applied physics notes for civil branch by prof. atul waghmare

8/11/2019 Applied Physics Notes for Civil branch by Prof. Atul Waghmare

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CHAPTER 1

MOTION

[A] Rectilinear and Angular Motion

Q. State three equations of motions in case of rectilinear motion.

Ans: The three equations of rectilinear motion are

1. V = u + at

2.

S = ut +1

2 at2

3. V2 = u2 + 2as

Where, u = Initial velocity

V = Final velocity

t = Time taken by a particle to change velocity from u to v

s = Distance travelled in time t

a = Uniform acceleration

Q. State the equation of distance travelled by a particle (body) in nth

second.

Ans: The equation of distance travelled by a body in nth second is

Snth = u +

2 (2n – 1)

Where, u = Initial velocitya = Uniform acceleration

Snth = Distance travelled in time n.

If the body freely falling under gravity then the equation of distance

travelled by the body in nth second will be

Snth = u +

2 (2 n – 1)



2

Q. State the equations of motion for a body falling freely due to gravity

along with the meaning of each symbol.

Ans: The equations of motion when a body falling freely due to gravity is as

follows.

1.

V = u + gt

2. S = ut +1

2 gt2

3. V2 = u2 + 2gs

Where, u= Initial velocity

V = Final velocity

g = Gravitational acceleration

t = Time taken

s = Vertically downward distance covered by body.

Q. State the equations of motion when a body moves vertically upward

(against gravity)Ans: The equations of motion when a body moves (through) vertically upward are

as follows

1. V = u – gt

2.

S = ut – 1

2 gt2

3. V2 = u2 – 2gs

Where, u = Initial velocity

V= Final velocity

g = Gravitational acceleration

t = Time taken

s = vertically upward distance coved by body



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Q. Define following terms

1. Uniform velocity

2. Uniform acceleration

3. Uniform retardation

Ans: A) Uniform velocity

“ If a body covers equal displacement in equal interval of time, then it is said

to be in uniform velocity.”

For e.g. A car moving with same speed (10 m/s) in the same direction.

B) Uniform acceleration

“If the acceleration of a body is uniform in magnitude & direction with

respect to time then it is called as uniform acceleration.”

For e.g. A ball falling down.

C) Uniform retardation

“If the acceleration of a body is negative & uniform in magnitude &

direction with respect to time then it is called as uniform retardation.”

For e.g. A ball moving up.


1. Angular displacement

2. Angular velocity

3. Angular acceleration

Ans:

1.

Angular displacement:

“It is defined as the angle through which radius vector turns when the

particle in circular motion moves from one position to other .”

It is denoted by . Its S.I. unit is radian (rad).

Unit: m/s or cm/s

Unit: m/s2 or cm/s2



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2.

Angular velocity:

“The rate of change of angular displacement with respect to time is called

as angular velocity of a particle”.

It is denoted by = /t

Its S.I. Unit is red/sec.

3.

Angular acceleration

“The rate of change of angular velocity with respect to time is called as

angular acceleration of a particle.”

It is denoted by ∝ Its S.I. Unit is red/s2

∝ = /t

Q. State the relation between linear velocity & angular velocity.

Ans: The relation between linear velocity and angular velocity is given by.

V = r

Where‟ V = Linear velocity

r = Radius

= Angular velocity

Q. Derive the relation between linear velocity and angular velocity.

Ans: Consider a particle particle performing (undergoing) uniform circular

motion. It moves from point A to point B in time T shown in figure.



5

We have,

Linear velocity =

V = s/t Were, S = Linear displacement

∴ = /t (∴ s = )

( = angular displacement)

= x /t (∴

= )

∴ V =

Where, = Radius of circle

= Angular velocity

Thus, linear velocity is radius times the angular velocity.

Q. State the three equations of angular motion along with meaning of each

symbol.

Ans: The three equations of angular motion are as follows

1. I = I + ∝ t

2. = I t +1

2 ∝ t

2

3. F2= I

2 + 2 ∝

Where, = Angular displacement, α= Angular accleration

I = Initial angular velocity

F = Final angular velocity

t = Time taken by a particle to change velocity from

I to F



6

Q. State the equation of angular distance travelled by a body in nth

second.

Ans: The equation for angular distance travelled by a particle in nth second is

given by,

nth = 1 +∝

2 (2n – 1)

Where,

nth = Angular distance travelled by a particle body in nth second

1 = Initial angular velocity

∝ = Constant angular acceleration



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[B] KINETICS


1. Momentum

2. Impulse

3. Impulsive force

Ans:

1. Momentum:

Momentum of a body is “the amount of motion that a body or a particle

has.” OR

It is defined as “the amount (quantity) of motion possessed by movingbody”.

Mathematically, momentum is the product of mass & velocity. It is

represented by

Momentum = Mass x Velocity

P = mxv

Its S.I.unit is N sec or Kgm/sIt is vector quantity

2. Impulse:

It is defined as the “change in momentum”

Impulse = change in momentum

Impulse = mv- mu

Where, m= Mass of a body

u= Initial velocity

v= Final velocity

Its S.I.unit is kg m/s or N sec,

It is vector quantity.



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3. Impulsive force:

Impulsive force is defined as “the force which acts over short time;

producing rapid change in the motion of a body.”

Mathematically, impulsive force is the rate of change of impulse withrespect to time.

Impulsive force =

=(−)

=

(−)

Its S.I.unit is N

Q. State Newton’s First law of motion OR State law of inertiaAns: Statement: It states that “every body continues in its state of rest or of

uniform motion in a straight line, unless, it is acted upon by some external

unbalanced force”

This means in the absence of unbalanced force, every object has a tendency

to resist any change in its state of rest of motion.

This tendency is inertia so this law is also known as law of inertia.

Examples: 1. A car moving with uniform velocity continues its motion until

we apply breaks

2. A book kept on table, remains at rest until we apply force

Q. State and explain Newton’s second law of motion.

Ans: Statement: It states that the “rate of change of momentum of a body is

directly proportional to the applied force & takes place in the direction of

force.”

Thus,−

∝ F

(−)

∝ F or

(−)

= Constant x F

ma = F [−

= a & assuming proportionality constant = 1]



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Where, F = Force

mv = Final momentum

mu = Initial momentum

t = Timem = Mass

a = Acceleration

u = Initial velocity

v = Final velocity

Q. State Newton’s third law of motion (Law of action reaction).

Ans: Statement: “For every action there is equal and opposite reaction.”

The means that for every force there is a reaction force that is equal in size

but opposite in direction.

Ex: Action: While swimming our body push the water backwards.

Reaction: The water pushes our body forward.

Action: When we hit the ball on wall

Reaction: It bounces back

Q. Explain recoil of a gun? OR What is recoil velocity of a gun & derive

the equation of recoil of gun.

Ans: “If a bullet is fired from a gun then bullet moves (shoot) out with a large

velocity & at the same time gun moves back (jerk) with the little velocity.”

“This backward movement (jerk) is known as recoil of the gun” and “the

velocity with which the gun moves backward is called the recoil velocity of a

gun”

1.

Before firing

Let Mg be the mass of the gun & M b that of the bullet



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2.

After firing

Let „Vg‟ be the velocity of the gun & „V b‟ that of the bullet.

But, according to law of conservation of momentum.

Initial momentum of Final momentum of

gun & bullet gun & bullet

But, initial momentum of the gun & bullet is equal to zero because they are

initially at rest.

Therefore,

Final momentum after firing = M b V b + Mg (- Vg) = 0

The negative sign indicates that the gun move in the backward direction

∴ M bV b – Mg Vg = 0

∴ M bV b = Mg Vg

∴ V b=

Mg Vg

Mb

This is the equation of recoil velocity.



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[C] WORK, POWER &ENERGY

Q. Define Work, Power & Energy. Write their S.I. Unit.

Ans: Work: “Work is defined as the product of force acting on the body & the

displacement produced.”

Work = Force x Displacement

The S.I. Unit of work is Newton meter

M.K.S. unit is joule

C.G.S.unit is erg.

Power: “ Power is defined as the rate (capacity) of doing work ”

Power =

The S.I. Unit of power is “watt”

Energy: “The energy of a body is defined as its capacity to do work ”

S.I. unit of work is joule or erg

All forms of energy are transferable

E.g. Light energy can be converted into electrical energy.

Q. Define potential energy & kinetic energy with equation.

Ans: Potential energy:“The energy possessed by a body due to its position is

called as potential energy.”

Potential energy = weight x height

∴ P. E. = mgh

S. I. unit Nm or joule



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Kinetic energy: “The energy possessed by a body virtue of its motion is

called as kinetic energy”.

K.E. = 12 mv2

Where, m = Mass of body

v = Velocity

S.I. unit Nm or joule

Q. State & explain work energy principle.

Ans: Statement:“ It states that the work done by a system of force acting on a

body between any two points is equal to the change in kinetic energy of a

body between these same two points.”

Consider a body of mass m

Let, F be the force on it

V1 be the velocity at position P1

V2 be the velocity at position P2

d. be the displacement of a body

∴ K.E. at P1 = 1/2 mv12

∴ K.E. at P2 = 1/2 mv22

Change in kinetic energy = 1/2 mv22 – 1/2mv1

2

= 1/2 m (v22 – v2

1)

Work done = Force X Displacement= [F] x d

= (m x a) x d

Using equations of motion, it can be proved that

Work done = change in kinetic energy



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Q. Define following terms.

1. Projectile motion

2. Trajectory

3. Angle of projection

4. Time of flight5. Range of projectile

6. Maximum height of projectile

Ans:

1. Projectile motion:“ Projectile motion of a body thrown (projected) in air at

angle [less than 900 ] with the horizontal.”

Ex: 1. Motion of football kicked in air. 2. Motion of cricket ball a batsman

hits a six

2. Trajectory:“The path along which projectile moves is called as trajectory.”

3.

Angle of projection:“ It is defined as it is the angle made by the velocity of

projection with the horizontal at the original point „O‟”

Angle of projection

= tan-1

(

)

Where = Angle of projection with horizontal at the origin

H = Maximum height of a projectile

R = Horizontal range of projectile

4.

Time of flight: “The time taken by the projectile to reach the ground from

the time it was projected is known as time of flight”.

It is given by

T =

Where, T = Time of flight

u = Initial velocity

θ = Angle of projection

g = Acceleration due to gravity.



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5. Range of projectile:“The total horizontal distance covered by a projectile

is called as a range (R).”

R = (V

2

sin2) / g (If = 45

0

C sin2 = 1)R = Range of a projectile

V = Velocity of projectile

= Angle of projection

g = gravitational acceleration

Maximum Height of projectile: The maximum vertical distance covered

by a projectile from the ground level is called height of a projectile

R = (V2 sin2θ) /2 g

R = Range of a projectile

V = Velocity of projectile

θ = Angle of projection

g = gravitational acceleration



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[D] CIRCULAR MOTION & PROECTILE MOTION

Q. Define

1. Circular motion

2. Uniform circular motion

3. Centripetal (radial) acceleration

Ans:

1. Circular motion:“Circular motion is defined as the motion of a particle

along the circumference of circle.”

Ex: a) Moon revolving around the earth

b) Electron revolving around the nucleus of an atom.

2. Uniform circular motion: “ It is defined as a motion of particle along the

circumference of circle with constant speed.”

Ex: a) Moon revolving around the earth

b) Electron revolving around the nucleus of an atom.

3.

Centripetal (radial) acceleration:“The acceleration in uniform circular

motion which is directed along the radius & towards the centered of a circle

is called as centripetal acceleration”.

Q. Define centripetal force. Give two examples

Ans: Centripetal force: “ It is defined as the force acting along the radius

towards the centre of the circular path which keeps the particle in uniform

circular motion.” Ex: Electron revolving around the nucleus - In this case the

electrostatic force of attraction between electron & proton plays the

role of centripetal force.

Moon revolving around the earth - The gravitational force of

attraction between earth and moon plays the important role



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Q. Define Centrifugal force. Give two examples. Write applications of

centrifugal force (circular motion)

Ans: Centrifugal force: “ It is defined as the force acting on a particle performing

uniform circular motion which is directed away from the centre and alongthe radius of a circular path.”

Ex: 1. Person sitting in merry go round or giant wheel, experiences outward

pull.

2. Motor cyclist driving in a artificial death well in a circus experiences

outward pull because of his high speed

Applications of Centrifugal Force:

1.

Centrifugal force used in banking of roads to avoids skidding of speedy

vehicle along a curved road.

2. Centrifuged pump is used to transfer liquid; centrifugal force is used in

this process.

3. A cream separator is used in the dairy works on the principle of centri

fugal force.

4. Centrifugal blower is used to blow air.

5. Centrifugal force can be used to generate artificial gravity for rotating

space stations6. Centrifuges are used in science & industry to separate substances.

Q. Differentiate between Centripetal force & Centrifugal force.

Ans:

Centripetal force Centrifugal force

1 It is the force acting on a particlein uniform circular motion which is

along the radius & towards thecentre of circular path

It is the force acting on particle performing uniform circular motion

which is along the radius & away fromthe circular path

2 This is a real force (has physical

existence)

This is an imaginary (pseuelo) force.

3 This force is acting towards the

centre

This force acting away from the centre



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CHAPTER 2

NON-DESTRUCTIVE TESTING OF MATERIALS

Q. Define ultrasonic waves & states its frequency range.

Ans: Definition:

“The ultrasonic‟s are the sound waves having frequency more than 20 kHz.”

The frequency range of UV- waves is more than 20 kHz.

Q. State the range for infransonic, sonic & ultrasonic waves

Ans: Type of wave Range of frequency

1.

Infrasonic Less than 20HZ

2.

Sonic Between 20 HZ to 20 kHz

3.

Ultrasonic More than 20 kHz

Q. State any four properties of Ultrasonic waves. (Any Four)

Ans: Properties of Ultrasonic Waves:

1. Frequency of these sound waves is more than 20khz

2.

Ultrasonic waves has shorten wavelength3. Ultrasonic sound waves carry high amount of sound energy

4.

The speed of propagation of ultrasonic waves increases with increase in

frequency

5. Ultrasonic waves show negligible diffraction

6. Ultrasonic waves travel over long distance without considerable loss

7. Ultrasonic waves undergo reflection and refraction at the separation of

two media

8.

If ultrasonic waves are passed through a fluid, then temperature of thefluid increases

9. Ultrasonic waves travel with constant speed through a homogeneous

material

10. Ultrasonic waves possess certain vibrations which are used as good

massage action in case of muscular pain



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Q. State engineering, scientific and medical applications of ultrasonic

waves.

Ans: Applications of ultrasonic waves:

a)

Engineering Applications:1.

Ultrasonic waves are used for cleaning or washing of delicate parts of

machine like watches, electronic components etc.

2.

Ultrasonic waves are used in Non Destructive Testing to detect flaw in

material without damaging.

3. Ultrasonic waves are used in drilling holes in the metal.

4. Ultrasonic waves are used in cutting metal parts in industry.

b)

Scientific Applications:

1.

Ultrasonic waves are used to determine the depth of sea.

2. Ultrasonic waves are used in sound navigation and ranging (SONAR.)

Because of high directionality, the ultrasonic waves are used to locate

obstacle and to determine their distance in sea.

3. Ultrasonic waves are used to produce some alloys like lead aluminum

which cannot be produced by conventional techniques.

c)

Medical Applications:

Ultrasonic waves are used by doctors to monitor the development of

unborn babies, diagnosis of heart disease and other disorders in patient

Q. Explain the production of ultrasonic waves using piezoelectric method.

Ans: Principle: When the electric field is applied across the crystal its dimension

changes and when alternating PD is applied across crystal then the crystal

sets into elastic vibrations

Piezoelectric effect:

When mechanical pressure is applied; the opposite charges are developed on

the parallel faces of certain crystals like quartz, tourmaline, barium titanate

and Rochelle salt etc.



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Method:

1.

A chip of piezoelectric crystal like quartz is placed between two plates as

shown in fig.

2.

A suitable Hartely oscillator is connected across it.

3. The electric oscillations along the electric axis produce mechanical

vibrations along the mechanical axis.

4.

Then the frequency of oscillator is increased.

5.

At a particular frequency of oscillator the oscillator frequency becomes

equal to natural frequency of vibration of crystal.

6. Then the crystal thrown into resonance vibrations & ultrasound is given out

(produced)

Q. What is meant by non destructive testing (NDT)?

Ans: “Non destructive testing is the method of examining the materials for

internal as well as surface discontinuities without the destruction of

material.”



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Q. State the advantage of non destructive testing (NDT) methods.

Ans: Advantages of NDT:

1. Material can be used for its intended purpose after testing.

2.

It is used for quality control tool, because material can be tested for all possible parameters.

3.

It is used for quality assurance tool; because after testing material one can

check its suitability at a particular place.

4. The testing of material is possible before its manufacturing.

5. The testing is possible after manufacturing.

6. The testing of material is possible during servicing of components

7.

Hundred percent examination of material is possible.

Q. State the limitations (Disadvantages) of Non- destructive testing. (NDT)

Ans: Limitations of NDT:

1. Minimum two methods for complete examination of the material are

required. With only one method, testing for all parameters of materials is

not possible.

2.

Trained and certified persons are authorised to conduct the test (level I,

II, and III) as per American Society for Non-destructive Testing (ASNT).

3. Cost of equipments is high and thus testing charges are more as

compared to destructive testing.

4. Qualitative testing is possible; however, quantitative testing is difficult.

5.

NDT interpretations are relative. One should know the standard results

first.

Q. Name the popular NDT methods used in mechanical industry.

Ans:

1. Visual testing (optical testing) (VT)

2. Liquid penetrate testing (LPT)

3. Eddy current testing (ET)

4.

Radiography testing(RT)



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a. X-ray radiography

b.

Fluoroscopy

c.

Gamma ray radiography

5. Ultrasonic scanning and flaw detector method(UT)

6.

Magnetic particle testing (MPT) 7. Thermography method

Q. Explain the criteria for selection of NDT method OR State any four

factors on which NDT method can be selected OR On which factors

selection of NDT method depends.

Ans: Following parameters are considered while selecting NDT methods

1. Codes or standard requirements

2.

Specifications of the material to be tested, for example, nature ofmaterial, its size & shape.

3. Types of disorders to be detected also depends on nature of disorders

4.

Testing also depends on manufacturing process of material to be tested

5.

It is also depending on the instruments available for testing

6.

Total cost required to test the material

Q. Describe Liquid penetrate testing (LPT) with its

1. Principle

2. Experimental procedure.

Ans: Principle:“ It works on the principle of capillarity

Experimental procedure:

Following steps are involved in liquid penetrant test or inspection.

1. Surface penetration:

Initially the surface of specimen is cleaned. Because the presence of

flakes, dirt, grease etc on the surface of work pieace prevents penetrant to

slip into the cracks. This give wrong information



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2. Application of dye penetrant: Suitable fluorescent dye is mixed in

penetrant so that its viscosity remains low. This dye penetrant is applied

evenly on specimen. Due to capillary action the penetrant goes into the

surface open discontinuities. It takes some time. In general case this

dwell time is 20-30 minutes.

3. Removal of excess penetrant: After dwell time is over, the excess

penetrant is removed from the surface carefully.

4.

Application of developer: A thin layer of developer is applied over the

surface. The role of developer is to pull the trapped penetrant out of the

crack this provides good visibility of crack.

5.

Inspection and evaluation of defect: Surface of the specimen is seen

under white light or ultraviolet or laser light. The crack can be visualized

under light.



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6. Post cleaning: After inspection the surface of the specimen is cleaned &

the specimen can be used for its intended purpose.

Q. What is role of developer in LP testing?

Ans: The role of developer is to pull the trapped penetrant out of the crack this

provides good visibility of crack

Q. Write advantages and disadvantages of LPT.

Ans: Advantages of LPT:

1.

Cheaper method2. No power required

3.

All types of material can be examined

4.

It is independent on size and shape

5.

Method is suitable for wide and varied application.

Disadvantages of LPT:

1. It is used only for porous materials

2.

It is used only for detecting surface defects3. It‟s running cost is more

4. It is slow process

Q. State applications of LPT in industry.

Ans: Applications of L.P.T.:

1.

Rotor discs and blades are checked by LPT2. Aluminum casting and forgings, Pistons and cylinder heads in

automotive industry are checked by LPT.

3. Moulded plastic parts can be checked.

4. Bogie frames of railway locomotives can be checked.

5. Electrical ceramic parts can be checked.



24

Q. State the principle, procedure of ultrasonic testing.

Ans:

Principle: When ultrasonics waves are introduced into a material, it gets

reflected, transmitted, scattered from surface or flaw

Procedure:

1. At first, the ultrasonic equipment is calibrated before testing the specimen

by using the standard blocks as per recommendation.

2. Then the ultrasonic signals are generated from the transducer.

3. These generated ultrasonic signals can pass through specimen

4. Then the reflected signals from the crack flow are received and amplified

5. Then the amplified signal is applied to the cathode ray tube (CRT)

6. CRT displays the details of specimen (material) i.e. the material is OK or

defective

Q. Name different types of ultrasonic testing methods

Ans: Types of ultrasonic testing

1. Transmission ultrasonic testing method

2. Pulse echo UT method

Q. State advantages of ultrasonic testing.

Ans: 1. It has high sensitivity and reliability

2. It can be used for testing all types of material

3. The equipment is portable and harmless

4. The result are instantaneous



25

Q. State the application of ultrasonic testing.

Ans: Application of UT:

1. This method is used to detect flaws in all metals, rubber, tyres, concrete,

wood, composite, plastic.2.

It is used to test casting, forging, welding, rolling, heat treatment.

3.

It is used to detect crack (flaws) in Air craft.

4.

It is used in tunnel inspection.

5. It is used in bridge inspection

6. It is used to detect surface discontinuities.

7. It is used in monitoring of thermal and atomic power plant

8.

It is used to determine thickness.

Q. If the crack is on the surface of job or near the surface which of the

following method is used. Why? UT, LPT, NDT

Ans: If the crack is on the surface of job or near the surface, then LPT method is

useful because.

1. It is cheaper method

2. No power required;

3.

By using this method all types of material can be examined



26

CHAPTER 3

ACOUSTICS AND INDOOR LIGHTING

[A] ACOUSTICS

Q. Define Echo.

Ans: “The echo is defined as the same sound heard again after an interval of

1/10th of second due to reflection of the original sound from a surface which

is at a distance greater than 16.5 m from the source of sound.”

Ex. A doctor stethoscope is a good example of multiple reflections of sound.

Q. Define Reverberation time.

Ans: “The time for which the sound persists in a hall even after the soure is cut

off is called as reverberation time.”

Reverberation time also depends upon the type of sound produced, e.g.

human speech, musical sound, noise, etc.

Q. State Sabine’s formula with usual meaning in acoustics

Ans: The reverberation time (t) given by Sabine‟s formula,

t = 0.164V/A

Where, V = Total volume of the hall in m3

A = Total absorption of sound in the hall = ∑ a.s

i.e. t = 0.164V/∑ a.s



27

Q. Distinguish between echo and reverberation.

Ans:

Q. State any four factors affecting acoustical planning of building and state

how they are to be adjusted for good acoustics.

Ans: To have best sound effects in a hall the following factors should be

considered

1. Intensity and Loudness: In order to hear sound clear in the hall, intensity

and loudness of sound produced should be sufficiently high. Intensity should be uniform throughout the hall.

a.

The length of hall should be more as compared to its breadth

b.

The wall surface behind the loudspeaker should form large parabola

2. Echoes: The repetition of sound due to reflection from an obstacle is called

echo. Echoes spoil the quality of sound.

Echoes can be avoided by using sufficient sound absorbing material in the

hall.

3.

Reverberation: The persistence of sound due to multiple refletion in hall

even after the source is cut off is called reverberation.

Reverbation creates confusion and affect quality of sound. To reduce the

effect, sufficient sound absorbing material should be used.

Echo Reverberation

1 The repetition of sound due toreflection from an obstacle is calledecho

It is the persistence of sound due tomultiple reflections in a hall even afterthe source of sound is cut off

2 Distance between the source and

obstacle is large

Distance between the source and the

surface is small

3 Reflected sound is clear Reflected sound is not clear



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4. Reverbation time: The time for which sound persist in a hall even after the

source of sound is cut off is called as reverbation time. Reverberation time of

good hall must be less than 2 second

5.

External noise: The outside noise can mix with the sound of speech ormusic in the hall and create confusion for the audience. This can be

decreased by making the hall soundproof cabins for machinery and type

writters

Q. State the condition for good acoustics in auditorium. Or State the

requirements of good acoustics of building.(Any four)

Ans: The conditions (Requirements) for good acoustics:

1.

The sound produced should be clear uniformly distributed throughout the

hall.

2. The sound produced should be heard at all points in the hall with

sufficient intensity.

3. There should not be focusing of sound

4.

There should not be any dead spots or silence zones in the hall.

5.

Reverberation should be minimum.

6.

Echo should be absent

7. There should not be overlapping of sound waves.

8. Echelon effect should be absent

9. The external noise should not enter the hall



29

[B] INDOOR LIGHTING

Q. Define luminous flux. State its SI unit

The amount of light which flows from a source per second is called as

luminous flux. Unit - lumen

Q. Define Luminous intensity (I) or illumination power. State its S.I. unit.

Ans: Luminous intensity or Illumination power: “The luminous flux emitted

per unit solid angle by a source in a particular direction is called as

luminous intensity or illumination power.”

Unit is lumen/steradian or Candela (cd )

I =

I =∅

Q. Define Illumination or Intensity of illumination of surface (E).

Ans: Illumination or intensity of Illumination of surface: “The Luminous flux

per unit area of a surface is called as illumination or intensity of

illumination of surface.”

E =

E =∅

Its unit is Lumen per square meter (Lumen/m2)



30

Q. State and explain inverse square law of illumination with diagram.

Ans: Inverse Square Law of illumination: The intensity of illumination of a

surface due to a point source of light is inversely proportional to the square

of distance of the surface from the source

E 1/r2

Let S be a point source which is emitting light in all directions and Q be the

amount emitted in unit time.

Let, A and B spherical surfaces with centre S.

r 1 – radius of surface A and r 2 - radius of surface B

The intensity of illumination E1 on surface is given by

E1 =

i.e. E1 = Q

4r 12

Similarly, the intensity of illumination E2 on surface B is given by,

E2 =

. E2 = Q

4r 22

∴ E2 r 22

r 12

∴ E 1/r2



31

Q. Draw a ray diagram of Bunsen’s Photometer. Write principle,

construction and working

Ans: Principle of Bunsen Photometer:

The Bunsen photometer is based on the principle of photometry i.e. theequivalence of luminance for two surfaces, Mathematically, E1 = E2

Construction: Bunsen photometer consist of paper screen centrally

mounted in box. The paper screen is spotted with grease so that spot become

semi transparent. Two plane mirror are kept in an inclined position on eitherside of the screen as shown in fig. The box is provided with two coaxial

windows. The box mounted on a vertical stand of adjustable height.

Working: A source of light of illuminating power I1 is placed at a distance

of r 1 from screen. Another source is placed on other side of screen. The

position of second source is adjusted in such way that the image of central

grease spot, seen in the two mirrors appears equally bright. The distance of

the second source from the screen is measured as r 2.

Then compare the luminous intensities of the source by relation

I1 / I2 = r1/ r2



32

Q. Write any two application of Bunsen photometer

Ans: Applications of Bunsen Photometer

1. If luminous intensity of one source is given then luminous intensity of

other source can be determined2.

It is used to compare the illuminating power of two given sources

Q. State (Name) different lighting schemes

Ans: Different lighting schemes

1. Direct lighting system

2. Indirect lighting system

3. Semi indirect lighting system

Q. State and explain factors affecting indoor lighting system

Ans: Factors affecting indoor lighting system are

1. Efficiency of source

2. Utilisation factor

3. Maintenance factor

4. Space height ratio

Efficiency of source: In light sources like electric bulbs, electrical energy is

converted to light energy. However, the total electrical energy is not

converted to light energy. A fraction of the electrical energy is consumed in

heating.

The ratio of the luminous flux obtained from a light source to the electrical

energy utilized is called efficiency of the source.

Utilisation factor: The whole amount of light emitted by the source doesnot reach the work plane. A fraction of light gets absorbed by the walls.

Ceiling and remaining part of light reaches the work plane.

The ratio of luminous flux obtained on the work plane to the luminous flux

emitted by the light source is called utilization factor.



33

Maintenance factor: When the walls, ceilings of the room are cleaned

properly and nicely painted, the illumination of hall/room increases.

The ratio of illuminance obtained under existing conditions to the

illuminance that will be obtain when everything is clean is called maintaince

factor.

Space height ratio: Good lighting system requires uniform distribution of

light on the working area

The ratio of horizontal distance between two consecutive lamps to their

mounting height from the work plane is called space height ratio.

Q. Which lighting system is preffered in workshop? Why?

Ans: Indirect lighting system is used in workshops. Because to avoid shadow

effect and glaring effect

Q. State the factors which are important while selecting a lighting scheme.Ans: While selecting lighting scheme following factors are to be considered.

1.

The particular working place and its available area.

2. The timing for working and availability of natural light.

3. The specific requirement such as commercial purpose or day to day

work.

4.

The economy and provision of budget.

5.

Whether it is a public or private place.



34

CHAPTER 4

MODERN PHYSICS

[A] PHOTO-ELECTRICITY

Q. Define Photoelectric effect.

Ans: “When light of suitable frequency is incident on metal surface, electrons are

emitted from it. This effect is called as photoelectric effect.

Q. State Planck’s hypothesis.

Ans: According to Planck‟s hypothesis, “Emission and absorption of energy can

occur only in discrete amounts of bundles.” Planck called this discrete

amount or bundle of energy as quanta or photon. The energy of each photon

is E=hv. So for emitted or absorbed energy,

E = nhv

Where, h is Planck‟s constant, has a value of 6.626 x 10-34 J s

v is the frequency,

n is an integer = 1, 2, 3, ----

Q. What is photon? State the properties of photon

Ans: Photon: It is small pocket or bundle of energy.

The properties of Photon are as follows

1. It is a mass less particle

2. It is electrically neutral

3. It moves with speed of light

4.

Photon do not ionise



35

Q. Define a) Stopping Potential b) Photoelectric Work Function c)

Threshold frequency d) Threshold Wavelength.

Ans: a) Stopping Potential: “It is reverse or negative potential which reduces

the photoelectric current to zero. It is denoted by V s.”

b) Photoelectric Work Function: “ It is the energy required to detach the

electron from the metal.”

It is given by w 0 = hv 0

c) Threshold frequency: “ It is the minimum frequency of incident light at

which emission just begins.

It is given by v 0 = c/ v 0

d) Threshold wavelength: “It is the maximum wavelength of incident light

at which emission just begins

It is given by 0 =

Q. State the properties (Characteristics) of Photoelectric effect.

Ans: Following are the characteristics of photoelectric effect

1.

A metal emits electrons only when the incident (light) radiation has

frequency greater than critical frequency (v0)

2.

A photoelectric current is directly proportional to the intensity of light

and independent of frequency

3.

The velocity of photoelectron is directly proportional to the frequency of

light and independent of intensify.



36

4. For a given metal surface, stopping potential is directly proportional to

the frequency

5.

This process is instantaneous. i.e. the emission of photoelectrons starts at

the moment light is incident on the metal surface.

Q. Derive Einstein’s photoelectric equation.

Ans: 1. According to Planck‟s Hypothesis, Energy is radiated and also travels in

the form of bundles or quanta known as photons

2. Each photon carries an energy h v

3. When a photon of incident light radiation interacts with an electron insidean atom, the whole amount of energy is absorbed by the electron.

4. The electron uses part of the incident energy, to make the electron free

from the metal (work function) and the rest part of the energy is converted

into kinetic energy

Thus, hv = 0 +1

2 mv2

max

hv = hv 0 +1

2 mv2

max

∴ 1

2 mv2

max = hv – hv0

mv

2max = h (v – v 0)

Where,1

2mv2

max is the maximum kinetic energy of electrons,

h is Planck‟s constant,

v is the frequency of incident light,

v0 is the threshold frequency

This is the required Einstein‟s Photoelectric equation



37

Q. State the Einstein’s Photoelectric Equation and explains the significance

of each term involved in it.

Ans: The Einstein‟s Photoelectric Equation is given by

mv

2max = h (v – v 0)

Where,1

2mv2

max is the maximum kinetic energy of electrons,

h is Planck‟s constant,

v is the frequency of incident light,

v0 is the threshold frequency

Q. Why electrons are not emitted from the surface of metal plate, if

frequency of incident radiation is less than threshold frequency v 0

Ans: 1. According to Einstein‟s Photoelectric Equation

12 m v2 max = h(v-v0)

2. If v < v0, then,12 mv2

max becomes negative

3. But for the emission of photoelectrons the kinetic energy must be positive

4. Thus, due to negative value of kinetic energy the photoelectrons are not

emitted from the surface of metal plate

Q. Define Photoelectric cell. State it’s any two applications.

Ans: “A Photocell is an electronic device in which light energy gets converted

into an electrical energy.” It is used to produce a current or voltage when

exposed to light or other electromagnetic radiation.



38

The two applications of Photoelectric cell are-

1. They are used in exposure meters

2. They are used in burglar alarms

Q. Explain the construction and working of photoelectric cell.

Ans: Principle:“The working principle of photoelectric cell is photoelectric

effect. It is an electronic device which converts light energy into an

electrical energy.”

Construction:

1.

A photocell is an evacuated tube consisting of two electrodes i.e. cathode

and anode

2. Cathode is concave in shape called an emitter, made from a material that

emits electrons easily. Whereas, anode is in the form of thin rod as shown

in the fig

3.

Cathode is connected to negative terminal of battery. An ammeter is

connected in the circuit to measure the current flowing through the

photoelectric cell



39

Working:

1. When light is allowed to fall on cathode it emits photoelectrons.

2. The photoelectrons are attracted by anode

3.

The photoelectric current starts flowing through the circuit and themilliammeter shows the deflection.

Q. Draw the diagram of Photoelectric cell.

Ans:

Q. State the applications of photoelectric cell. (Any four)

Ans: The applications of photoelectric cell are

1.

It is used in Burglar alarm

2.

It is used in fire alarm

3. It is used in Lux meter to determine the intensity of light

4.

It is used in Exposure meter

5. It is used in automatic control of traffic signal

6.

It is used in automatic counting objects

7.

It is used in automatic shutting and opening doors

8.

It is used in detecting flaws in metals

9. It is used in television sets



40

Q. State the principle of light dependent resistor (LDR) or photo resistor.

Ans: Principle: “The electrical resistance of LDR decreases as the intensity of

incident light increases.”

Q. State the applications of LDR.

Ans: Application of LDR:

1.

It is used for detecting ships and air craft‟s by the radiations given out.

2. It is used as flame, smoke and burglar detectors.

3. It is used as a automatic lighting controls for street light.

4.

It is used in Camera for exposure control.5. It is used in Xerox machine- to controls the density of toner.

6.

It is used in Colorimetric test equipment.

7.

It is used as automated real view mirror etc.

Q. Draw the symbol of LDR.

Ans:



41

[B] X-RAYS

Q. What are X-rays?

Ans: X ray is electromagnetic radiations of short wavelength ranging from 0.01

A0 to 100A0.

1 A.U.= 1 Angstron unit = 10-10 metre.

Q. State the properties of X-rays.(Any four)

Ans: Properties of X-rays:

1.

X rays are electromagnetic radiations of very short wavelength

2. X rays travels with speed of light

3. X rays are electically neutral

4. X rays can ionises gas

5. X rays can damage or kill living cells

6.

X rays are invisible

7.

X rays can affect photographic plate8. X rays are not deflected by electric or magnetic field

Q. State the formula for minimum wavelength of X-rays also state the

meaning of symbols used in it

Ans:

= /. Where, = minimum wavelength of x rays

h= planks constant, V= applied voltage

C = velocity of light, e =charge of electron



42

Q. Explain the production of X-rays using Coolidge X-ray tube.

Ans: X ray is produced whenever fast moving electrons strike a high atomic

weight solid (tungsten) in vacuum.

1.

The Coolidge X-ray tube is shown in figure

2. It consists of a highly evacuated hard glass tube containing cathode (K)

and the target (T) attached to anode (A).

Working:

1. When the cathode is heated by electric current it produces electrons due to

thermionic emission.

2. The beam of electron is then focussed on the anode (Target).

3. The electrons from cathode are then accelerated by applying high voltage

between cathode and anode using step up transformer.

4. When these fast moving electrons are suddenly stopped by tungesten

anode, they lose their kinetic energy and X rays are produced from the

target.

5. Some amount of kinetic energy is converted to large amount of heat.



Q. State the applications of X-rays.

Ans: Applications of X-rays:

A) Industrial [Engineering] applications

1. X rays are used to detect defects within metals, machine parts, and

castings etc.

2.

X rays are used to detect manufacturing defect in rubber tyres or tennis

ball in quality control.

3.

X rays are used to detect cracks in the wall

4. X rays are used to detect the cracks in the body of aeroplane or motor car

5. X rays are use to distinguish real diamond from duplicate one

6.

X rays are used to detect smuggling gold at airport and ship yard

B) Scientific Research applications:

1. X rays are used to investigate the structure of the atom.

2. X rays are used for analyzing the structure of complex organic molecules.

3. X rays are used in determining the atomic number and identification ofvarious chemical elements.

C) Medical Applications:

1. X rays are used in detecting fractures in bones

2. X rays are used to cure skin diseases and destroy tumors.

3. X rays are used to detect bullet position inside the body.

4. X rays are used to cure diseases like cancer

applied physics notes for civil branch by prof. atul waghmare

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