume stress = P

iii. Shearing stress = F

A

iv. Breaking stress

= Breaking force

Area of cross-section=

F

A

2. Strain:

i. Longitudinal strain = L

l

ii. Volume strain = V

V

iii. Shearing strain = h

x

3. Hooke’s law:

Coefficient of elasticity E = Stress

Strain

4. Young’s modulus: Y = 2

MgL

r l

5. Elongation in the wire due to load:

l = 2

MgL

r Y

6. Bulk modulus:

K = Bulk stress

Bulk strain =

PV

V

7. Modulus of rigidity for small angle:

= F

A =

Fh F

Ax A tan

8. Relation between moduli of elasticity:

i. 9 3 1

Y K

ii. 9K

Y3K

9. Poisson’s ratio:

= L D

D

l

= L r

r

l

(In magnitude)

1. Define Young’s modulus, bulk

modulus, rigidity modulus of elasticity. State relation between them.

[W-06, 09, 13; S-07, 12, 15] [4 M] Ans: Refer 1.6.(a), 1.6.(b), 1.6.(d), 1.6.(e) 2. Define factor of safety, elastic limit.

[W-06, 11] [2 M] Ans: Refer 1.8.(c), 1.5.(a) 3. Define stress, strain, restoring force,

deforming force. [W-07, 11] [4 M] Ans: Refer 1.3, 1.4, 1.2(b), 1.2 (a) 4. State whether rubber is more elastic or

steel, why? [W-07; S-10] [2 M] Ans: Refer 1.6.(a) Knowledge bank 5. State and explain Hooke’s law. Define

modulus of rigidity. Write down its SI unit. [S-08] [4 M]

Ans: Refer 1.5.(b), 1.6.(d) 6. Explain elastic limit and yield point on

stress-strain diagram. [S-08] [2 M] Ans: Refer 1.5.(a), 1.8 7. State Hooke’s law of elasticity and

define Young’s modulus of elasticity. [W-08] [2 M]

Ans: Refer 1.5.(b), 1.6.(a) 8. Define deforming force and restoring

force. State SI unit of stress. [W-08; S-12] [4 M]

Ans: Refer 1.2.(a), 1.2.(b), 1.3 9. Define stress and Young’s modulus of

elasticity. [S-09] [2 M] Ans: Refer 1.3, 1.6.(a) 10. Describe the behavior of wire under

continuously increasing load using stress-strain curve.

[S-09, 10; W-11] [4 M] OR

Explain behavior of wire under continuously increasing load.

[S-15] [4 M] Ans: Refer 1.8.(a)


25


MSBTE Numericals

11. Define stress, strain and Hooke’s law. [W-09] [2 M]; [S-10] [4 M]

Ans: Refer 1.3, 1.4, 1.5.(b) 12. Define elasticity and plasticity.

[W-10; S-15] [2 M] OR

State elasticity and plasticity property. [S-13] [2 M]

Ans: Refer 1.2.(c), 1.2.(d) 13. State and explain three types of strain.

[W-10] [4 M] Ans: Refer 1.4 14. State and explain Hooke’s law of

elasticity. Hence define elastic limit. [W-10] [4 M]

Ans: Refer 1.5 15. What is elasticity property and elastic

body? [S-11] [2 M] Ans: Refer 1.2.(c) 16. Using the behaviour of uniform cross

section material wire under continuously increasing load by stress-strain curve, define [S-11] [4 M]

i. Elastic-stress limit ii. Yield stress iii. Breaking stress Ans: Refer 1.5.(a), 1.8.(a), 1.8 17. State types of stress. State SI unit of

stress. [S-11] [4 M] Ans: Refer 1.3 18. Define factor of safety. [S-12] [2 M] Ans: Refer 1.8.(c) 19. State Hooke’s law. Explain stress-strain

diagram. [S-12] [4 M] Ans: Refer 1.5.(b), 1.8 20. State any two factors affecting elasticity.

[W-12, 14] [2 M] Ans: Refer 1.2.(f) 21. Define tensile, volumetric, shear strain

and breaking stress. [W-12] [4 M] Ans: Refer 1.4, 1.8

22. Give relation between Bulk modulus of elasticity and compressibility.

[S-13] [2 M] Ans: Refer 1.6.(c) 23. Define: [S-13] [4 M] i. Yield point ii. Ultimate stress iii. Breaking stress iv. Factor of safety in Elasticity Ans: Refer 1.8, 1.8.(b),1.8, 1.8.(c) 24. Define: [W-13] [2 M] i. Stress ii. Strain Ans: Refer 1.3, 1.4 25. Define Poisson’s ratio. [W-13] [2 M] Ans: Refer 1.7 26. State elastic body and plastic body.

[S-14] [2 M] Ans: Refer 1.2.(c),1.2.(d) 27. Define [W-14] [4 M] i. Elastic limit ii. Yield point iii. Poisson’s ratio iv. Factor of safety. Ans: Refer 1.5.(a), 1.8, 1.7, 1.8.(c) 28. Define three modulii of elasticity, Y, K

and . [S-14] [4 M] Ans: Refer 1.6.(a), 1.6.(b), 1.6.(d) 29. State Hooke’s law. [S-15] [2 M] Ans: Refer 1.5.(b) 1. Calculate stress if a load of 10 N is

attached to the lower end of wire of radius 1 mm. [W-14] [2 M]

Solution: Given: F = 10 N, r = 1 mm = 103 m, A = r2 = (103)2 = 106 m2 To find: Stress

Formula: Stress = F

A

26

Basic Science Physics (F.Y.Dip.Sem.-1) MSBTEPublications Pvt. Ltd. Target Calculation: From formula,

Stress = 6

10

3.142 10

= 3.183 106 N/m2 Ans: The stress of wire is 3.183 106 N/m2. 2. A material wire elongates by 1 % of

its original length when loaded. Calculate tensile strain for the wire.

[S-14] [2 M] Solution:

Given: l = 1 % of L = L

100

To find: Tensile strain

Formula: Tensile strain = L

l


Tensile strain = L /100

L

= 1

100= 0.01

Ans: Tensile strain for the wire is 0.01. 3. A wire of length 1.5 m extends by

1.5 mm when a force is applied to it. Calculate the stress produced in it.

(Given: Y = 2 1011 N/m2) [S-07] [4 M]

Solution: Given: l = 1.5 mm = 1.5 103 m, L = 1.5 m, Y = 2 1011 N/m2 To find: Stress (F/A)

Formula: Y = FL

Al


F

A=

Y

L

l

= 11 32 10 1.5 10

1.5

F

A= 2 108 N/m2

Ans: The stress produced in wire is 2 108 N/m2.

4. A wire gets elongated by 0.002 mm, when a stress of 2000 N/m2 is applied. Find Young’s modulus of elasticity of material of wire if the original length of wire is 4 m. [W-12] [4 M]

Solution: Given: l = 0.002 mm = 2 106 m,

F

A= 2000 N/m2, L = 4 m

To find: Young’s modulus (Y)

Formula: Y = FL

A l


Y = 6

2000 4

2 10

= 4 109 N/m2

Ans: Young’s modulus of elasticity of wire is 4 109 N/m2.

5. A wire of diameter 2 mm and of length 2.5 m extends by 1.5 mm by applying a force of 15 N. Find Young’s modulus of wire.[S-08] [4 M]

Solution: Given: D = 2 mm, r = 1 mm = 103 m, L = 2.5 m, l = 1.5 mm = 1.5 103 m, F = 15 N To find: Young’s modulus (Y)

Formula: Y = 2

FL

( r ) l


Y = 3 2 3

15 2.5

3.142 (10 ) 1.5 10

= 7.957 109 N/m2 Ans: Young’s modulus of the wire is

7.957 109 N/m2. 6. A wire of diameter 3 mm and length

4 m extends by 2.5 mm when a force of 10 N is applied. Find the Young’s modulus of material of wire.

[S-15] [4 M] Solution: Given: D = 3 mm, r = 1.5 mm = 1.5 103 m,


27

Publications Pvt. Ltd. Target L = 4 m, l = 2.5 mm = 2.5 103 m, F = 10 N To find: Young’s modulus (Y)

Formula: Y = 2

FL

( r ) l


Y = 3 2 3

10 4

3.142 (1.5 10 ) 2.5 10


2.263 109 N/m2. 7. Calculate Young’s modulus of

elasticity for a wire having length 100 cm and diameter 5 mm. The wire elongates by 2 mm when subjected to a load of 10 N. [S-13] [4 M]

Solution: Given: L = 100 cm = 1 m, D = 5 mm, r = 2.5 mm = 2.5 103 m, F = 10 N, l = 2 mm = 2 103 m To find: Young’s modulus (Y)

Formula: Y = 2

FL

r l


Y = 3 2 3

10 1

3.14 (2.5 10 ) 2 10

= 2.548 108 N/m2 Ans: Young’s modulus of elasticity for a

given wire is 2.548 108 N/m2. 8. Calculate Young’s modulus of

elasticity for material wire 2 m long, 0.4 mm diameter, if weight applied is 100 N which elongates the wire by 0.001 mm. [S-14] [4 M]

Solution: Given: L = 2 m, D = 0.4 mm, r = 0.2 mm = 2 104 m, F = 100 N, l = 0.001 mm = 106 m To find: Young’s modulus (Y)

Formula: Y = 2

FL

r l


Y = 4 2 6

100 2

3.142 (2 10 ) 10

= 1.591 1015 N/m2 Ans: Young’s modulus of elasticity of wire is

1.591 1015 N/m2. 9. A wire extends by 0.02 cm when loaded

with 2 kg. If its length is 2 m and radius 0.05 cm, calculate Young’s modulus.

[W-07] [4 M] Solution: Given: l = 0.02 cm = 2 104 m, M = 2 kg, L = 2 m, r = 0.05 cm = 5 104 m To find: Young’s modulus (Y)

Formula: Y = 2

MgL

r l


Y = 4 2 4

2 9.8 2

3.142 (5 10 ) 2 10


2.495 1011 N/m2. 10. A copper wire of length 5 m and cross

section 5 105 m2 stretches by the same amount as a steel wire of length 6 m and cross-section 4 105 m2 under given load. Find the ratio of Young’s modulus of copper to steel.

[S-12] [4 M] Solution: For copper, L1 = 5 m, A1 = 5 105 m2 For steel, L2 = 6 m, A2 = 4 105 m2 Let load be Mg and extension produced

be l then, Young’s modulus for copper

YCu = 1

1

Mg L

A l and

Young’s modulus for steel

Yst = 2

2

Mg L

A l

28

Basic Science Physics (F.Y.Dip.Sem.-1) MSBTEPublications Pvt. Ltd. Target Taking ratios,

Cu

st

Y

Y= 1 1

2 2

Mg L / A

Mg L / A

l

l

= 1 2

2 1

L A

L A =

5

5

5 4 10

6 5 10

Cu

st

Y

Y =

4

6 =

2

3

= 0.667 Ans: The ratio of Young’s modulus of copper

to steel is 0.667. 11. An unknown weight attached to the

lower end of wire of length 4 m, radius 0.7 mm, extends it by 0.8 mm. If Y = 2 1011 N/m2, find the unknown weight. [S-10] [4 M]

Solution: Given: L = 4 m, r = 0.7 mm = 7 104 m, l = 0.8 mm = 8 104 m, Y = 2 1011 N/m2 To find: Weight (Mg)

Formula: Y = 2

(Mg) L

r l


Mg = 2r Y

L

l

4 2 4 113.142 (7 10 ) 8 10 2 10

4

Mg = 61.58 N Ans: The unknown weight is 61.58 N. 12. Find the weight attached to the lower

end of wire having length 150 cm, radius 0.3 mm and extension produced is 0.6 mm, if Young’s modulus of wire is 2 × 1011 N/m2.

[W-14] [4 M] Solution: Given: L = 150 cm = 1.5 m, r = 0.3 mm = 3 104 m l = 0.6 = 6 104 m, Y = 2 1011 N/m2

To find: Weight (Mg)

Formula: Y = 2

(Mg)L

r l


Mg = 2r Y

L

l

=11 4 2 42 10 3.142 (3 10 ) 6 10

1.5

Mg = 22.62 N Ans: The weight attached is 22.62 N. 13. A weight exerts force of 100 N on a

wire of cross-sectional area 0.02 cm2. Find the extension produced if the length of wire is 5 m.

(Given Y = 2 1011 N/m2) [W-08; S-09] [4 M]

Solution: Given: F = 100 N, L = 5 m, Y = 2 1011 N/m2, A = 0.02 cm2 = 2 102 cm2 = 2 106 m2 To find: Extension (l)

Formula: Y = FL

Al


l = FL

YA

=11 6

100 5

2 10 2 10

= 1.25 103 m

l = 1.25 mm Ans: The extension produced in wire is 1.25 mm. 14. An aluminium wire 3 mm in diameter

and 4 m long is used to support a mass of 50 kg. What is the elongation of the wire? Young’s modulus of aluminium is 7 1010 N/m2.

[W-09, 13] [4 M] Solution: Given: D = 3 mm, r = 1.5 mm = 1.5 103 m, M = 50 kg, Y = 7 1010 N/m2,

L = 4 m


29

Publications Pvt. Ltd. Target To find: Elongation (l)

Formula: Y = 2

MgL

r l


l = 2

MgL

r Y

= 3 2 10

50 9.8 4

3.142 (1.5 10 ) 7 10

= 3.96 103 m l = 3.96 mm Ans: The elongation of wire is 3.96 mm. 15. A wire has diameter 1 mm and length

2 m. It is stretched by force of 2 kg weight. What is the increase in length of wire?

(Given: Y = 2 1011 N/m2, g = 9.81 m/s2) [S-12] [4 M]

Solution: Given: D = 1 mm, r = 0.5 mm = 5 104 m, M = 2 kg, L = 2 m, Y = 2 1011 N/m2, g = 9.81 m/s2 To find: Elongation (l)

Formula: Y = 2

MgL

r l


l = 2

Mg L

r Y

= 4 2 11

2 9.81 2

3.142 (5 10 ) 2 10

= 2.498 104 m = 0.2498 mm l 0.25 mm Ans: The increase in length of wire

(elongation) is nearly 0.25 mm. 16. A wire of length 1 m extends by 1 mm

when stretched by a load of 1 kg. Find the area of cross-section of the wire.

(Given: Y = 2 1011 N/m2 and g = 9.81 m/s2). [W-10] [4 M]

Solution: Given: L = 1 m, l = 1 mm = 103 m, M = 1 kg, Y = 2 1011 N/m2, g = 9.81 m/s2

To find: Area of cross-section (A)

Formula: Y = Mg L

Al


A = Mg L

Yl

= 11 3

1 9.81 1

2 10 10

A = 4.905 108 m2 Ans: The area of cross-section of wire is

4.905 108 m2. 17. A weight exerts force of 100 N on steel

wire of cross-sectional area of A m2. If the extension produced is 2 mm, for 5 m length of wire then find the diameter of wire. (Y = 2 1011 N/m2)

[S-11] [4 M] Solution: Given: F = 100 N, A = A m2, l = 2 mm = 2 103 m, L = 5 m, Y = 2 1011 N/m2 To find: Diameter (D)

Formulae: i. Y = FL

Al

ii. A = 2

D

2


A = FL

Yl

= 11 3

100 5

2 10 2 10

A = 1.25 106 m2 ….(1) From formula (ii),

D = 4A

= 64 1.25 10

3.142

….[From (1)] = 1.261 103 m D = 1.261 mm Ans: The diameter of the wire is 1.261 mm.

30

Basic Science Physics (F.Y.Dip.Sem.-1) MSBTEPublications Pvt. Ltd. Target 18. If the change in volume of a gas

having original volume 100 cc is 0.001 cc at NTP, find Bulk modulus of elasticity. [W-12] [2 M]

Solution: Given: V = 0.001 cc, V = 100 cc At NTP, P = 76 cm of Hg P = hg = 76 cm 13.6 gm/c.c. 980 cm/s2 = 1.013 106 dyne/cm2 To find: Bulk modulus (K)

Formula: K = P V

V


K = 61.013 10 100

0.001

= 1.013 1011 dyne/cm2 Ans: Bulk modulus of elasticity is

1.013 1011 dyne/cm2. 1. Distinguish between elasticity and

plasticity. Ans: Refer 1.2.(e) 2. Distinguish between deforming force

and stress. Ans: Refer 1.2.(a), 1.3 3. Define compressibility. State its unit and

dimensions. Ans: Refer 1.6.(c) 4. Distinguish between Young’s modulus,

bulk modulus and modulus of rigidity. Ans: Refer 1.6.(f) 5. What are elastomers? Ans: Refer 1.8.(a) 1. A material wire elongates by 6% of its

original length when loaded. Calculate tensile strain for the wire.

2. Two wires of 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.

3. A steel wire of length 5.0 m and cross-

section 3.0 105 m2 stretches by the same amount as a copper wire of length 3.0 m and cross-section 4.0 105 m2 under a given load. What is the ratio of Young’s modulus of steel to that of copper?

4. For a cube of each side 5 cm, the upper

face is displaced by 0.65 cm due to a force of 0.25 N. Calculate shearing stress and shearing strain.

5. The bulk modulus of water is

3 106 N/m2. How much change in pressure is needed to compress it by 0.2% ?

6. The bulk modulus of a certain liquid is

3 106 N/m2. Find its compressibility. 7. A tangential force of 2100 N is applied

on a surface area 3 106 m2 which is at 0.1 m from fixed face. The force produces a shift of 0.7 mm of upper surface with respect to bottom. Calculate modulus of rigidity.

8. Young’s modulus of the material of a

wire is 7.1 1010 N/m2 and its bulk modulus is 7.7 1010 N/m2. Find the modulus of rigidity of the material of the wire.

9. A wire of length 3 m and diameter

1 mm is stretched so that its length increases by 2 mm. Find the decrease in the diameter of the wire if Poisson’s ratio for the material of the wire is 0.36.

10. A metal bar can withstand maximum

force of 24 106 N. If the bar has area of cross-section 0.04 m2, find the stress.

Additional Theory Questions

Problems for Practice

Contents

Chapter No. Topic Page No.

Unit - I: Properties of Solids 1

1 Properties of Solids 2

Unit - II: Properties of Liquids 32

2 Fluid Friction 33

3 Surface Tension 51

Unit - III: Thermal Properties of Matter 68

4 Modes of transformation of heat 69

5 Gas Laws 87

Unit - IV: Optics 106

6 Refraction of Light 107

Unit - V: Wave Motion 134

7 Wave Motion 135

8 Resonance 152

Model Question Papers

Model Question Paper I 167

Model Question Paper II 169

Model Question Paper III 171

MSBTE Question Papers

Question Paper – Summer 2014 173

Question Paper – Winter 2014 175

Question Paper – Summer 2015 177


31

Publications Pvt. Ltd. Target 11. An unknown weight is attached to the

lower end of wire of length 6 m, radius 0.7 mm, extends it by 0.8 mm. If Y = 1.8 1011 N/m2, find the unknown weight.

12. A metal block of volume 1 m3 is

subjected to change of pressure of 5 atmospheres. If the change in volume of the block, is 1.2 104 m3, calculate the bulk modulus.

(1 atmosphere = 105 N/m2) 13. A wire of length 6 m extends by 1 mm

when stretched by a load of 10 kg. Find the area of cross-section of the wire.

(Given: Y = 2 1011 N/m2, g = 9.81 m/s2) 14. A weight exerts force of 360 N on a

wire of cross-sectional area 0.02 cm2. Find the extension produced if the length of wire is 4 m.

(Given: Y = 2 1011 N/m2). 15. An unknown weight attached to the

lower end of wire of length 2 m and radius 0.1 mm, extends it by 0.2 mm. If Y = 2 1011 N/m2, find the unknown weight.

16. A weight exerts force of 200 N on steel

wire. If the extension produced is 2 mm, for 10 m length of wire, then find the diameter of wire. (Y = 2 1011 N/m2).

Answers to Practice Problems

1. 0.06 2. 8 : 1 3. 20 : 9 4. 100 N/m2, 0.13 5. 6 103 N/m2

6. 3.33 107 m2/N 7. 1 1011 N/m2 8. 2.6368 1010 N/m2

9. 24 108 m 10. 6 108 N/m2

11. 36.95 N 12. 4.167 109 m2

13. 29.43 107 m2

14. 3.6 103 m 15. 0.6284 N 16. 2.52 mm

Documents

Best of luck to all the aspirants! - Target · PDF fileNo part of this book may be reproduced or transmitted in any form or ... (Reference of answer to each question is ... Pascal’s