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15O102

ENGINEERING PHYSICS LABORATORY

S1 B.E./B.Tech. (Common to all Branches)

Year of Release: 2015

BANNARI AMMAN INSTITUTE OF TECHNOLOGY

SATHYAMANGALAM – 638 401

DEPARTMENT OF PHYSICAL SCIENCES

LAB MANUAL



15O102

ENGINEERING PHYSICS LABORATORY

S1 B.E./B.Tech. (Common to all Branches)

Prepared by Approved by

Mr. J. Vivekanandan

Dr. N. Pongali Sathyaprabu

Mr. A. Mahudeswaran

Dr. K. Sadasivam

Year of Release: 2015

BANNARI AMMAN INSTITUTE OF TECHNOLOGY

SATHYAMANGALAM – 638 401

DEPARTMENT OF PHYSICS

LAB MANUAL



R 2015: 15O102: Engineering Physics Laboratory

List of Experiments as per the syllabus

1.

Determine the moment of inertia of the disc and calculate the rigidity modulus of agiven wire using torsion pendulum (symmetrical masses method).

2. Find the elevation of the given wooden beam at the midpoint by loading at the ends

and hence calculate the Young’s modulus of the material.

3. Find the depression at the midpoint of the given wooden beam for 50g, 100 g, 150 g,

200 g and 250 g subjected to non-uniform bending and determine the Young’s

modulus of the material of the beam.

4.

Determine the coefficient of viscosity of the given liquid by Poiseulle’s method.

5.

Form the interference fringes from the air wedge setup and calculate the thickness of

the given wire.

6. By applying the principle of diffraction, determine the wavelength of given laser and

the average particle size of lycopodium powder using laser source.

7.

Determine the

(i) wavelength of ultrasonics in a liquid medium,

(ii) velocity of ultrasonic waves in the given liquid

(iii) compressibility of the given liquid using ultrasonic interferometer.



List of experiments

S. No. Experiment Page No.

1 Rigidity Modulus - Torsion Pendulum 6

2 Young’s Modulus - Uniform Bending 16

3 Young’s Modulus - Non-uniform Bending 26

4 Coefficient of Viscosity -Poiseuille’s Method 36

5 Thickness of a Thin Wire - Air Wedge 44

6 Wavelength of Laser and Particle Size – Diode LASER 52

7 Adiabatic Compressibility – Ultrasonic Interferometer 62



Department of Physical Sciences, Bannari Amman Institute of Technology, Sathyamangalam 5

Figure 1.1 Torsional Pendulum

Torsional Pendulum Torsional Pendulum Torsional Pendulum(without masses) (with masses at d1

close to the axis)

(with masses at d2

away from the axis)




Expt. No. : Date:

AIM

To find the moment of inertia of the disc and the rigidity modulus of the material of the

suspension wire subjected to torsional oscillations.

GENERAL OBJECTIVE

To assess the shear elastic behavior of a given material using torsional pendulum

SPECIFIC OBJECTIVES

1. To measure the time period of the torsional pendulum

2. To calculate the moment of inertia of the disc

3. To measure the radius of the wire using screw gauge

4. To determine the rigidity modulus of the wire using the formula

APPARATUS REQUIRED

Metallic disc

Brass/steel wire

Symmetrical masses

Stop clock

Metre scale

Screw gauge

Stand

FORMULA

1. Moment of inertia of the disc

)()(2

2

1

2

2

2

02

1

2

2T T

T d d m I

(kg m2)

Rigidity Modulus - Torsional Pendulum




TABLE – I

To find 2

OT

L and

)( 21

22

20

T T

T

S.No

Length of

the

suspension

wire L

(10-2m)

Distance

between

the masses(10-2m)

Time for 10 oscillations(s)Period

T

(s)

L / T02

10-2

(m/s2 )

T02

------------

( T22 - T1

2 )Trial I Trial II Mean

1 53.4

No mass

d1 = 2.5

d2 = 5.5

65

66

70

63

64

70

64

65

70

T0 = 6.4

T1 = 6.5

T2 = 7.0

1.3037 6.0682

2 72.8

No mass

d1 = 2.5

d2 = 5.5

75

77

82

77

79

84

76

78

83

T0 = 7.6

T1 = 7.8

T2 = 8.3

1.2604 7.1752

Mean 2

OT

L = 1.2821× 10-2 m/s2

Mean)( 2

1

2

2

20

T T

T

= 6.6217




2. Rigidity modulus of the material of the wire

2

0

4

8

T

L

r

I n

(N/m2)

Symbol Explanation Unit

I Moment of inertia of the disc kg m2

m Value of one of the two masses placed on the disc kg

d1 Closest distance between the center of mass and the wire m

d2 Farthest distance between the center of mass and the wire m

T0 Time period without any mass placed on the disc s

T1 Time period when two masses are placed at a distance d1 s

T2 Time period when two masses are placed at a distance d2 s

n rigidity modulus of the material of the suspension wire N/m2

L length of the suspension wire m

r radius of the suspension wire m

PREREQUISITE KNOWLEDGE

1. Torsional oscillation

It is the periodic oscillation produced by twisting the wire

2. Torsional pendulum

A body suspended from a rigid support by means of a thin elastic wire is called a

torsional pendulum

3. Rigidity modulus

Rigidity modulus is defined as ratio of shearing stress to shearing strain

4. Moment of Inertia

It is a measure of a body’s tendency to resist the angular acceleration5. Torque

The product of moment of inertia and angular acceleration (or) the rate of change of

angular momentum




LEAST COUNT FOR SCREW GAUGE

Least Count (LC) =Ph

h l

Pitch = Distance moved

Number of rotations given

=5

5

= 1mm

LC =1

100 = 0.01mm

TABLE - II

To measure the radius of the wir e (r ) using screw gauge

Zero Error (ZE) : 10 division Zero Correction (ZC) : 0 .10 mm

S. No.

Pitch Scale

Reading

PSR

( 10-3 m)

Head Scale

Coincidence

HSC

(divisions)

Observed Reading

OR = PSR + ( HSC × LC )

(10-3 m)

Correct Reading

CR = OR ± ZC

( 10-3 m)

1.

2.

3.

4.

5.

0

0

0

0

0

47

46

47

47

46

0.47

0.46

0.47

0.47

0.46

0.57

0.56

0.57

0.57

0.56

Mean diameter (d) = 0.566 × 10-3 m

Radius (r = d/2) = 0.283 × 10-3 m




PROCEDURE

1. One end of the material of the wire is clamped using a vertical chuck. A metallic disc

is attached to the other end of the wire.

2. The length of the suspension wire is fixed to a particular value. The disc is slightly

twisted so that the disc executes torsional oscillations.

3. The time taken for ten oscillations is noted using a stop clock. Two trials are taken for

each length. The mean time period T0 is found.

4. Now two equal masses are placed on either side of the center of the disc close to the

suspension wire.

5. The closest distance d1 from the center of the mass to the center of the suspension

wire is found.

6. Now the disc is made to execute torsional oscillations. The time taken for 10

oscillations is measured and the time period T1 is calculated.

7. Now the two equal masses are placed at the edges of the disc. The farthest distance d2

from the center of the mass to the center of the suspension wire is found.

8. Now the disc is made to execute torsional oscillations. The procedure is repeated and

the time period T2 is calculated.

9.

The radius of the wire(r) is found using a screw gauge.




OBSERVATION

CALCULATION

Moment of inertia of the disc

)()(2

2

1

2

2

2

02

1

2

2T T

T d d m I

kg m2

Value of one of the masses placed on the disc m = 50 ×10-3 kg

The closest distance between the suspension wire

and the center of the mass d1 = 2.5 ×10-2 m

The farthest distance between the suspension wire

and the center of the mass d2 = 5.5 ×10-2 m

Radius of the suspension wire r = 0.283 × 10-3 m

Mean value of L/T02 = 1.282 ×10-2 m/s2

Mean value of T02 /( T2

2 - T12) = 6.621




RESULT

The moment of inertia of the disc I = 1.58 10-3 kg m2

The rigidity modulus of the material of the given wire n = 7.97 10 10 N/m2

APPLICATIONS

Torsional pendulum clocks, shafts in automobiles

VIVA VOCE QUESTIONS

1. Why is it called torsion pendulum?

2. What happens to the period of oscillation when the length of suspension wire is

decreased?

3. Define moment of inertia.

4. If the radius of the wire is doubled, what happens to the moment of inertia?

5. Is it possible to determine the rigidity modulus other than solids?

STIMULATING QUESTIONS

1. Why is a hollow shaft stronger than a solid shaft of same mass and material?

2. Identify the role of moment of inertia of the clothes from the beginning to end ofcycle in washing machine




Rigidity modulus of the material of the wire

2

0

4

8

T

L

r

I n

N/m2




Figure 2.1 Young’s modulus- Uniform bending

Figure 2.2 Model Graph





Least Count (LC) =Ph

h l


Number of rotations given =

5

5 = 1mm

LC =1

100 = 0.01mm

TABLE – I

To determine the thickness (d) of the beam using screw gauge

Zero Error (ZE) : 12 divisions Zero Correction (ZC) : 0 .12 mm

S. No.

Pitch Scale

Reading

PSR

10-3 m

Head Scale

Coincidence

HSC

divisions

Observed Reading

OR = PSR + ( HSC LC )

(10-3 m)

Correct Reading

CR = OR ±ZC

( 10-3 m)

1

2

3

4

5

5

5

5

5

5

79

45

54

61

74

5.79

5.45

5.54

5.61

5.74

5.91

5.57

5.66

5.73

5.86

Mean (d) = 5.746 x10-3 m




FORMULA

Young’s modulus of the material of the beam

3

2

2

3

sbd

MgaLY (N/m2)


Y Young’s modulus of the material of the beam N/m2

M Load applied kg

L Distance between the knife edges m

a Distance between the load and the nearest knife edge m

g Acceleration due to gravity m /s2

b Breadth of the beam m

d Thickness of the beam m

s Elevation produced for ‘M’ kg load m

Unit Equivalent Units

N/m2 kg m-1 s-2 1 Pa


1. Elastic materials

Materials which can completely regain their original condition of shape and size on

removal of deforming forces are said to be elastic

2. Plastic materials

Materials which retain the deformed nature even after the removal of deforming

forces are said to be plastic3. Hook e’s law

Within the elastic limit, the stress is directly proportional to the strain




LEAST COUNT FOR VERNIER CALIPER

Least Count (LC) = Value of 1 Main Scale Division (MSD)/ Number of

divisions in the vernier

10 MSD = 1 cm

Value of 1 MSD = 1/10 cm = 0.1 cm Number of divisions in the vernier = 10

LC = 0.1/ 10 = 0.01 cm

TABLE - II

To determine the breadth (b) of the beam using vernier caliper

LC = 0.01 cm Zero Error (ZE): Nil Zero Correction (ZC): Nil

S. No.

Main Scale

Reading

MSR

(10-2 m)

Vernier Scale

Coincidence

VSC

(divisions)

Observed Reading

OR = MSR + (VSC × LC)

(10-2 m)

Correct

Reading

CR = OR ± ZC

(10-2 m)

1

2

3

4

5

2.9

2.9

2.9

2.9

2.9

3

5

4

2

6

2.93

2.95

2.94

2.92

2.96

2.93

2.95

2.94

2.92

2.96

Mean (b) = 2.94 x 10-2 m




PROCEDURE

1. The given beam is supported on two knife edges separated by a distance ‘L’

A pin is fixed vertically at the mid-point.

2. Two weight hangers are suspended, one each on either side of the knife edges so that

their distances from the nearer knife edge are equal. The beam is brought to the

elastic mood by loading and unloading it several times.

3. With the dead load ‘W’, the pin is focused through microscope. The microscope is

adjusted so that the horizontal crosswire coincides with the tip of the pin. The

microscope reading is taken.

4. The load is changed in steps of 0.05 kg and in each case the microscope reading is

taken during loading and unloading. The readings are tabulated. The elevation at the

mid-point for ‘M’ kg is calculated.

5. The distance between the knife edges (L) is measured using a metre scale. The

breadth (b) and thickness (d) of the beam are found using vernier caliper and screw

gauge, respectively.




LEAST COUNT FOR TRAVELLING MICROSCOPE



20 MSD = 1 cm

Value of 1 MSD = 1/20cm = 0.05 cm

Number of divisions in the vernier = 50

LC = 0.05/50= 0.001 cm

TABLE -III

To f ind elevation ‘s’

LC = 0.001 cm *TR= MSR + (VSC LC)

Mean (s) = 0.051 × 10-2 m

*Note: Total Reading (TR) = Main Scale Reading (MSR) + (VSC LC)

Load

M

(10-3 kg)

Microscope readingElevation

‘s’ for M

kg

( 10-2 m )

Loading UnloadingMean

( 10-2 m )MSR

( 10-2 m )

VSC

(div)

TR

( 10-2 m )

MSR

( 10-2m )

VSC

(div)

TR

( 10-2 m )

W

W + 50

W + 100

W + 150

W+ 200

7.6

7.65

7.70

7.75

7.80

8

19

9

13

10

7.608

7.669

7.709

7.763

7.810

7.6

7.65

7.70

7.75

7.80

4

10

5

9

10

7.604

7.660

7.705

7.759

7.810

7.606

7.665

7.707

7.761

7.810

0.059

0.042

0.054

0.049




RESULT

The Young’s modulus of the material of the given beam Y = 0.826 × 1010 N/m2

APPLICATIONS

AFM probe, wings of air craft, helicopter rotator, marine fittings, designing of bridges,

bicycle frames and wind mill turbine blades

VIVA VOCE QUESTIONS

1. What is the effect of temperature on elastic modulii?

2. Which dimension among breadth, thickness and length has significant role in

Young’s modulus? Why?

3. How do you ensure in your experiment that the elastic limit is not exceeded?

4. What kind of elasticity is observed in (a) suspension bridge (b) an automobile tyre?


1. Bridges are declared as unsafe after long use. Reason out.

2. Which one is more elastic, foam or steel? Justify your answer.




OBSERVATION

CALCULATION


3

2

2

3

sbd

MgaLY (N/m2)

Mass for the elevation M = 50 ×10-3 kg

Distance between the two knife edges L = 80 ×10-2 m

Acceleration due to gravity g = 9.8 m / s2

Breadth of the beam b = 2.94 × 10-2 m

Thickness of the beam d = 5.764 × 10-3 m

Elevation produced for ‘M’ kg of load s = 0.051 × 10-2 m

Distance between one of the knife edges and the

adjacent weight hanger a = 5 × 10-2 m




Figure 3.1 Young’s modulus - Non-uniform bending

Figure 3.2 Model Graph




Expt. No. : Date:

AIM

To find the Young’s modulus of the given material of the beam by non-uniform bending.

GENERAL OBJECTIVE

To evaluate the elastic behavior of the given wooden beam by pin and microscope

experimental method and to find its Young’s modulus

SPECIFIC OBJECTIVES

1.

To measure the thickness and breadth of the given wooden beam using screw gaugeand vernier caliper, respectively

2. To determine the depression of the given wooden beam loaded at its midpoint by

non-uniform bending method

3. To find the slope from the graph drawn between the load versus depression

4. To calculate the Young’s modulus of the wooden beam from the mean depression

and slope obtained from table and graph, respectively

5. To analyze the elastic behavior of the given wooden beam from the results obtained

APPARATUS REQUIRED

Wooden beam

Weight hanger with slotted weights

Knife edges

Travelling microscope Vernier caliper

Screw gauge

Metre scale

Young’s Modulus - Non-uniform Bending





Least Count (LC) = Pitch

Number of head scale divisions


Number of rotations given =

5

5 = 1mm

LC =1 mm

100 = 0.01mm

TABLE -I

To determine the thickness (d) of the beam using screw gauge

Zero Error (ZE) : 12 div Zero Correction (ZC) : 0 .12 mm

S. No.

Pitch Scale

Reading

PSR

( 10-3 m)

Head Scale

Coincidence

HSC

(div)

Observed Reading

OR = PSR + ( HSC

LC )

(10-3 m)

Correct Reading

CR = OR ± ZC

( 10-3 m)

1

2

3

4

5

5

5

5

5

5

79

45

54

61

74

5.79

5.45

5.54

5.61

5.74

5.91

5.57

5.66

5.73

5.86

Mean (d) = 5.746 x10-3 m




FORMULA


3

3

4 sbd

MgLY (N/m2)


Y Young’s modulus of the material of the beam N/m2

M Load applied kg

L Distance between the knife edges m

g Acceleration due to gravity m /s2

b Breadth of the beam m

d Thickness of the beam m

s Depression produced for ‘M’ kg load m

Unit Equivalent Units

N/m2 kg m-1 s-2 1Pa


1. Stress

Stress is a dimension quantity defined as force per unit area.

2. Strain

Strain is the relative change in shape or size of an object due to externally applied

forces. It is dimensionless quantity and has no units.

3.

Young’s modulus Young’s modulus is defined as the ratio between linear stress and linear strain.




LEAST COUNT FOR VERNIER CALIPER



10 MSD = 1 cm

Value of 1 MSD = 1/10 cm = 0.1 cm


LC = 0.1/ 10 = 0.01 cm

TABLE-II

To determine the breadth (b) of the beam using vernier caliper

LC = 0.01 cm Zero error (ZE) : Nil Zero Correction (ZC): Nil

S. No.

Main Scale

Reading

MSR

(10-2 m)

Vernier Scale

Coincidence

VSC

(divisions)

Observed Reading

OR =MSR + (VSC×LC)

(10-2 m)

Correct Reading

CR = OR ± ZC

(10-2 m)

1

2

3

4

5

2.7

2.7

2.7

2.7

2.7

3

5

4

2

6

2.73

2.75

2.74

2.72

2.76

2.73

2.75

2.74

2.72

2.76

Mean (b) = 2.74×

10

-2

m




4. Uniform and non-uniform bending

In uniform bending, the beam is elevated due to load, and non-uniform bending, the

beam is depressed due to load.

In uniform bending, every element of the beam is bent with the same radius of

curvature whereas in non-uniform bending, the radius of curvature is not the same for

all the elements in the beam.

PROCEDURE

1. The given beam is supported on two knife edges separated by a distance ‘L’

A pin is fixed vertically at the mid-point. A weight hanger is suspended at the mid-

point of the beam. The beam is brought to the elastic mood by loading and unloading

it several times.

2. With the dead load ‘W’, the pin is focused through microscope. The microscope is

adjusted so that the horizontal crosswire coincides with the tip of the pin. The

microscope reading is taken.

3. The load is changed in steps of 0.05 kg and in each case the microscope reading is

taken during loading and unloading. The readings are tabulated. The depression at the

midpoint for ‘M’ kg is calculated.

4. The distance between the knife edges (L) is measured using a metre scale. The

breadth (b) and thickness (d) of the beam are found using vernier caliper and screw

gauge, respectively.







20 MSD = 1 cm



LC = 0.05/50= 0.001 cm

TABLE -III

To find depression‘s’


Mean (s) = 0.105 × 10-2 m


Load

M

(10-3 kg)

Microscope readingDepression

‘s’ for M

kg

( 10-2 m )

Loading UnloadingMean

( 10-2 m )MSR

( 10-2 m )

VSC

(div)

TR

( 10-2 m )

MSR

( 10-2m )

VSC

(div)

TR

( 10-2 m )

W

W + 50

W + 100

W + 150

W+ 200

6.80

6.70

6.60

6.50

6.35

6

14

28

40

36

6.806

6.714

6.628

6.540

6.386

6.80

6.65

6.55

6.45

6.35

6

34

40

44

36

6.806

6684

6.590

6.494

6.386

6.806

6.699

6.609

6.517

6.386

0.107

0.09

0.092

0.131




RESULT

The Young’s modulus of the material of the given beam Y = 1.149 x 1010 N/m2

APPLICATIONS

AFM probe, wings of air craft, helicopter rotator, marine fittings, designing of bridges,

bicycle frames and wind mill turbine blades.

VIVA VOCE QUESTIONS

1. Define elastic limit.

2. When a beam is loaded at its midpoint, it is then said to be under non-uniform

bending. Why?

3. Differentiate between elasticity and plasticity.

4. Give the significance of neutral axis.




OBSERVATION

CALCULATION

Young’s Modulus of the material of the beam

3

3

4 sbd

MgLY (N/m2)

Mass for the depression M = 50 ×10-3 kg

Distance between the two knife edges L = 80 ×10-2 m

Acceleration due to gravity g = 9.8 m / s2

Breadth of the beam b = 2.74 × 10-2

m

Thickness of the beam d = 5.746 × 10-3 m

Depression produced for ‘M’ kg of load s = 0.105 × 10-2 m





1. What happens to the Young’s modulus of the material if its dimension is

increased?

2. Defense force is not allowed to do march past on the bridges. Reason out

3. Load vs depression plots for copper and steel are given. Which material is

stiffer? Justify.




Figure 4.1 Experimental arrangement of Poiseuille’s Method

Figure 4.2 Calculation of radius of the capillary tube




Expt. No: Date:

AIM

To determine the coefficient of viscosity of the given liquid by Poiseuille’s method.

GENERAL OBJECTIVE

To assess the viscous behavior of the given liquid by Poiseuille’s method

SPECIFIC OBJECTIVES

1. To determine the driving height of the liquid level

2. To find the time taken for the uniform flow of given volume of the liquid

3. To measure the internal radius of the capillary tube

4. To calculate the coefficient of viscosity of the given liquid using the formula

APPARATUS REQUIRED

Burette

Rubber tube

Capillary tube

Pinch cock

Traveling microscope

Stop clock

Metre scale

Beaker

Viscosity of a Liquid -Poiseuille’s Method




TABLE - I

To fi nd time of f low (t) and height

TABLE - II

To find ‘ ht ’

The height of the axis of the capillary tube from the table (h0) = 11 × 10-2m

S. No.Volume of the liquid

(10-6 m3 )

Time of

flow

t (s)

h1

(10-2 m)

h2

(10-2 m)

h=[(h1+h2 )/ 2]-h0

(10-2 m)

ht

(10-2 ms )

1

2

3

4

5

0 – 10

10 – 20

20 – 30

30 – 40

40 – 50

28

36

41

54

76

85

75

65

55

45

75

65

55

45

35

69

59

49

39

29

1932

2124

2009

2106

2204

Mean ‘ht’ = 2075×10-2 ms

Burette reading

(10-6 m3 )Time taken (s)

Height of burette reading from the

table

H (10-2

m)0

10

20

30

40

50

0

28

64

105

159

235

85

75

65

55

45

35




FORMULA

Coefficient of viscosity of the given liquid

LV

ht gr

8

)(4 (Nsm-2)


Density of the given liquid kg m-3

g Acceleration due to gravity ms-2

r Internal radius of the capillary tube m

L Length of the capillary tube m

h Driving height of the liquid m

t Time taken for the flow of liquid s

V Volume of the liquid collected m3


1. Streamline flow

It is defined as the flow of a fluid in which velocity is constant or varies in a regular

manner.

2. Shear stress

It is defined as the force applied parallel to the liquid layer surface per unit area.

3. Velocity gradient

The difference in velocity between adjacent layers of the fluid is known as a velocity

gradient.

4.

Density It is defined as the mass of the liquid per unit volume.







20 MSD = 1 cm



LC = 0.05/50= 0.001 cm

TABLE - III

To fi nd the in ternal radius of the capill ary tube


Position

Microscope Reading Diameter

(2r) = R 1 R 2

10-2m MSR

10-2m

VSC

(div)

TR

10-2m

Left R 1 9.5 30 9.53

0.10

Right R 2 9.6 30 9.63

Top R 1 12.9 40 12.94

0.10

Bottom R 2 13 40 13.04

Mean diameter (2r) = 0.10 × 10-2 m

Radius (r) = 0.05 × 10-2 m





5. Coefficient of viscosity

It is defined as the ratio of applied shear stress to velocity gradient in a fluid flow.

PROCEDURE

1. The burette is fixed vertically in the stand and filled fully with the liquid for which

the viscosity is to be measured.

2. At the lower end of the burette, a capillary tube is attached using a rubber tube.

3. The capillary tube is placed on a table such that the tube is in horizontal position.

This arrangement allows the liquid to flow freely through the capillary tube

without the influence of gravity.

4. The knob in the bottom of the burette is opened and the water is allowed to drain

through the capillary tube. When the liquid level reaches zero mark level, the stop

clock is started

5. The time taken to reach 10, 20,…..50 cc is noted. Then the time interval for each

10 cc, namely 0-10, 10-20, ………., 40-50 is found and tabulated. The height (H)

of each marking namely 0, 10----50 cc is measured from the table. Also the height

(h0) of the axis of the capillary tube from the table is found. Then the actual height

of each marking is obtained using the relation (H – h0)

6. The driving height h = [(h1+h2)/2]-h0 for every 10 cc namely 0-10, 20-30---40-50

is calculated by taking the height of initial marking as h1and final marking as h2

for each range.

7. The mean value of (ht/V) is calculated. The diameter of the capillary tube is

measured using a travelling microscope and then radius (r = diameter/2) is

calculated from it.

8. Substituting the values in the given formula, the coefficient of viscosity can be

calculated.




OBSERVATION

CALCULATION

Coefficient of viscosity of the given liquid

LV

ht gr

8

)(4 (Nsm-2)

Density of the given liquid = 1000 kg m-3

Acceleration due to gravity g = 9.8 m/s2

Inner radius of the capillary tube r = 0.05 × 10-2 m

Length of the capillary tube L = 35 × 10-2 m

Volume of the liquid V = 10 × 10-6 m3

Mean value of ht = 2075 × 10-2 ms




RESULT

The co-efficient of viscosity of the given liquid = 1.4 ×10-3 Nsm-2

APPLICATIONS

Flow rates of liquids in pipes, selection of proper lubricant oils, paper coating processes,

atomization of fuel oils to droplets in boilers for efficient burning, smooth application of paints in walls and flow behavior of adhesives.

VIVA VOCE QUESTIONS

1. Define co-efficient of viscosity.

2. Point out the fluids having viscosity less than and greater than that of water.

3. Comment the various factors that affect co-efficient of viscosity.

4. Why is viscous force dissipative?

5. Compare streamline flow with turbulent flow.


1. The inter molecular forces in oil are less than water but still the viscosity of oil is

more than water. Justify.

2. If the temperature increases, the viscosity of liquid decreases whereas the viscosity of

gases increases. Comment.




Figure 5.1 Air wedge arrangement

Figure 5.2 Fringe pattern




Expt. No. : Date:

AIM

To calculate the thickness of a thin wire by forming interference fringes using an air wedge

arrangement.

GENERAL OBJECTIVE

To measure the thickness of a given thin wire by air wedge method

SPECIFIC OBJECTIVES

1. To form an interference pattern between two glass plates using air wedge setup

2.

To calculate the band width of the interference pattern from the microscope readings

3. To measure the length of the air wedge using a scale

4. To determine the thickness of a given material using formula

APPARATUS REQUIRED

Travelling microscope

Optically plane glass plates

A thin wire

Sodium vapour lamp

Reading lens

Scale

Thickness of a Thin Wire – Air Wedge







20 MSD = 1 cm

Value of 1 MSD = 1/20 cm = 0.05 cm


LC = 0.05/50= 0.001 cm

TABLE - I

To determine the fr inge width (

)


Order of

Fringes

Microscope reading

Width for 5 fringes

(10-2 m)Fringe width

(10-2 m)MSR

(10-2 m)

VC

(div)

TR

(10-2 m)

n

n+5

n+10

n+15

n+20

n+25

n+30

n+35

n+40

13.0

12.85

12.7

12.55

12.4

12.25

12.15

12.0

11.9

14

23

12

29

24

15

13

30

32

13.014

12.873

12.712

12.579

12.424

12.265

12.163

12.030

11.932

0.141

0.161

0.133

0.155

0.159

0.102

0.133

0.098

0.028

0.032

0.026

0.031

0.032

0.020

0.026

0.019

Mean ( ) = 0.029×10-2 m





OBSERVATION

Length of the air wedge L = 5.7x 10-2 m

Wavelength of the sodium light = 5893 x 10 – 10 m

Band width = 0.029 x 10-2 m

CALCULATION

Thickness of the given wire

t = L/2 (m)




2.

The distance between the wire and the tied end (L) is measured using a scale.

3.

Light from a sodium vapour lamp is incident on a plane glass plate inclined at 45 to

the horizontal.

4.

The reflected light from the plane glass plate is incident normally on the optically

plane glass plates forming the air wedge and reflected back.

5. The reflected light from the air-wedge is viewed through the eye-piece of a

microscope. The microscope is moved up and down and adjusted for clear

interference fringes of alternate dark and bright.

6. The microscope is fixed so that the vertical cross-wire coincides with the dark band

(say nth band) and the reading is noted.

7.

The microscope is moved across the fringes and readings are noted when the vertical

cross-wire coincides with the (n+5)th, (n+10)th….. dark bands.

8. The observed readings are tabulated and the band width () is calculated.

9. The thickness of the given wire/thin-sheet is calculated using the formula.

RESULT

The thickness of a thin wire using air wedge method (t) = 5.79 x 10-5 m

APPLICATIONS

Testing the flatness of a plane surface, measuring the thickness of a thin material




VIVA VOCE QUESTIONS

1.

What is monochromatic light? Give an example.

2. What is the condition for the occurrence of interference phenomenon?

3.

When the length of the air-wedge is increased, what happens to the fringe width?

4. Why the glass plate used in the pathway of the light source should be inclined exactly

at 45°?

5.

Bright and dark fringes are formed alternatively in interference pattern. Justify.

6. What happens to the fringe width, if the thickness of the material is increased?

7.

Why do we get straight line fringes in an air wedge?


1. Can we use the polychromatic light instead of monochromatic light in air wedge

method?

2.

Is there any loss of energy in interference phenomenon?




Figure 6.1 Determination of wavelength

Figure 6.2 Particle size determination by LASER




Expt. No.: Date:

AIM

To calculate the wavelength of the given laser using grating and particle size of the given

lycopodium powder using laser diffraction method.

GENERAL OBJECTIVE

To evaluate the wavelength of the laser using a grating and to find the particle size of

lycopodium powder

SPECIFIC OBJECTIVES1.

To obtain diffraction spots on the screen using grating and the laser source

2.

To measure the distance between centre spot and first order spot by varying the

distance between the grating and screen

3. To find the angle of diffraction (sin ) using the formula

4. To calculate the wavelength of laser source using the formula

5. To obtain diffraction pattern of lycopodium powder using laser source

6. To measure the radius of the first and second order rings for various screenglass

plate distances and calculate the particle size

APPARATUS REQUIRED

Laser source

Glass plate with lycopodium powder

Grating

Stand

Screen

Metre scale

Wavelength of Laser and Particle Size – Diode LASER




TABLE - I

To determine the wavelength of the given laser ‘λ’

Mean sin = 0.39

S. No.

Distance between the

grating and the centre spot

AB(10-2m)


centre spot and first

order spot

BC

(10-2m)


first order spot (BC)

and the grating slit

AC=√ (AB2+BC2)

(10-2m)

Sin =

BC/ AC

1

2

3

4

5

6

6.5

7

8

8.5

14

15.1

16.3

17.4

18.5

15.23

16.44

17.73

19.15

20.36

0.38

0.39

0.39

0.41

0.41




FORMULA

Wavelength of the given laser

mN

sin (m)

Particle size ‘a’ of the given powder

r

Dma

22.1 (m)


Wavelength of the laser light m

N Number of lines per metre length of the given grating lines/ m

a Particle size of the given powder m

m Order of diffraction no unit

D Distance between the screen and the glass plate m

r Radius of the ring m

Angle of diffraction degree


1.

LASER

It is an acronym of Light Amplification by Stimulated Emission of Radiation.

2. Properties of LASER

High intensity

Directionality

Monochromaticity

Coherence

3.

Diffraction

Bending of light around the edges of an obstacle




TABLE - II

To determine the particle size ‘a’

Mean (a) = 8.98 × 10-6m

Order

m

Distance between screen

and glass plate

D (10-2

m)

Radius of the

ring

r(10-2 m)

Particle

size

a (m )

121

1.7 9.79510-6

2 3.6 9.25110-6

126.4

3 6.91810-6

2 4.7 8.90810-6

128.5

2.4 9.41610-6

2 5 9.04010-6




4.

Optical grating

It is a glass plate consisting of alternate ruling and slit

5. Pythagoras theorem

It states that the square of the hypotenuse (the side opposite the right angle) is equal

to the sum of the squares of the other two sides

PROCEDURE

I. To determine the wavelength of the given laser source

1.

An optical grating of known N value is fixed on the grating mount that is placed on a

wooden stand.

2. Laser beam from the given semiconductor laser source is made to fall normally on the

fixed grating.

3. Now, the grating diffracts laser beam. A screen is kept on the other side of the grating

to obtain the diffraction spots.

4. The distance between the grating and screen (AB) is fixed.

5. The distance between the centre spot and first order diffraction spot (BC) on either

side of the screen is measured.

6.

The above procedure is repeated for different values of AB.

7.

The distance between the diffracting slit and the first order diffraction spot is

calculated using AC=√(AB2+BC2)

8.

From the values of BC and AC, sin is calculated and the mean value is found.

9. The wavelength of the laser is calculated using the formula.

II. To determine the particle size of the given powder

1.

The lycopodium powder dispersed in a transparent thin glass plate is kept vertically

using a stand in between the laser source and screen.

2. The laser beam incident on the glass plate undergoes diffraction by the particles.




OBSERVATION

Number of lines per metre in the given grating N = 6 x 105 lines/ m

Order of the diffraction m = 1

CALCULATION

1. Wavelength of the given laser

mN

sin (m)




3. By adjusting the distance of the glass plate from the screen, a clear concentric ring

pattern is obtained. The ring pattern is due to the diffraction of the laser light by the

powder particles.

4. The centre of the ring pattern is marked on the screen. The radius of the first order

ring (m = 1) is measured and the measurement can be made for other orders

(m = 2, 3, 4,…….) also.

5.

The readings are taken for different values of D (distance between screen and glass

plate).

RESULT

1.

Wavelength of the laser = 6506 x 10-10 m

2. The average particle size of the given powder a = 8.898 x 10-6 m

APPLICATIONS

Photography (3D view-Hologram), Medicine (treatment of a detached retina), Computer

(Printers), Meteorology (laser interferometer to measure length), Weapons (LIDAR - Light

Detection and Ranging), Industry (To weld or melt the materials)

VIVA VOCE QUESTIONS

1. Define coherence.

2. Identify two types of coherence.

3.

Can we use X-rays instead of laser source in this experiment?




2. Particle size of the given lycopodium powder

r

Dma

22.1 (m)




4. List six applications of LASER.

5. Why do we get rings in the particle size determination instead of spots?

6. Is there any other method to measure the particle size?


1.

Can we use the sodium vapour lamp as a source to measure particle size?

2. While increasing the distance between the screen and glass plate, what happens to

the particle size?




Figure 7.1 Ultrasonic Interferometer

Figure 7.2 Maxima and minima in the ammeter reading




Expt. No.: Date:

AIM

To find the compressibility of the given liquid using ultrasonic interferometer

GENERAL OBJECTIVE

To study the compressibility of the liquid by measuring the wavelength and velocity of

ultrasonic waves in the liquid using ultrasonic interferometer

SPECIFIC OBJECTIVES

1. To generate high frequency ultrasonic waves in the liquid using piezoelectric

oscillator

2.

To form standing waves between quartz crystal and reflector plate

3. To measure the distance ‘d’ between two successive maxima or minima

4. To find the wavelength and velocity of the ultrasonic waves in the liquid

5.

To evaluate the compressibility of the liquid from the formula

APPARATUS REQUIRED

Ultrasonic interferometer

Sample liquid (Water/Kerosene)

High frequency generator

FORMULA

1.

Wavelength of the ultrasonic wave

λ = 2d (m)

2. Velocity of ultrasonic waves in a given liquid ν = f λ (m/s)

3. Compressibility of the liquid

2

1

Adiabatic Compressibility – Ultrasonic Interferometer

(m2/N)




TABLE – I

To calculate ‘d ’

Least Count (LC) = 0.01 mm

Frequency of the ultrasonic waves = 2 106 Hz

TR = PSR + (HSC × LC)

Mean d = 0.313 x 10-3 m

S. No.

Order of

maxima

or

minima

Micrometer reading for n

maximum deflections(n+15)th reading –

nth reading) = 15 d

(10-3 m)

d

(10-3 m)PSR

(10-3 m)

HSC

(div)

TR

(10-3 m)

1. n 0 0 0 4.73 0.315

2. n+5 1.5 8 1.58 4.72 0.314

3.

n+10 3.0 15 3.15 4.73 0.315

4. n+15 4.5 23 4.73 4.75 0.316

5.

n+20 6.0 30 6.30 4.77 0.318

6. n+25 7.5 38 7.88 4.66 0.310

7.

n+30 9.0 48 9.48 4.62 0.308

8.

n+35 11 7 11.07 4.61 0.307

9. n+40 12.5 4 12.54

10.

n+45 14 10 14.10

11.

n+50 15.5 18 15.68





λ Wavelength of the stationary ultrasonic wave m

d Distance between two successive maxima or minima m

v Velocity of ultrasonic wave m/s

f Frequency of the ultrasonic wave Hz

ρ Density of the given liquid kg/m3

β Adiabatic compressibility of the given liquid m2/N


1.

Ultrasonic waves

Ultrasonic waves are the sound waves of frequency above audible range (i.e.)

above 20000 Hz.

2.

Properties of ultrasonic waves

highly energetic

travel through long distances

undergo reflection, refraction and absorption similar to ordinary sound waves

produce stationary wave pattern in liquids of suitable dimension and behave as

an acoustical grating

generate heat in materials for a longer time of exposure

3. Ultrasonic interferometer

An ultrasonic interferometer is a simple and NDT device to determine the

ultrasonic velocity in liquids with a high degree of accuracy.

4.

Standing wave

It is a wave in a medium in which each point on the axis of the wave has an

associated constant amplitude.

5.

Adiabatic compressibility

Compressibility is a measure of the relative volume change of a fluid or solid as a

response to a pressure (or mean stress) change.




CALCULATION

1.

Wavelength of the ultrasonic wave

λ = 2d (m)

2. Velocity of ultrasonic waves in a given liquid

ν = f λ (m/s)

3. Compressibility of the liquid

2

1

m2/N




PROCEDURE

1. Ultrasonic interferometer is used to determine the velocity of ultrasonic waves in

liquids. It consists of a high frequency generator and a measuring cell.

2. The high frequency generator is used to excite the quartz crystal fixed at the bottom

of the measuring cell. The measuring cell is a double walled cell to maintain the

temperature of the cell at a constant value.

3.

The measuring cell is connected to a high frequency generator. The cell is filled with

the given liquid and the frequency of the generator is set at a desired value.

4. Then ultrasonic waves are reflected back from the movable plate, and standing waves

are formed between the quartz crystal and the reflector plate.

5.

The micrometer screw is moved till the anode current reaches maximum.

Microammeter readings are noted for ‘n’ number of maxima / minima.

6. The distance ‘d’ between two successive maxima and minima is obtained from the

readings taken.

7.

The wavelength of the ultrasound is calculated using the ‘d’ value and hence the

velocity of the ultrasonic wave can be calculated using the known frequency.

8. The compressibility of the given liquid is determined by knowing the density of the

given liquid.




RESULT

1.

The wavelength of ultrasonic wave is λ = 0.626 x 10-3 m

2.

The velocity of ultrasonic waves in the given liquid

medium is ν

= 1255.6 m/s

3. The adiabatic compressibility of the given liquid β = 7.323 x 10-10 m2/N

APPLICATIONS

Detection of flaws in metals, SONAR for detection of submarines, iceberg and other objects

in ocean, soldering and metal cutting, diagnostics applications such as detection of tumors

and defects in human body, ultrasonic cleaner and humidifier.

VIVA VOCE QUESTIONS

1. Why ultrasonic waves are not audible to humans?

2. Are ultrasonic waves electromagnetic waves? Comment.

3. Define acoustic grating.

4.

List two methods to produce ultrasonic waves.

5.

Name two technological importance of measuring the velocity and adiabatic

compressibility of liquids.


1.

How do bats locate the prey and objects?

2.

Recognize how dolphins communicate with each other.

3. Can we calculate adiabatic compressibility for a solid?

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