TDC-IISc, Kudapura
Physics Experiment
PU College Teachers Training Program
PHYSICS
EXPERIMENTS
TALENT DEVELOPMENT CENTRE
INDIAN INSTITUTE OF SCIENCE, KUDAPURA
Challakere, Chitradurga District,
Karnataka-577536
TDC-IISc, Kudapura
Physics Experiment
Contents
1.
Measurement of linear dimensions and density of materials
2.
Acceleration due to gravity - Simple Pendulum
3.
Compound pendulum
4.
Projectile motion
5.
Determination of Spring Constant
6.
Determination of Young’s Modulus – Searle’s method
7.
Determination of Moment of Inertia and Rigidity modulus
8.
Equilibrium of concurrent coplanar forces
9.
Moment bar: Equilibrium of parallel forces
10.
Determination of Solar Constant
11.
Verification of Conservation of Energy
12.
Coefficient of viscosity – Stoke’s method
13.
Determination of surface tension of water – capillary rise
14.
Determination of velocity of sound - Resonance Column
15.
Sonometer - Verification of Laws of Transverse Vibrations of Stretched String
16.
Resistivity of metals - Metre Bridge
17.
Determination of Absolute Zero – Ideal gas law
18.
Temperature coefficient of resistance  of metal
19.
Temperature Dependence of Resistance of semiconductor – Energy Bandgap Eg
20.
Heat Transfer
21.
Thermal expansion
22.
Specific heat capacity of metals
23.
Newton’s law of cooling
24.
Hydrogen spectra and determination of Rydberg constant
25.
Focal Length of lenses and mirrors
26.
Refractive index of glass - Lateral Shift
27.
Refractive index of glass Prism - Pin Method
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Physics Experiment
28.
Refractive index of liquid - Travelling microscope
29.
Refractive index of glass Prism - Spectrometer
30.
Refractive index of liquid - Spectrometer
31.
Determination of Specific rotation - Polarization
32.
Determination of wavelength of laser using Transmission Grating
33.
Determination of wavelength of laser using Reflection Grating
34.
Determination of grating constant - spectrometer - Sodium lamp
35.
Newton’s rings
36.
Comparison of Magnetic Moment of two bar magnets - Deflection
Magnetometer
37.
Determination of BH - Tangent Galvanometer
38.
Determination of M of bar magnet by Mapping of Magnetic lines of force
39.
Strength of Magnetic field due to a Solenoid and Cylindrical Magnet
40.
Verification of Ohm’s Law
41.
Diode Characteristics
42.
Zener diode characteristics
43.
Half wave and Full wave Rectifier Circuits
44.
Transistor Characteristics
45.
Transistor (Common Emitter) Amplifier Circuit
46.
Measurement of self inductance of inductor
47.
Series resonance of LCR circuit
48.
Parallel resonance of LCR circuit
49.
Charging and discharging of capacitor
50.
Verification of truth table of logic gates
51.
Amplitude Modulation - Demonstration
52.
Determination of e/m of an electron
53.
Determination of Planck’s constant - Photoelectric Effect
Physical constant, standard values and units
TDC-IISc, Kudapura
Physics Experiment
Experiment 1a. Density of solids
Aim: Determination of density of solids.
You are given a digital vernier caliper.
Measure the dimensions of the object, find the volume and mass and finally the density of the given
material/object.
Formula:
Volume of Rectangular block =
𝑙 𝑖𝑛 𝑐𝑚  𝑏 𝑖𝑛 𝑐𝑚  ℎ 𝑖𝑛 𝑐𝑚 = ________cm3
3
Volume of Sphere =
4 d 
  =
3 2
____________cm3
2
Density
g/cm3
Density
kg/m3
Solid
Cylinder
Rectangular
Block
d 
Volume of Solid Cylinder =    h = ____________ m3
2
Observations:
Mean
Volume Mass g
Object
Dimensions
Trial
mm
cm
mm
cm3
1
Length
L
2
1
Breadth
B
2
1
Height
h
2
1
Diameter
Sphere
d
2
1
Height
h
2
1
Diameter
d
2
Similarly make columns and determine the density of all the solids given to you.
Experiment 1b. Screw Gauge
Aim: To determine the dimensions of a solid using Screw gauge and density of a given material.
Principle:
pitch of the screw
Least count =
mm.
no: head scale div:
Observation:
Object
Dimension
Trial
mm
Mean
mm
cm
Volume
(cm3)
Mass
(g)
1
2
Cylinder
1
length
2
1
Steel ball
diameter
2
3
Similarly make columns and determine the density of all the solids given to you.
Result:
diameter
Density of the given solid _____ is ___________g/cm3 =___________kg/m3.
Density
(g/cm3)
Density
(kg/m3)
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Physics Experiment
Experiment 1c. Density of a Solid by Archimedes method
Aim: Density by measuring the volume displaced by solids as in experiment 1a and 1b and also
irregular shaped solids such as small stones.
Take a given solid and weigh it accurately in a digital balance.
Mass of the solid, m = __________ g
Take water in the 5 ml measuring cylinder and measure the volume.
Volume V1 = __________cm3
Drop the given solid material slowly into the cylinder and read the volume again.
Volume V2 =__________ cm3
Volume of the solid = (V2- V1) cm3;
Density = mass (g) / Volume in cc = ________ g/cc
Solid
taken
Mass (g)
Volume
V1(ml=cc=cm3)
Volume V2 V2 – V1 Density
(cm3)
(cm3)
(g/cm3)
Density reported in the literature
(g/cm3)
1
2
3
4
Experiment 1d. Density of liquids
Temperature of the liquid
=_________ͦ C.
Take 50 ml beaker.
Mass of the beaker
=
g (A) Add 10 ml of a liquid from a pipette.
Mass of liquid + Beaker
=
g (B).
Mass of 10 ml of liquid =
(B) – (A) = _____g.
Therefore Density of liquid =
_______ g/cc.
Repeat the experiment 4 times and find accurate density of water, alcohol, methanol, CCl4.
Compare the results with the standard value from Clark’s table.
Experiment 2. Acceleration due to gravity - Simple Pendulum
Aim:
To determine the acceleration due to gravity (g) at the place.
Apparatus:
Metallic bob with hook, clamp stand, split halves of a cork with plane faces, fine cotton thread
about 150 cm in length, digital Vernier calipers, 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 analyzed properly one can obtain the value
of g accurately.
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Physics Experiment
Formula:
 L
2 
T 
(a) g  4 2 
(ms-2)
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.
Clamp
ObservationTable 1: Determination of the length of the
Simple Pendulum.
Stand
l
Inextensible
thread
Bob
2r
Diameter
Trial
Mean r
of
the
Number
2r
in m
bob, 2r
1
2
Figure: Simple Pendulum
Table 2: Determination of g.
Serial
Trial
L(cm)
number
No
1)
1.
2)
1)
2.
2)
1)
3.
2)
1)
4.
2)
1)
5.
2)
1)
6.
2)
1)
7.
2)
1)
8.
2)
n
Tn(s)
T
Tn
n
Mean
T(s)
 L
g  4 2  2 
T 
(ms-2)
n = Number of oscillations, Tn = Time for n number of oscillations, T 
Acceleration due to gravity g, is = ………… ms-2
Determination of the length of Second’s Pendulum:
L
T
T2
Tn
= Period of oscillation,
n
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Physics Experiment
* Plot the graph between L – T and L-T2 graph.
* Find the length of the seconds Pendulum. [Pendulum having time period of two seconds is called
seconds pendulum.]
Questions:
1. What is the value of g at the Pole and at the Equator of the Earth?
2. What is the value of g at the moon’s surface?
3. Is there any other method to determine the g at a place? If so, mention them.
4. Why does the value of g vary from place to place?
5. What is the relation between g (acceleration due to gravity) and G (universal
gravitational constant)? What is the value of g on the surface of a planet with twice
the mass of the earth and four times the radius of the earth?
6. Can we use extensible thread in simple pendulum?
7. What will be the time period of simple pendulum if its length is infinity?
Experiment 3. Compound pendulum
Aim
To determine acceleration due to gravity and radius of gyration of a compound pendulum bar about its
center of mass.
Apparatus
A bar pendulum, stop clock, meter scale.
Description
A compound bar pendulum AB is a metallic, thick rectangular bar. A number of small circular holes
of about 1 cm diameter are drilled along the length of the bar at equal distance (about 2 cm) from each
other. The bar pendulum can be suspended vertically from each of these holes through a horizontal
edge K.
Procedure
The bar pendulum is suspended by the horizontal edge passing
through the first hole from one end A (say). With the help of a
pointer, the position of rest of the pendulum is noted. The bar is set to
small oscillations about the equilibrium point. Leaving first three or
four oscillations, the time taken for twenty oscillations with two trials
is noted. The distance of horizontal edge from the top end A is found
out. The experiment is repeated by suspending the bar in each hole and the distance
from A to horizontal edge is measured. After crossing center of gravity, the bar is
suspended upside down, but the distance of horizontal edge is measured from same
end A of the bar. Observations are tabulated.
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Physics Experiment
Period of oscillation
Time for 20 oscillations
No. of hole Distance
from
from end A
end A (cm)
Trail 1
Trail 2
Mean
Mean
period T
(s)
A graph is drawn taking distance l of horizontal edge along x-axis and period of oscillation T along yaxis. The graph is symmetrical about the line passing through the center of gravity C parallel to y-axis.
It consists of two similar curves on either side of C. A line PQRS is drawn, parallel to x-axis, cutting
the curve at four points P, Q, R and S. The points P and R lying on either side of C, correspond to the
center of suspension and center of suspension and center of oscillation respectively. Similarly, other
pair of points is Q and S. Hence, the length L of equivalent simple pendulum is
L = (PR + QS)/2
The other such lines are drawn and corresponding periods T are noted and tabulated.
Determination of g
Period T (s)
PR (cm)
QS (cm)
Length of equivalent
pendulum:
L=(PR+QS)/2 L/T2
(cm)
Mean L/T2 =
The acceleration due to gravity
ms-2
𝐿
𝑇2
is determined and mean value is take. From the graph, we not that PM=SM=l1 and RM=QM=l2 and
hence the radius of gyration about the axis passing through C is given
𝐾 = √𝑙1 𝑙2
The mean value of K is calculated.
𝑔 = 4𝜋 2
S.No.
Result:
(i)
(ii)
l1=PS/2 (cm)
Acceleration due to gravity =
Radius of gyration
=
l2=QR/2 (cm)
ms-2
m
𝐾 = √𝑙1 𝑙2
𝑐𝑚
TDC-IISc, Kudapura
Physics Experiment
Experiment 4. Projectile Motion
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.
There are three kinds of projectile motion.
1) Horizontal projectile motion: The projectile launched from a height in the horizontal
direction is called horizontal projectile. The range of the projectile depends on initial velocity
u , the height (h) from which projectile is launched and g. The expression for R can be written
as
2ℎ
𝑅 = 𝑢√
𝑔
R
2) Ground to ground projectile: The projectile launched from the ground reaches the same
horizontal level as its initial position, after its flight in air (shown in the figure below). This is
called ground to ground projectile.
3) Projectile launched from a height at an angle:
u
ux
h
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Physics Experiment
The projectile motion of this kind can easily analyzed by considering two one dimensional motion
separately along vertical and horizontal directions. The motion along vertical direction can be
1
described using equationℎ = 2 𝑔𝑡 2 − 𝑢𝑦 𝑡 . Where, t is the time of flight of the projectile. The motion
along horizontal can described using the equation𝑅 = 𝑢𝑥 𝑡.
EXPERIMENT to find the Muzzle speed
1. Choose one corner of a table to place the projectile launcher. Make sure a distance of about 2m is
clear on the floor around the table. Measure the height of the launcher.
2. Clamp the launcher to the corner of the table using the Universal Table Clamp
3. Adjust the angle of the launcher to 0o. This is a case of horizontal projectile. Place carbon paper on
the ground such that the projectile falls on it.
4. Align the projectile launcher and launch the ball and measure the range using the mark made by
the carbon paper on the floor.
5. Using equation 1 estimate the speed of the projectile.
Height (h) =___________
Range Velocity
EXPERIMENT SETUP - LAUNCHING AT AN ANGLE (GROUND TO GROUND
PROJECTILE
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 200. 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).
Data Table
Velocity =
Angle
(degrees)
20
30
40
45
50
60
70
80
Horizontal
Range (m)
experiment
Horizontal
range
(theory)
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Physics Experiment
QUESTIONS
1. Sketch the trajectory of your projectile when it was shot at an angle of 60 degrees. Draw 5
qualitative horizontal and vertical velocity vectors at different locations on your sketch. Make sure
the lengths of the vectors represent the relative magnitudes of the velocities. In other words, low
velocities should be represented by short arrows and long arrows should represent large velocities.
2. 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.
3. What angle corresponds to the maximum range? Explain why this particular angle produces the
maximum range.
Experiment 5. 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
Observation Table:
Trial
No.
Load,
F
( 10-39.81 N)
Load
Increasing
(cm)
Pointer reading
Load
Decreasing
(cm)
Extension
x ( 10-2 m)
Average
(cm)
1.
x0
x0- x0=
2.
x1
x1- x0=
3.
x2
x2- x0=
4.
x3
x3- x0=
5.
x4
x4- x0=
6.
x5
x5- x0=
7.
x6
x6- x0=
18
16
12
-2
Extension (x10 m)
14
10
8
6
4
2
0
0
100
200
300
-3
4. Plot F - x Graph.
Load ( x 9.81x10 N)
400
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Physics Experiment
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.
Experiment 6. Young’s Modulus – Searle’s Method
Aim: To determine the Young’s Modulus of the material of the given wire using Searle’s apparatus.
Apparatus: Searle’s apparatus, given wire, meter scale, 0.5 kg weights, screw gauge.
Formula:
Young’s modulus of the material of the wire
𝑌=
𝑀𝑔𝑙
(𝑃𝑎𝑠𝑐𝑎𝑙)
𝜋𝑟 2 𝑥
Where M = load (in kg)
g = acceleration due to gravity (in ms-2)
l = length of the wire (in m)
r = radius of the wire (in m)
x = extension for the load M (in m)
Procedure:
The Searle’s apparatus is hung by the two long wires from a rigid clamp on the ceiling. The
experimental wire connects the frame in which the micrometer works. The weight hanger with a load
W is attached to this frame. The dead load is attached to the frame connected to the other wire.
The screw gauge is adjusted to make the two frames to be at the same level. The PSR and HSR are
noted. Now a weight 0.5 kg is added to the weight hanger. The frame is brought to horizontal by
adjusting the screw head. As before the PSR and HSR are noted. The procedure is repeated several
times for loading and then while unloading the weights.
The diameter and hence the radius of the wire is found out using the screw gauge. The length of the
given wire is measured using a meter scale.
A graph is drawn with load on the x axis and extension on the y axis.
Hence the Young’s modulus of the given wire is calculated using the formula given.
Observations:
Length of the wire
: ______m
Diameter of the wire
: ________10-3m.
Radius of the wire
: ________10-3m.
Micrometer reading
Mean
Extension
Sl
Load
loading
unloading
-3
no
(kg)
10 m
10-3m
PSR
HSR PSR+HSRLC PSR HSR PSR+HSRLC
1.
x0 =
--------2.
x1 =
x1- x0 =
3.
x2 =
x2-x0 =
4.
x3 =
x3- x0 =
5.
x4 =
x4-x0 =
6.
x5 =
x5- x0 =
Result:
Young’s modulus of the material of the wire = ____________Pascal
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Physics Experiment
Experiment 7. Moment of Inertia and Rigidity Modulus
Aim:
i.
ii.
To determine the rigidity modulus of the material of the given wire
To determine the moment of Inertia of a disk and dumb bell
Theory:
Consider a wire securely fixed on both ends. If the wire is twisted, it will exert a restoring torque when
trying to return to its original untwisted position. For small twists, the restoring torque is proportional
to the angular displacement of the wire.
=-
(1)
The proportionality constant, , depends on the properties of the wire and is called the torsion spring
constant. It is given by
𝜋𝜂𝑟 4
𝜅=
(2)
2𝑙
Where r and l are the radius and length of the wire.  is the rigidity modulus of the wire.
When the object attached to the wire is twisted and released, the objects executes simple harmonic
motion with a period T, given by
𝑰
𝑇 = 2𝜋√𝜅
(3)
The moment of Inertia of a disk is given by
1
𝐼 = 2 𝑀𝑅 2
(5)
Where M is the mass of the disk and R is the radius of the disk.
The moment of inertia of the dumbbell about the axis passing through its centre of mass is given by
1
4
𝐼 = 2 ( 𝑀𝐿2 +
1
𝑀𝑙 2
12
1
4
+ 𝑀(𝑟12 + 𝑟22 )) +
1
𝑚𝐿2
12
(6)
Where l is the length of the cylinder, r1 and r2 are the inner and outer radius of the cylinder, M is the
mass of the cylinder. L and m is the length and mass of the brass rod.
Procedure
Part A: Determination of the rigidity modulus of the given wire
1. Measure the length and radius of the wire.
2. Measure the radius and mass of the disk.
3. Calculate the moment of inertia of the disk.
4. Set the disk into torsional oscillations and measure the time for 20 oscillations.
5. Determine the period of oscillation, T.
6. Find torsional spring constant,  using eq (3) and rigidity modulus using eq (2).
Part B: Determination of the moment of inertia of the dumbbell
1. The dumbbell is attached to the same kind of wire of same length in the adjacent setup.
2. Set the dumbbell into torsional oscillations
3. Measure the time for 20 oscillations.
4. Estimate the time period and hence the moment of inertia of the dumbbell.
5. Calculate the moment of inertia of the dumbbell by measuring the dimensions and mass of
dumbbell using eq (6) and compare with the experimental value.
Result:
Rigidity modulus of the given wire
Calculated value of Moment of inertia of the dumbbell
Experimental value of Moment of inertia of the dumbbell
:____________GPa
:____________kg.m2
:____________ kg.m2
Experiment 8. Equilibrium of concurrent coplanar forces
Aim: To verify
 law of parallelogram of forces
 converse of triangle of forces
 Lami’s theorem
Apparatus: Drawing board with pulleys attached, weight hangers, slotted weights and thread
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Physics Experiment
Principle:
Law of Parallelogram of forces:
states that “If two forces at a point are represented in magnitude and direction
by the two adjacent sides of a parallelogram drawn from a point, then the
resultant is represented both in manitude and direction by the diagonal of the
completed parallelogram drawn from the same point.” After getting the
tracing of the three strings on the paper points A, B and C are marked on the
lines so that OA, OB and OC represent the forces P, Q and R respectively.
The parallelogram OAC’B is completed. The side OC’ is measured. OC’ is
found to be equal to OC and angle COC’ is found to be 180. This means OC’
represents a force R’ which is equal and opposite to R. Since R is the
equilibriant of P and Q, R’ represented by the diagonal OC’ should be the
resultant of P and Q. Thus the law of parallelogram of force is verified. The
experiment is repeated for different values of P, Q and R.
Converse of the law of triangle of forces:
If three forces at a point are at equilibrium and a triangle is constructed by drawing lines parallel to the
lines of action of the forces then the sides of triangles taken in order are proportional to the forces to
which they are parallel.
DE, EF and FD are drawn parallel to OM, ON and OL respectively so that
they form triangle DEF. sides DE, EF and FD are measured. The ratios
𝑃 𝑄
𝑅
, and 𝐷𝐹 are calculated and found to be equal verifying the converse
𝐷𝐸 𝐸𝐹
of the law of triangle of forces. The experiment is repeated for different
values of P, Q and R.
Lami’s theorem: If three forces acting at a point are in equilibrium then
each force is proportional to the sine of the angle between the other two.
Having obtained the trace of knot and of the strings OL, OM and ON angles ,  and  are measured.
𝑃
𝑄
𝑅
The ratiossin 𝛼, sin 𝛽 and sin 𝛾 are calculated. The
experiment is repeated
for different values of P, Q and R.
Procedure: The drawing board is fixed vertically. Two frictionless pulleys P1 and P2 are fixed to the
top corner of the drawing board. A light inextensible thread is used to hang the weight hanger P and Q
through the pulleys. Another thread is tied to the thread connecting the pulleys to form a common
point O. A drawing sheet is fixed on the board. The weight R is disturbed and the position of O and
the threads are noted when it comes to rest.
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Physics Experiment
Observation:
Law of Parallelogram of forces
P
Q
Trial no
kg.wt
kg.wt
1.
2.
3.
Converse of the law of triangle of forces:
DE
Trial
P
Q
R
No
kg.wt
kg.wt
kg.wt
 10-2m
1.
2.
3.
Lami’s theorem:
Trial
P
No
kg.wt
1.
2.
3.
Q
kg.wt
R
kg.wt
OC
 10-2m
R
kg.wt

OC’
 10-2m
Angle
COC’
DE
 10-2m
DE
 10-2m
𝑃
𝐷𝐸
𝑄
𝐸𝐹
𝑅
𝐹𝐷


𝑃
sin 
𝑄
sin 
𝑅
sin 
Result:
Within the limits of experimental errors the law of parallelogram of forces, converse of law of triangle
of forces and Lami’s theorem are verified.
Experiment 9. Moment bar: Equilibrium of parallel forces
Aim: To verify the conditions of equilibrium of parallel forces.
Apparatus: Two spring balances, two weight hangers, slotted weights, scale.
Principle: Sum of clockwise moments = sum of anticlockwise moments , moments taken about any
point in their plane.
Procedure:
The notation is followed as depicted in the figure above. The
weights W1 and W2 are adjusted along the scale until the scale is
horizontal. The distances x, x1, x2, y1 and y2 and the readings P
and Q from the springs are noted. The experiment is repeated for
different values of W1 and W2 and the results are
tabulated.
Observation: x=________cm
Forces acting
Up
Trial
Distances
down
y1
P
Forces
Q
W1
W2
y2
x1
x2
P+Q
W1+W
+W2
Moments
(P y1) +
(Q  y2)
(W1 x1) +
(W2  x2) +
(W  x)
1.
2.
3.
4.
Result: Verified that for coplanar forces under equilibrium, the Clockwise moment is equal to
anticlockwise moments.
TDC-IISc, Kudapura
Physics Experiment
Experiment 10. 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.
 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.
TDC-IISc, Kudapura


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 gives-
slope 


Physics Experiment
SA
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 Month
(degree)
36.39
July
28.71
August
18.14
September
7.28
October
6.78
November
10.79
December
Zenith
(degree)
10.13
6.9
12.34
23.34
32.0
37.35
angle
Observation table:
Time
Temperature
Solar Constant, S
Solar Constant without Luminosity of the
atm. absorption, S0
Sun, L
Experiment 11. 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.
Formula:
1
1 2 1
K2 
2
2
mgh  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.
TDC-IISc, Kudapura
Physics Experiment
m
Timer sensor
h
s, t
Figure: 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.
Note:
K2 2
K2 1
K2

for
solid
sphere;
for
solid
cylinder;

 1* for solid cylinder.
r2 2
r2
r2
5
*assuming inner diameter and outer diameter of the cylinder is almost same.
Experiment 12. Coefficient of viscosity – Stoke’s method
Aim: To find the coefficient of viscosity of given fluids.
Procedure:
1. Measure the density of the fluid fl.
2. Measure the density of the ball B.
3. Fill the fluid inside the tube and identify two points, one at top and
other at bottom, separated by distance L.
4. Take the ball and release it at the surface of fluid slowly.
5. Using the stopwatch record the time when it reaches the top point. Call
this as ‘initial time’.
6. Record the time when the ball reaches the bottom point. This is ‘final
time’.
7. Calculate the difference in final and initial time. This is t.
8. Calculate terminal velocity 𝑣, using the formula 𝑣 = 𝐿/Δ𝑡
9. Use the terminal velocity to calculate coefficient of viscosity 𝜂 using
equation (1).
TDC-IISc, Kudapura
Physics Experiment
Formula:
𝜼 = 𝟐𝒓𝟐 𝒈
𝝆𝑩 − 𝝆𝒇𝒍
(1)
𝟗𝒗
Where,
𝜂 = 𝐶𝑜𝑒𝑓𝑓𝑖𝑐𝑖𝑒𝑛𝑡 𝑜𝑓 𝑣𝑖𝑠𝑐𝑜𝑠𝑖𝑡𝑦
𝑟 = 𝑅𝑎𝑑𝑖𝑢𝑠 𝑜𝑓 𝑡ℎ𝑒 𝑏𝑎𝑙𝑙
𝑔 = 𝐴𝑐𝑐𝑒𝑙𝑒𝑟𝑎𝑡𝑖𝑜𝑛 𝑑𝑢𝑒 𝑡𝑜 𝑔𝑟𝑎𝑣𝑖𝑡𝑦
𝜌𝐵 = 𝐷𝑒𝑛𝑠𝑖𝑡𝑦 𝑜𝑓 𝑡ℎ𝑒 𝑚𝑎𝑡𝑒𝑟𝑖𝑎𝑙 𝑜𝑓 𝑡ℎ𝑒 𝑏𝑎𝑙𝑙
𝜌𝑓𝑙 = 𝐷𝑒𝑛𝑠𝑖𝑡𝑦 𝑜𝑓 𝑡ℎ𝑒 𝑓𝑙𝑢𝑖𝑑
𝑣 = 𝑇𝑒𝑟𝑚𝑖𝑛𝑎𝑙 𝑣𝑒𝑙𝑜𝑐𝑖𝑡𝑦
Observations :
Sr
N
o
𝜌𝑓𝑙
(kg/m3
)
Weight of
ball
W
Diameter of
ball
D
Density of the
ball
𝜌𝐵
(kg/m3)
Time (s)
t
Length
L
(m)
𝑣
(m/s)
Coefficient
of viscosity
𝜂
(kgm-1s-1)
1
2
Result:
Viscosity of the given fluid: __________ kgm-1s-1
Experiment 13. Surface Tension of Water by Capillary Rise
Aim: To determine the surface tension of water by capillary rise method.
Apparatus: A capillary tube of about 15 cm length, travelling microscope, water taken in beaker, an
index needle and supporting stands.
Formula
Surface tension of water
=
T
hgr
2
Nm-1
where h = capillary rise (m),
g = acceleration due to gravity (ms-2),
tube (m), = density of water (kg /m3)
r = radius of the capillary
Procedure:
To find the capillary rise (h):
Capillary tube is cleaned with dilute acid. It is clamped vertically by a stand and its bottom
end is dipped into water taken in a beaker. The water will be found to raise into the tube and stand at a
height. A needle P is mounted vertically from a separate stand and is so adjusted that the tip of the
needle just touches the surface of the water in the beaker.
The microscope M is focused on the lower meniscus of the water in the tube. The horizontal
cross wire is made to be tangential to the lower meniscus. The reading on the vertical scale is noted.
Then the beaker is removed without disturbing the needle. Then the microscope is next focused at the
tip of the needle by lowering the microscope till the image of the tip just touched the horizontal cross
wire. The reading on the vertical scale is again noted. The difference between the two readings gives
the capillary rise h.
The observations are repeated a few times by keeping the water at various levels in the beaker.
The mean capillary rise h is calculated.
To determine the radius of the capillary tube (r):
The capillary tube is held horizontally in front of the microscope with the help of a stand. The
end face of the capillary tube is viewed. The microscope is adjusted so that the bore of the tube is
TDC-IISc, Kudapura
Physics Experiment
clearly seen in the field of view. The vertical cross wire is made tangential to one side of the bore. The
reading on the horizontal scale is noted. The vertical cross wire is moved to the diametrically opposite
side of the bore and the reading is again noted.
Similar readings were taken for the horizontal cross wire also and the mean difference of these
two readings gives the mean diameter of the capillary. So the radius is found.
𝑟
𝑟(ℎ+ )𝜌𝑔
The surface tension of water is calculated using the formula 𝑇 = 2 cos3 𝜃
For narrow capillary tube r≪h and in case of water =0 therefore cos 0 = 1
𝑟ℎ𝜌𝑔
Hence the formula becomes 𝑇 = 2 Nm-1.
Sl.
No
(i)To find the capillary rise (h) :
Least count = 0.001 × 10-2 m
Correct reading = MSR + (VSR  LC)
Reading of the meniscus in the capillary
tube (h1)
VSR
Correct reading
MSR 
10-2 m
 10-2 m
Reading of the tip of the needle (h2)
MSR  10-2 m
Capillary rise
ℎ = (ℎ1 − ℎ2 )  10-2 m
VSR Correct reading
10-2 m
Mean height of the capillary rise, h =
 10-2 m
(ii) To find the radius of the capillary tube (r) :
Least count = 0.001 × 10-2 m
Correct reading = MSR + (VSR x LC)
Scale on
Reading when cross wire is
Reading when cross wore is tangential Diameter of the bore
which the
tangential to the bore on one side
to the bore on the other side
𝑑 = 𝐷1 ~ 𝐷2 x 10−2 𝑚
-2
reading is
MSR x
VSR
Correct reading MSR  10
VSR Correct reading
−2
taken
10-2 m
𝐷1  10 𝑚
𝐷2  10−2 𝑚
m
Horizontal
Vertical
Mean diameter of the capillary tube, d =
𝑑
Radius of the capillary tube, r = =
2
Result:
Surface tension of water: ____________ Nm-1
 10-2m
 10-2m
TDC-IISc, Kudapura
Physics Experiment
Experiment 14. Determination of velocity of sound - 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 
1
T
273
l2 cm
l1 cm
Water level
First maximum
Second maximum
Figure: Resonance column, measurement of first and second resonance.
Observations:
1st Resonating length
Frequency of l1
Tuning fork
Trial1 Trial2 Mean
f (Hz)
(cm)
(cm)
l1 (cm)
2nd Resonating length
l2
Trial1
(cm)
Trial 2 Mean
(cm)
l2 (cm)
V  2 f l2  l1 
V
V0 
1
m/s
m/s
Mean velocity of sound in air at room temperature = ………….. m/s
Mean Velocity of sound in air at 00C = ……………m/s
Question:
1. In which parameters velocity of sound dependent upon?
2. By this experiment can we find the frequency of unknown tuning fork?
3. Will the velocity of sound in gasses change when its pressure changes?
Experiment 15. Sonometer – Verification of Laws of Transverse
Vibrations of Stretched String
Aim: To verify the first and second laws of transverse vibrations of a stretched string using sonometer,
i.e., to show (i) nl = constant and (ii)
√𝑇
𝑙
= constant
Apparatus: Sonometer, tuning forks of known frequency, rubber block, slotted weights, paper riders
(small V shaped paper bits).
T
273
TDC-IISc, Kudapura
Physics Experiment
TDC-IISc, Kudapura
Physics Experiment
Procedure:
Verification of I Law:
The sonometer wire is kept under tension by a suitable load, say 2 kg. A small paper rider is
place on the wire between the movable bridges. The stem of an excited tuning fork of known
frequency (n) is placed on the sonometer box. By adjusting the positions of the bridge gently, the
length of the vibrating segment is changed till the paper rider flutters violently and is thrown off. The
length of string between the movable bridges , ‘l’ gives the resonating length.
Keeping the tension constant the experiment is repeated with the tuning forks of different
frequencies and corresponding vibrating lengths of the wire are found out as before. The values are
tabulated and the product ‘nl’ is found to be a constant verifying the first law.
Verification of II Law:
The sonometer wire is kept under tension by a load of 1 kg. Using a tuning fork of known
frequency, the resonating length (l) is found out as explained earlier. By increasing the load in steps of
0.5 kg, the corresponding resonating lengths are found out for the same fork. The tension of the wire T
is calculated in each case using the relation T = mg where g is the acceleration duet to gravity. For the
√𝑇
same tuning fork and the same sonometer wire, the value of 𝑙 is found to be constant, verifying the
second law.
Verification of I Law: Tension (T) =
 9.8 N
S.N.
Frequency of the tuning fork (n)
Resonating length (l)
nl Hz m
Verification of II Law: Frequency of the tuning fork (n) =
S.N
Load (m) kg
Tension (T) N
√𝑇 𝑁1/2
Hz.
Resonating length 10-2m
√𝑇
𝑙
𝑁1/2 𝑚-1
Result:
The first and second laws of transverse vibrations of stretched strings are verified.
Experiment 16. Resistivity of metals - Meter Bridge
Aim: To find the resistance and resistivity of a given wire (unknown resistance) using meter bridge.
Formula:
Resistance of the given wire:
R
S l
100  l
Resistivity of the material of the wire:

d 2 R
4L
TDC-IISc, Kudapura
Physics Experiment
Unknown wire, R
Standard resistance box, S
100-l
l
G
battery
key
Figure: Circuit connection of Metre Bridge Experiment
Determination of diameter of the wire ‘d’, using screw gauge
Material
Diameter,
d (mm)
Diameter,
d (m)
Kanthal
Iron
Copper
Determination of unknown resistance, R
Material
Kanthal
Iron
Copper
Trial
S()
Balancing
length l(cm)
𝑅(Ω) =
𝑆×𝑙
(100 − 𝑙)
Mean R
()
1
2
3
1
2
3
1
2
3
Resistivity of the wire:
Material
Kanthal
Iron
Copper
Questions:
Resistance,
R ()
Unknown wire
Diameter,
Length,
L (m)
d(10-3m)
Resistivity,
calculated
(m)
Resistivity (m)
(Reported in literature
at 20 C)
139×10−8
10×10−8
1.68×10−8
TDC-IISc, Kudapura
Physics Experiment
1. Describe the principle of Metre Bridge experiment?
2. Why do we use small currents in Metre Bridge experiment?
3. Arrange the elements in increasing order of their electrical resistivity.
Elements - Au, Cu, Ag, Pt, Al.
4. Arrange the elements in increasing order of their conductivity.
Elements - Au, Cu, Ag, Pt, Al.
5. Which one among the above wires measured in the experiment forms a good heating element?
Experiment 17. Determination of Absolute Zero - Ideal Gas Law
Aim:
To study the relationship between pressure and temperature of given gas under constant volume and to
find absolute zero.
Apparatus: Gas law set up box, digital thermometer, liquid nitrogen and a digital stop watch.
Procedure:
Dip the pressure filled gas law box into the liquid nitrogen and allow to cooled up-to the liquid
nitrogen temperature. Take out the gas law box. Record the pressure with respect to temperature and
time. Record the pressure up to room temperature. Plot the graph of pressure against temperature.
Observation table:
Temperature
Pressure
Result:
Absolute Zero temperature: ________C.
Experiment 18. Temperature coefficient of resistance ‘’ of metal
Aim: To study the temperature variation of resistance of a metal.
Apparatus:
Cu coil, hot water, beaker, thermometer and a multimeter.
Dip the copper coil in the hot water bath and measure the resistance of the coil as the water cools as a
function of temperature till room temperature.
Or dip the copper coil in liquid nitrogen till temperature reaches -196 ͦ C and allow it to warm to room
temperature
TDC-IISc, Kudapura
Physics Experiment
Observation :
Temperature
Resistance
30 0C
Plot a graph of resistance(Y) vs temperature(X).
Calculate the temperature coefficient of resistance of copper.
(𝑅2 −𝑅1 )
𝛼 = (𝑇 −𝑇
where R0 is the resistance at 0 ͦ C and R1and R2 are the resistances at T1 and T2
)𝑅
2
1
0
temperatures respectively. Use the plot to find these values.
Compare the α value obtained with the values in Clark’s table.
Result:
Questions:
1. Does the resistance of copper increases with temperature ?
2. Does the resistance of copper increases linearly with temperature ?
3. If you cool copper wire to absolute zero temperature will the resistance go to zero Ohms ?
Experiment 19. Temperature Dependence of Resistance of semiconductor –
Energy Bandgap Eg
Aim: To examine the temperature variation of resistance of a semiconductor
Apparatus:
Thermistor, hot water, beaker, thermometer and a multimeter
Procedure:
Dip the thermistor in the hot water bath and measure the resistance as it cools as a function of
temperature till it reaches room temperature.
Observation :
Temperature
( ͦ C)
Resistance,R ()
Temperature,
T( K)
Plot a graph of R vs T.
Plot a graph of loge(R) vs 1000/T.
and calculate the slope and hence calculate the Eg.
R=R0exp (Eg/2kT)
ln(R)=ln(R0) + Eg/(2kT)
ln(R)
1000/T
TDC-IISc, Kudapura
Physics Experiment
k=85.6  10-6 eV.K-1
Questions:
Discuss the behavior of the graph.
Experiment 20. 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:
1. Pour 350-400ml of water to the beaker.
2. Heat the water up to a constant temperature, say 90 0C .
3. Mark two positions to all the rods at same length from one end (say at 4cm and 20 cm
from one end).
4. Pick any two rods and hold by your bare hand at 20 cm the mark point.
5. Dip the two rods up to the mark (4 cm) into the hot water. As shown in figure.
6. Wait for some time. After a while your hand will feel heat.
7. Identify which hand feels heat first.
8. The rod which was held by your hand that feels first heat sensation has higher thermal
conductivity.
9. 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 : experimental set up for heat transfer
Observation Table:
Given materials
Thermal conductivity
In ascending order
Brass, Copper, Aluminium, Steel, ………..
Experiment 21. Thermal expansion
Aim: determination of linear coefficient of thermal expansion of solids
Introduction: Under the application of heat, oscillation of atoms increases with greater amplitudes
which in turn increases the average separation between the atoms in the metal and thus metals gets
expansion and vice versa. The change in the length under the application of heat is called linear
expansion and is different from one metal to another. The quantity that defined the change in length
per 0C is known as linear coefficient of thermal expansion denoted by α. Since the change in length of
the metal due to thermal expansion is very small, one can use the single slit experiment to determine
the α precisely. It is designed in such a way that whenever there is a change in the length of the metal
due to heat, there will be a same amount of change in the slit width and thus the fringe widths.
TDC-IISc, Kudapura
Physics Experiment
Apparatus: diode laser, metal rod (1cm diameter, 5-6 cm length), sharpener blades, water bath with
heater (or mica band heater), measuring tape, digital slide caliper, digital thermometer.
Formula: Under the condition that θ very small (i.e. θ ≈ 00) then,

1
D
1
  
L0 T2  T1   x 2 x1 
Here, D = distance between slits and the screen, λ = wavelength of laser light, L 0 = length of the road
at absolute zero temperature, xi= mth order fringe width at temperature Ti ( i= 1 or 2).
Movable frame
Blades
Non conducting frame
Metal rods
Procedure:
1. Figure shows the schematic diagram of the experiment.
2. Measure the length of the metal rod.
3. Allow the laser light to passes through the single slit and observe a diffraction pattern. Adjust
the slit width so that at least 4 bright fringes can be seen on the screen. Note the fringe width
x i with corresponding temperature, T0. Rise the temperature of the water bath and thus the
temperature of the metal. As the temperature of the metal increases correspondingly the slit
width also increases. Once the slit with increases, the fringe width also changes. Note the
change in the fringe width with respect to temperature. Take the readings of fringe width at
40, 45, 50, 55, 60, 65, 70 0C.
4. CARE MUST BE TAKEN THAT DURING THE EXPERIMENT NO HEAT SHOUD
FALLD ON THE BLADES USED IN MAKING THE SLITS.
1
1
  versus Ti  T1 , where i = 1, 2, 3, ….. trials.
 x i x1 
5. Plot 
1
1
  
 x i x1  which is equal to D . i.e., m  D
6. Take the slope, m 
L 0
Ti  T1
L 0
7. Since Li  L0 1  Ti  , one can derive  
Ti  Ta .
8. Using expression at 6, one can determine α.
Dm
. It is given that Li  La when
La  DmTa
TDC-IISc, Kudapura
Physics Experiment
Observation table:
La  ................ at Ta  ....................
Trial number
Temperature
Change
in mth
order
Temperature
Fringe width
Ti 0C
Ti  T1 0C
1
70
2
65
3
60
4
55
5
50
6
45
7
40
x i in mm
1
1
  
 x i x1 
Result:
Temperature coefficient of thermal expansion for the ……Copper/Aluminium.. is =
Experiment 22. Determination of Specific Heat Capacity of Fe, Cu and Al
Aim: To determine specific heat capacity of solids by the method of mixtures.
Apparatus: Calorimeter, stirrer, oven, thermometer, metal pieces and water.
Procedure:
Measure a clean, dry calorimeter (glass beaker) mass in electronic balance after that tare it to zero.
Add 20 ml of water to calorimeter and find the mass M1 of 20 ml water. Note down the initial
temperature T1 of water. You are given Fe, Cu and Al rods; take the masses M2 of each one. Put these
into a constant temperature water bath for 30 minutes. Note down the temperature T 2 of water bath in
which rods are heated. After that take the piece and dip into the calorimeter. Find the maximum rising
temperature T3.
Let the specific heat capacity of given solid be s. The specific heat capacity of water is 4190 J kg-1K-1.
Heat gained by water by increasing the temperature from initial temperature (T1) to maximum rising
temperature T3 is given by:
Heat gained = (water mass (M1) x water specific heat of water)  (T3– T1)
Heat lost by the rod when it is cooled from water bath temperature (T2) to maximum rising
temperature T 3 is given by:
Heat lost = metal mass (M2)  metal specific heat of metal (s)  (T2-T3)
By the principle of method of mixtures,
Heat gained = Heat lost
Therefore,
(Water mass (M1)  water specific heat of water)  (T3– T1)
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Physics Experiment
= metal mass (M2)  metal specific heat of metal (s)  (T2-T3)
Please Note: We have neglected heat gain by calorimeter (beaker).
Find the specific heat capacity of the solid.
Repeat this experiment four times for one metal. Find sp. Heat capacity of Al, Cu and Fe.
Experiment 23. Newton’s law of cooling
Aim:
To verify Newton’s law of cooling.
Apparatus:
Calorimeter, Thermometer, Hot water, stopwatch.
Principle:
According to Newton’s law of cooling the rate of heat loss of a hot body is proportional to the
difference in temperature between the hot body and its surroundings.
𝑑𝑄
ie 𝑑𝑡 ∝ −(𝜃𝑏𝑜𝑑𝑦 − 𝜃𝑠𝑢𝑟 )
or
𝑙𝑜𝑔𝑒 (𝜃𝑏𝑜𝑑𝑦 − 𝜃𝑠𝑢𝑟 ) = −𝑘𝑡 + 𝐶
Procedure:
The room temperature is noted. The calorimeter is filled with hot water. The temperature is noted and
the timer is started. The variation in temperature is noted as function of time. The observations are
tabulated and a graph is plotted with 𝑙𝑜𝑔𝑒 (𝜃𝑏𝑜𝑑𝑦 − 𝜃𝑠𝑢𝑟 ) on the Y axis and time on the X axis. A
straight line is obtained.
Observations:
𝜃𝑠𝑢𝑟 =______C
Sl no
Time(s)
𝜃𝑏𝑜𝑑𝑦 (C)
𝜃𝑏𝑜𝑑𝑦 − 𝜃𝑠𝑢𝑟 (C)
𝑙𝑜𝑔𝑒 (𝜃𝑏𝑜𝑑𝑦
− 𝜃𝑠𝑢𝑟 )
Result:
Newton’s law of cooling is verified.
Experiment 24. Hydrogen Spectra and determination of
Rydberg Constant
Aim: Determination of Rydberg constant R and energy level diagram of hydrogen.
Principle: Rydberg constant is one of the fundamental constants given by the equation 𝐸𝑛 =
𝑅
−𝑍 2 𝑛2 𝑤here R = Rydberg constant, 𝐸𝑛 = Energy level of Hydrogen with principal quantum number,
n=1,2… and Z is atomic number( 1 for H, 2 for He1+)
TDC-IISc, Kudapura
Physics Experiment
In a hydrogen discharge tube, electron occupied in the ground state n=1of Hydrogen atom is
excited to the higher energy levels. They come back to n = 1,2… states giving rise to Lyman, Balmer,
Paschen series. Balmer lines are in visible region of the electromagnetic spectrum.
Procedure:
In the given spectrometer observe and find the wavelengths of transitions
n=3  2, n= 4  2, n = 5  2 emission lines.
1
1
Energy of emission lines are given by 𝐸𝑛=32 = 𝑅 [22 − 32 ]
n=∞
Find the value of R in eV.
E = h = 6.62310-34  c/ Joules.
c= 3108 m/s;  - wavelength in m.
1 ev = 1.602  10-19 Joules.
-E
n
n =1
Wavelength of Balmer and Lyman lines are given below from accurate spectroscopic measurements.
Balmer Series
Lyman Series
656.112
nm
121.502 nm
n=32
n=21
486.009 nm
102.518 nm
n=42
n=31
433.937 nm
97.202 nm
n=52
n=41
410.070 nm
94.824 nm
n=62
n=51
93.730 nm
n=61
(a) Find the value of Rydberg constant R in eV from the Balmer and Lyman emission lines.
(b) Draw the hydrogen energy levels for n = 1, 2, 3,4,5,6…∞ to the state.
(c) Find the ionization energy of Hydrogen when the electron is in n= 2 state.
(d) Calculate Energy levels of He1+ ion for n=1, 2, 3, 4, 5, 6 and plot them.
TDC-IISc, Kudapura
Physics Experiment
Experiment 25a. Focal length of Convex lens
Aim:
To determine the focal length of a given convex lens.
To find the magnification in each cases.
Apparatus:
Convex lens, Light source, screen and optical bench.
Formula:
(a)
1 1 1
 
f u v
𝑢×𝑣
or 𝑓 = (𝑢+𝑣 )
f= focal length (cm), u= object position (cm), v = image position (cm).
u
v
Object
Image
Lens
find position
for clear image
Variable position
Of lens
Fix Position
Figure. Measurement of object and image position
Procedure:
1. Focus an object which is at infinite distance on the screen.
2. Measured the distance between lens and the position of the clearest image. Let this length be
denoted by F.
Observation : Determination of focal length.
Lens 1 (Distant object method): F = ……..….cm
Position of Object,
u (cm)
Position of
the Image
v (cm)
Magnification
𝑣
𝑚=−
𝑢
Nature of
the Image
Focal
length,
f (cm)
more than 2F
u =…………
at 2F
u =…………
between F and 2F
u =……..…
Make similar tabular column for other given convex lenses also.
4. ResultExperimental observed focal length of the lens is = ………….m.
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?
Average
focal
length,
fav (cm)
TDC-IISc, Kudapura
Physics Experiment
Experiment 25b Determination of focal length of
concave lens using convex lens
Aim: To determine the focal length of the given concave lens.
Focal length of convex F1 =
cm (using distant object method)
Focal length of convex and concave lens:
Sl. No
Size of image
Distance between
the lenses and
object ‘u’ (cm)
Distance between
the combination
and screen(cm)
𝐹2 =
𝑢×𝑣
(𝑐𝑚)
𝑢+𝑣
U > 2F
Mean F2 =
cm
Determination of focal length of concave lens F3:
𝐹 ×𝐹
Formula 𝐹3 = 𝐹1−𝐹2
1
2
F3 focal length of concave lens =
Sl. No
Size of image
cm
Distance between
the concave lens
and screen(S1) ‘u’
cm
U > 2F
Result:
Focal Length of concave lens:
10-2m
Distance between
the concave lens
and screen (S2)’v’
cm
𝐹2 =
𝑢×𝑣
(𝑐𝑚)
𝑢−𝑣
TDC-IISc, Kudapura
Physics Experiment
Experiment 25c. Focal length of concave mirror
Aim: To find :
o the focal length of the given concave mirror by
(1) distant object method (2) u-v method and (3) u-v graphical method.
o the radius of curvature of given concave mirror.
o the magnification in each case.
Procedure:
(1) Distant object method:
The given concave mirror is placed on the stand and placed near an open window
facing a distant object. A white screen is adjusted in front of the mirror without blocking the
light rays from the distant object (the screen should be placed in an angular direction to the
mirror) till a clear inverted image is formed. The distance between the mirror and the screen
gives an approximate value of the focal length ’f’.
(2) u-v method:
The concave mirror on the stand is placed in front of the illuminated object at a
distance greater than its focal length found in distant object method. The screen is adjusted to
and fro in front of the mirror till a clear inverted image is formed on the screen. The distance
between the mirror and object is measured as ‘u’ and the distance between the mirror and the
screen is measured as ‘v’. The focal length of the given mirror is calculated using formula.
f = uv/u+v
Experiment is repeated by changing the value of (4 readings greater than 2f value for
diminished image and 4 readings less than 2f value-not less than ‘f’ value from previous
method-for enlarged image)and measuring ‘v’ in each case, the readings are tabulated. f is
calculated in each case and the average value of the focal length is found out.
(3) u-v graphical method:
Using the values of ‘u’ and ‘v’ obtained above, a graph is drawn taking ‘u’ along the
X-axis and ‘v’ along the Y-axis with the same scale and symmetrical origin for the both the
axes. We get a hyperbola curve. A straight line is drawn through the origin making an angle of
45° with anyone of the axes. This cuts the curve at a point P. from P, perpendiculars PA and
PB are drawn to X&Y axes. Now OA or OB gives 2f. Hence f is found.
(i) Distant object method:
The distance between the mirror and screen =
(This value directly given the focal length)
Focal length of the concave mirror f =
(ii) u – v method :
x 10-2m
x 10-2m
TDC-IISc, Kudapura
Physics Experiment
SL. NO
Distance between
the mirror ans
object ‘u’ x 10-2m
1.
2.
3.
4.
1.
2.
3.
4.
Size of the image
Distance between
the mirror and
screen ‘v ‘x 10-2m
Focal length of
the mirror
𝑢𝑣
𝑓 = 𝑢+𝑣 x 10-2m
Magnified
U < 2f
Diminished
U < 2f
Mean f =
(iii) u-v graphical method
Result:
The focal length of the given concave mirror by
(1) distant object method =
10-2m
(2) u-v method
=
10-2m
(3) u-v graphical method =
10-2m
x 10-2m
TDC-IISc, Kudapura
Physics Experiment
Experiment 26. Refractive index of glass slab – Lateral shift
Aim: To 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.
Formula: Lcalc  t 
sin i  r 
cos r 
n
sin i 
sin  r 
i
r
t
Lmeas
Where Lcalc =calculated lateral shift,
t = thickness of the glass slab, n = refractive index, i= angle of incident and r = angle of refraction.
Figure : Ray diagram when a ray of light passing from rarer medium to a denser medium
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
r
n
i-r
Lcalculated
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?
Lmeasured
TDC-IISc, Kudapura
Physics Experiment
Experiment27. Refractive index of Prism – pin method
Aim: To find the refractive index of the prism by finding the angle of minimum deviation.
Principle: Refractive index of the material of the prism is given by
𝑛=
𝐴 + 𝐷𝑚𝑖𝑛
sin (
)
2
𝐴
sin ( 2 )
where n = refractive index, A= Angle of the prism, Dm is the angle of minimum deviation.
A
i
D
Figure 19: Ray diagram for prism using pin method
Observation:
Angle
of Angle
of Angle
of
minimum Refractive index of
incident, i
Deviation, D
Deviation (from graph) Dmin the prism
40
44
48
52
56
60
64



Plot the i - D graph.
Find the angle of minimum deviation from the curve.
Find the refractive index of the prism.
Question:
1. Why angle of minimum deviation takes place in this experiment?
TDC-IISc, Kudapura
Physics Experiment
Experiment 28. Refractive index of liquid using Travelling microscope
Aim: To determine the refractive index of water by shift method using a traveling microscope.
Apparatus:
Travelling microscope, beaker, pin, 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.
Eye piece
Eye piece
Eye piece
Focusing
knob
Focusing
knob
R1
Objective
R2
Focusing
knob
Objective
Objective
Saw
dust
Traveling microscope
coin
without water
coin
with water
with
water and
coin
saw dust
Figure 14. Position of real and apparent measurement
Procedure: Measurement of Ri’sDetermination of R1
2. Least count (LC) of a traveling microscope is determined by using the formula-
LC 
Value of 1 MSD
Total number of VSD
3. The traveling microscope is set for vertical traverse. The axis of the microscope is also made
vertical.
4. The microscope is focused on a coin which is at the bottom of a beaker.
5. The main scale reading (MSR) and the coinciding vernier scale division (VSD) are noted.
6. The total reading R1 is calculated using the relationR1 = MSR + (CVD x LC)
Determination of R2
7. Pour the water into the beaker to a height about 3-4 cm. Consequently, the pin is out of focus.
8. 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.
9. Repeat the steps 4 and 5, to calculate R2.
Determination of R3
10. A small quantity of dust is sprinkled on the surface of the water.
11. Microscope is now focused on the saw dust.
12. 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
R3
TDC-IISc, Kudapura
Physics Experiment
Table 1: Determination of Refractive Index of water.
Trial
no.
R1
MSR
CVD
R2
R1
MSR
CVD
R3
R2
MSR
CVD
R3
n
R3  R1
R3  R2
Mean R.I.
(n)
1
2
3
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?
Why the apparent depth is lesser than the real depth?
Experiment 29. Refractive Index of Glass Prism - Spectrometer
Aim
To determine the angle of the given glass prism and the angle of minimum deviation of
different colours using a spectrometer and hence to determine the refractive index of given glass prism
and liquids.
Apparatus:
Spectrometer, Mercury lamp, glass prism, hollow prism, liquids, spirit level and reading lens.
Formula:
𝑅 −𝑅
i) Angle of the prism, 𝐴 = 1 2 2 (degree)
Where R1 = total reading in one vernier for the reflected image on one refracting face, (degree)
R2 = Total reading in the same vernier for the reflected image on the other refracting face,
(degree)
ii) Angle of minimum deviation, D = R3-R4 (degree)
R3=total reading in the same vernier for direct ray position (without the prism)
R4 = total reading in one vernier for minimum deviation position (degree)
iii) Refractive index of the prism, 𝑛 =
𝐴+𝐷
)
2
𝐴
sin( )
2
sin(
TDC-IISc, Kudapura
Physics Experiment
R1
R2
Procedure:
i) Focusing the eyepiece on the cross wire:
The telescope is turned towards a white wall and the eyepiece is adjusted until the cross wires are seen
distinctly.
ii) Adjusting the telescope for parallel rays:
The telescope is turned towards a distant object and is adjusted till a well-defined image is seen.
The telescope should not be disturbed hereafter.
iii) Adjusting the collimator for parallel rays
The slit is illuminated by Mercury lamp and the collimator is adjusted for a clear image of the
slit.
iv) Leveling the prism table.
The prism table is made horizontal by using a spirit level and adjusting the three leveling
screws.
To find the angle of the prism (A):
The prism is fixed on the prism table such that the opaque rectangular base faces the telescope.
The two reflected images from the two refracting faces of the prism are located with the naked
eye. The telescope is brought to one side and clamped. The reflected image is made to coincide
with the cross wire by adjusting the tangential screw of the telescope. Care is taken to see that
the vernier clamping screw is tight so that the vernier scale does not move. The readings of the
circular main scale and vernier scale are taken on both the verniers. The total reading is
calculated for each vernier. Let them be R1 for both the cases. Similarly R2 also is taken. From
this the angle of the prism A is calculated.
To find the angle of minimum deviation (D):
The prism is placed at the centre of the prism table and the refracted image is observed. The
prism table is rotated slowly increasing the angle of incidence. The refracted image moves
towards the direct ray position and the angle of deviation decreases. The telescope is also moved
to follow the image. At once stage the image stops turns back and moves in the opposite
direction. The position of the image where it turns back is the minimum position. When the
image turns the telescope is adjusted so that the vertical cross wire coincides with the image at
this position. The reading of the verniers are noted as R3.
The prism is removed and the image is viewed directly through the telescope. The telescope is
adjusted and fixed at the position when the vertical cross wires coincides with the direct image.
The readings of the two verniers are noted as R4. The difference R3 - R4 gives the angle of
minimum deviation D.
The refractive index of the material of the given glass prism and liquid is given by
TDC-IISc, Kudapura
Physics Experiment
𝑛=
𝐴+𝐷
sin ( 2 )
𝐴
sin ( 2 )
Angle of the prism
Vernier 1
Reflection
MSR
Face 1(R1)
Face 2(R2)
Angle of the
prism
VC
Vernier 2
Total
reading
MSR
𝑅1 − 𝑅2
𝐴=(
)=
2
Total
reading
VC
𝑅1 − 𝑅2
𝐴=(
)=
2
Mean angle of the prism = ________
Angle of minimum deviation for different lines
Vernier 1
Lines
Direct (R3)
Violet(R4)
Blue(R4)
Green(R4)
Red(R4)
MSR
VC
Total
reading
Vernier 2
Minimum
deviation,
D
R3 - R4
===
MSR
VC
Total
reading
Minimum
deviation,
D
R3 - R4
===
Mean D for
each color
===
Result
Angle of the prism
Angle of the minimum deviation for different colors
1. Violet
2. Blue
3. Green
4. Red
Refractive index of the prism for various colors
1. Violet
2. Blue
3. Green
4. Red
:
:
:
Experiment 30. Refractive Index of liquid - Spectrometer
Aim:
To determine the refractive index of the given liquid in hollow prism.
Apparatus:
Spectrometer, Mercury lamp, hollow prism, liquids, spirit level and reading lens.
Formula:
𝑅 −𝑅
i) Angle of the prism, 𝐴 = 1 2 2 (degree)
Where R1 = total reading in one Vernier for the reflected image on one refracting face, (degree)
R2 = Total reading in the same Vernier for the reflected image on the other refracting face,
(degree)
ii) Angle of minimum deviation, D = R3-R4 (degree)
R3=total reading in the same vernier for direct ray position (without the prism)
R4 = total reading in one Vernier for minimum deviation position (degree)
TDC-IISc, Kudapura
Physics Experiment
iii) Refractive index of the prism, 𝑛 =
𝐴+𝐷
)
2
𝐴
sin( )
2
sin(
R1
R2
Procedure:
Follow the instructions given in previous experiment.
Observations:
Angle of the prism
Vernier 1
Reflection
Total
MSR
VC
reading
Face 1(R1)
Face 2(R2)
𝑅1 − 𝑅2
Angle of the
𝐴
=
(
)=
prism
2
Vernier 2
MSR
Total
reading
VC
𝑅1 − 𝑅2
𝐴=(
)=
2
Mean angle of the prism = ________
Angle of minimum deviation for different lines
Vernier 1
Lines
MSR
Direct (R3)
Violet(R4)
Blue(R4)
Green(R4)
Red(R4)
VC
Total
reading
Vernier 2
Minimum
deviation,
D
R3 - R4
===
MSR
Total
reading
VC
Result
Angle of the prism
Angle of the minimum deviation for different colors
5. Violet
6. Blue
7. Green
8. Red
Refractive index of the prism for various colors
:
:
:
Minimum
deviation,
D
R3 - R4
===
Mean D for
each color
===
TDC-IISc, Kudapura
5.
6.
7.
8.
Physics Experiment
Violet
Blue
Green
Red
Experiment 31. Determination of Specific rotation
Aim:
To determine the specific rotation of given sugar solution.
Apparatus:
Laser source, Two Polarizers with rotating mounts, one used as a polarizer and another as an analyzer.,
Rectangular container to hold sugar solution, Sugar solution of different concentration.
Formula:
𝜃 = 𝑆𝑙𝑐
Where  = angle of rotation of the plane of polarization
S= specific rotation
l= length of the liquid column(m)
c=concentration of the optically active liquid (kg.m-3)
Procedure:
1) Pour sugar solution in to the glass container till the laser beam passes through it.
2) The light beam coming out of the glass container may be laterally shifted. Hence some
adjustment in the direction may be required.
3) The intensity of light is increased due to rotation of plane of polarization. Turn the analyzer
till the light intensity is minimum and note down the angle 2 of the analyzer.
4) The difference in the angle 2 and 1 gives the optical rotation .
5) Measure the inner length l of the rectangular container.
6) Repeat the experiment for at least four concentration of sugar solution.
7) Using the values of ‘’ and ‘C’, Plot an appropriate graph and find the slope value of the
linear line to obtain the specific rotation of sugar solution.
Observations
S.No.
-3
Conc (kg.m )
1 (degree)
2(degree)
=(2-1)
(degree)
Result:
Specific rotation of the sugar solution S =__________________ rad.m2.kg-1
𝑆=
𝜋
 (180)
𝑙C
TDC-IISc, Kudapura
Physics Experiment
Experiment 32. Determination of wavelength using Transmission Grating
Aim: Determination of wavelength of given laser using transmission grating.
Apparatus required: Gratings and its stand, Laser source, Meter Scale, Stand and graph paper.
Formula:
m  d sin θm
where m = order,
d = distance between the two adjacent grating slits,
 m = angle between mth order fringe and the 0th order fringe to the grating.
Diagram:
d
Grating lines
Grating
Figure 16a. Grating.
2nd order
2x2
st
1 order
θ2
Laser
Light
2x1
1
0th order
D
Grating
Screen
Figure 16b. Grating Diffraction.
Calculation1.
2.
d
1
number of lines per unit length
(Convert in meter)
D = distance between the grating and the screen (in cm).
TDC-IISc, Kudapura
Physics Experiment
3. Measurement of angle  m :
1st order fringe
2nd order fringe
Grating
1 =
2x1
Cm
x1
cm
x1
D
1 x 
tan  1 
D
1
= 2x2
d sin 1 cm
2 
x2
cm
x2
D
x 
1
tan  2 
D
 
2 =
d sin  2
2
Result: Wavelength of the laser light source is,
exp t 
1  2
2
 ..............
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?
TDC-IISc, Kudapura
Physics Experiment
Experiment 33. Diffraction due to Reflection Grating
Aim:
To observe the diffraction pattern of laser using a scale and hence to determine the
wavelength
Figure:
ym
m
1
i
y1
0
y0
β1

Diffraction spots
β0
d
Z0
𝑑
2 −𝑦 2
𝑦𝑚
0
)
𝑚
𝜆 = (2𝑍2 ) (
0
Procedure:
A ruler is placed on a table and the laser beam is aligned such that the beam is incident at the grazing
angle as shown in the figure. A screen is placed at a distance of ~ 2 m to observe the diffraction spots.
First, the position of the direct beam in the absence of ruler is marked. The position of these diffraction
spots are measured from the midpoint of the position of the direct beam and the specularly reflected
𝑦 2 −𝑦 2
Sl no
1.
2.
3.
4.
5.
beam position. From the average value of ( 𝑚 0 ), wavelength of the laser is calculated.
𝑚
Observations:
2
(𝑦𝑚
Position of the
− 𝑦02 )
2
2
Spots
(𝑦𝑚
− 𝑦02 )
𝑦𝑚
m
spots, ym
𝑚
ym
(cm2)
(cm2)
(cm)
(cm2)
0
y0
1
y1
2
y2
3
y3
4
y4
Result:
Diffraction pattern due to reflection from the ruler was observed.
Mean wavelength of the laser used = ______________ nm
Experiment 34. Determination of grating constant
Aim: To determine the grating constant of the given grating.
Apparatus: Sodium vapour lamp, Spectrometer, reading lens, grating.
Formula:
𝑠𝑖𝑛(𝜃) = 𝑁𝑚𝜆
N = Number of lines per unit length
m = order of the spectra
 = wavelength of the light used.
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Physics Experiment
Procedure:
Follow the instructions given for experiment “Refractive Index of Glass Prism – Spectrometer” for the
initial arrangement of the spectrometer. Mount the given transmission grating on the prism table.
Observations:
1=60’=3600’’
1’=60’’
1 Main Scale Division
30 𝑚𝑖𝑛𝑢𝑡𝑒
Least count, LC = 𝑛𝑢𝑚𝑏𝑒𝑟 𝑜𝑓 𝑑𝑖𝑣: 𝑜𝑛 𝑡ℎ𝑒 𝑣𝑒𝑟𝑛𝑖𝑒𝑟=
= 1 𝑚𝑖𝑛𝑢𝑡𝑒=1’
30
𝑁
=
Vernier 1
Order
line
MSR
Direct
(R1)
===
st
1st
order
m=1
2nd
order
m=2
1 line
589 nm
2nd line
589.6 nm
1st line
589 nm
2nd line
589.6 nm
VC
Total
reading
 0=

sin 𝜃
𝑚𝜆
Vernier 2
MSR
VC
===
𝑁
Total
reading

0=
===
 1=
1~0=
1=
1~0=
 1=
1~0=
1=
1~0=
 2=
2~0=
2=
2~0=
 2=
2~0=
2=
2~0=
=
Mean N :__________
Result:
Average value of N :________
Experiment No: 35 Newton’s Rings
Aim
To determine the radius of curvature of a given convex lens using Newton’s rings experiment.
Apparatus
Sodium vapour lamp, a short focus convex lens, two plane glass plates, a vernier microscope.
Procedure
A large focal length convex lens L is placed on a glass plate P, kept on the bed plate of microscope.
Rays of light from sodium vapour lamp S, incident horizontally on a glass plate G inclined at 45 ͦare
reflected vertically downward and are incident normally on the air film enclosed between the lens and
glass plate. Due to interference between the light reflected from top and bottom surface of air film, the
alternate dark and bright concentric rings can be observed through the microscope. At the point of
contact of lens with the plate, the thickness of air film is zero. Therefore, the center of concentric rings
appears dark. When moves away from the point of contact, the thickness of air film increases
symmetrically and hence, alternate bright and dark rings are obtained. These rings are called newton’s
rings.
Let the first clear dark ring be nth ring. The microscope is moved slowly
to the left side to cover, say, twenty one dark rings. The rings are
counted as n, n+3, n+6 upto n+21. The vertical cross wire is made
tangent to (n+21)th dark ring and the reading in horizontal scale is noted.
The microscope is moved in the same direction to the right of the rings
system. Similarly readings are taken by moving the
vertical crosswire tangential to the (n+21)th, (n+18)th, …
sin 𝜃
𝑚𝜆
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Physics Experiment
nth rings on the right side. The difference in readings between two sides of a particular ring gives the
diameter of that ring.
L.C. = 0.001 cm
Microscope reading
(d2n+m – dn2)
Order of ring
dn × 10-2m
dn2 × 10-4m-2
(m=12)
Left side (cm)
Right side (cm)
n
x1
n+3
x2
n+6
x3
n+9
x4
n+12
x5
x5 – x1
n+15
x6
x6 – x2
n+18
x7
x7 – x3
n+21
x8
x8 – x4
Mean value = × 10-4m-2
Let dn and dn+m be the diameter of nth and (n+m)th dark Newton’s rings respectively. Wavelength λ is
5893× 10-10m of sodium vapour lamp and R is the radius of the curvature of the lens, then
𝑑𝑛2 = 4𝑅𝑛𝜆
2
𝑑𝑛+𝑚
= 4𝑅(𝑛 + 𝑚)𝜆
𝑅=
2
− 𝑑𝑛2 )
(𝑑(𝑛+𝑚)
4𝑚𝜆
Result
The radius of curvature of the given convex lens =
m
Experiment 36. Comparison of Magnetic Moment of two bar magnets Deflection Magnetometer
Aim : To compare the Magnetic moments of two bar magnets (M1/M2) using Deflection
Magnetometer(DM).
Principle : Tangent Law
Null Deflection method:
M 1 d13

M 2 d 23
Equal Distance method:
M 1 Tan 1 

M 2 Tan  2 
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:
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Physics Experiment
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:
d1magnet A
(in cm)
Trial
d2 - magnet B (in cm)
1
2
3
Mean d2
4
M1  d1 
 
M 2  d2 
3
1
2
3
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 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)
L
1 (for magnet A)
R
L
R
L
2 (for magnet B)
R
L
R
Mean 1
tan(1)
Mean 2
tan(2)
1)
2)
3)
Trial #
d(in cm)
1)
2)
3)
tan(1)
tan(2)
M1  tan1 


M 2  tan 2 
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Physics Experiment
Result:
Mean M1/M2 = _______
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?
Experiment 37. Determination of BH - Tangent Galvanometer
Aim: To determine the Horizontal component of earth’s magnetic field BH at a place using Tangent
Galvanometer.
Formula:
BH 
I
K
tan  
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
K=reduction factor
I = current through Tangent Galvanometer
Initial Arrangement:
1) The Tangent galvanometer is made horizontal with the help of spirit level and leveling
screws.
2) The compass box is rotated till the 90 - 90 line is parallel to the plane of the coil.
3) The coil is rotated until the aluminium pointer reads 0 - 0.
4) The coil is now set in the magnetic meridian.
TG
Commutator
Battery
Key
A
Figure: Circuit connection for Tangent galvanometer
Rheostat
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Physics Experiment
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)
Deflections observed (degrees) 30° < 𝜃 < 60°
1
2
3
4
Mean 
tan()
𝐾=
𝐼
(𝐴)
tan 𝜃
1
2
3
4
Mean value of K
Radius of the coil
Number of turns of the coil used
Horizontal component of the earth’s magnetic field, BH
= ___________ A.
= ___________ m.
= ___________.
= __________ T.
Questions:
1) How does BH vary with Latitude?
2) What will happen if the coil of the tangent galvanometer is along east-west direction?
Experiment 38. Mapping of magnetic lines of Force
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.
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Physics Experiment
N –pole of the magnet pointing geographic North :
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 : 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 : 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?
TDC-IISc, Kudapura
Physics Experiment
Experiment 39. Strength of Magnetic field due to a Solenoid and
Cylindrical Magnet
Objective: To study the magnetic field along the axis of
1) A finite solenoid and compare it with the theory
2) A Cylindrical magnet and find the dependence with distance.
Apparatus:
1) Solenoid of 1000 turns and cylindrical magnet.
2) Acrylic rod having Hall probe IC and 9V battery.
3) Another acrylic rod for winding the pickup coil and Enameled wire.
4) Digital Multimeter (DMM).
5) DC regulated power supply
Introduction: A solenoid is a helical coil which can produce magnetic field when an electric current
is passed through it. It is used as an inductor in many electronic circuits. The solenoid without any
magnetic material in its core is called air core solenoid. The magnetic field produced depends on the
number of turns per unit length, current in the solenoid and geometry of the coil. In the present
experiment, we will study the magnetic field produced by an air core solenoid along its axis.
Hall probe IC: The probe is a semiconductor device in which a small voltage is generated,
proportional to the component of magnetic field applied perpendicular to its plane. The voltage is
amplified and read on a voltmeter. The Hall effect sensor has three leads. The middle lead is common
or ground terminal. Its first and common lead are connected to the 9V battery. A battery connector is
provided to make the necessary connections. The third lead and common lead are connected to a
multimeter used as a voltmeter with 20V dc range. The crocodile connectors are to be used to make
this connection. Red and black colours of the crocodile connectors should be connected to positive and
negative terminals of the voltmeter respectively.
Initially in the absence of the magnetic field, the multimeter displays a voltage V0 (nearly equal to
4.10V) volts. This may vary from probe to probe. Hence note V0 for your instrument. In the presence
of a magnetic field B, the multimeter reading changes to V. The strength of the field is obtained using
the relation
B  0.14  V in tesla
(1)
where V  V  V0 volts. V can be either positive or negative depending on the direction of B
Theory:
Part A: Consider an air core solenoid of n number of turns per unit length (N/L), length L and radius r0
as shown in the figure 1.
The magnetic field at a distance x from one end of the solenoid is given by
𝐵=
𝜇0 𝑛𝐼
(cos𝜃1
2
− cos𝜃2 )
(2)
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Physics Experiment

1
x

2
Figure 1
Part B: The axial magnetic field due to a cylindrical magnet of magnetic moment M at distance d
from the center of the magnet can be written as (d > length of the magnet)
B
2 0 M
4 d n
(3)
where the index n is expected to be an integer.
Experimental Setup:
1) The magnetic Hall probe IC is fixed to a long acrylic rod that can be moved along the axis of a
cylindrical holder (figure 3). When the cylindrical holder is fixed to a clamp it can help in
conveniently moving the Hall probe along a fixed direction.
Figure 3
2) Place the Solenoid below the Hall probe assembly as shown in the figure 4. The solenoid has
700 turns. Connect the solenoid to the DC regulated power supply. Apply voltage such that 0.5
A current flows through the solenoid.
3) Connect the Hall probe IC to 9V batter using the battery connector. The output wires must be
connected to DMM. Check the output of the IC. It should be around 4.00V. If it is lesser than
3.5 V change the battery.
Procedure:
1) Place the solenoid so that the Hall probe assembly can easily move along its axis. Let the top
end of the solenoid be taken as the origin of the axis.
2) Measure the position of the probe on the axis of the solenoid. The position can be measured by
noting the length of the rod above the cylindrical holder (You may follow your own procedure).
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Physics Experiment
3) Measure the magnetic field at this position (Follow similar procedure as in Experiment on
magnet-magnet interaction). Measure the magnetic field in the region above, below and inside
the solenoid.
4) Measure the inner and outer radius of the solenoid and estimate its average. This can be taken as
the effective radius of the solenoid. Estimate the theoretical values of the magnetic field at the
above positions and compare with the experiment.
5) Plot a graph of Magnetic field as a function of position on the axis.
6) Note the region on the axis of the solenoid where the magnetic field is constant.
R1 = __________
R2= __________
Average radius R0 =__________________
Ambient voltage of the Hall sensor, V0=_______________
1
2
….
….
….
…..
….
n
Procedure: Magnetic field due to cylindrical magnet.
The same setup can be used for measuring the axial magnetic field due to a cylindrical
magnet.
1) Measure the ambient voltage (V0) of the Hall probe IC in the absence of the magnet.
2) Measure the magnetic field using the Hall probe IC at different position on the axis of the
magnet. The IC gives the voltage which can be used to estimate the magnetic field by the
method described in the introduction. The magnet position can be changed using the nut and
bolt assembly provided with the acrylic tube.
3) Measure the magnetic field at least for eight different positions between 1.5 cm to 3.5cm.
4) Plot a suitable graph to determine n in equation (3).
5) Using the data collected also determine the value of magnetic moment M of the magnet.
1
2
….
…
….
n
Questions:
1) What is the region on the axis of solenoid where the magnetic field is uniform?
2) Consider two cylindrical magnets separated by distance d (center of one magnet to that of
another). The force is F. What will the force if the separation is doubled.
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Physics Experiment
Experiment 40. Verification of 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:
1. Ohm’s Law: R 
V
I
2. Equivalent resistance, R s for resistors R1 and R 2 connected in series , Rs R1  R2
3. Equivalent resistance R p for resistors R1 and R 2 connected in parallel, R p 
R1 R2
R1  R2
V
mA
V
mA
I
R
R1
Power
Supply
R2
Power
supply
Figure 4a: Ohm’s Law
Figure4b: Series Resistance Connection
I1
Power
Supply
V
1
I2
R1
R2
Figure4c: Parallel Connection of Resistance
Observations:
Part 1: Ohm’s law
Trial
For resistor R1
V1
I1
R1
For resistor R2
V2
I2
1
2
3
4
Mean R1
Mean R2
R2
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


Physics Experiment
Plot Voltage-current graph.
Compare the resistance value obtained from slop, calculated value and color code value.
State whether Ohm’s law is verified or not.
Part 2: Series Connection
R1=_______ , R2=_______.
Trial
Vs
(Volts)
Is
(A)
Rs 
Vs
Is

1
2
3
4
Mean Rs=

Compare the theoretical value of R s (using formula) and experimental value of R s .
Part 3: Parallel connection:
R1=_______ , R2=_______.
Trial
V(Volts)
I1(A)
I2(A)
Rp 
V
I1  I 2
1
2
3
4
Mean Rp=

Compare the theoretical value of R p (using formula) and experimental value of R p .
Questions:
Which of the following materials are conducting? glass, stone, Copper wire, Aluminium rod, rubber,
plastic, wood.
Experiment 41. Diode Characteristics
Aim:
1. To draw the forward and reverse bias characteristics curve of a semiconductor diode.
2. Determination of knee voltage and bulk 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 a non-linear curve. There
is a large flow of current in forward bias mode after exceeding the knee voltage, while negligible
current (nA) flows in reverse bias mode.
Apparatus: Diode, 0-20 Volt DC power supply, digital voltmeter, digital milli-ammeter and
micro-ammeter and resistor. Use
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Physics Experiment
Formula for calculating bulk resistance
R = Diode bulk resistance =
1
slope
forward bias  I V  characteristic
μA
mA
V
V
Figure : Reverse Bias
Figure : Forward Bias
Table:
Trial
number
1
2
3
4
5
6
7
8
9
10
1.
2.
3.
4.
5.
6.
Forward Bias
Voltage V
Current mA
0.1
0.2
0.3
0.4
0.5
0.55
0.60
0.65
0.70
0.75
Reverse Bias
Voltage V
Current µA
1.0
2.0
3.0
4.0
5.0
6.0
7.0
8.0
9.0
10.0
Plot voltage versus current graph.
Find the slope from the forward bias characteristics curve.
Find the bulk resistance.
bulk resistance = inverse of the slope in I-V characteristics curve.
Find the knee voltage.
Connect an LED and record it’s forward current verses diode voltage.
Questions :
1.
2.
3.
4.
5.
What is a semiconductor diode?
What is the barrier potential in a silicon diode?
Give important uses of diode.
Name few pentavalent and trivalent elements used for doping.
In which bias, the diode has high resistance?
Experiment 42. Zener diode characteristics
Aim:
1) Determination of forward and reverse bias characteristics of a zener diode.
2) To determine the break down voltage.
Introduction: Zener diode is a heavily doped p-n junction diode which is made to conduct heavily in
breakdown region. It works as a normal diode in forward bias mode, while in reverse bias, it acts as a
voltage regulator. In this experiment we study both forward and reverse bias characteristics. We also
find the breakdown voltage and understand how it acts as a voltage regulator.
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Apparatus : Zener diode, 0-20 Volt DC power supply, digital voltmeter, digital milli-ammeter and
resistor. Use the kit for the experiment.
mA
Vreversed
Vforward
Ireverse (mA)
V
Figure : Reverse Bias circuit
Iforward (mA)
Circuit Diagram:
Figure : I-V characteristics curve
Observation Table:
Trial
number
1
2
3
4
5
6
7
8
9
10
Forward Bias
Voltage V
Current mA
0.1
0.2
0.3
0.4
0.5
0.55
0.60
0.65
0.70
0.75
Reverse Bias
Voltage V
Current mA
1. Take Readings in steps of 1.0V till the breakdown voltage. Near the breakdown voltage, take
the readings in steps of 0.1V. (Example, a 5V zener will start breaking down at 4.5V. So go in
steps of 1.0V till 4V and 0.1V steps upto 5.5V)
2. Plot voltage versus current graph.
3. Find the break down voltage.
Questions:
1. What is the difference between general rectifier diode and Zener diode?
2. Give few applications of Zener diode.
3. Why the power dissipation of zener diode is important in designing a regulator?
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Experiment 43. Rectifier circuits
Aim:
1. To construct a half wave, a full wave and a bridge rectifier circuit.
2. Measure RMS, Vdc with and without capacitor using digital multimeter for three AC input
voltages.
3. Observe voltage waveforms on oscilloscope.
4. Measure Peak voltage and the AC ripple using oscilloscope
.
Introduction: The electronic gadgets generally work on low dc voltage. The voltage supplied to
household is 220V ac. Thus we need to convert high voltage AC to low voltage DC. The high AC
voltage is first stepped down to low voltage AC using a transformer. Then it is converted to DC using
a rectifier circuit. A single diode converts only one half wave of AC input, a two diode circuit fully
converts both half waves into dc. A bridge rectifier circuit needs four diodes connected in a bridge
fashion. A capacitor smoothens the fluctuating dc voltage, which appears across the load resistor.
Apparatus: Diodes, multi-tapped transformer, digital voltmeter, Resistor, capacitor, circuit unit,
patch cords.
𝑉𝐷𝐶(ℎ𝑎𝑙𝑓 𝑤𝑎𝑣𝑒) =
𝑉𝐷𝐶(𝑓𝑢𝑙𝑙 𝑤𝑎𝑣𝑒) =
𝑉𝐷𝐶(𝑏𝑟𝑖𝑑𝑔𝑒) =
15 V
~
D
R
10K
𝑉𝑝𝑒𝑎𝑘 (𝑉𝑟𝑚𝑠 √2 − 𝑉𝑑 )
=
𝜋
𝜋
2𝑉𝑝𝑒𝑎𝑘 2(𝑉𝑟𝑚𝑠 √2 − 𝑉𝑑 )
=
𝜋
𝜋
2𝑉𝑝𝑒𝑎𝑘 2(𝑉𝑟𝑚𝑠 √2 − 2𝑉𝑑 )
=
𝜋
𝜋
+
V
_
C
100 μF
~
D
0
V
R
10K
V
- 15
V
0V
Figure Half wave rectifier
+15
V
C
100μF
μF
Figure Full wave rectifier
Input ac Voltage
Output dc voltage without C
Output dc voltage with C
Figure. H.W. Rectifier Input – Output waveform
Input ac Voltage
Output dc voltage without C
Output dc voltage with C
Figure F.W. rectifier input-ouput waveform.
TDC-IISc, Kudapura
Physics Experiment
Bridge Rectifier Circuit
Figure Bridge Rectifier Circuit.
Observation Table:
Trial No
VRMS
VDC without
capacitor
VDC with
capacitor
VDC
Calculated
1
2
3
Half Wave
Rectifier
Ripple Voltage VP-P
Ripple Frequency
1
Full Wave/
Bridge
2
Rectifier
3
Ripple Voltage VP-P
Ripple Frequency
Oscilloscope observation:
No. of trials
X-axis
No. of Div
Time/Div
Total Time
Y-axis
No. of Div
Volt/Div
Total Voltage
Frequency
1
2
Questions:
1.
2.
3.
4.
Explain a rectifier circuit. What are it’s applications?
Why VDC with capacitor is more than VDC without capacitor?
What is the difference between a full wave rectifier and a bridge rectifier?
Write basic electrical components with a circuit you need to design a 5V and 100 mA
DC power supply.
TDC-IISc, Kudapura
Physics Experiment
Experiment 44. Transistor characteristics
Aim:
1. To study the input and output characteristics 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-base junction should be forward biased while collector-base
junction should be in reverse bias. One of the most important applications of a transistor is the
amplification of a signal. In this experiment we study the input and output characteristics of a
transistor. We will find the current gain α and β in a CE mode.
Apparatus: A transistor, two variable dc power supply (0-5V, and 0-20V), two dc ammeter (0-200μA;
0-100mA), digital voltmeter, circuit board with a base resistor of 50kΩ, a collector resistor of 1kΩ and
a NPN transistor.
Formula:
 
I C
I B

I C 2  I C1
I B 2  I B1
;
 

1 
Circuit Diagram:
R
Ic
Ib
μA
mA
C
VBB
VBE
B
VCC
VCE
E
Figure . Common Emitter NPN transistor characteristics circuit diagram.
VCE = 1 V
IB =80 μA
μA
μA
I =μA
40μA
IC
(mA)
IB
(μA)
0
IB =120 μA
VBE
K
Figure 4b .Input Characteristics
B
μA
VCE (V)
Figure 4c. Output characteristics
TDC-IISc, Kudapura
Physics Experiment
Table- Input characteristics.
Observation
VCE = 1V
Number
VBE (V)
IB (μA)
1
0.1
2
0.2
3
0.3
4
0.4
5
0.5
6
0.55
7
0.60
8
0.65
9
0.70
10
0.75
VCE = 10V
VBE (V)
IB (μA)
0.1
0.2
0.3
0.4
0.5
0.55
0.60
0.65
0.70
0.75
4. Plot VBE (V) versus IB (μA) graph.
Table- Output characteristics.
Observation
IB = 20 μA
Number
VCE (V)
IC (mA)
1
2
..
..
..
..
IB = 40 μA
VCE (V)
IC (mA)
5. Plot VCE (V) versus IC (mA) graph.
Questions :
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 do the current gain α and β of a transistor suggest?
How do you amplify an ac signal using a transistor?
Experiment 45. Transistor (Common Emitter) Amplifier Circuit
Aim:
1. To study the Amplifier characteristics of a given transistor.
2. To determine various voltages and plot the Q point.
3. To connect a sinewave input and observe the amplified output on oscilloscope.
4. To calculate the Voltage Gain of the amplifier.
Introduction: Transistor common emitter amplifier circuit consists of a voltage divider bias, an input
coupling and an output coupling capacitors for coupling the AC signal to input and output
respectively. To create an ac ground at the emitter, a bypass capacitor across the emitter resistor is
used.
Vcc
R1
R2
R1=10k
R2=2.2k
RC=3.3k
RE=1k
RL=100k
C1, C2=10µF
CE=100µF
TDC-IISc, Kudapura
Physics Experiment
Apparatus: A transistor, circuit board, resistors, capacitors, dc power supply, digital voltmeter, sine
wave generator, oscilloscope.
 Construct the CE emitter amplifier circuit as per the circuit.
 Measure the following transistor voltages: VB, VC, VE, VCC, VBE and VCE.
 Draw the load line and mark the Q point.
 Calculate IC and power dissipation of the transistor as follows:
IC=(VCC-VC) / RC
Power Dissipation PD = VCE x IC
 Set about 20mV sine wave signal on the function generator and connect it to the input
coupling capacitor as Vin.
 Measure the output Vout across the load resistor .
 Calculate the voltage gain using the following equation:
Voltage gain = ouput voltage / input voltage
Gain in dB = 20 log (Voltage gain)
VB (V)
VC (V)
Vin =
VE (V)
VCC (V)
VBE (V)
Voltage Gain =
Vout =
VCE (V)
IC (mA)
PD
(mW)
Voltage Gain in dB =
Experiment 46 Measurement of self inductance of an inductor
Aim:
1. To find out the self inductance of a coil.
Introduction: In a series RL circuit, we know that the effective impedance is given by,
Z  R2  X L2
Where R is the internal resistance of the coil.
Also we know that
Z
Vrms
I rms
We can measure R through a multimeter and note down its value. Now, if we measure I for the given
applied V and for a particular frequeny, we can theoretically calculate Z for each frequency applied.
Since we know Z and R the only unknown term is XL and hence rearranging the terms, we get
X L  Z 2  R2
Connect the coil as given in the figure 8a. Keep the applied voltage at some particular value and
change the frequency for a particular range. Note down the Irms and calculate Z and L in turn, for each
case. After 10 observations, take average of L.
Apparatus: unknown inductor, variable amplitude since wave generator, digital voltmeter, digital
ammeter, bread-board and connecting wires.
Figure . AC circuit : Inductor
TDC-IISc, Kudapura
Physics Experiment
Observations:
Observation
number
1
2
3
4
5
6
7
8
Vrms (V)
Irms (A)
Frequency (Hz)
Z (Ω)
L (H)
Experiment 47. Series resonance of an RLC circuit
Aim:
1. To find out the resonant frequency of a series RLC circuit and check with its theoretical
value.
2. Plotting the series current vs frequency curve.
Introduction: In a series RLC circuit, resonance occurs when inductive reactance equals the
capacitive reactance. Since the phase of these two reactances are opposite in nature, they cancel out
and the effective impedance equals the series resistance of the circuit. Also since the impedance goes
to minimum, the rms current through the circuit goes to its maximum value at resonance.
In other words, at resonance, XL = XC
wL = 1/wC
2πfL = 1/2πfC
f 
1
2 LC
Also we know that the series impedance of an RLC circuit is
Z  R 2  ( X L  X C )2
At resonance, XL = XC. So, Z  R
Connect the RLC components as given in the figure 6a. Keep the applied voltage at some particular
value and change the frequency. Note down the Irms for each case. After 10 observations, plot Irms vs
frequency. The plot should be a bell shaped curve as shown in figure 6b.
Now theoretically calculate the resonant frequency and compare it with the experimental results.
Apparatus: Known values of RLC, variable amplitude sine wave generator, digital voltmeter, digital
ammeter, bread-board and connecting wires.
Figure 6a. Series RLC circuit
Figure 6b. Irms vs f
TDC-IISc, Kudapura
Observation
number
1
..
10
Physics Experiment
Vrms (V)
Irms (A)
Frequency (Hz)
Experiment 48. Parallel resonance of an RLC circuit
Aim:
1. To find out the resonant frequency of a parallel RLC circuit and check with its theoretical
value.
2. Plotting the series current vs frequency curve.
Introduction: In a parallel RLC circuit, resonance occurs when inductive reactance equals the
capacitive reactance. Since the phase of these two reactances are opposite in nature, they cancel out
and the effective impedance equals the series resistance of the circuit. So at resonance the current is
only passing through the R and the currents in L branch and C branch cancel out.
In other words : At resonance, XL = XC
wL = 1/wC
2πfL = 1/2πfC
f 
1
2 LC
Connect the RLC components as given in the figure 7a. Keep the applied voltage at some particular
value and change the frequency. Note down the Irms for each case. After 10 observations, and also
plot Irms vs frequency. The plot should be something like figure 7b.
Now theoretically calculate the resonant frequency and compare it with the experimental results.
Apparatus required: Known values of RLC, variable amplitude since wave generator, digital
voltmeter, digital ammeter, bread-board and connecting wires.
Figure 7a. Parallel RLC circuit
Figure 7b. Irms vs f
Observation
number
1
2
3
4
5
..
10
Vrms (V)
Irms (A)
Frequency (Hz)
TDC-IISc, Kudapura
Physics Experiment
Experiment 49. Charging and discharging of capacitor
Aim:
1. To find out the charge-discharge characteristics of an electrolytic capacitor
2. Determination of time constant, charge stored and voltage across the capacitor.
3. Plotting the charging and discharging curve.
Introduction: A capacitor when charged with a voltage source, charges exponentially as shown in the
figure 5b. A similar exponential decay is observed when the fully charged capacitor is discharged. To
observe the charging characteristics, a large time constant RC is required. With the given values of
1000µF and 100KΩ, RC=100s. Note the voltmeter reading every 10s after closing the switch and plot
the voltage verses time. To do the discharging study, connect the resistor parallel to the capacitor and
close the switch. Note the voltmeter reading every 10s and plot the voltage verses time.
Note: 1. The capacitor must be fully discharged before the experiment.
2. The readings must be observed for atleast 4RC time constant value.
Apparatus: Electrolytic capacitor 1000µF, resistor 100kΩ, 5 Volt DC power supply, digital voltmeter,
digital timer, start-stop switch. Use the kit supplied.
Figure 5a. Charging circuit Figure 5b. Charge characteristics curve of capacitor
Observation
Number
1
2
..
..
..
..
Figure . Discharging circuit
Charging a Capacitor
Time (sec)
Voltage (V)
Figure . Discharging curve of capacitor
TDC-IISc, Kudapura
Physics Experiment
Observation
Number
1
2
..
..
..
..
Discharging a Capacitor
Time (sec)
Voltage (V)
Result
Time constant of the RC circuit = ______s
Experiment 50 Logic Gates verification
Aim:
1. To study the truth table for the logic gates OR, AND, NOT, NOR and NAND.
2. To build the OR, AND, NOT and XOR logic gates using NAND gates (IC 74LS00 two
input quad NAND).
Introduction: A simple two input logic gates can be constructed using diodes, transistor and resistors,
switches and LEDs. Similarly, using the universal gate concept, all types of logic gates can be built.
For this, uncommitted two input NAND gates from the IC 74LS00 are used. Finally the truth table is
drawn for four input combination for all the gates and verified.
Building OR, AND, NOT and XOR logic gates using universal Gates (NAND)
2 input Exclusive OR gate
Observation:
 Connect the NAND gates to form other types of gates.
 Use the switches to input the Logic 0 and Logic 1 to the gates.
 Verify the result by writing the Truth Table.
 Think of real life examples of application of gates.
TDC-IISc, Kudapura
Physics Experiment
Experiment 51. Amplitude modulation - Demonstration
Aim:
To see the demo experiment of amplitude modulation and demodulation on the kit and perform the
following operations:








Connect the oscilloscope to carrier signal and measure it’s frequency and amplitude.
Set 1kHz sinewave signal as modulating signal using the function generator.
Connect the oscilloscope to measure the amplitude and the frequency of the modulating
signal.
Connect the oscilloscope to modulated output and observe the wave shape.
Vary the modulating signal amplitude and observe the depth of modulation.
Vary the modulating signal frequency and observe the modulated output.
Connect the modulated output to the AM demodulator.
Connect the oscilloscope to demodulated output and measure the frequency and amplitude.
Experiment 52. Determination of e/m of electron
Aim:
To determine the ratio of electronic charge ‘e’ to the mass ‘m’of electron by Thomson’s
method.
Apparatus:
CRT, magnetic compass, High Voltage Power Supply (HVPS), 2 permanent magnet, U
shaped wooden stand (U stand)
Procedure:
1. Earth’s magnetic meridian is marked on the table using a magnetic compass.
2. The U shaped wooden stand is placed with its arm perpendicular to the magnetic meridian and
the CRT parallel to the magnetic meridian.
3. The HVPS is kept away from the CRT. The CRT is connected to the HVPS and the brightness
and focus controls are adjusted to get a bright spot at centre of the screen.
4. The 2 bar magnets are kept on either side of the ‘U stand’ at equal distance ‘D’ from the CRT.
5. The deflection of the spot ‘y’ on the screen is noted.
6. The Y deflection voltage applied to bring the spot back to its origin is noted as V.
7. This is repeated for different positions of the magnet ‘D’.
8. The magnetic field is reversed by exchanging the position of the bar magnets and the
experiment is repeated.
9. The CRT is removed without disturbing the ‘U stand’ and the magnetic compass is placed
with its pointer reading 90 -90 along the meridian and 0-0 perpendicular to the meridian.
10. The bar magnets are placed at the same distances ‘D’ as in the previous part of the experiment
and the deflections ‘’ are noted.
11. The magnets are reversed and the experiment is repeated.
𝑒
1
𝑉𝑦
=
(
)
2
𝑚 𝐾𝐵𝐻 (tan 𝜃)2
K = 12.3 cm  3.1 cm  2.8 cm
= 106.7 cm3
= 106.7  10-6 m3
𝐵𝐻 = 40𝜇𝑇 = 40 × 10−6 𝑇
Observations:
Sl no
Deflection
of the spot,
y (cm)
Position of
the magnet,
D (cm)
Deflecting
voltage
V (Volts)
Deflection
Vy
V.cm
1
2
(tan )2
=(1+2)/2
1.
2.
3.
4.
5.
6.
7.
8.
9.
10.
𝑀𝑒𝑎𝑛
𝑉𝑦
(tan 𝜃)2
= _______V.cm. = ______10-2 V.m.
Result The average value of e/m of electron : ___________ C.kg-1
Experiment 53. Determination of Planck’s constant - Photoelectric Effect
Aim: To determine the Planck’s constant using photodiode.
Apparatus: Light source, color filters, photoelectric effect setup with voltmeter and ammeter.
h
Formula:
eV

 e  slope ;
h= planck’s constant, e = electronic charge, ν = frequency.
Connecting
board
Photocell
Ray of light
A
-
V
+
Light
source
Color filters
(a) Electrical connection
(b) Experimental set up we have seen.
Figure 17: Photoelectric effect
Procedure:
1. Connect the electrical connection as shown in figure (a).
2. The bulb is allowed to warm up for 10 minutes with the metal box lid removed.
𝑉𝑦
(tan 𝜃)2
3. A color filter (a color glass disc) say, orange color, is inserted in the color filter window
provided between photocell and light source.
4. Measured/noted the stopping voltage, V.
5. Repeat the steps (1-4) for different color filters.
Table 1: Determination of Planck’s constant.
Color
Wavelength
(x 10-9m)
Frequency
(x 1014 Hz)
Photo current,
I = ? when V = 0
Blue
Green-1
Green-2
Orange
Red
1. Plot frequency - Stopping potential graph, and find the slope.
2. Plot frequency - Photo current graph.
3. Planck’s Constant as,
he
V

 e  slope ;
e =1.602 x 10-19C
Discuss the following:
1.
2.
3.
4.
Give few applications of photoelectric effect.
Which nature of light is used to explain photoelectric effect?
Why we are using color filters in this photoelectric effect?
From the graph, try to find the threshold frequency.
Stopping potential
V=? when I= 0
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
4. Electric charge, e
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
Value
Physical parameter
Units
1. Electric charge
Coulomb, C
2. Electric current
Ampere, A
Am-1
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
Hz or s-1
2.9979 x 108 ms-1
6.626 x 10
-34
Js
4π x 10-7 Hm-1
1.602 x 10-19 C
7
-1
1.0974 x 10 m
331.3 ms-1
1.485- 1.925
1.3330
Thermal Conductivity of Selected materials:
Material
Thermal Conductivity
1. Copper
385 Js-1m-1K-1
2. Aluminium
205 Js-1m-1K-1
-1 -1 -1
3. Brass
109 Js m K
4. Steel
50.2 Js-1m-1K-1
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. micro, μ =10-6
3. milli, m = 10-3
4. killo, k = 103
Tesla, T
V
Ohm, Ω
Ωm
m
IISc Press
Talent Development Centre
Indian Institute of Science, Kudpura
Challakere, Chitradurga District, Karnataka- 577536