HIGH SCHOOL SCIENCE TEACHERS TRAINING PROGRAM
PHYSICS
EXPERIMENTS
TALENT DEVELOPMENT CENTRE
INDIAN INSTITUTE OF SCIENCE, KUDAPURA
Challakere, Chitradurga District,
Karnataka-577536
Contents
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Name of the Experiment
To estimate linear dimensions and density of material
Using vernier calliper
Regular Shape
Using screw gauge
Any Shapes
Using Archimedes’ principle
Simple Pendulum : To estimate the value of g at the a place and
behavior of time period corresponds to its Length.
Spring Constant: To verify Hooks’ law and determination of
Spring Constant
Conservation of Energy
Projectile motion
Resonance Column: To determine the velocity of sound at 0 0C
Temperature dependence of resistance of a semiconductor
Diode Characteristics
Zener Diode Characteristics
To study full and Half wave rectifier Circuit
Transistor Characteristics
Temperature coefficient of resistance of a metal
Ohm’s Law: To verify Ohms’ law and
Metre Bridge: To determine the resistivity of a conductor
Mapping of Magnetic Field lines
Deflection Magnetometer
Tangent Galvanometer
Laws of reflection
Laws of refraction and Lateral Shift
Refractive index of liquid using shift method
Focal Length
Refractive index of a Prism-Pin Method
Diffraction through Grating
Photoelectric Effect
H2 Lamp Spectroscopy
Solar Constant
Heat Transfer
To determine the latent heat of fusion of ice
To determine the specific heat of a liquid
To determine the specific heat of a solid
Mixing of liquids of different temperatures
Experiment. Dispersion of light
Physical constant, standard values and units
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TDC-IISc, Kudapura
Physics Experiment
1. Experiment: To estimate linear dimensions and density of material
Aim:
1. To measure the linear dimensions and mass of a given object.
2. To estimate the density in g/cc and kg/m3.
Introduction: Measurements of dimension and mass of an object are the first excise for the
higher secondary students. From these measurements it is possible to estimate the density of
material which is the most basic property of matter. In this experiment we will estimate the
density of some solid objects made of different material. In part A and B we will determine
the density of regularly shaped objects and in part C we will find the density of irregularly
shaped objects.
Formula: Density 
Mass
Volume
4 d
(a)Volume of Rectangular block = l  b  h ; (b)Volume of Sphere =
π 
3  2 
2
d
(c) Volume of Solid Cylinder = π  h
 2
3
Where l = length, b = breadth, h = height and d= diameter.
Part A: Vernier Calliper
Apparatus: Digital Vernier Caliper, Solids of Different shapes of different materials
(aluminum, teflon, brass, copper, steel, graphite).
Figure 1: Digital Vernier calliper
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TDC-IISc, Kudapura
Physics Experiment
Observation Table:
Object
Shape
Aluminum
Rectengular
Block
Brass
Cylinder
Copper
Cylinder
Steel
Cylinder
Dimension
Trails
1st
2nd
Mean
Mean
in cm
Volume
Mass
Density
Length
Breadth
Height
Diameter
Height
Diameter
Height
Diameter
Height
Part B: Screw Gauge:
Figure 2: Screw Gauge
Formula:
1. Least Count LC  
Pitch of the Screw
number of division on head scale
2. TR = PSR + (HSR-ZE) x LC
TR= Total Reading; PSR= Pitch Scale Reading; HSR= Head Scale Reading; ZE = Zero Error
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TDC-IISc, Kudapura
Physics Experiment
Observations:
1. Pitch = ……… mm.
Object
Trial
Steel
Ball
1
2
1
2
wire
2. Least count =……… mm.
PSR
(mm)
HSR
TR
(mm)
Mean
(mm)
3. Zero error = ………….. mm.
Mean
(cm)
Volume
(cc)
Mass
(g)
Density
(g/cc)
Results
Diameter of the steel ball = ………. mm
Density of the steel ball = ……….. mm
Diametre of the wire = …………….. mm
Part C: Density of solids of any shape:
In this case the volume of the object cannot be determined using vernier calipers and
screw gauge. We will find the volume of the object using Archimedes principle and then its
density.
Buoyancy method:
Archimedes principle is the supreme principle to determine the density of materials in any
shape. It states that the buoyant force experience experienced by a submerged object is equal to the
weight of the liquid displaced by the object. Because of buoyant force weight of an object inside
the liquid feels lighter. If 𝑀𝑜 is the mass of the object in air, ML is the mass of the object when
it is submerged to a liquid of density DL, then the Density of the object D is
𝐷=
𝐷𝐿 × 𝑀𝑜
𝑀𝐿
Procedure:
a. Measure the mass of the object= Mo
b. Take a beaker and fill it up with water up to 2/3 of its height. Put the water filled beaker on an
electronic balance. Make its mass be zero by pressing ‘TARE’ on the electronic balance.
c. Tie the object using a weightless string.
d. Submerge the object into the water (by holding other part of the string) such that the object
should not touch the bottom of the beaker.
e. Measure the mass of the object when it is submerged state. Mass is showing on the weighing
balance. This is the apparent mass of the object. Apparent mass = ML.
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TDC-IISc, Kudapura
Mass of the object in
air, 𝑀𝑜
Physics Experiment
Setting mass of the
water+beaker to ZERO
Apparent mass of the
object, 𝑀𝐿
Table of Observations:
Object
Mass of the object
in air, 𝑀𝑜
Mass of the object in
liquid, 𝑀𝐿
Density of the
liquid 𝐷𝐿
Density of the
Object
Questions:
1. Find the percentage difference in the estimated and standard values for the density of
the object.
2. Suggest a method to improve the measurement.
3. Arrange the following materials in descending order with respect to their density.
(Materials- Au, Pt, Os, Water, Na, K, Zn).
4. What is the least count of
a. A meter scale = _______________; Vernier caliper = ______________
b. Measuring jar = ______________ ; Your wrist watch = _____________
5. The least count of the instrument represent in many cases the error in the
measurement. In the case of screw gauge what is the error?
6. What is fractional error and percentage error in measuring the diameter of the steel
ball.
7. Suppose if we measured the steel ball with Vernier calipers the percentage error in the
measurement would be.
8. Density of an object is 2.3 kg/m3. Will it float or sink to the water?
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TDC-IISc, Kudapura
Physics Experiment
Density for few materials
Object
Aluminum
Copper
Brass
Stainless Steel
Iron
Osmium
Density
Kg/m3
2712
8960
8525
7480-8000
7850
22500
Object
Gold
Water, liquid, 100 0C
Water, 4 0C
Water, liquid, 0 0C
Water, ice, 0 0C
Teflon
Density
Kg/m3
19300
958.40
999.98
999.84
916
2200
2. Experiment: Simple Pendulum
Aim: To determine the acceleration due to gravity (g) at the place and behavior of time
period corresponds to its Length.
Apparatus:
Metallic bob with hook, clamp stand, split halves of a cork with plane faces, fine cotton
thread about 200 cm in length, digital Vernier callipers, stop clock, meter scale.
Introduction: Simple pendulum is the one of the earliest experiment done in the history of
Science (1584 AD). Galileo used the hanging lamp in the church as a simple pendulum and
studied the variation of its time period (using his pulse) on the length of the string, mass and
size of the bob. Now we use this experiment as one of the most important methods for
determination of the acceleration due to gravity (g). When the experiment is done carefully
and analysed properly one can obtain the value of g accurately.
 L
Formula: g  4 2  2  (ms-2)
T 
T = period of oscillation of the bob (s); L = length of the simple pendulum (m) = [distance
from the point of suspension to the centre of gravity of the bob.
ObservationTable 1: Determination of the length of the Simple Pendulum.
Length of the
Diameter of the
Mean
r
Trial Number
string
bob, 2r
2r
m
l
1
2
Length of the
Simple Pendulum
L= l + r m
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TDC-IISc, Kudapura
Physics Experiment
Figure 4: Simple Pendulum
Selecting small angle of oscillations
Table 2: Determination of g.
Trial
No.
L
(m)
Total time
Number of
for n
oscillations
oscillation
n
Tn
Time
period
T
T n
n
Mean T
 L
g  4 2  2  ms-2
T 
1
2
1
2
1
2
1
2
n = Number of oscillations, Tn = Time for n number of oscillations, T 
Acceleration due to gravity g, is = …………
6
Tn
= Period of oscillation,
n
TDC-IISc, Kudapura
Physics Experiment
Table 3: Determination of the length of Second’s Pendulum:
L
T
T2
What you have to do1. Plot the graph between T vs L
2. Plot the graph between T2 vs L graph.
3. Find the length of the seconds Pendulum.
[Pendulum having time period of two seconds is called seconds pendulum.]
Questions:
1.
2.
3.
4.
5.
What is the value of g at the Pole and at the Equator of the Earth?
What is the value of g at the moon’s surface?
Is there any other method to determine the g at a place? If so, mention them.
Can we use extensible thread in simple pendulum?
What will be the time period of simple pendulum if its length is infinity?
3. Experiment: Spring Constant
Aim: Determination of spring constant of a given spring.
Introduction: When a spring is extended or compressed from its natural length it develops a
restoring force proportional to the extension (or compression). The constant of
proportionality is called the spring constant. Greater the spring constant it requires more
force to extend (or compress) it. In this experiment we will find the spring constant of a given
spring by measuring the extension for a given load (force).
Formula:
⃗𝑭 = −𝑘𝒙
⃗
where F = applied force, k = spring constant, x = displacement. “-ve sign says that the restoring
force is opposite to the direction of the extension/applied force.”
𝑘=
𝒇𝒐𝒓𝒄𝒆 𝒂𝒑𝒑𝒍𝒊𝒆𝒅
𝒄𝒉𝒂𝒏𝒈𝒆 𝒊𝒏 𝒍𝒆𝒏𝒈𝒕𝒉
=
𝑭
𝒙
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TDC-IISc, Kudapura
Physics Experiment
Observation Table:
Trial
No.
1
2
..
..
F = Load
x 9.8x 10-3
N
Pointer reading
Load Increasing
x 10-2 m
Load Decreasing
x 10-2 m
Average
x 10-2 m
x = Extension
(Change in
length)
(x 10-2 m)
0
M1
M2
4. Plot x vs F Graph.
18
16
12
-2
Extension (x10 m)
14
10
8
6
4
2
0
0
100
200
300
400
-3
Mass ( x 9.81x10 N)
5. Find the slope.
6. Spring constant, k =
1
(Nm-1)
slope
Question:
1. What is the physical significance of the spring constant?
2. When F =- kx breaks down?
3. In shock absorbers springs are used. How will this help.
4. Experiment: Conservation of Energy
Aim: Verification of conservation of energy and determination of frictional loss.
Apparatus:
Solid sphere, solid cylinder, hollow cylinder, adjustable inclined plane, digital stop
clock, digital balance and vernier caliper.
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TDC-IISc, Kudapura
Physics Experiment
Formula: An object of mass m is at rest at height h from the ground level. The mass object is
rolling without slipping on an incline plane from rest and continue to roll to horizontal plane.
The kinetic energy of the mass object when the object is just at the horizontal plane is =
1 2 1 2
mv  I and the potential energy at the height h is = mgh.
2
2
If the energy is conserved then
1 2 1 2 1 2 K2 
mv  I  mv 1  2 
2
2
2
r 

g= acceleration due to gravity, m = mass of the object, v = velocity of the object, I = moment of
inertia, K = radius of gyration, r = radius of the object under motion.
mgh 
m
Timer sensor
h
s, t
Figure 5 : Conservation of Energy
Table:
Object
Height, h
Distance,
s
Time, t
Total
K.E.
P.E.
K .E.
P.E.
Solid
Sphere
Solid
Cylinder
Hollow
Cylinder
4. ResultFrictional loss = potential energy- total kinetic energy.
Questions:
1. In this experiment, is K.E.=P.E? Support your answer.
2
2
K
1
K
K2 2
*
 for solid sphere; 2  for solid cylinder; 2  1 for hollow cylinder.
2
r
2
r
r
5
*assuming inner diameter and outer diameter of the hollow cylinder is almost same.
Note:
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TDC-IISc, Kudapura
Physics Experiment
5: Experiment: Projectile motion
Aim: To determine the angle of projection for which range is maximum.
INTRODUCTION
The projectile motion known to the mankind from the times of Archimedes is an example for
two dimensional motion. The motion occurs in a vertical plane defined by the direction of
launch. In the simplest case (when air resistance is neglected and motion occurs close to the
surface of earth) the projected body experience uniform accelerated motion along vertical
direction and uniform velocity motion along horizontal direction. The horizontal range (R),
time of flight (T), maximum height (H) attained by the projectile are some of the parameters
of interest. In the following experiment we will try to explore the dependence of these
parameters on the initial conditions (speed, angle and height of launch) and acceleration due
to gravity (g). Using the appropriate relations we will also find the value of g. Understanding
of projectile motion is very useful in sports like Javelin throw, shot put, long jump, high jump
etc.
Procedure:
1. Place the launcher at one corner of a table and clamp it.
2. Make sure that the table is long enough for the projectile with maximum range falls on it.
3. Keep track of the location on the table where the ball lands. The launcher should be set to
the level of the table by moving the semicircular frame on the stand.
4. Place carbon paper on the table such that the ball falls on it.
5. Place the steel ball on the launcher and rotate it to an angle of 300. Launch the ball and
measure the range.
6. Repeat the experiment for the angles listed below and compare it with theory (Using the
initial velocity and the angle, the horizontal range can be calculated).
Table of observation:
Angle
30
35
40
45
Horizontal range
Angle
Horizontal
range
50
55
60
65
Maximum range ______ is obtained when the projectile is projected at the angle of _____.
QUESTIONS
1. Imagine two balls at the same height. At the same instant, one is dropped and the other is
fired horizontally. Which ball would hit the ground first? Use the force diagrams and
vectors drawn above to explain your answers.
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TDC-IISc, Kudapura
Physics Experiment
6. Experiment: Resonance Column
Aim: To find the velocity of sound in air at room temperature using resonance column.
Formula: 1. Velocity of sound at a temperature (T0C) in air medium is V  2 f l2  l1  ;
V
2. The velocity of sound at 00C in air is V0 
T
1
273.16
l2 cm
l1 cm
Water level
First maximum
Second maximum
Figure 6: Resonance column, measurement of first and second resonance
Observations:
Frequency
of Tuning
fork
f (Hz)
1st Resonating length
l1 (cm)
Trial Trial Mean
1
2
l1
2nd Resonating length
l2 (cm)
Mean
Trial 1 Trial 2
l2
V  2 f l2  l1 
V
V0 
1
m/sec
T
273.16
m/sec
Velocity of sound in air at ____ 0C = ………….. ms-1 and at 0 0C = _________ ms-1
Question:
1. In which parameters velocity of sound dependent upon?
2. By this experiment can we find the frequency of unknown tuning fork?
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TDC-IISc, Kudapura
Physics Experiment
7. Experiment: Temperature dependence of resistance of a semiconductor
Aim: To examine the variation of resistance of a semiconductor due to variation of
temperature.
Introduction: Based on the electrical properties, materials can be devided into three classes(1) metal- good conductor of electricity; (2) insulator- Bad conductor of electricity and (3)
semiconductor- electrical conductivity between that of a conductor such as copper and that of
an insulator such as glass. Semiconductors are the basic blocks of todays’ electronics. Unlike
metals, the electrical resistivity of a semiconductor material decreases with increasing
temperature.
Apparatus: A semiconductor, a metal, water heater, beaker, thermometer and a multimeter.
1. Observation Table:
Observation
Number
1
2
3
..
..
..
..
..
..
Semiconductor
Temperature
Resistance
0
90 C
Room temperature
* Plot (Resistance vs Temperature).
Questions:
1. Why resistance of a semiconductor decreases with increase in temperature?
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TDC-IISc, Kudapura
Physics Experiment
8. Experiment: Diode Characteristic
Aim: 1. To draw the forward and reversed bias characteristics curve of a semiconductor
diode.
2. Determination of knee voltage and forward bias resistance.
Introduction: Diode is a semiconductor device which allows easy flow of current only in one
direction (unidirectional device). It consists of a junction formed by a p-type and n-type
semiconductor. The relation between the current flow and applied voltage is different from
that in the case of conductors. There is a large flow of current when it is connected in forward
bias (Voltage exceeding the knee voltage) while negligible current flows in reverse bias. In
this experiment we will study the variation of current in both Forward and reverse bias.
Apparatus required: Diode, 0-20 Volt DC power supply, digital voltmeter, digital
milliammeter, resistor, circuit unit, patch cords.
Formula- R = Forward bias resistance =
Figure 8a: Forward Bias
1
slope forward bias I V  characteri stic
Figure8b: Reverse Bias
Figure8c: A typical I-V
characteristic
of
a
semiconductor diode
Procedure :
1. Plot voltage versus current graph.
2. Find the slope from the forward bias characteristic curve.
3. Find the forward bias resistance.
4. Forward bias resistance = inverse of the slope in I-V characteristic
5. Find the knee voltage.
13
Table of Observation:
Forward Bias
Reverse Bias
Trial
number
Voltage
Current
Voltage
Current
1
0.1
1
2
0.2
2
3
0.3
3
4
0.4
4
5
0.5
5
6
0.55
6
7
0.60
7
8
0.65
8
9
0.70
9
10
0.75
10
11
0.80
11
Result:
Forward Biased Resistance= ______________
Question :
1. What is a semiconductor diode?
2. What are the applications of the diode?
3. Are the semiconductors used in diodes intrinsic semiconductors? Explain.
4. Name few elements for p-type and n-type doping?
5. In which bias, the diode has high resistance?
Circuit Diagram:
mA
mA
Vreversed
V
14
9b. Reverse Bias
Ireverse (mA)
Vforward
V
Figure 9a: Forward Bias
Iforward (mA)
9. Experiment: Zener diode characteristic
Aim: 1) Determination of forward and reversed bias characteristic of a zener diode .
2) To determine the break down voltage.
Introduction: Zener diode is also a p-n junction diode but with low breakdown voltage.
When worked in reverse bias in the breakdown region, it acts as a voltage regulator. In
this experiment we study the forward and reverse bias characteristics. We also find the
breakdown voltage and understand how it acts as a voltage regulator.
Apparatus required: Zener diode, 0-20 Volt DC power supply, digital voltmeter, digital
milliammeter, resistor, circuit unit, patch cords.
9c. A typical I-V characteristic
of a Zener diode.
Observation Table:
Trial No
1
2
3
4
5
6
7
8
9
10
11
Forward bias
Voltage
Current
0.1
0.2
0.3
0.4
0.5
0.55
0.60
0.65
0.70
0.75
0.80
Reverse Bias
Voltage
Current
1
2
3
4.5
4.6
4.7
4.8
4.9
5.0
.
5.5
1. Plot voltage versus current graph.
2. Find the break down voltage.
Question:
1. What is the difference between general rectifier diode and Zener diode?
2. Give few applications of Zener diode?
3. When Zener diode works in breakdown voltage will it not get damaged?
10. Experiment: To study full and Half wave rectifier Circuit
Aim:
1. To draw the waveform of a half wave and full wave rectifier circuit.
2. Compare the Output VDC.
Introduction: The electronic gadgets generally work on low dc voltage. The voltage
supplied to house hold is 220V ac. Thus there is a requirement of converting high voltage
AC to low voltage DC. The high AC voltage is first step down to low voltage AC with the
help of a transformer. Then it is converted to DC using a rectifier circuit. It consists of
diode (or diodes) since it allows current to flow in only one direction. When only one
diode is sued the rectifier is called half wave while if two diodes are used then it is called
full wave rectifier. However in both cases only fluctuating DC is obtained. In order to get
a constant DC , a capacitor is used in parallel to the load resistor.
Apparatus required: Diodes, ±15 Volt power supply, digital voltmeter, Resistor,
capacitor, circuit unit, patch cords.
Formula-
VDC (half wave) =
(𝑽𝑹𝑴𝑺 −𝑽𝒅 )√𝟐
𝝅
;
VDC(Bridge Rectifier) = 2 [
(𝑉𝑅𝑀𝑆 −2𝑉𝑑 )√2
𝜋
]
15
Circuit Diagram:
Figure 10a:
Half wave rectifier
10b. Full wave rectifier
Input ac Voltage
Input ac Voltage
Output dc voltage without C
Output dc voltage without C
Output dc voltage with C
Output dc voltage with C
10c. Half Wave Rectifier
Input-Output wave-form
10d. Full Wave Rectifier
input – Output wave-form
Observation Table:
Trial No
Half Wave
Rectifier
1
2
3
Bridge
Rectifier
1
2
3
VRMS
VDC
without
capacitor
VDC with
capacitor
Question:
1. What are the uses of rectifier circuits?
2. Why VDC with capacitor is more than VDC without capacitor?
3. To design a 5V and 5 mA DC power supply, write basic electrical
components you need.
16
VDC
Calculated
11. Experiment: Transistor characteristics
Aim: 1. To study the input and output characteristic of a given transistor.
2. To determine its α and β.
Introduction: Transistor is a two junction three terminal device. Transistor comes in two
configuration; npn and pnp. The emitter and base is generally connected in forward bias
while collector and base in connected in reverse bias. One of the most important
applications of a transistor is power amplification of a signal. In this experiment we study
the input and output characteristic of a transistor. We will find the current amplification
parameter α and β in CB and CE mode.
Apparatus required: A transistor, two variable dc power supply (0-5V, and 0-20V), two
dc ammeter (0-1000μA; 0-100mA), digital voltmeter, circuit board with a base resistor of
5.1kΩ, patch cords.
I
I  I C1


Formula:   C  C 2
;
I B
I B 2  I B1
1 
Circuit Diagram:
R
Ic
Ib
μA
mA
C
VBB
VBE
B
VCC
VCE
E
Figure11a. Common Emitter NPN transistor characteristic circuit diagram.
VCE = 1 V
IB =80 μA
μA
μA
I =μA
40μA
IC
(mA)
IB
(μA)
0
IB =120 μA
B
μA
K
VCE (V)
11b .Input Characteristics
11c. Output characteristic
VBE
17
Table- Input characteristic.
Observation
Number
1
2
..
..
..
..
VCE= 1V
VBE (V)
0.1
0.2
0.3
0.4
0.5
0.51
0.52
0.53
0.54
0.55
0.56
IB (μA)
3. Plot VBE (V) versus IB (μA) graph.
Table- Output characteristic.
Observation
Number
1
2
..
..
..
..

IB = 20 μA
VCE (V)
IC (mA)
0.1
0.2
0.3
0.4
0.5
….
1.0
2.0
3.0
IB = 40 μA
VCE (V)
IC (mA)
0.1
0.2
0.3
0.4
0.5
….
1.0
2.0
3.0
Plot VCE (V) versus IC (mA) graph.
Question :
1.
2.
3.
4.
5.
What is a transistor?
What is the difference between NPN and PNP transistor?
What are the uses of a transistor?
What is the physical meaning of α and β of a transistor?
In case of step up transformer we find output voltage greater than the input
voltage. Then what is the difference between a transistor and a transformer.
_______________________________________________________________________________
18
12. Experiment: Temperature coefficient of resistance of a metal
Aim: To examine the variation of resistance of a metal due to variation of temperature.
Introduction: Each conducting materials like Cu increases its resistance while increasing
its temperature. Near room temperature most of the metals changes its resistance linearly
with temperature. At low enough temperature and enough high temperatures, resistance
of the conducting meterials does not changing its resistance linearly with temperature.
Apparatus: A metal, water heater, beaker, thermometer and a multimeter.
1. Observation Table:
Observation
Number
1
2
3
..
..
..
..
..
..
Metal
Temperature
90 0C
Resistance
Room temperature
* Plot (Resistance vs Temperature).
Questions:
1. Why resistance of a metal increases with increase in temperature?
19
13. Experiment: Ohm’s law
Aim: 1. To verify Ohm’s law.
2. To verify law of combination of resistors in series and parallel.
Introduction: The current flowing through a conductor is proportional to potential
difference across it. For a given potential difference the current flow depends on the
property of the conductor which is measured in terms of either resistance or conductance.
This is known as Ohm’s law. It is important to note that the Ohm’s law is valid only in the
case of conductors when the temperature and other parameters are kept constant. In this
experiment we will investigate the variation of the current with potential difference
across it and find the resistance of the conductor. Also we will find the effective resistance
when resistors are connected in parallel and series combination
Formula:
V
I
Equivalent resistance for series connection, Rs R1  R2
RR
Equivalent resistance for parrallel connection R p  1 2
R1  R2
Ohm’s Law: R 
V
V
mA
mA
I
R
R1
Power
Supply
Figure13b: Series Resistance
Connection
Figure 13a: Ohm’s Law
I1
Power
Supply
V
I2
R1
R2
Figure13c: Parallel Connection of Resistance
Observations:
1) Ohm’s law
Trial
V1
For resistor R1 ()
I1
R1
V2
For resistor R2 ()
I2
R2
1
2
3
Mean R1
20
R2
Power
supply
Mean R2
2) Series Connection: R1=_______ , R2=_______.
Trial
Vs
(Volts)
Is
(A)
Rs 
Vs
Is

1
2
3
4
Mean Rs=
3) Parallel connection: R1=_______ , R2=_______.
Trial
V(Volts)
I1(A)
I2(A)
Rp 
V
I1  I 2
1
2
3
4
Mean Rp=
I.
Plot the graph of Current vs Voltage for R1 and R2. Checked whether Ohm’s law is verified
or not.
II.
Check whether R1 andR2 values calculated using colour codes and experimental are equal
or not.
III.
Plot the current vs voltage graph from series and parallel combination resistance
observation table.
IV.
Check whether the calculated value of Rs, and Rp, using theoretical, experimental and
graphical are equal or not.
Questions:
1. Can Ohm’s Law be applied to all the materials?
2. Is resistance of a metal constant with respect to temperature?
3. How is electrical resistance developed in a conductor?
4. At a given condition which metal has the highest resistance?
5. What are the conductance of the given resistors?
21
14. Experiment: Meter Bridge
Aim: To find the resistance and resistivity of a given wire (unkown resistance) using
meter bridge.
S l
Formula :
Resistance of the given wire: R 
100  l
d 2 R
Resistivity of the material of the wire:  
4L
Unknown wire, R
Standard resistance box, S
l
100-l
G
battery
key
Figure 14: Circuit connection of Metre Bridge Experiment
Determination of diameter of the wire ‘d’, using screw gauge
LC =
Pitch
 __________ mm ; Zero Error, ZE = ___________
Total no of HSD
Material
Cu
Fe
Kanthal
Trials
PSR(mm)
HSR
D=PSR+(HSR-ZE)x LC
Mean
Diameter,
D (m)
1
2
1
2
1
2
Determination of unknown resistance, R
22
Material
Trial
Cu
1
2
3
S()
Balancing
length l(cm)
R
S l
100  l
Mean R
Fe
Kanthal
1
2
3
1
2
3
Resistivity of the wire:
Material
Resistance,
R
Length,
L
Diameter,
D
Resistivity
(Calculated)
Resisvity (Reported
in literature)
Cu
Fe
Kanthal
Questions:
1. Which principle is used in Metre Bridge experiment?
2. Why do we prefer to use small currents Metre Bridge experiments?
3. Arrange the elements in descending order with respect to their electrical resistivity.
Elements- Au, Cu, Ag, Pt, Al.
4. Which alloy forms a good heating element?
_______________________________________________________________________________
15. Experiment: Mapping of magnetic field lines
Aim: Mapping of Magnetic lines of Force of a Bar Magnet to find its Magnetic Moment.
Introduction: A null or neutral point is that point at which the Earth’s magnetic field is
nullified by the field due to bar magnet. Before starting the experiment, make sure that
the null points are within the given A3 sheet. Don’t keep any magnetic materials in the
immediate vicinity of the bar magnet while doing this experiment.
Procedure :
1. Fix the drawing sheet firmly on the board using board pins.
2. Draw two long perpendicular lines passing through the centre of the drawing sheet.
3. Place the magnetic compass at the centre of the sheet.
4. Align one of the line along the magnetic N-S direction.
5. Place the bar magnet at the centre of the drawing sheet.
6. Draw the outline of the bar magnet.
7. Mark the magnet’s poles and the geomagnetic N-S direction.
8. Using the magnetic compass start marking the field direction without disturbing the
board or the bar magnet. (Use pencil only).
9. Continue till Null point is located properly.
10. Once null point is located, note the corresponding ‘d’ values.
N –pole of the magnet pointing geographic North :
23
When a short magnet is placed with its axis is the meridian with its north pole
pointing north of the earth, two null points are obtained on the equatorial.
In this case,
M
4d 3B

H
0
Where 0 = permeability of free space = 4 x 10-7 H.m-1, d = null point distance, BH =
horizontal component of earth’s magnetic field.
N
null
point
N
d
S
S
Figure 15a: Magnet north pointing geographic North
N-pole of the magnet pointing geographic south :
When a short magnet is placed in the magnet meridian with the south pole of the
magnet pointing north of the earth, two null points are obtained on the axial line of the
magnet.
In this case,
M
4d3B
2
H
0
Null point
N
S
d
N
d
Figure 15b: Magnet north pointing geographic South
Questions:
1. What will be happened to shape of the magnetic lines of forces if it is plotted only
the magnetic lines of forces due to the earth’s magnet?
2. Why magnetic lines of forces never intersect each other?
24
16. Experiment: Deflection Magnetometer
Aim: To compare the Magnetic moments of two bar magnets (M1/M2) using Deflection
Magnetometer(DM).
Principle : Tangent Law
Equal Distance method:
M 1 Tan 1 

M 2 Tan  2 
Null Deflection method:
3
1
3
2
M1 d

M2 d
Initial Adjustments:
The magnetometer is adjusted for tan A position as follows:
1) The DM board is placed along the east west direction.
2) All magnets and magnetic materials are removed from the working table.
3) The compass box is rotated until the pointer reads 0-0 and parallel to the scale.
4) The centre of the given magnets are marked carefully.
5) Mark the given 2 magnets as A and B.
Procedure:
Null Deflection method:
The magnet A with magnetic moment M1 is placed on one of the arms of the DM
board such that its axial line passes through the centre of the DM needle. The distance d1
from the needle to the center of the magnet is noted. The magnet B is placed on the other
arm of the DM such that its axial line also passes through the centre of the needle. The
position of the magnet B is adjusted until the pointer reads 0-0. The distance d2 of the
center of magnet B to the needle is noted. In this position the fields B1 and B2 due to
magnet A and B cancel each other reducing the deflection to zero. The expt is repeated by
reversing the poles of the magnets A and B and by interchanging the arms of the DM. The
ratio of the magnetic moments of the two magnets A and B is calculated using the
formula given in the previous section.
Table – Null Deflection Method:
Trial
d1magnet
A
(in cm)
d2 - magnet B (in cm)
1
2
3
4
Mean d2
M1  d1 
 
M 2  d2 
3
1
2
Equal Distance method:
In this method, only one magnet is used at a time. Place the magnet A at a fixed
distance from the centre of the magnetic compass needle. Note the deflection as 1 and 2.
Reverse the magnet and keep at the same distance on the same arm. Note the deflection
25
as 3 and 4. Repeat the expt on the other arm of the DM and take the readings as 5, 6, 7
and 8. Take the average of these  values. Repeat the expt for different distances.
Repeat the expt for the magnet B also. The ratio M1/M2 is calculated using the formula in
the previous section.
Table - Equal Distance method
Trial #
d(in cm)
Trial #
d(in cm)
1
1 (for magnet A)
2
3
4
1
2 (for magnet B)
2
3
4
tan(1)
Mean 
tan(1)
Mean 
tan(2)
M1  tan1 


M 2  tan 2 
tan(2)
Questions:
1. What is the use of Deflection magnetometer?
2. Why a short magnet is used in deflection magnetometer?
3. Can you use deflection magnetometer at the magnetic poles of the Earth?
17. Experiment: Tangent Galvanometer
Aim: To determine the Horizontal component of earth’s magnetic field B H at a place
using Tangent Galvanometer.
Formula:
K
I
tan  
K=reduction factor;
I = current through Tangent
Galvanometer
26
BH 
0 nK
2r
0 = Permeability of free space
= 4 x 10-7 H.m-1.
n = number of turns used. r = radius of the coil
Initial Arrangement:
The electrical connections are made as shown in the circuit diagram. The Tangent
galvanometer is made horizontal with the help of spirit level by adjusting the leveling
screws. The compass box is rotated till the 90 - 90 line is parallel to the plane of the coil.
The coil is rotated until the aluminium pointer reads 0 - 0. The coil is now set in the
magnetic meridian.
TG
Commutator
Battery
Key
A
Rheostat
Figure 17: Circuit connection for Tangent galvanometer
Procedure:
The power supply is switched ON and the rheostat is adjusted for a suitable current such
that the deflection in the TG lies between 30 and 60. The current I and the deflections 1
and 2 are noted. The current through the Tangent Galvanometer is reversed and 3 and 4
are noted. The mean value of  is calculated. The reduction factor K of the TG is
calculated. The procedure is calculated for different values of the current, I. The values
are tabulated and the mean value of K is obtained.
The radius of the coil ‘r’ and the number of turns ‘n’ is noted. BH is calculated using the
given formula.
Precaution:
1. Don’t keep any magnets or magnetic materials near TG.
2. Keep the rheostat away from the TG.
Observation Table-
Trial
Current,
I(A)
1
Deflections observed (degrees)
2
3
4
Average
tan()
K
I
tan ( )
1
2
3
4
Radius of the coil = ___________ m.
Number of turns of the coil used = ___________.
Horizontal component of the earth’s magnetic field, BH = __________ T.
27
Questions:
1) How does BH vary with Latitude?
2) What will happen if the coil of the tangent galvanometer is along east-west
direction?
18: Experiment : Laws of reflection
Aim: To verify the laws of reflection of light using a plane mirror
Apparatus Required
Plane mirror, wood board, white sheet, protractor, pins and ruler.
Concept:
It is known that the incident light, reflected light and normal line are lie in the same plane
and coincide at point of incident. Laws of reflection also states that the angle of incidence
(θi) is equal to the angle of reflection (θr), that is, θi = θr.
Procedure
1. Fasten a sheet of paper to a drawing board or flat surface into which pins can be
pressed easily. Draw a thin line (AB) in the middle of the paper. Also draw a normal line
̅̅̅̅̅̅̅ ⊥ 𝑁𝐶
̅̅̅̅ ). Draw an incident ray at θi to the
(CN) at right angles to this line i.e., (𝐴𝑁𝐵
normal, ( say, θi = 30ᵒ)
2. Press pin1 and pin 2 into the paper at the positions shown in the figure above.
3. Stand the mirror upright with its reflecting surface faces the line pin1 and pin 2 are
pressed.
̅̅̅̅̅̅̅.
4. The painted surface of the plane mirror should be exactly on the line 𝐴𝑁𝐵
5. With your eye at bench level, look into the mirror and find a position where the image
of pin2 covers pin1.
6. Now press in first pin3 and pin4 so that they in turn cover the images of pin1 and pin 2.
Pin3 and pin4 will be in line with the images of pin1 and pin2.
7. Mark the position of Pin3 and pin4. After marking, removed all the pins.
8. Dwar a line by joining the positions of the pin 3 and pin 4.
9. Measure the angle between normal and line joining by pin 3 and pin 4. This angle is
angle of reflection, r.
Repeat the experiment for the other angles of incidences, 40ᵒ, 50ᵒ, 55ᵒ, 60ᵒ and 65ᵒ.
28
Observation Table:
S.No
Angle of
incidence θi
Angle of
reflection θr
Difference
θi ~ θr
Is normal, incident ray,
reflected ray meet at a point?
1
2
3
4
5
Observation:
1. 1st law of reflection is verified.
2. 2nd Law of reflection is verified.
19. Experiment: Laws of refraction and determination of Lateral Shift
Aim: To verify the laws of refraction and find
i)
the lateral shift of a light ray when passing through a glass slab.
ii)
the refractive index of the material of the glass slab.
sin i  r 
sin i 
n
cos r 
sin  r 
Where Lcalc =calculated lateral shift, t = width of the glass slab, n = refractive index,
i= angle of incident and r = angle of refraction.
Formula: Lcalc  t 
i
r
t
Lmeas
Figure 19: Ray diagram when a ray of light passing from rarer medium to a denser medium
29
Procedure:
Place the glass slab on a sheet of paper and draw the outline. Remove the slab and
draw the normal and the incident ray with certain angle ‘i’ on one face using a protractor
and scale. Insert two pins vertically on the incident ray. Place the glass slab. Insert two
pins on the other side of the glass slab such that all the pins appear to lie in a straight line.
Remove the glass slab and the pins and join the marks made by the pin to draw the
emergent ray. The perpendicular distance between the incident ray and the emergent ray
is noted as Lmeas. The lateral shift Lcalc is calculated using the given formula and compared
with Lmeas.
The angle of refraction ‘r’ is measured and the refractive index is calculated.
Observation table:
Trial
1
2
3
4
i
30
40
45
55
r
n
i-r
Lcalc (cm)
Lmeas (cm)
Question:
1. Why the direction of ray and the incident ray is parallel in this experiment?
2. What is the refractive index of crown glass and flint glass with respect to air?
20. Experiment: To determine the refractive index of a liquid using Shift Method
Aim: To determine the refractive index of a liquid by shift method using a traveling
microscope.
Apparatus:
Travelling microscope, beaker, pin/coin, water, reading lens and saw dust.
Formula:
(a)
n
Real depth
R  R1
 3
Apparent depth R3  R2
R1, R2, R3 are the readings on the micrometer eyepiece at different conditions.
R1 = position of the image of the object (pin/coin) when no liquid is in the container.
R2 = position of the image of the object (pin/coin) when few liquid is in the container.
R3 = position of the image of the saw dust floating on the liquid surface.
30
Eye piece
Eye piece
Eye piece
Focusing
knob
Focusing
knob
R1
Objective
Focusing
knob
R2
R3
Objective
Objective
Saw
dust
coin
Traveling
microscope
coin
without water
coin
with water
with water and
saw dust
Figure 20. Position of real and apparent measurement
Procedure: Measurement of Ri’sDetermination of R1
1. Least count (LC) of a traveling microscope is determined by using the formulaValue of 1 MSD
LC 
Total number of VSD
2. The traveling microscope is set for vertical traverse. The axis of the microscope is
also made vertical.
3. The microscope is focused on a coin which is at the bottom of a beaker.
4. The main scale reading (MSR) and the coinciding vernier scale division (VSD) are
noted.
5. The total reading R1 is calculated using the relationR1 = MSR + (CVD x LC)
Determination of R2
6. Poured the water into the beaker to a height about 3-4 cm. Consequently, the pin is
out of focus.
7. The microscope carrier is moved up until the pin gets focused. Care must be taken,
not to disturbed focusing screw of the microscope while trying to see pin.
8. Repeat the steps 4 and 5, to calculate R2.
Determination of R3
9. A small quantity of dust is sprinkled on the surface of the water.
10. Microscope is now focused on the saw dust.
11. Repeat the steps 4 and 5, and calculate R3.
Repeat the experiment for different water levels.
ObservationsRi = MSR + (CVD x LC)
where MSR=Main Scale Reading, CVD = Coinciding Vernier Division., LC=least Count
31
Table 1: Determination of Refractive Index of water.
R1
Trial
no. MSR CVD
1
2
3
R1
R2
MSR CVD
R3
MSR CVD
R2
R3
n
R3  R1
R3  R2
Mean R.I.
(n )
Result1. Refractive index of water, n =……….
2. Compare your experimental value of R.I. with standard value of R.I. of water
Question:
1. Is there any other method(s) to determine the R.I of water? If so, give the names of
those experiments.
2. Can this method be employed to determine the R.I. of any material, especially liquids?
3. Why the apparent depth is lesser than the real depth?
21: Experiment: Focal Length of a given lens
Aim: To determine the focal length of a given convex lens.
Apparatus: Convex lens, Light source, screen and optical bench.
Formula: Depending on the position of the object and focal length of the lens, the size and
position of image varies. The relation between focal length, image distance, and object
distance is given by the formula (a) and the magnification is given by formula (b)
1 1 1
v
 
(a)
(b) M 
f v u
u
f= focal length, u= object position, v = image position.
u
v
Image
find position
for clear image
Object
Lens
Variable position
Of lens
Figure 21. Measurement of object and image position
32
Fix Position
Procedure:
1. Keep the lens in the lens bench.
2. Focus an object which is at infinite distance on the screen.
3. Measured the distance between lens and the position of the clearest image. Let this
length be denoted by F.
4. Keep the lens, object (light source) and the screen on the optical bench as shown in
figure.
5. Keep the object at any place such a way that the distance between object and lens
more than 2F.
6. Find the position of the image when object position is more than 2F.
7. Note these object position (in distances) and image position.
8. Also note that the nature of the image.
9. Calculate the focal length using the given formula.
10. Repeat the steps (5-9) for different image positions say,
(i) u = 2F.
(ii) F  u  2F.
11. Find the mean focal length.
ObservationTable : Determination of focal length.
Lens 1: F = ……..….
Position of
Object
u (cm)
Position of
the Image
v (cm)
Nature of
the Image
Magnification
Focal
length,
f (cm)
Average
focal
length,
fav (cm)
Focal
length,
f (cm)
Average
focal
length,
fav (cm)
more than 2F,
u =…………
at 2F,
u =…………
between F and
2F,
u =……..…
Lens 2: F = …………
Position of
Object
u (cm)
Position of
the Image
v (cm)
Nature of
the Image
Magnification
more than 2F,
u =…………
at 2F,
u =…………
between F and
2F,
u =……..…
33
4. ResultExperimental observed focal length of the lens is = ………….
Questions:
1. Compare F and fav.
2. In this method can we see the image on the screen when object is less than F?
3. With this method can we determine focal length of any given lens?
22. Experiment: Refractive index of a Prism-Pin Method
Aim: To find the refractive index of the prism by finding the angle of minimum deviation.
Formula: Refractive index of the material of the prism is given by
 A  Dm 
sin 

2


n
 A
sin  
2
where n = refractive index, A= Angle of the prism, Dm is the angle of minimum deviation.
A
i
D
Figure 22: Ray diagram for prism using pin method
Observation Table:
Angle of
incident, i
40
44
48
52
56
60
64
34
Angle of
Deviation, D
Angle of minimum
Deviation (from graph)
Refractive index
of the prism



Plot the D vs i graph.
Find the minimum deviation from the curve.
Find the refractive index of the prism.
Question:
1. Why angle of minimum deviation takes place in this experiment?
23. Experiment: Diffraction through Grating
Aim: Determination of wavelength of laser Light source using grating diffraction.
Apparatus required: Gratings and its stand, Laser source, Meter Scale, Stand and graph
paper.
Formula:
m  d sin θm ,
where m = order number,
d = distance between the two adjacent grating slits,
 m = angle between mth order fringe and the 0th order fringe to the grating.
Diagram:
Figure 23a. Gratting.
Figure 23b. Grating Diffraction.
Calculation1.
d
1
number of lines per unit length
(Convert in meter)
35
2.
D = distance between the grating and the screen (is in cm, do not convert in meter).
3. Measurement of angle  m :
Grating
1st order fringe
2x1
Cm
x1
cm
x1
D
1 =
tan
1 x1 
 
D
2nd order fringe
1 =
d sin 1
2 
2x2
cm
x2
cm
x2
D
x 
1
tan  2 
D
 
Result: Wavelength of the laser light source is,
 
exp t  1 2  ..............
2
Error calculation:
  exp t
error %  real
 100 %
real
Question;
1. From this experiment, what can you tell that to get more accurate value of
wavelength?
2. Which property of light is used in diffraction phenomenon?
3. What is mean by resolving power of grating?
4. What will happen if we used a non-monochromatic light source instead
monochromatic one?
36
2 =
d sin  2
2
24. Experiment: Photoelectric Effect
Aim: To determine the Planck’s constant using photodiode.
Apparatus: Light source, color filters, photoelectric effect setup with voltmeter and
ammeter.
eV
Formula:
h
 e  slope ;

h= planck’s constant, e = electronic charge, ν = frequency.
Connecting
board
A
-
Photocell
Ray of light
V
+
Light
source
Color filters
(b) Experimental set up we have seen.
(a) Electrical connection
Figure 17: Photoelectric effect
Procedure:
The bulb is allowed to warm up for 10 minutes with the metal box lid removed. A
color filter (a color glass disc) say, orange color, is inserted in the color filter window
provided between photocell and light source. Measured/noted the stopping voltage, V.
Repeat the steps (1-4) for different color filters.
Table 1: Determination of Planck’s constant.
Color
Wavelength Frequency
(x 10-9m)
(x 1014 Hz)
Photo current,
I = ? when V = 0
Stopping potential
V=? when I= 0
1. Plot Stopping potential versus frequency, and find the slope.
3. Planks Constant as,
4. Calculate error as-
he
V

 e  slope ; e =1.602 x 10-19C
6.626  10 34  hexpt
Error % 
 100 %
6.626  10 34
Question:
1. Give few applications of photoelectric effect.
2. Which nature of light is used to explain photoelectric effect?
3. Why we are using color filters in this photoelectric effect?
4. From the graph, try to find the threshold frequency.
37
25. Experiment: H2 Lamp Spectroscopy
Aim: 1. To measure the wavelength of the Balmer series of visible emission lines from the
hydrogen.
2. To determine the Rydberg Constant using Bohr model formulation.
Apparatus:
Spectrometer, a grating, hydrogen discharge lamp, power supply, a magnifying
glass, a small night light, spirit level and a black cloth to block out stray light.
Formulae1.
mλ  d sinθ
where n is an integer given by m = 0, 1, 2, 3, ….
2.
Rydberg formula1
1
1
 R 2  2 

 n f ni 
where R is the Rydberg constant, ni and nf are integers, 1,2,3,4,…. up to infinity,
with ni > nf. For hydrogen atom, nf = 2 corresponds to the Balmer series.
a. H2 Lamp-Spectrometer
b. Image formation.
c. Spectrum
d. Alignment and Measurement.
Figure 25: H2 Lamp Spectrometer and spectrum
38
ObservationValue of 1 MSD = S = …….
Total number of VSD = N = ……….
S
Least Count, LC   ………
N
Total reading, say, θ ox = MSD + (CVD x LC);
x = R or L
Table 1: Observation table for hydrogen Balmer series.
Trial
no
Color
ni
m
v
Blue
Cyan
red
violet
Blue
cyan
red
▀
▀
▀
▀
▀
▀
▀
▀
6
5
4
3
6
5
4
3
1
1
1
1
2
2
2
2
θ oR
θ oL
θ oL
Corrected
(If
needed)

d sin 
m
1

1 1 
  2
 4 ni 
1 1 
1
* Draw the graph   2  versus .

 4 ni 
* Slope = R =……………….
Accepted value of Rydberg constant, R = 10.9737 x 106 m-1.
% difference 
Accepted value  Measured value
Accepted value
 100%
Questions:
2
1. In the analysis of your results, suppose you had plotted 1/λ versus 1/n i instead of (¼ 2
1/ni ). Would your data give a straight line and would its slope give the same results
2.
3.
for R? How would the intercept change?
Is this experiment tells wave nature or particle nature of light?
How?Can this Rydberg formula be true for any other hydrogen like chemical
elements, e.g. He+, Li2+, Be3+ etc.?
39
26. Experiment: Solar constant
Aim: Measure Solar constant and hence the luminosity of the Sun
Apparatus: Aluminium (Al) Plate blackened on one surface, K-type thermocouple
connected to a multimeter, PVC pipe used as holder for Al plate and stop clock
Introduction: The solar constant (S) is the amount of energy incident normally per unit
area per unit time on the surface of earth. This can be determined by exposing a metal
plate normal to solar radiation and measuring its raise in temperature. Once S is obtained
the Luminosity of Sun can be estimated.
Theory:
Solar Constant (S):
The Al plate exposed to Sun rays for duration of time t receives heat energy equal to
Q  StA ------------ (1)
Where, S is the solar constant and A is the surface area of the blackened surface of the Al-plate.
The heat energy results in the rise in the temperature of the plate. If the specific heat of Al is C and
the mass of the plate is m, the rise in temperature can be written as
 
Q
-------------------(2)
mC
Substituting for Q from equation 1 in equation 2
 
SAt
mC
Measuring  and t, Solar constant, S can be determined.
Luminosity of Sun (L):
The amount of energy emitted by Sun in one second in all the direction. The relation
between Luminosity and Solar constant can be obtained in the following way:
Consider a sphere of radius R equal to distance between Sun and earth (Astronomical Units). All
the energy emitted by Sun should pass through the surface of this sphere. In one second the Sun
emits energy equal to Luminosity of Sun. Thus on the surface of the earth the amount energy
received per unit area per second (Solar constant) should be given by
S0 
L
4R 2
If S0 is known then L can be estimated. The mean value of R =1.49X1011 m
Note:
units.
Specific heat of Al = 913 Jkg-1 per degree centigrade. All measurement should be in S.I.
Experimental setup:
 The PVC pipe is fixed to the retort -stand and placed outside in the Sun. Adjust the
orientation of the pipe such that the solar rays are normal to the surface of the Al
plate to be placed latter. If the rays are incident normally the shadow of the pipe on
the floor will be shortest.
 The Al plate is placed inside the pipe such that the blackened side faces the open
end of the pipe.
40

A thermocouple is attached to the center of the back side of Al plate. The terminal
of the thermocouple is connected to multimeter with the selector knob turned to
the position indicating the temperature measurement.
Procedure:
 The Al plate must be at least 50 C lesser than the ambient (room) temperature.
Measure the mass and area of the blackened surface of the Al plate before placing
it in the PVC pipe. Note down the time at which the experiment is started. This
will help in finding the zenith angle of the Sun from the data table books.
 Note down the initial temperature of the Al plate. Start the stop watch.
 Start noting down the temperature of the Al plate, every 30s till the temperature
rises by 200C.
 Plot a graph of Change in temperature Vs time. The slope of the straight line graph
SA
givesslope 
mC
 Knowing the mass, specific heat and surface area of the of the Al plate, Solar
constant can be determined.
 The solar constant measured on the surface of the earth (S) should be corrected for
the atmospheric absorption. If the S0 is the value of solar constant without the
atmospheric absorption then the relation between S and S0 is given be
S  S0asecz 
Where, a is the coefficient of absorption due to earth’s atmosphere, whose average value
can be taken as 0.7 and z is the Zenith angle. At 12:00 noon the Zenith angle in different
months are given the table below
Month
January
February
March
April
May
June
Zenith angle
(degree)
36.39
28.71
18.14
7.28
6.78
10.79
Month
July
August
September
October
November
December
Zenith angle
(degree)
10.13
6.9
12.34
23.34
32.0
37.35
Observation table:
Time
Temperature
Solar Constant,
S
Solar Constant without
atm. absorption, S0
Luminosity of
the Sun, L
41
27. Experiment: Heat Transfer
Aim: Comparison of the thermal conductivity of given materials
Apparatus: Metal rods of same shape and size (1ft in length), 500 ml glass beaker and
a heater.
Procedure:
4. Pour 350-400ml of water to the beaker.
5. Heat the water up to a constant temperature, say 90 0C .
6. Mark two positions to all the rods at same length from one end (say at 4cm
and 20 cm from one end).
7. Pick any two rods and hold by your bare hand at 20 cm the mark point.
8. Dip the two rods up to the mark (4 cm) into the hot water. As shown in
figure.
9. Wait for some time. After a while your hand will feel heat.
10. Identify which hand feels heat first.
11. The rod which was held by your hand that feels first heat sensation has
higher thermal conductivity.
12. Repeat the procedure 4-8 for all the materials given to you and compare them.
hand
B
D
A
C
Hot water
Constant Temp.
Fire
Figure 21: experimental set up for heat transfer
Observation Table:
Given materials
Thermal conductivity
In ascending order
42
Brass, Copper, Aluminium, Steel, ………..
28. Experiment: Latent Heat of Fusion of ice
Aim: To determine latent heat of fusion of ice.
Introduction: if we supply heat to a solid ice then it can convert from solid state to liquid
state without changing its temperature. The heat energy released or absorbed by an object
during a thermodynamic process without changing its temperature is called latent heat,
like ice’s change of phase from solid to liquid at 0 0C, change phase from liquid water to
steam water at 100 0C. Latent heat of fusion or enthalpy of fusion is the latent heat of the
object when the object undergoes its phase changes from solid to liquid at constant
temperature.
Procedure:
1. Measure a clean, dry calorimeter (glass beaker/Cu) mass = Mc.
2. Add 40-50 ml of water to calorimeter and find the mass of water,
Mw = (Mass of water + Mass of Calorimeter) –Mass of calorimeter
3. Measure temperature of water = Tw.
4. Take two cubes of dry, clean ice and measure ice temperature = Ti.
5. Add these two cubes of ice to the water inside the beaker.
6. Stir the mixed solution until the solution become constant temperature. Measure this
constant temperature= Tf .
7. Measure the mass of the total system = Ms
Ms = (Mass of Calorimeter + Mass of water + Mass of ice )
8. Mass of the ice = Mi,
Mi = (Mass of Calorimeter + Mass of water + Mass of ice) – (Mass of Calorimeter + Mass of water)
= Ms – (Mc+ Mw)
Formulaa. Heat gain by (ice to melt+ melt ice to rise up its temperature from Ti to Tf)
= 𝑀𝑖 × 𝐿𝑖 + 𝑀𝑖 × 𝑆𝑤 (𝑇𝑓 − 𝑇𝑖 )
b. Heat loss by (water + calorimeter) from 𝑇𝑤 to 𝑇𝑓 .
= 𝑀𝑤 × 𝑆𝑤 (𝑇𝑤 − 𝑇𝑓 ) + 𝑀𝑐 × 𝑆𝑐 (𝑇𝑤 − 𝑇𝑓 )
By principle of mixing,
Heat Loss = Heat Gain
𝑴𝒊 × 𝑳𝒊 + 𝑴𝒊 × 𝑺𝒘 (𝑻𝒇 − 𝑻𝒊 ) = 𝑀𝑤 × 𝑆𝑤 (𝑇𝑤 − 𝑇𝑓 ) + 𝑀𝑐 × 𝑆𝑐 (𝑇𝑤 − 𝑇𝑓 )
Li= Latent heat of fusion of ice
𝑀𝑤 × 𝑆𝑤 (𝑇𝑤 − 𝑇𝑓 ) + 𝑀𝑐 × 𝑆𝑐 (𝑇𝑤 − 𝑇𝑓 ) − 𝑴𝒊 × 𝑺𝒘 (𝑻𝒇 − 𝑻𝒊 )
𝐿𝑖 =
𝑴𝒊
Literature data:
Specific heat of Water, Sw = 4186 𝐽𝐾𝑔−1 𝐾 −1;
SC = 670 (crown), SC = 500 (flint),
−1 −1
SC = 840 (ordinary) 𝐽𝐾𝑔 𝐾 and
SC = 3860 𝐽𝐾𝑔−1 𝐾 −1 for copper.
Observation Table:
Trial
1
2
Mcalorimeter
Mwater
Mice
Twater
Result: Latent heat of fusion of ice is ……………
Tice
Tfinal
Li
% error = ………………
43
29. Experiment : Measurement of Specific Heat Capacity of a liquid
Aim: To measure the specific heat capacity of water
Introduction: When a current I is passing through a resistor for a period of time. The
electrical energy can be converted in to heat energy following Joules heating effect. If the
resistor is inside a liquid medium, then, this heat energy form electricity can in turn rise
the temperature of the liquid.
Apparatus: Heater Coil, Constant Current Supply, Digital Thermometer, Beaker, Stop
Watch, Digital balance.
Procedure
1. Find the mass of the 100 ml empty beaker (calorimeter) = Mc.
2. Fill 80 ml water into a beaker and measure the mass of water = Mw.
3. Measure the initial temperature T1 of the water.
4. Measure the current I from the power supply and Resistance R of the coil by multimeter.
5. Put the coil (heater) into the water and then switch on the power supply for 15 minutes.
Measure the time t by a stop-watch.
6. Stir the water continuously measure the corresponding temperature.
7. Record the final temperature T2 of water.
Formula:
Energy supplied by heater = heat gained by water + Heat gained by calorimeter
I2  R  t = Mw × S × ΔT + (Mc  Sc ΔT)
𝐼 2 × 𝑅 × 𝑡 − 𝑀𝑐× 𝑆𝑐 × ∆𝑇
𝑆=
𝑀𝑤 × ∆𝑇
where
ΔT = change in the temperature =(T2-T1) ;
Sc = Specific Heat of Calorimeter ( = 840 J/Kg K for ordinary glass) .
Observation Table:
Trial
1
2
Mc
Mw
I
R
t
T1
T2
ΔT
S
Result: The specific heat capacity of water is _________ J/kg K.
The standard value of the specific heat capacity of __________ is _________ J/kg K.
44
30. Experiment:
Determination of Specific Heat Capacity of Metals
Aim: To determine specific heat capacity of solids by the method of mixtures.
Introduction: Specific Heat capacity of an object is the amount of heat energy supplied to
per unit mass of the object to raised its temperature by 1 0C. Its SI unit is J/K and it is
defined as the amount of heat energy required to raise 1 kg mass of the object by 1 K.
Procedure:Take a clean, dry glass beaker. Mass of the beaker = (Mg). Add 20 ml of water
to the beaker and find the mass Mw of 20 ml water. Note down the initial temperature T1
of water. You are given Fe, Cu and Al rods. Find the mass MM metal. Put these into a
constant temperature water bath for 20 minutes. Note down the temperature T2 of the
water bath in which the rods are immersed. Take metal and dip into the beaker
containing 20ml of water. Find the maximum temperature T3 reached.
Let SM = specific heat capacity of metal. Specific heat capacity of water (Sw) = 4186 J kg-1K-1
and ordinary glass beaker (Sb) = 840 J kg-1K-1.
a. Heat gain by (water + beaker) to rise up its temperature from T1 to T3
= (𝑀𝑤 × 𝑆𝑤 + 𝑀𝑏 × 𝑆𝑏 ) × (𝑇3 − 𝑇1 )
b. Heat loss by metal when it cooled from 𝑇2 to 𝑇3 .
= 𝑀𝑀 × 𝑆𝑀 (𝑇2 − 𝑇3 )
By the principle of method of mixtures, Heat gained = Heat lost.
Therefore,
Mw x Sw x (T3 – T1) + Mg × Sb x (T3 – T1) = MM × SM × (T2 – T3)
Therefore specific heat capacity of metal Al.
𝑆𝑀 =
(𝑀𝑤 × 𝑆𝑤 + 𝑀𝑔 × 𝑆𝑏 )× (𝑇3 – 𝑇1 )
𝑀𝑀 × (𝑇2 – 𝑇3 )
J kg-1K-1
Similarly find the specific heat capacity for Cu and Fe.
Observation Table
Metal
Mass of
metal, 𝑀𝑀
Temp of
metal,
𝑇2
Mass of
water,
𝑀𝑤
Mass of
beaker,
𝑀𝑤
Temp of
water, 𝑇1
Final Temp.
of water, 𝑇3
Specific heat of Metal,
𝑆𝑀
Results:
i.
The specific heat capacity of Al metal = ……….. J kg-1K-1
ii.
The specific heat capacity of Cu metal = ……….. J kg-1K-1
iii.
The specific heat capacity of Fe metal = ……….. J kg-1K-1
45
31. Experiment. Mixing of liquids of different temperatures
Aim: To determine the temperature of mixed solution when two solutions of different
temperatures mixed.
Apparatus: Calorimeter, stirrer, thermometer, waters of different temperatures.
Procedure:
1. Measure a clean, dry calorimeter (glass beaker/Cu) mass = Mc
2. Add 20-50 ml of water “a” to calorimeter and find the mass of water = Ma
3. Ma = (Mass of beaker, Mc + Mass of water, Ma) – (Mass of beaker, Mc)
4. Measure temperature of water “a” = Ta.
5. Take 40-60 ml of water “b” of higher temperature than water “a”.
6. Measure temperature of water = Tb.
7. Meaure the mass of water “b” = Mb
8. Mb = (mass of calorimeter +mass of water “a” + Mass of water “b”)
– (mass of calorimeter + mass of water “a”)
9. Stir the mixed solution until the solution become constant temperature. Measure this
constant temperature= Tfinal .
10. Compare this experimental final temperature and calculated temperature.
Formulaa. Heat gain by cold water (water a) to rise its temperature up to Tf.
= 𝑀𝑎 × 𝑆𝑤 × (𝑇𝑓𝑖𝑛𝑎𝑙 − 𝑇𝑎 ) + 𝑀𝑐 × 𝑆𝑐 × (𝑇𝑓𝑖𝑛𝑎𝑙 − 𝑇𝑎 )
b. Heat loss by hot water
= 𝑀𝑏 × 𝑆𝑤 × (𝑇𝑏 − 𝑇𝑓𝑖𝑛𝑎𝑙 )
Heat Loss = Heat Gain gives𝑀𝑎 × 𝑆𝑤 × 𝑇𝑎 + 𝑀𝑐 × 𝑆𝑐 × 𝑇𝑎 + 𝑀𝑏 × 𝑆𝑤 × 𝑇𝑏
𝑇𝑓𝑖𝑛𝑎𝑙 =
𝑀𝑎 × 𝑆𝑤 + 𝑀𝑏 × 𝑆𝑤 + 𝑀𝑐 × 𝑆𝑐
Literature data:
Specific heat of Water, Sw = 4186 𝐽𝐾𝑔−1 𝐾 −1, Sc = 840 𝐽𝐾𝑔−1 𝐾 −1 ( for ordinary glass)
Observation Table:
Mass
Temp
cold
of cold
Trial
water,
water,
Ma
Ta
1
2
46
Mass
hot
water
Mb
Temp
of hot
water,
Tb
Mass of
Calorimeter
Mc
Final temp
Tfinal
Calculated
Final Temp
Tf inal
Experimental
32. Experiment. Dispersion of light
Aim:
(a) To observed the dispersion of light.
(b) To determine the wavelength and energy associated by different colors of light.
Procedure:
(i) Set the spectrometer properly.
(ii) Observed the different colors (violet, green and red).
(iii) Coincide the vertical cross wire with the time.
(iv) Measure the corresponding wavelength.
Formula: Energy (E) associated with a photon of wavelength () and frequency (ν) is
ℎ𝑐
𝐸 = ℎ𝜈 =
𝐽
𝜆
ℎ𝑐
𝐸 = ℎ𝜈 =
𝑒𝑉
𝑒𝜆
Where ℎ = 𝑃𝑙𝑎𝑛𝑐𝑘 ′ 𝑠𝑐𝑜𝑛𝑠𝑡𝑎𝑛𝑡 = 6.626 × 10−34 𝐽. 𝑠
𝑒 = 𝑒𝑙𝑒𝑐𝑡𝑟𝑜𝑛𝑖𝑐 𝑐ℎ𝑎𝑟𝑔𝑒 = 1.6 × 10−19 𝐶
𝑐 = 𝑠𝑝𝑝𝑒𝑑 𝑜𝑓 𝑙𝑖𝑔ℎ𝑡 𝑖𝑛 𝑣𝑎𝑐𝑢𝑢𝑚 = 2.9979245 × 108 𝑚𝑠 −1
Color of the light
Wave length
Frequency

ν
Energy at Joules
ℎ𝑐
𝐸 = ℎ𝜈 =
𝜆
Energy in eV.
ℎ𝑐
𝐸=
𝜆𝑒
47
Physical Constants, Standard values and Units
Thermal Conductivity and specific heat capacity of selected materials:
Thermal Conductivity
Specific heat capacity
Material
-1
-1 -1
Js m K
J kg-1K-1
Copper
385
385
Aluminium
205
913
Brass
109
380
Steel
50.2*
452*
Iron
34.6- 80.4
444 - 500
* It changes depending on composition.
Physical Constants, Standard values and Units
Physical Constant,
Standard Parameters
1. Speed of Light, c
2. Planck’s
constant, h
3. Permeability of a
vacuum, 0
Value
Physical parameter
Units
2.9979 x 108 ms-1
1. Electric charge
Coulomb, C
6.626 x 10-34 Js
2. Electric current
Ampere, A
1.361 kWm-2
3. Magnetic field
strength
4. Magnetic flux
density
5. Potential difference
6. Electric resistance
7. Resistivity
8. Wavelength
3.839 x 1026 W
9. Frequency
4π x 10-7 Hm-1
4. Electric charge, e
1.602 x 10-19 C
5. Rydberg constant
6. Velocity of Sound
at 0 0C
7. Solar Constant
8. Luminosity of the
Sun
9. Refractive index
Glass (crown & flint)
10. Refractive index
water
1.0974 x 107 m-1
331.3 ms-1
Am-1
Tesla, T
V
Ohm, Ω
Ωm
m
Hz or s-1
1.485- 1.925
1.3330
Resistivity (at 20 0C) and Temperature Coefficient of resistance of selected materials::
1. Copper
1.68 x 10-8 Ωm
0.0039 K-1
2. Aluminium
2.82 x 10-8 Ωm
0.0039 K-1
3. Nicrome
100-150 x 10-8 Ωm 0.0004 K-1
4. Steel
16- 74 x 10-8 Ωm
5. Kanthal
139-145 x 10-8 Ωm
Multiplication factor
1. nano, n = 10-9
2. milli, m = 10-3
micro, μ =10-6
kilo, k = 103
This manual is compiled by Dr. K. J. Singh
48
IISc Press
Talent Development Centre
Indian Institute of Science, Kudpura
Challakere, Chitradurga District, Karnataka- 577536
49