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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
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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
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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.
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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
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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)
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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
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Department of Physical Sciences, Bannari Amman Institute of Technology, Sathyamangalam 7
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
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Department of Physical Sciences, Bannari Amman Institute of Technology, Sathyamangalam 8
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
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Department of Physical Sciences, Bannari Amman Institute of Technology, Sathyamangalam 9
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
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Department of Physical Sciences, Bannari Amman Institute of Technology, Sathyamangalam 10
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.
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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
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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
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Department of Physical Sciences, Bannari Amman Institute of Technology, Sathyamangalam 13
Rigidity modulus of the material of the wire
2
0
4
8
T
L
r
I n
N/m2
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Figure 2.1 Young’s modulus- Uniform bending
Figure 2.2 Model Graph
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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 – 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
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FORMULA
Young’s modulus of the material of the beam
3
2
2
3
sbd
MgaLY (N/m2)
Symbol Explanation Unit
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
PREREQUISITE KNOWLEDGE
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
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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
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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.
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LEAST COUNT FOR TRAVELLING MICROSCOPE
Least Count (LC) = Value of 1 Main Scale Division (MSD)/ Number of
divisions in the vernier
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
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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?
STIMULATING QUESTIONS
1. Bridges are declared as unsafe after long use. Reason out.
2. Which one is more elastic, foam or steel? Justify your answer.
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OBSERVATION
CALCULATION
Young’s modulus of the material of the beam
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
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Figure 3.1 Young’s modulus - Non-uniform bending
Figure 3.2 Model Graph
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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
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LEAST COUNT FOR SCREW GAUGE
Least Count (LC) = Pitch
Number of head scale divisions
Pitch = Distance moved
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
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FORMULA
Young’s modulus of the material of the beam
3
3
4 sbd
MgLY (N/m2)
Symbol Explanation Unit
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
PREREQUISITE KNOWLEDGE
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.
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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
2.7
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
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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
mid- point 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.
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LEAST COUNT FOR TRAVELLING MICROSCOPE
Least Count (LC) = Value of 1 Main Scale Division (MSD)/ Number of
divisions in the vernier
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 find depression‘s’
LC = 0.001 cm *TR= MSR + (VSC LC)
Mean (s) = 0.105 × 10-2 m
*Note: Total Reading (TR) = Main Scale Reading (MSR) + (VSC LC)
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
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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.
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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
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STIMULATING QUESTIONS
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.
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Figure 4.1 Experimental arrangement of Poiseuille’s Method
Figure 4.2 Calculation of radius of the capillary tube
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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
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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
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FORMULA
Coefficient of viscosity of the given liquid
LV
ht gr
8
)(4 (Nsm-2)
Symbol Explanation Unit
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
PREREQUISITE KNOWLEDGE
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.
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LEAST COUNT FOR TRAVELLING MICROSCOPE
Least Count (LC) = Value of 1 Main Scale Division (MSD)/ Number of
divisions in the vernier
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 fi nd the in ternal radius of the capill ary tube
LC = 0.001 cm *TR= MSR + (VSC LC)
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
*Note: Total Reading (TR) = Main Scale Reading (MSR) + (VSC LC)
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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.
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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
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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.
STIMULATING QUESTIONS
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.
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Figure 5.1 Air wedge arrangement
Figure 5.2 Fringe pattern
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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
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LEAST COUNT FOR TRAVELLING MICROSCOPE
Least Count (LC) = Value of 1 Main Scale Division (MSD)/ Number of
divisions in the vernier
20 MSD = 1 cm
Value of 1 MSD = 1/20 cm = 0.05 cm
Number of divisions in the vernier = 50
LC = 0.05/50= 0.001 cm
TABLE - I
To determine the fr inge width (
)
LC = 0.001 cm *TR= MSR + (VSC LC)
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
*Note: Total Reading (TR) = Main Scale Reading (MSR) + (VSC LC)
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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)
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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
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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?
STIMULATING QUESTIONS
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?
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Figure 6.1 Determination of wavelength
Figure 6.2 Particle size determination by LASER
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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
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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)
Distance between the
centre spot and first
order spot
BC
(10-2m)
Distance between the
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
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FORMULA
Wavelength of the given laser
mN
sin (m)
Particle size ‘a’ of the given powder
r
Dma
22.1 (m)
Symbol Explanation Unit
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
PREREQUISITE KNOWLEDGE
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
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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
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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.
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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)
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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?
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2. Particle size of the given lycopodium powder
r
Dma
22.1 (m)
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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?
STIMULATING QUESTIONS
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?
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Figure 7.1 Ultrasonic Interferometer
Figure 7.2 Maxima and minima in the ammeter reading
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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)
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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
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Symbol Explanation Unit
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
PREREQUISITE KNOWLEDGE
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.
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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
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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.
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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.
STIMULATING QUESTIONS
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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