Practical-I MCT 3

Exp. No.1.1
Flywheel- Moment of inertia
Aim: To find the moment of inertia of a fly wheel.
Apparatus: The flywheel, weight hanger with slotted weights, stop clock, metre scale etc.
Theory: A flywheel is an inertial energy-storage device. It absorbs
mechanical energy and serves as a reservoir, storing energy during the
period when the supply of energy is more than the requirement and
releases it during the period when the requirement of energy is more than
the supply. The main function of a fly wheel is to smoothen out variations
in the speed of a shaft caused by torque fluctuations. Many machines have
load patterns that cause the torque to vary over the cycle. Internal
combustion engines with one or two cylinders, piston compressors, punch
presses, rock crushers etc. are the systems that have fly wheel.
A flywheel is a massive wheel fitted with a strong axle projecting on
either side of it. The axle is mounted on ball bearings on two fixed
supports as shown in fig.b. There is a small peg inserted loosely in a hole
on the axle. One end of a string is looped on the peg and the other end
carries a weight hanger. A pointer is arranged close to the rim of the flywheel. To do the
experiment, the length of the string is adjusted such that when the descending mass just touches
the floor, the peg must detach the Pointer
axle. Now a line is drawn on the
rim with a chalk just below the Peg
pointer. The string is then attached
to the peg and the wheel is rotated
for a known number of times ‘n’
such that the string is wound over Axle
‘n’ turns on the axle without
overlapping. Now the mass m is at
a height ‘h’ from the floor. The Wheel
m
mass is then allowed to descend
down. It exerts a torque on the h
axle of the flywheel. Due to this
torque the flywheel rotates with an
angular acceleration. Let  be the
angular velocity of the wheel Fig.b
when the peg just detaches the axle and W be the work done against friction per one rotation,
then by law of conservation of energy,
1 2 1 2
mgh = Iω  mv  nW (1)
2 2
Let N be the number of rotations made by the wheel before it stops. Since the kinetic
energy of rotation of the flywheel is completely dissipated when it comes to rest, we can write,
1
NW = Iω2
2

3
4 MCT Practical I

Iω 2
Or, W = (2)
2N
Using eqn.2 in eqn.1,
1 2 1 Iω2 1 2 n 1
mgh = Iω  mv  n
2
= Iω 1    mr 2ω2
2 2 2N 2  N 2
Nm  2gh 2 
 I =  r  (3)
N  n  ω2 
where, ‘r’ is the radius of the axle. To determine ‘’ we assume that the angular retardation of
the flywheel is uniform after the mass gets detached from the axle. Then,
Total angular displacement
Average angular velocity =
Time taken
ω0 2πN
=
2 t
4πN
  = (4)
t
Procedure: To start with the experiment one end of the string is looped on the peg and a
suitable weight is placed in the weight hanger. The fly wheel is rotated ‘n’ times such that the
string is wound over ‘n’ turns on the axle without overlapping. The flywheel is held stationary at
this position. The height ‘h’ from the floor to the bottom of the weight hanger is measured. The
flywheel is then released. The mass descends down and the flywheel rotates. Start a stop watch
just when the peg detaches the axle. Count the number of rotations ‘N’ made by the wheel during
the time interval between the peg gets detached from the axle and when the wheel comes to rest.
The time interval ‘t’ also is noted. The experiment is repeated for same ‘n’ and same mass ‘m’.
The average value of ‘N’ and ‘t’ are determined. The moment of inertia ‘I’ is calculated using
equations (3) and (4). The entire experiment is repeated for different values of ‘n’ and ‘m’ and
the average value of I is calculated.
 Ensure that the length of the string is such that when the mass just touches the floor the
peg gets detached from the axle.
 In certain wheels the peg is firmly attached to the axle. In such case, one end of the string
is loosely looped around the peg such that when the mass just touches the floor the loop
gets slipped off from the peg.
 ‘m’ is the sum of mass of weight hanger and the additional mass placed on it.
Observation and tabulation
To determine the radius of the axle using vernier calipers
Value of one main scale division (1 m s d) = ………. cm
Number of divisions on the vernier scale, x = ……….
Value of one main scale division 1msd
Least count, L. C = = = ……. cm
Number of divisions on the vernier scale x
Practical-1 MCT 5

Trial No. MSR VSR D = M S R + V S RL C Mean diameter D

cm cm cm

D = …….. cm = ……… m
Diameter of the axle
D
Radius of the axle r = = …….. m
2
Determination of moment of inertia
No. of rotations of Time interval in
Number of windings of string
Height from the floor to the
Mass suspended at one end of

bottom of weight hanger ‘h’ m

the wheel after the between the detachment

detachment of the of the peg and when the
peg from the axle wheel comes to stop, ‘t’
‘N’ sec.
4πN I
the string ‘m’ kg

ω kg.m2
on the axle ‘n’

t
Mean Mean t
1 2 N 1 2 sec

Result
Moment of inertia of the given flywheel, I = ………. kg.m2
6 MCT Practical I

Exp.No.1.2
Compound pendulum- To find ‘g’ and radius of gyration
Aim: To determine (a) the value of acceleration due to gravity ‘g’ at the given place by using a
compound pendulum, (b) the radius of gyration and hence the moment of inertia of the
compound pendulum about an axis passing through its centre of mass.
Apparatus: The compound pendulum, stop watch, etc.
Theory
A compound pendulum, also known as a physical pendulum, is a body of any arbitrary
shape pivoted at any point so that it can oscillate in a plane when its centre of mass is slightly
displaced on one side and is released.
In the figure S is the suspension centre and G is the centre of gravity of the body. Let the
vertical distance SG be l when the body is in its normal position of rest. If the body is oscillated
through an angle θ about an axis passing through S and perpendicular to the vertical plane of the
body, its centre of gravity takes the position G’. The torque acting on the body due to its weight
mg is given by,
 = Mglsinθ
The negative sign indicates that the torque acts opposite to the direction of increase of θ. If I is
the moment of inertia of the body about the axis of rotation, then the torque is also given as,
d 2θ
 = I = I 2
dt Reaction of
dθ 2
weight mg
i.e. I 2 = Mglsinθ
dt
If the angular displacement θ is very small, sinθ = θ . Then the
equation of motion becomes, S
d 2θ Mgl
+ θ = 0 (1) θ
dt 2 I
l
Eqn.1 shows shat the motion of the pendulum is simple
S
Mgl
harmonic with an angular frequency, ω0  . Its period of
I A lsinθ G'
oscillation is given by,
G
2π I mg
T = = 2π (2) O'
ω0 Mgl
O
I
Now we define L = (3) S
Ml
Fig.a
L
Then, T = 2π (4)
g
where, L is called the length of an equivalent simple pendulum.
Practical-1 MCT 7

If K is the radius of gyration of the compound pendulum about an axis through the centre of
mass, the moment of inertia is,
ICM = MK2 (5)
Applying the parallel axes theorem the moment of inertia around the pivot is given by,
I = ICM + Ml2 = MK2 + Ml2 = M  K 2  l 2  (6)

Hence from eqn.2 we get,

K2  l 2 L
T = 2π = 2π (7)
gl g

I K2  l 2
where, L = = (8)
Ml l
Thus, if we know the radius of gyration of an irregular body around an axis through the centre of
mass, the time period of oscillation of the body for different points of pivoting can be calculated.
Fig.b shows the graph between the time period T in the Y axis and the distance of the point of
suspension (axis of rotation) from one end of the bar in the X axis.
Centres of suspension and oscillation are mutually interchangeable: In fig.a consider the
point O on the line joining the centre of suspension ‘S’ and centre of gravity G at a distance
 K2  K2
  l  from ‘S’ or from G . This point is called the centre of oscillation. An axis passing
 l  l
through the centre of oscillation and parallel to the axis of suspension is called axis of oscillation.
K2
Let S G = l1 and G O = l2= .
l1 Y
Let T1 be the time period with ‘S’ as
point of suspension. Now we find out
Time period

the period of oscillation T2 with O

as point of suspension. Then,

4π 2  K 2 
T2 =   l2 
g  l2 
L
2
K
But, l2 = (9) L
l1
T A
P E Q R F S
4π  K
2 2
 Tm
Then, T2 =   l1 
g  l1  O E' G F' X
Distance of knife edge from one end of the bar
= T1 Fig.b

Thus the axes of suspension and oscillation are interchangeable. And if ‘L’ is the distance
between them we can write,
8 MCT Practical I

K2 K2
L =  l1 =  l2 (10)
l1 l2
L
And T = T1 = T2 = 2π
g
Thus by knowing L and T value of acceleration due to gravity g can be obtained as,
4π 2 L
g = (11)
T2
To determine L and K: Draw the graph between the time period T in the X axis and the
distance of the point of suspension (axis of rotation) from one end of the bar as shown in fig.b.
From the graph, for a given T,
A
PR  QS
L = (12)
2
By eqn.8, K = l1l2 = PA  AR = QA  AS
PA  AR  QA  AS
Thus, K = (13)
2
Procedure: In our experiment we use a symmetric compound pendulum as shown
in fig.c. The compound pendulum is suspended on a knife edge passing through the
first hole near one of the ends, say, A. The pendulum is pulled aside slightly and is
released so that the pendulum oscillates with small amplitude. The time for 20
oscillations is determined twice and the average is calculated. From this, the period
of oscillation T of the symmetric pendulum is found out. Similarly, the time periods
of the pendulum by suspending the pendulum in successive holes till the hole near
the other end B. (For holes beyond the centre of gravity, the pendulum gets
inverted). The distances ‘x’ from the end A to the edge of the holes at which the
knife edge touches are measured by a metre scale.
The centre of gravity of the bar is determined by balancing it on a knife edge. B
The position of centre of gravity from the end A is also measured. The mass of the Fig.c
bar (including the knife edge if it is attached to the bar) is measured using a balance.
A graph is drawn taking the distances ‘x’ of the holes from the end A along the X-axis and
the time periods T along the Y-axis as shown in fig.b.
To determine the length of the equivalent simple pendulum and the radius of gyration K
about the axis passing through the centre of gravity from the graph, draw lines parallel to the X-
axis for particular values of T. Determine PR and QS and from these L is calculated. Also
determine PA, AR, QA and AS and from these K is calculated. Finally, using eqn.11 the value of
acceleration due to gravity ‘g’ is calculated and the moment of inertia of the bar about an axis
through the centre of mass (centre of gravity) using eqn.5. We can also calculate the moment of
inertia of the bar about an axis at a distance ‘a’ from the end A and perpendicular to the bar by
applying the parallel axes theorem, I  MK 2  Ma 2 .
 Distances ‘x’ from the end A depends on how the knife edge is fixed in the holes. It may
be the top end, bottom end or centre of the hole. The inversion of the bar also is taken
into account in this case.
Practical-1 MCT 9

Observation and tabulation

Mass of the bar, M = …….. kg.
Position of centre of gravity G from the end A = ……. m
Distance ‘x’ from Time for 20 oscillations in sec. Period T
the end A in metre 1 2 Mean sec

To determine acceleration due to gravity (Observations from graph)

Sl.No. T PR QS PR  QS L 2 L  L
sec m m L  m 2
ms Mean 2
ms 2 g  4π 2  2  ms 2
2 T T T 

To find radius of gyration and moment of inertia (Observations from graph)

Sl.No. T PA AR QA AS PA  AR  QA  AS Mean K ICM  MK 2
sec m m m m K m m kg.m2
2

Result
Acceleration due to gravity at the place, ‘g’ = ……… ms 2
Radius of gyration about an axis through the centre of mass, K = ………. m
Moment of inertia about an axis through the centre of mass, ICM = ………. kg.m2
10 MCT Practical I

Exp.No.1.3
Surface Tension by capillary rise method
Aim: To determine the surface tension of the given liquid by capillary rise method.
Apparatus: Beaker with the given liquid, capillary tube, travelling microscope, etc.
Theory
T

Pointer
Travelling microscope

hl hw

Fig.a liquid water

When a capillary tube of inner radius ‘r’ is dipped in
a liquid of surface tension ‘T’ the liquid rises through the
tube to a certain height. This is known as capillary rise. If 
Fig.b
is the angle of contact of the liquid with the capillary tube,
the height of the capillary rise is such that the upward surface tension force is equal to the weight
of the liquid column in the tube. Let ‘h’ be the height from the liquid surface in the beaker to the
liquid meniscus in the capillary tube. Then,
1
2πrT cosθ = πr 2 hρg  πr 3ρg (1)
3
The second term in the R H S is the weight of the liquid in the meniscus portion, which is
negligibly small.
 r
 h   rρg
T = 
3
(2)
2 cos θ
We usually use the capillary rise method to find out the surface tension of liquid that wets the
glass and have negligibly small angle of contact. Thus,
 r
 h   rρg
T = 
3
(3)
2
The density of the liquid can be determined by Hare’s apparatus as shown in the fig.b. Let
hw is the height of the water column and hl is the height of the liquid column, then,
Atmospheric pressure, H = P + hwwg = P + hlg
Practical-1 MCT 11

i.e. hww = hl

h Height of water column
 = w ρw = 1000 kgm 3 (4)
hl Height of liquid column
Procedure: The capillary tube and the pointer are arranged as shown in the fig.a. Raise and
lower the beaker and check that the meniscus also raises or lowers correspondingly. Otherwise,
clean the tube and is again checked that the tube is completely wet with the liquid. The pointer is
arranged such that its tip just touches the liquid surface in the beaker. Then the travelling
microscope is focused to see the liquid meniscus. (It is better to place the microscope close to the
tube and is then pulled back till the tube is seen clearly and then it is raised or lowered to see the
meniscus). The horizontal wire of the microscope is made to coincide with the meniscus and the
readings on the vertical scale are noted. Now the beaker is removed and the microscope is
adjusted to see the tip of the pointer. By adjusting the vertical tangential screw the tip of the
pointer is made to coincide with the horizontal cross wire. The reading on the vertical scale is
noted. The difference between the two vertical scale readings gives the capillary rise. The
experiment is repeated after the beaker is placed in another level.
To find the diameter of the bore of the capillary tube it is arranged horizontally. The
travelling microscope is adjusted to see the bore clearly. The horizontal tangential screw of the
microscope is adjusted such that the vertical cross wire is tangential to
the left side of the bore. The reading on the horizontal scale is noted. The
vertical wire is then made to coincide with the right side of the bore and
the reading on the horizontal scale is again noted. The difference
between the readings gives the diameter in the horizontal direction.
Similarly by adjusting the vertical tangential screw the horizontal cross
wire is made to coincide with the top and bottom and the corresponding
readings on the vertical scale are noted. The difference between the D
readings gives the vertical diameter. The average diameter and hence the Fig.c
radius of the bore is calculated.
The density of the given liquid is determined by Hare’s apparatus. The beakers containing
the given liquid and water are arranged as shown in fig.b. The height of the liquid column and
the height of the water column are determined and the density is calculated using eqn.4.
Finally, the surface tension of the liquid is calculated using eqn.3.
 The interior of the tube must be clean. It is free from any surface contamination. When
the beaker is raised or lowered the liquid meniscus also is raised or lowered
correspondingly. Otherwise clean the tube.
 Ensure that there are no air bubbles inside the capillary tube.
 Ensure that the pointer just touches the water surface before taking the meniscus reading.
 Ensure that while using Hare’s apparatus hw is the height from the water surface in the
beaker to the meniscus and not the scale reading against the meniscus. Also hl is the
height from the liquid surface to the meniscus.
Observation and tabulation
Value of one main scale division (1 m s d) = …….. cm
Number of divisions on the vernier scale, n = ……..
Value of one main scale division 1msd
Least count ( LC) = = = …… cm
Number of divisions on the vernier n
12 MCT Practical I

Determination of capillary rise

Trial Reading against the liquid meniscus Reading against the tip of the pointer Capillary
No. MSR VSR Total reading MSR VSR Total reading rise
cm ‘a’ cm cm ‘b’ cm h = ab
cm
1
2
3
4
Mean ‘h’ = ……. cm

Determination of radius of the capillary tube

Reading corresponding to Reading corresponding to Diameter
Mode Left/top Right/bottom ‘D’
M S R V S R Total reading MSR VSR Total reading cm
cm cm cm cm
Horizontal
Vertical
Mean ‘D’ = ……. cm

Determination of density of liquid using Hare’s apparatus

Density of water, w = 1000 kg/m3
Trial No. Height of liquid column, Height of water column, hw
hl cm hw cm ρ ρ w kg/m3
hl
1
2
3
4
5
Mean  = …………… kg/m3

 r
 h   rρg
Surface tension of the given liquid, T =  3
= ………….
2
= ……… N/m

Result
The surface tension of the given liquid = …….. N/m
Practical-1 MCT 13

Exp.No.1.4
Young’s modulus of the material of bar-Non-uniform bending
(using pin & microscope)

Aim: To determine the Young’s modulus of the material of a bar by subjecting it to non-
uniform bending and measuring the depression at centre of the bar by using pin and microscope.
Apparatus: A long uniform bar, two knife edges, a travelling microscope, pin, weight hanger
and slotted weights, etc.
Theory: Let a beam AB be supported by two knife-edges K1 and K2 and loaded at the middle C
with a weight W = Mg as shown in Fig.(a). The length of the beam between the knife-edges is l
and the reaction at each knife-edge is W/2, acting upwards. The depression is maximum at the
middle. Let this maximum depression be δ. Since the middle of the beam is almost horizontal,
the beam may be considered to be equivalent to two inverted cantilevers CA and CB, each of
length l 2 and carrying an upward load W/2. Therefore, the maximum depression δ of C below
the knife-edges is equivalent to the elevation of A and B from the lowest position C.
Now, consider a vertical section P, distant x from C. Then, the moment of the deflecting
couple on the section PB is

W W l 
.PB =   x W/2 W/2
2 2 2 
l
In the equilibrium condition, this deflecting couple is A C B
balanced by the bending moment.
YI W l  K1 K2
=   x (1)
R 2 2 
W
where, Y is the Young’s modulus, R is the radius of Fig.a
curvature at any point on the bent beam and I is the
geometrical moment of inertia. For a beam with W/2
Y axis

rectangular cross-section, the geometrical moment of

bd 3 l/2
inertia I = , where b is the breadth and d is the B
12
thickness of the bar. For a circular beam of radius r, it is, P 
π r4 y
I =
4 C W/2 X axis
If y is the elevation of the section P above C, the x
radius of curvature of the neutral axis at this section is Fig.b
given by,
1 d2 y
=
R dx 2
Substituting this value of 1/R in eqn.1,
14 MCT Practical I

d2 y W l 
YI =   x
dx 2 2 2 
d2 y W l 
Or, =   x (2)
dx 2 2YI  2 
dy W l x2 
On integration we get, =  x   + C1
dx 2YI  2 2 
dy
where C1 is the constant of integration. Since x = 0 and = 0 at C (i.e. at l = 0), C1 = 0.
dx
dy W l x2 
Therefore, =  x  
dx 2YI  2 2 
Integrating the expression again, we get
W  l x 2 x3 
y =    + C2
2YI  2 2 6 
where, C2 is a constant of integration. Since y = 0 at x = 0, C2 = 0.
l
At the free end, x = and if the corresponding elevation y = δ, we can write,
2
W  l l2 l3 
 =    
2YI  2 8 48 
Wl 3 Mgl 3
Or,  = = (3)
48YI 48YI
bd 3
For a beam of rectangular cross-section, I =
12
Wl 3 Mgl 3
Then,  = = (4)
4bd 3Y 4bd 3Y
Mg  l 3 
Or, Y =   (5)
4bd 3  δ 
π r4
For a beam of circular cross-section, I = and hence,
4
Wl 3 Mgl 3
 = = (6)
12Yπr 4 12Yπr 4
Mg  l 3 
Or, Y =   (7)
12πr 4  δ 

Procedure: The given bar is supported symmetrically on two knife edges, such that the length
of the bar in between the knife edges is l, say 40 cm, as shown in fig.c. (If a metre scale is used
as the bar, we can place the bar on the knife edges such that they are at 30 cm and 70 cm marks).
The weight hanger is suspended at the midpoint of the bar (in between the knife edges). A pin is
Practical-1 MCT 15

fixed vertically at the midpoint of the bar. A travelling microscope is focused such that the
horizontal wire is at the tip of the pin.
Now the bar is B
brought into an elastic
mood by loading and l
unloading it step by
step several times. A
sufficient dead load is
A
placed in the weight
hanger. Let ‘w0’ be
the mass of weight W
hanger and the Fig.c
additional dead load
placed in it. The microscope is focused such that the tip of the pointer is at the horizontal wire.
(The microscope is initially placed very close to the pin. It is then pulled back till the pin is seen
clearly. Then the rack and pinion arrangement is adjusted to see the pin very clearly. The
microscope is raised or lowered by adjusting the main screw and tangential screws to make the
horizontal wire to coincide with the tip of the pin). The reading on the vertical scale is taken.
Now the slotted weights are added to the weight hanger in steps of mass ‘m’. In each case the
microscope is made to coincide with the tip of the pin and the readings are taken. Then the mass
in the weight hanger is unloaded in steps and again the readings are noted. From these readings
l3
the average depression  is found out for a particular mass M, say M = 4m =200 gm, and is
δ
l3
calculated. The experiment is repeated for different values of ‘l’ and the mean value of is
δ
determined.
The breadth ‘b’ of the bar is determined with a vernier calipers and the thickness ‘d’ by a
screw gauge. The Young’s modulus of the bar is calculated using eqn.5.
Observation and tabulation
To find breadth of the bar using vernier calipers
Value of one main scale reading of the vernier calipers (1 m s d) = …….. cm
Number of divisions on the vernier scale n1 = ……..
1msd
Least count (LC) of the vernier calipers = = …… cm
n1
Trial No. MSR VSR b = M S R + V S RL C Mean breadth b
cm cm cm
16 MCT Practical I

To find the thickness of the bar using screw gauge

Distance moved by the screw tip for 5 rotations of the head = ……… mm
Distance moved by the screw tip
Pitch of the screw, P = = ……… mm
Number of rotations of the head
Number of divisions on the head scale = ………
Pitch
Least count (L C) = = ……. mm
Number of divisions on the head scale
Zero coincidence = …….. ; Zero error = …….
Zero correction = ……..
Trial No. PSR Observed Corrected Thickness Mean d
‘x’ mm HSR H S R ‘y’ d  x  y  LC mm mm
1
2
3
4
5

l3
To find
δ
Value of one main scale reading of the microscope (1 m s d) = …….. cm
Number of divisions on the vernier scale n = ……..
1msd
Least count (LC) of the travelling microscope = = …… cm
n
Mass for which depression is calculated, M = 4m = 0.2 kg.
Microscope readings
Suspended Load

the mass M=4m

Depression for
Mean reading

loading Unloading  l3 
‘’ in cm
Mean 
‘l’ in m

 
Length

δ
cm
cm
MSR

MSR
VSR

VSR
Total

Total
cm

cm

cm

cm

m2
in kg.

W0
W0 + m
W0 + 2m
W0 + 3m
W0 + 4m
W0 + 5m
W0 + 6m
W0 + 7m
Practical-1 MCT 17

W0
W0 + m
W0 + 2m
W0 + 3m
W0 + 4m
W0 + 5m
W0 + 6m
W0 + 7m
W0
W0 + m
W0 + 2m
W0 + 3m
W0 + 4m
W0 + 5m
W0 + 6m
W0 + 7m
l3
Mean = …….m2
δ
Mg  l 3 
Young’s modulus of the material of the bar, Y =   = ……
4bd 3  δ 

= …….. Nm2
Result
Young’s modulus of the material of the bar, Y = ……. Nm2
18 MCT Practical I

Exp.No.1.5
Young’s modulus of the material of a bar -Uniform Bending
(Using optic lever, telescope and scale)
Aim: To determine the Young’s modulus of the material of the given bar by subjecting it to
uniform bending and by measuring the elevation using an optic lever, scale and telescope
arrangement.
Apparatus: A long uniform bar, two knife edges, an optic lever, scale and telescope
arrangement, weight hanger and slotted weights, etc.
Theory
Consider a beam supported symmetrically on two knife edges A and B and with a length l
between the knife edges. The beam is loaded with equal weights W = Mg at the ends at equal
distances p from the knife edges, as shown in Fig.(a). The bar is bent uniformly since it is loaded
symmetrically at both ends. Let δ be the elevation of the midpoint O of the bar when it is loaded.
Consider the equilibrium of one half of the bar, say OC. The only external forces acting on
this section of the beam are the load W acting vertically downwards at C and its reaction W
acting vertically upwards at the knife edge A. The distance between these two forces is p. These
two forces constitute a couple, whose moment is given by W.p. In the equilibrium condition, this
YI
moment is balanced by the bending moment .
R
YI
 W.p = (1)
R
The bar bends into the arc of a circle as shown in Fig. (b). If R is the radius of curvature of the
neutral surface and δ is the elevation,
W W
l l l2
 2R  δ  δ = . or 2Rδ  δ2 = p O p
2 2 4
Since δ2 is negligibly small compared to 2Rδ, we C
l
D
can write
l2 l2 A B
2Rδ = or R = W W
4 8δ
Fig.a
Substituting for R in eqn.1,
O
YI 8YI 
W.p = = δ C D
2
l 8δ l2 l/2 l/2

Wpl 2
  =
2R

8YI
bd 3
For a bar of rectangular cross section, since I = Fig.b
12
Wpl 2 12Wpl 2 3Wpl 2
 = = =
bd 3 8Ybd 3 2bd 3Y
8Y
12
Practical-1 MCT 19

3Wpl 2 3Mgpl 2
Or, Y = = (3)
2bd 3δ 2bd 3δ
π r4 2 M
Similarly, for a bar of circular cross section, I = M
4
Wpl 2 Wpl 2
= =
π r4 2πr 4 Y A B
8Y B C
4 A C
Wpl 2 Mgpl 2 B B
Or, Y = = (4) a
N A N
2πr 4 δ 2πr 4 δ A C C
Principle of optic lever: In this experiment we A
determine the elevation ‘’ by using an optic lever, scale A  N
a
and telescope arrangement. The principle behind it is s2
that, if the mirror turns through an angle  the reflected s
ray turns through an angle 2. The optic lever consists of s1
a triangular frame with three legs and a mirror strip is 2
fixed perpendicularly on it as shown in fig.c. The optic Fig.c D M
lever is placed with its front leg ‘A’ at the midpoint of
the experimental bar arranged on the knife edges. The back legs B and C rest on another bar
placed behind the experimental bar. A scale and telescope is arranged at a distance D, say 1 m,
from the mirror of the optic lever such that the image of the scale is obtained on the cross wire of
the telescope. Let s1 be the scale reading that coincides with the horizontal wire. Then the bar is
loaded symmetrically. Due to the elevation  of the bar, the optic lever and hence its mirror turns
through an angle ‘’. Since the reflected ray turns through an angle 2, we get another scale
reading s2 that coincides with the horizontal wire of the telescope. Then,
Shift in scale reading when the bar is loaded, s = s2  s1
Let ‘a’ be the length of the line joining the front leg and the midpoint of the line joining the back
legs of the optic lever.
δ
Angle turned by the optic lever due to the elevation  of the bar,  =
a
s
Angle turned by the reflected ray from the mirror of the optic lever, 2 =
D
δ s
i.e. =
a 2D
as
 = (5)
2D
Using eqn.5, eqns.3 and 4 become,
3MgD  pl 2 
For rectangular bar, Y =   (3a)
abd 3  s 
MgD  pl 2 
For cylindrical bar, Y =   (4a)
πr 4a  s 
20 MCT Practical I

Procedure: The given bar is

supported symmetrically on two
knife edges with length of the p
bar in between the knife edges
A
is ‘l’. For uniform bending of
the bar, equal weights are l p
suspended at equal distances, W
‘p’ from the knife edges. An B
optic lever is arranged behind
W
the bar such that its front leg is
at the midpoint of the bar and
the two back legs are on another
bar arranged as shown in the
fig.d. (Ensure that the two bars
do not touch each other). The
scale and telescope arrangement Fig.d
is placed in front of the bar with
the distance in between the
scale and the mirror D is greater
than or equal to 1 m. The
telescope can be focused as follows.
 Looking above the edges of the telescope with one eye (other eye closed) the stand on
which the telescope fixed is moved sideways till the image of the scale is seen clearly
on the mirror of the optic lever.
 Looking through the edges in the right side of the telescope with one eye, adjust the
leveling screws (vertical and sidewise) till the telescope is exactly towards the mirror
strip of the optic lever.
 Now looking through the eyepiece, the telescope is focused by adjusting the rack and
pinion arrangement till the clear image of the scale is seen exactly on the cross wires of
the telescope.
 Before starting to take reading, ensure that it is possible to get scale readings for the
minimum and maximum loads. The scale is raised or lowered if needed.
Now the bar is brought to elastic mood by loading and unloading in steps for several times.
Now suitable dead loads are placed in the weight hangers. The scale reading that coincides with
the horizontal cross wire is noted. Increase the slotted weights in the weight hangers in steps of
mass ‘m’ and in each case the scale reading is noted. The scale readings are also taken during the
unloading of the slotted weights. The average shift in scale reading for particular loads, say M =
4m in each weight hanger, is determined. The entire experiment is repeated for different values
of l.
The breadth of the bar is determined by a vernier calipers and its thickness by a screw
gauge. The length ‘a’ between the front leg and the midpoint of the line joining the back legs is
determined as follows. The optic lever is pressed on a paper so that the impressions of the three
legs are obtained on the paper. Construct the triangle with these impressions. Find out the
midpoint N of the line joining the back legs (refer fig.c). Then measure the distance ‘a’ between
this point and the point corresponding to the front leg.
Practical-1 MCT 21

Observation and tabulation

Mass used to increase the load in steps, m = …….. kg.
Mass for which the elevation is calculated, M = 4m = …….. kg.
Length P D Suspended Telescope reading in cm Shift in scale reading Mean  Dpl 2 
l in m m m load in kg loading Unloading Mean ‘s’ for a mass M=4m ‘s’ in  
 s 
cm m m3
W0
W0 + m
W0 + 2m
W0 + 3m
W0 + 4m
W0 + 5m
W0 + 6m
W0 + 7m
W0
W0 + m
W0 + 2m
W0 + 3m
W0 + 4m
W0 + 5m
W0 + 6m
W0 + 7m
W0
W0 + m
W0 + 2m
W0 + 3m
W0 + 4m
W0 + 5m
W0 + 6m
W0 + 7m
Mean …….
To find breadth of the bar using vernier calipers
Value of one main scale reading of the vernier calipers (1 m s d) = …….. cm
Number of divisions on the vernier scale n = ……..
1msd
Least count (LC) of the vernier calipers = = …… cm
n
Trial No. MSR VSR b = M S R + V S RL C Mean breadth b
cm cm cm
22 MCT Practical I

To find the thickness of the bar using screw gauge

Distance moved by the screw tip for 5 rotations of the head = ……… mm
Distance moved by the screw tip
Pitch of the screw, P = = ……… mm
Number of rotations of the head
Number of divisions on the head scale = ………
Pitch
Least count (L C) = = ……. mm
Number of divisions on the head scale
Zero coincidence = …….. ; Zero error = …….
Zero correction = ……..
Trial No. PSR Observed Corrected Thickness Mean d
‘x’ mm HSR H S R ‘y’ d  x  y  LC mm mm
1
2
3
4
5

To determine ‘a’ B

A N
a=…

Distance between the front leg and the midpoint of the line joining the back legs,
a = …….. m

3Mg  Dpl 2 
Young’s modulus of the material of the bar, Y =   = ………….
abd 3  s 
= ……… Nm2
Result
Young’s modulus of the material of the bar, Y = ……… Nm2
Practical-1 MCT 23

Exp.No.1.6
Torsion pendulum
(Moment of inertia of a disc and rigidity modulus)
Aim: To determine the moment of inertia of a given disc and the rigidity modulus of the
material of the wire used to suspend the disc by the method of torsional oscillations.
Apparatus: The torsion pendulum consisting of the suspension wire and the heavy disc, two
identical masses, stop watch etc.
Theory: A heavy body, say a disc, having a moment of inertia I is suspended by a metallic wire
whose one end is fixed on a rigid support. The body is twisted slightly by applying a torque and
is released. Then the body executes torsional oscillations. This arrangement is called a torsion
pendulum. Using the law of conservation of energy we can show that the torsional oscillations
are simple harmonic and find out the period of oscillations.
The total energy of the system is equal to the sum of the kinetic energy of rotation of the
body and the work done in twisting the wire.
2
1 2 1 2 1  dθ  1
i.e. E = Iω + Cθ = I   + Cθ 2
2 2 2  dt  2
dE r
Since the total energy is conserved = 0
dt
1  dθ   d 2θ  1  dθ 
Thus, 0 = 2I    2  + 2Cθ  
2  dt   dt  2  dt 
 dθ 
Dividing throughout by   we get, l
 dt  l

dθ 2
I 2 + Cθ = 0
dt
d 2θ C
i.e. + θ = 0
dt 2 I
R θ
d 2θ C Fig.b
Or, 2
=  θ
dt I Fig.a
That is, the angular acceleration is proportional to
angular displacement from the equilibrium position and is opposite to it. Hence the oscillations
of the torsion pendulum are simple harmonic. Comparing with the standard equation for a simple
d2 y C
harmonic motion 2 + ω2 y = 0 we get, 2 =
dt I
2π 2π I
Thus the period of oscillation , T= = = 2π
ω C C
I
π nr 4
where, C = is the couple per unit twist of the suspension wire and I the moment of inertia
2l
of the suspended body.
24 MCT Practical I

Determination of rigidity modulus of a wire

Method (1): To determine the rigidity modulus of the material of a wire a torsion pendulum is
arranged. It consists of a heavy disc suspended by a thin uniform wire whose rigidity modulus is
to be determined. Length between the chucks is adjusted to a suitable value as shown in the fig.a.
Now the disc is rotated through a small angle and is released. The period of oscillation T 0 is
determined. The radius of the wire ‘r’, mass of the disc ‘M’ and the radius of the disc ‘R’ are also
determined. The rigidity modulus of the material of the wire is calculated as follows.
I
The period of oscillation T0 = 2π
C
I
Squaring and rearranging we get, C = 4π 2 2
T0
Substituting for couple per unit twist C we get,
π nr 4 I
= 4π 2 2
2l T0
8πI  l 
 Rigidity modulus n =   (1)
r 4  T02 
MR 2
where, I is the moment of inertia of the disc that can be calculated using I = (2)
2
Method (2) - using identical masses: In this method two identical masses (say, cylindrical in
shape) of mass ‘m’ and moment of inertia I0 about its own axis, are placed at equal distances d1
from the suspension wire as shown in fig.c. Let I be the moment of inertia of the disc and I1 be
the moment of inertia of the system. Applying parallel axes theorem, the moment of inertia of the
identical masses about the axis through the suspension wire is 2I0  2md12 . Hence the moment of
inertia of the system,
I1 = I  2I0  2md12 (3)
If the identical masses are placed at distances d2 from the wire, the
moment of inertia of the system is given by,
I2 = I  2I0  2md 22 (4)
I2  I1 = 2m  d 22  d12  (5)

Let T0 be the period of oscillations of the torsion pendulum without the

identical mass and T1 and T2 be the corresponding periods with identical l
masses at distances d1 and d2, respectively. Then,
I
T02 = 4π 2 (6)
C
d1 d1
I
T12 = 4π 2 1 m m
C
CT12
Or, I1 = (7) Fig.c
4π 2
Practical-1 MCT 25

CT22
And, I2 = (8)
4π 2

2  2
T 2  T12 
C
I2  I1 = (9)
4π
From eqns. 5 and 9,
C 2m  d 22  d12 
= (10)
4π 2  T22  T12 
Using eqn.10 in eqn.6, we get,
T02
I = 2m  d 22  d12  (11)
 T22  T12 
I
2
8π 2 m  d 22  d12 
By eqn.6, C = 4π 2 =
T0 T 2
2  T12 

π nr 4 8π 2 m  d 22  d12 
i.e. =
2l  T22  T12 
16πm  d 22  d12   l 
 n =  2 2 
(12)
 T2  T1 
4
r
Procedure: A reference line is drawn on the disc along its diameter. The torsion pendulum is
set for a desired length ‘l’ in between the two chucks, one on the clamp and the other on the disc.
A pointer is arranged close to the disc. This helps to count the oscillations. The disc is twisted
slightly and is released. The pendulum executes torsional oscillations. The time for 20
oscillations is noted. This is done once again and the average time for 20 oscillations is
calculated. From this the average time period T0 is determined.
The two identical masses are now placed (on a diametrical line of the disc) at equal
distance d1 each from the centre of the disc (from the wire) and the time period T 1 of the new
oscillations is determined as above. Then the distance of the identical mass is changed to d 2 and
the corresponding time period T2 is determined.
The entire experiment is repeated for different lengths l. The mass of the identical masses
‘m’ and the mass of the disc ‘M’ are measured with a balance. Radius R of the disc is determined
by measuring its diameter by a scale. The radius ‘r’ of the suspension wire is determined
accurately using a screw gauge.
In method (1), the moment of inertia of the disc is calculated using eqn.2 and the rigidity
modulus of the material of the suspension wire by eqn.1.
In the method (2), the moment of inertia of the disc is calculated by eqn.11 and the rigidity
modulus by eqn.12.
Observation and tabulation
Mass of the disc, M = …….. kg
Mass of identical masses, m = …….. kg
Radius of the disc, R = …….. m
26 MCT Practical I

To determine ‘I’ and ‘n’ (d1 and d2 fixed)

l Distance of the Time for 20 oscillations t  l  T02  l 
T  2  2 2 
m mass ‘m’ from
the axis in m
‘t’ sec 20  T0   T22  T12   T2  T1 
1 2 Mean sec ms2 ms2
Disc alone T0 =
d1 = T1 =
d2 = T2 =
Disc alone T0 =
d1 = T1 =
d2 = T2 =
Disc alone T0 =
d1 = T1 =
d2 = T2 =
Disc alone T0 =
d1 = T1 =
d2 = T2 =
Disc alone T0 =
d1 = T1 =
d2 = T2 =
Disc alone T0 =
d1 = T1 =
d2 = T2 =
Mean
To measure the radius ‘r’ of the wire using screw gauge
Distance moved by the screw tip for 5 rotations of the head = ……… mm
Distance moved by the screw tip
Pitch of the screw, P = = ……… mm
Number of rotations of the head
Number of divisions on the head scale = ………
Pitch
Least count (L C) = = ……. mm
Number of divisions on the head scale
Zero coincidence = …….. ; Zero error = ……. ; Zero correction = ……..
Trial No. PSR Observed Corrected Thickness Mean d
‘x’ mm HSR H S R ‘y’ d  x  y  LC mm mm
1
2
3
4
5
Radius of the wire, r = d/2 = …….. mm
Practical-1 MCT 27

Calculations
Method 1
MR 2
Moment of inertia of the disc, I= = ………….. = ……… kg.m2
2
8πI  l 
Rigidity modulus of the material of the wire, n =   = ………..
r 4  T02 

= ………. Nm2
Method 2
T02
Moment of inertia of the disc, I = 2m  d 22  d12 
 T22  T12 
=
= ………. kg.m2
16πm  d 22  d12   l 
Rigidity modulus, n =  2 2 
=
 T2  T1 
4
r

= ………. Nm2
Result
Moment of inertia of the disc, I = ………. kg.m2
Rigidity modulus of the material of the wire, n = ………. Nm2
28 MCT Practical I

Exp.No.1.7
Rigidity modulus of a material-Static torsion
Aim: To find out the rigidity modulus of the material of a rod using static torsion apparatus.
Apparatus: The static torsion apparatus, mirror strip, scale and telescope arrangement, slotted
weights etc.
Theory: The static torsion A
apparatus consists of a l
heavy metallic frame that M
can be fixed on a table. The B
experimental rod is passed P
through the hole in the R
frame B. One end of the
experimental rod is rigidly
clamped at the frame A.
The other end P of the rod
is held tightly by the
chucks on a metallic wheel
having radius R. One end
of a metal wire is fixed on
a small peg on the wheel.
The wire can be wound
clockwise or anti-clockwise
over the wheel in the groove provided on it. The free end of the wire carries a weight hanger.
When a mass M is suspended on the wheel, the wheel and hence the rod get twisted through an
angle . If C is the couple per unit twist of the rod, we can write,
πnr 4θ
MgR = C =
2l
where, n is the rigidity modulus of the material of the rod, ‘r’ its radius and ‘l’ is the length of the
rod from the fixed end of the rod to the point at which the mirror is fixed. Then,
2MgR  l 
Rigidity modulus, n =   (1)
πr 4  θ 
The angle ‘’ is measured by an indirect method by using a scale and telescope, which employs
the principle that when the mirror turns through an angle , the reflected ray turns through an
angle 2. If ‘s’ be the shift in scale reading for a mass ‘M’,
s
2 = (2)
D
where, D is the distance between the mirror and the scale. Then,
4MgR  lD 
n =   (3)
πr 4  s 
Practical-1 MCT 29

Procedure: The given rod is clamped in the static torsion apparatus. The mirror strip M is
fixed at a distance ‘l’, say 20 cm, from the end A. The weight hanger carrying the dead load is
suspended at the free end of the metal wire wound clockwise on the wheel. The scale and
telescope arrangement is placed at a distance D, say 1 m, from the mirror. The telescope is
adjusted as mentioned in exp.No.5 so that the scale is seen clearly in the telescope. Then the
weight hanger is loaded and unloaded in steps several times so as to bring the rod in elastic
mood. Before starting to take the reading, check that we get the scale readings for the minimum
and maximum weight in the weight hanger.
To start to take reading, the weight hanger is loaded with the minimum weight W0. The
scale reading that coincides with the horizontal wire of the telescope is noted. Then the load is
increased in steps and in each case the coinciding scale reading is noted. After taking the reading
for maximum load, the load is decreased in steps and again the corresponding scale readings are
noted. Now the experiment is repeated after the metal wire is wound over the wheel
anticlockwise. The entire experiment is repeated for different values of ‘l’.
Using a piece of twine wound over the wheel, its circumference can be measured and from
it the radius R can be calculated. The radius ‘r’ of the rod is measured using a screw gauge.
Finally, the rigidity modulus is calculated using eqn.3.
Observation and tabulation
To find the radius of the wheel
Circumference of the wheel, L = ……. cm
L
Radius of the wheel, R= = …….. m
2π
To find the radius of the rod using screw gauge
Distance moved by the screw tip for 5 rotations of the head = ……… mm
Distance moved by the screw tip
Pitch of the screw, P = = ……… mm
Number of rotations of the head
Number of divisions on the head scale = ………
Pitch
Least count (L C) = = ……. mm
Number of divisions on the head scale
Zero coincidence = …….. ; Zero error = …….
Zero correction = ……..
Trial No. PSR Observed Corrected Thickness Mean d
‘x’ mm HSR H S R ‘y’ d  x  y  LC mm mm
1
2
3
4
5
Radius of the rod, r = d/2 = …….. mm
30 MCT Practical I

 lD 
To find out  
 s 
Mass used to increase the load in steps, m = …….. kg.
Mass for which the elevation is calculated, M = 4m = …….. kg.

Telescope reading in cm
Clockwise Anticlockwise
Mean shift

Shift for a mass

Shift for a mass

l D Suspended ‘s’ for the
 lD 
m m load in
Unloading
mass  

unloading
kg. M = 4m  s 
Loading

M = 4m

M = 4m
loading
m
Mean

Mean
metre

W0
W0 + m
W0 + 2m
W0 + 3m
W0 + 4m
W0 + 5m
W0 + 6m
W0 + 7m
W0
W0 + m
W0 + 2m
W0 + 3m
W0 + 4m
W0 + 5m
W0 + 6m
W0 + 7m
W0
W0 + m
W0 + 2m
W0 + 3m
W0 + 4m
W0 + 5m
W0 + 6m
W0 + 7m
Mean ……..
4MgR  lD 
Rigidity modulus of the material of the rod, n =   = ………….
πr 4  s 
= ………. Nm2
Result
Rigidity modulus of the material of the rod, n = ………. Nm2
Practical-1 MCT 31

Exp.No.1.8
Melde’s String- Frequency of a tuning fork
Aim: To determine the frequency of a tuning fork by Melde’s string arrangement set for (a)
transverse mode of vibration and (b) longitudinal mode of vibration.
Apparatus: An electrically maintained tuning fork, sufficient length of string, a light scale
pan, smooth pulley, weight box, common balance, etc.
Theory: When the entire string vibrates with one loop, the corresponding frequency of the
string is called its fundamental frequency. The theory of vibrations of a stretched string shows
that the fundamental frequency of transverse vibrations in a stretched string of length ‘l’ is

1 T
n = (1)
2l m
where, T is the tension on the string and ‘m’ is its linear density (mass per unit length).

Rh

Fig.a: Transverse mode

Rh L

Fig.b: Longitudinal mode

In the longitudinal mode of vibration, the fundamental frequency of vibration is given by,

1 T
n = (2)
l m
32 MCT Practical I

When the stretched string vibrates in unison with the tuning fork, the frequency of the tuning
fork N is same as the fundamental frequency of the stretched string. Thus for transverse
vibrations, if ‘l’ is the length of one loop, the frequency of the tuning fork,

1 T 1 Mg g M
N = n = = =   (3)
2l m 2l m 4m  l 2 
where, M is the sum of the mass of scale pan and the mass placed in it.
And for longitudinal mode of vibration, the frequency of the tuning fork,
1 T 1 Mg g  M 
N = n = = =   (4)
l m l m m  l 2 
where, M is the sum of the mass of scale pan and the mass placed in it.
If L is the length of ‘p’ loops in transverse mode, the length of one loop is given by,
L
l = (5)
p
And, if L is the length of ‘q’ loops in longitudinal mode, the length of one loop is given by,
L
l = (6)
q

Procedure
(a) Transverse mode: The apparatus is arranged and the connections are made as shown in the
fig.a. A suitable weight, say 1 or 2 gm, is placed in the scale pan. By adjusting the screw the
tuning fork is set into vibration. Place one of the two pointers at a well defined node and the
other pointer at another node. Count the number of loops ‘n’ in between the two pointers and
measure the length ‘L’ of the string in between the pointers.
(b) Longitudinal mode: In this case the arrangements are done as shown in the fig.b. Suitable
weights (500 mg or 600 mg) are placed in the scale pan and the tuning fork is set into vibration.
The number of loops and the length of the string in between the two pointers are measured.
 Adjust the total length between the tuning fork and the pulley by moving the tuning fork
back or forth so that the nodes and hence the loops are well defined. This adjustment is
needed since the string in between the two fixed ends (one at the tuning fork and the
other at the pulley) must contain integral number of loops.
 The mass M is the mass of the scale pan plus the mass placed in it.
 Use masses of the order of a few grams in the transverse mode and masses of the order of
milligrams in the case of longitudinal mode.
 Instead of the apparatus shown in the figure we may use an electromagnet with
alternating current and a strip of magnetic material to vibrate with the frequency of the
alternating current used. In this case the length of the strip is to be adjusted to get
oscillations.
Measurement of the mass of the scale pan and the linear density of the string: The mass of
the scale pan is determined by a common balance in the sensibility method. To find the linear
density (mass per unit length) ‘m’ of the string, take 10 metre length of the same string and find
out the mass of it using a common balance in the sensibility method.
Practical-1 MCT 33

Observations and tabulations

To find mass of the scale pan M0 and the linear density ‘m’
Load in the pans of the balance Turning points Resting Sensibility Correct weight
left Right (gm) Left (3) Right (2) point S
0.01 M=W+S(R1R0)
R1  R 2 gm
Nil Nil R0 = …
Scale pan W R1 = …
W + 0.01 R2 = …
Known length W R1 = …
(L1) of string W + 0.01 R2 = …

Mass of the scale pan, M0 = ………. gm = ………. kg

Mass of known length (L1 = …. metre) of the string, M1 = …….. kg
M1
Linear density, m = = ……. kg/metre
L1
To find frequency-Transverse mode
Trial Mass in the Total mass Number of Length of p Length of one M
No. scale pan suspended loops ‘p’ loops in cm loop l in cm 2
kg.m 2
l
x gm M = (M0+x) gm
1
2
3
4
5
Mean …………
g M
Frequency of the tuning fork, N =   = ……….. = ……… Hz
4m  l 2 
To find frequency-Longitudinal mode
Trial Mass in the Total mass Number of Length of q Length of one M
loops ‘q’ loops in cm loop l in cm kg.m 2
No. scale pan suspended l 2

x gm M = (M0+x) gm
1
2
3
4
5
Mean …………

g  M 
Frequency of the tuning fork, N =   = ……….. = ……… Hz
m  l 2 

Result
Frequency of the tuning fork, N = …….. Hz
34 MCT Practical I

Exp.No.1.9
Lee’s disc- Thermal conductivity of a bad conductor
Aim: To determine the thermal conductivity of a bad conductor by Lee’s disc method.
Apparatus: The Lee’s disc apparatus, bad conductor in the form of disc, two thermometers,
steam boiler, etc.

Steam
Thermometers
1 2+5 to
2
2  5
Brass disc Brass disc

Fig.a Fig.b

The Lee’s disc apparatus consists of a circular brass disc of about 8 to 12 cm diameter and
thickness about 1 to 2 cm. It is suspended on a stand as shown in the fig.a. A steam chamber of
the same diameter is used to heat the disc. The bad conductor whose thermal conductivity is to
be determined is taken in the form of a disc of the same diameter of the brass disc. It is kept in
between the brass disc and the steam chamber. There are holes provided on the steam chamber
and the disc to insert the thermometers.
Theory: The quantity of heat conducted per second through a conductor is proportional to the
area through which heat conducts and the temperature gradient. That is,

 θ θ 
Q  A 1 2 
 d 
 θ θ 
= λA  1 2  (1)
 d 
where,  is a constant for a particular material. The constant  is called the thermal conductivity
of the material. 1 and 2 are the temperatures on both sides of the bad conducting disc and ‘d’ is
its thickness. These temperatures, respectively, are the temperature of the steam chamber and the
brass disc near the bad conductor. At the steady state condition, the quantity of heat conducted
through the experimental disc is completely radiated from the brass disc.
The quantity of heat radiated by the brass disc is calculated as follows. The brass disc alone
is heated (after removing the bad conducting disc) to a temperature greater than the steady state
 dθ 
temperature 2 and is allowed to cool by radiation. Let   is the rate of cooling of the Lees
 dt θ2
disc at a temperature 2. Then the rate of loss of heat by the brass disc is proportional to the area
Practical-1 MCT 35

of its exposed region. When the disc is completely exposed for radiation we can write, the rate of
loss of heat per unit area
 dθ 
Mc  
Qtotal  dt θ2
=
Total area of the Lee's disc 2πr  2πrh
2

where, M is the mass, c is the specific heat capacity, r is the radius and h is the height of the
brass disc. At the steady state condition during the experiment, the exposed area of the brass disc
does not contain the upper face. Therefore, the rate of loss of heat from the disc during the
experiment is given by,
Q = Rate of loss of heat through unit areaexposed area of the disc
 dθ 
Mc  
 dt θ2  dθ   r  2h 
=   πr 2  2πrh  = Mc     (2)
2πr  2πrh
2
 dt θ2  2r  2h 
At the steady state condition, since the quantity of heat conducted through the experimental disc
is completely radiated from the brass disc, Q = Q. Thus from eqns.1 and 2,
 θ θ   dθ   r  2h 
λA  1 2  = Mc    
 d   dt θ2  2r  2h 
Mc  d   dθ   r  2h 
  =     
A  θ1  θ 2   dt θ2  2r  2h 
Mc  d   dθ   r  2h 
=      (3)
πr 2  θ1  θ 2   dt θ2  2r  2h 
 dθ 
  is the rate of cooling of the brass disc
 dt θ2
Temperature 

at the temperature 2. This can be determined

by finding the slope of the time-temperature
d
graph at the temperature 2. 2
Procedure: The diameter and the thickness
of the brass disc are measured by a vernier dt
calipers. Its mass M is measured by a balance.
The thickness ‘d’ of the experimental disc is
determined by a screw gauge.
The experimental arrangements are set
up as shown in the fig.a. The brass disc is Time t in seconds
Fig.c
suspended by a heavy retort stand. The
experimental disc and the steam chamber are placed on it. Thermometers are inserted in the holes
provided for that. Steam from a boiler is allowed to pass through the steam chamber till the two
thermometers show steady temperatures. Note the steady temperatures 1 of the steam chamber
and 2 of the brass disc. Then the experimental disc is removed and the steam chamber is kept in
contact with the brass disc. When the temperature of the brass disc is raised by about 8 or 10
degree, the steam chamber is removed and the brass disc is allowed to cool (as shown in fig.b).
36 MCT Practical I

When the temperature of the brass disc reaches 2+5, a stop watch is started. The time is noted at
regular intervals of temperature, say 0.5C (or 0.2C) till the temperature falls to 25. A graph
is plotted with time along the X axis and the temperature along the Y axis as shown in fig.c. To
 dθ 
find out   , draw a line parallel to the X axis at the temperature 2. At the point of
 dt θ2
intersection of this line with the curve, draw the tangent of the curve. Now construct a triangle as
shown in fig.c and the slope of the curve at 2 is determined.
 Do not stop the stop watch while taking the temperature-time observation. Count the time
in minutes and seconds.
 It should be remembered that the slope of the curve is determined at the steady state
temperature 2 of the brass disc.
Observation and tabulation
To find the thickness ‘d’ of the experimental disc using screw gauge
Distance moved by the screw tip for 5 rotations of the head = ……… mm
Distance moved by the screw tip
Pitch of the screw, P = = ……… mm
Number of rotations of the head
Number of divisions on the head scale = ………
Pitch
Least count (L C) = = ……. mm
Number of divisions on the head scale
Zero coincidence = …….. ; Zero error = ……. ; Zero correction = ……..
Trial No. PSR Observed Corrected Thickness Mean d
‘x’ mm HSR H S R ‘y’ d  x  y  LC mm mm
1
2
3
4
5
To find the radius ‘r’ and thickness ‘h’ of the brass disc using vernier calipers
Value of one main scale reading of the vernier calipers (1 m s d) = …….. cm
Number of divisions on the vernier scale n = ……..
1msd
Least count (LC) of the vernier calipers = = …… cm
n
Radius ‘r’
Trial M S R V S R D = M S R + V S RL C Mean diameter Mean radius
No. cm cm ‘D’ cm r = D/2 cm
Practical-1 MCT 37

Thickness ‘h’
Trial No. MSR VSR h = M S R + V S RL C Mean thickness ‘h’
cm cm cm

Mass of the brass disc, M = ………. kg

Specific heat capacity, c = ……….. J/kg.K
Steady temperature of the steam chamber, 1 = ………C
Steady temperature of the brass disc, 2 = ………C
 dθ 
To find  
 dt θ2
Temperature
Time
Temperature
Time
Temperature
Time

Calculation
 dθ 
Rate of cooling of brass disc at the temperature 2, =   = ……… K/s
 dt θ2

Mc  d   dθ   r  2h 
Thermal conductivity,  =     
πr 2  θ1  θ 2   dt θ2  2r  2h 

= ……………………………….
= ………. Wm1K1
Result
Thermal conductivity of the given ………… disc = ………. Wm1K1

Physical constants and data

Specific heat capacity of brass = 370 J/(kg.K)
38 MCT Practical I

Exp.No.1.10
Newton’s law of cooling- Specific heat of a liquid
Aim: To determine the specific heat capacity of a liquid by the method using Newton’s law of
cooling.
Apparatus: A spherical calorimeter, a thermometer, stop clock, given liquid, water, etc.
Theory: Newton’s law of cooling states that the rate of cooling of a body is proportional to the
mean difference of temperature between the body and
its surroundings. If 1 is the initial temperature of the
body, 2 is the temperature after a time ‘t’ seconds and
0 be the temperature of the surroundings we can write,
θ1  θ 2 θ θ
Rate of cooling,  1 2  θ0 (1)
t 2
Since the rate of cooling of the body is proportional to
its rate of loss of heat,
Mc  θ1  θ 2  θ θ 
= K  1 2  θ0  (2)
t  2 
where, M is the mass of the body, ‘c’ is its specific heat
capacity and K is a constant.
Let the calorimeter is first filled with hot water. If
‘tw’ is the time taken by the calorimeter and water to
cool from 1 to 2, Fig.a

mc cc  θ1  θ 2   m w c w  θ1  θ 2  θ θ 
= K  1 2  θ0  (3)
tw  2 
where, mc is the mass of calorimeter, cc is the specific heat capacity of the calorimeter, mw mass
of water and cw is specific heat capacity of water.
If the calorimeter is filled with the given hot liquid and is allowed to cool from the same
range of temperature and ‘tl’ be the corresponding time taken, we can write,
mc cc  θ1  θ 2   ml cl  θ1  θ 2  θ θ 
= K  1 2  θ0  (4)
tl  2 
where, ml is mass of liquid and cl is its specific heat capacity. From eqns.3 and 4,
mc cc  θ1  θ 2   ml cl  θ1  θ 2  mc cc  θ1  θ 2   m w c w  θ1  θ 2 
=
tl tw
 tl 
mccc  m w c w     mccc
 cl =  tw  (5)
ml
Practical-1 MCT 39

tl
Usually is determined by plotting
tw

Temperature
the cooling curves for water filled
calorimeter and liquid filled calorimeter
as shown in the fig.b.
Procedure: The mass mc of a clean
dry spherical calorimeter is determined Water
by a common balance. It is then almost
filled with hot water of temperature
nearly 90C. It is then suspended in air Liquid
as shown in fig.a. A sensitive
thermometer is inserted in the tl tw
calorimeter. When the temperature falls
to 80C, start a stop watch and the time
temperature observations are made at Fig.b Time
regular intervals of temperature or time.
(The time may be noted at a regular fall of temperature of 1C till the temperature falls to about
60C or the temperature may be noted at a regular interval of half a minute till the temperature
falls to 60C. The first method is advised since the time measurement is more sensitive than the
temperature measurement). Let the calorimeter is cooled to room temperature. Then the mass of
the calorimeter and water is determined. Let it be m2.
The water is poured out and the calorimeter is dried. It is then filled with the hot liquid and
the time-temperature observations are made for the same temperature range (80C to 60C) as in
the case of water. The calorimeter is again cooled to room temperature and the mass of
calorimeter plus liquid, m3, is determined.
The time-temperature observations are plotted on the same graph paper as shown in fig.b.
t
Find out l for different temperature ranges and its average is calculated. Finally, the specific
tw
heat capacity of the given liquid is calculated using eqn.5.
Observation and tabulation
Temperature 80 79 78 77 76 75 74 73 72 71 70
Time in minutes water
and seconds Liquid
Temperature 69 68 67 66 65 64 63 62 61 60
Time in minutes water
and seconds Liquid

Temperature 80 79 78 77 76 75 74 73 72 71 70
Time in seconds water
Liquid
Temperature 69 68 67 66 65 64 63 62 61 60
Time in seconds water
Liquid
40 MCT Practical I

tl
To find
tw
Sl.No Range of temperature Time of cooling in seconds tl
Water Liquid tw

Mean ………

To find masses of calorimeter, water and liquid

Load in the pans of the balance Turning points Resting Sensibility Correct weight
left Right Left (3) Right (2) point S
0.01 m = W+S(RR0)
W (gm) R 0  R1 gm
Nil Nil R0 = …
Nil 0.01 R1 = …
Empty calorimeter R=… m1 = …….
Calorimeter + water R=… m2 = …….
Calorimeter + liquid R=… m3 = …….

Mass of calorimeter, mc = m1 = …….. gm = ………. kg.

Mass of water, mw = m2m1 = …….. gm = ………. kg.
Mass of liquid, ml = m3m1 = …….. gm = ………. kg.
Specific heat capacity of (copper) calorimeter, cc = ………. Jkg1K1
Specific heat capacity of water, cw = ………. Jkg1K1

 tl 
mccc  m w c w     mccc
Specific heat capacity of liquid, cl =  tw 
ml
= ………………………..
= ……… Jkg1K1
Result
Specific heat capacity of liquid, cl = …….. Jkg1K1
*Standard data
Specific heat capacity of water = 4190 Jkg1K1
Specific heat capacity of copper = 385 Jkg1K1
Practical-1 MCT 41

Exp.No.1.11
Spectrometer - Refractive index of the material of a prism
Aim: To determine the refractive index of the material of the prism by finding out the angle of
minimum deviation.
Apparatus: Spectrometer, sodium vapor lamp, prism, reading lens etc.
Theory: For a given prism, corresponding to a
given angle of deviation there are two possible A
angles of incidence i1 and i2. These two angles
are such that if one of the angles is the angle of d
incidence, the other angle will be the angle of
i1 r1 i2
emergence. r2
Let i1 and i2 be the two angles of incidence
and r1 and r2 be the corresponding angles of
refraction for the given angle of deviation d. Fig:a
Then,
i1  i 2 = A + d (1)
r1  r2 = A (2)
d
Fig.b gives the variation of angle
of deviation d with angle of
incidence i. When the angle of
deviation is minimum, i1 = i2 = i,
r1 = r2 = r and d = D. Then, from
eqn.1 we get,
2i = A + D (3) D
AD i
i = (4) i1 i1=i2 i2
2
From eqn.2, Fig.b : i-d curve for an equilateral prism of  = 1.62
A
r = (5)
2
AD
sin  
Refractive index of the material of the prism,  =
sin i
=  2 
(6)
sin r A
sin  
2
Procedure: The following preliminary adjustments of the spectrometer are to be made.
1. Turn the telescope to the white wall. Hold the telescope with left hand firmly. By looking
through the eye piece, it alone is pushed in or pulled out with right hand till the cross wire
is seen clearly.
2. The telescope is then turned towards the distant object and the rack and pinion
arrangement is adjusted till the image of the distant object is formed clearly on the cross
wire.
42 MCT Practical I

3. The telescope is brought in a line with the collimator and sees the image of the slit. If
there is no image, check whether the slit is opened.
4. Looking through the telescope, the rack and pinion arrangement of the collimator is
adjusted till the image of the slit is seen clearly on the cross wire. (Usually the image is
blurred and spread. Focus the collimator till the image is not blurred and its width is
minimum).
5. Now adjust the width of the slit, if needed, to a minimum by rotating the slit width
adjusting screw.
Rack and pinion of collimator
Collimator
Slit width adjusting screw
Rack and pinion of telescope
Slit

Telescope

Prism table
Leveling screw of prism table
Ver I

Eye piece Vernier table

Ver II

Leveling screw Main screw of telescope

Main screw of vernier table
Fig:c Leveling screw

Tangential screw of vernier table

Tangential screw of telescope

6. The prism table is leveled either by observing the reflected images from both the sides of
the prism or by using a spirit level. In the former method, the prism is mounted on the
prism table with its base is parallel and
Collimator

close to the clamp. The prism table (or

vernier table) is rotated till the refracting
edge of the prism is towards the
collimator. Turn the telescope and the
reflected image from one of the faces of
the prism is observed. The two leveling A
screws on that side of the prism table are
adjusted so that the image is bisected by
the horizontal cross-wire. Now the
telescope is turned to the other side of the
prism and the reflected image from the B C
other face is viewed through the
telescope. Then the leveling screw on that
side of the prism table is adjusted till the
image is bisected by the horizontal wire. Reflected ray 2A Reflected ray
Fig:d
This process may be repeated once again.
Practical-1 MCT 43

To determine the angle of the prism A: After doing the preliminary adjustments, the telescope
is turned towards one side of the prism and the reflected image from that face is viewed through
the telescope (refer fig.d). The vernier table and the telescope are clamped by tightening their
main screws. The tangential screw of the telescope is adjusted till the reflected ray coincides with
the vertical cross-wire. The readings on both the verniers are noted. Now release the telescope
and is turned to the other side of the prism till the reflected image from that face is obtained in
the telescope. Clamp the telescope there. By adjusting its tangential screw, the reflected image is
made to coincide with the vertical cross-wire. The readings on both the verniers are again noted.
The difference between the corresponding vernier readings gives the angle of the prism.
To determine the angle of minimum deviation: The vernier table is released and is rotated
such that one of the refracting faces is towards the

Collimator
collimator (See fig.e and also fig.c in the next
experiment). Looking through the other face with one
eye (other eye closed) the vernier table is rotated till the

Incident ray
refracted image is seen. Find approximately the position
at which the image turns back. Now bring the telescope
in the line of the refracted ray and view the refracted
image. Looking through the telescope the vernier table is
slightly turned to and fro and finds the exact position at
B

Base of the prism

which the refracted image just turns back. Now the
prism is set for its minimum deviation position. The
vernier table and the telescope are clamped at this
position by tightening their main screws. Now adjust the A
tangential screw of the telescope so that the refracted
image coincides with the vertical cross wire. The C
readings on both the verniers corresponding to this
position are taken. The prism is now removed and the
telescope is brought in the line of the direct ray. After
Direct ray

D
clamping the telescope, its tangential screw is adjusted Refracted ray
such that the direct image coincides with the vertical
wire. The readings on both the verniers are again taken. Fig:e
The difference between the minimum deviation position
reading and the direct reading gives the angle of minimum deviation.
 If the prism table is not properly leveled one may not get the image in the field of view of
the telescope. In such a case the reflected images are seen directly with naked eye without
using telescope and the approximate leveling is to be done.
 The vernier table and the prism table are initially adjusted at the proper positions (both
the verniers are in a line perpendicular to collimator) so that the readings on both the
verniers are conveniently taken.
 Reading lens must be used to observe the vernier readings.
 Don’t forget to clamp the vernier table and the telescope after each adjustment. For
taking the direct reading, the prism must be removed carefully without any change in the
vernier table.
 The removal of the prism to take the direct reading can be avoided if initially the prism
table is adjusted to a height such that upper half the light from the slit passes above the
44 MCT Practical I

prism and the lower half of light passes through the prism so that we can simultaneously
see the direct image and the image formed by the prism.
Observation and Tabulation of data:
Value of one main scale division (1 m s d) = ……………
Number of divisions on the vernier n = ……………
Value of 1 m s d
Least count (L C) = = ……………
n
[One degree = 60 minute, (1 = 60 )]
To determine the angle of the prism A
Reading of the Ver I Ver II Mean
MSR VSR Total MSR VSR Total 2A A
Reflected image from first face
‘a’
Reflected image from second
face ‘b’
Difference between the above readings 2A = ab 2A = ab

To determine the angle of minimum deviation D

Reading of the Ver I Ver II Mean
MSR VSR Total MSR VSR Total D
Refracted image corresponding
to minimum deviation ‘x’
Direct image ‘y’
Difference between the above readings D = xy D = xy

Calculation
AD
sin  
Refractive index of the material of the prism,  =  2  = ……….
A
sin  
2
= ……….
Result
Angle of the prism A = ………….
Refractive index of the material of the prism,  = ………….
Standard data*
Refractive index against air for mean sodium line (589.3 nm)
Crown glass 1.48  1.61
Flint glass 1.53  1.96
Practical-1 MCT 45

Exp.No.1.12
Spectrometer-Dispersive power of a prism
Aim: To determine the dispersive power of the material of the prism for different pairs of
spectral lines.
Apparatus: Spectrometer, mercury vapor lamp, prism, reading lens etc.
Theory: For a given prism, corresponding to a
given angle of deviation there are two possible A
angles of incidence i1 and i2. These two angles
are such that if one of the angles is the angle of
incidence, the other angle will be the angle of d
emergence. i 1
r1 i2
r2
Let i1 and i2 be the two angles of incidence
and r1 and r2 be the corresponding angles of
refraction for the given angle of deviation d. Fig.a
Then,
i1  i 2 = A + d (1)
r1  r2 = A (2)
Fig.b gives the variation of angle of d
deviation d with angle of incidence
i. When the angle of deviation is
minimum, i1 = i2 = i, r1 = r2 = r and
d = D. Then, from eqn.1 we get,
2i = A + D (3)
A  D D
i = (4) i
2 i1 i1=i2 i2
From eqn.2,
A
r = (5) Fig.b : i-d curve for an equilateral prism of  = 1.62
2
Refractive index of the material of the prism,
 A  D 
sin  
 =
sin i
=  2  (6)
sin r  A 
sin  
 2 
The refractive index  of a material depends on the wavelength of the light. Hence it is a
dμ
function of wavelength . The dispersive power of the material is defined as . By Cauchy’s
dλ
relation,
B
Refractive index,  = A  2 (7)
λ
where, A and B are constants.
46 MCT Practical I

dμ 2B
Dispersive power,  = = 3 (8)
dλ λ
Since the dispersive power varies inversely with cube of the wavelength, it is different at
different wavelength region. If 1 and 2 are the refractive indices for the wavelengths 1 and 2,
it can be shown that the dispersive power of the material in that wavelength range as,
μ 2  μ1 μ  μ1
12 = , where, μ  2 (9)
μ 1 2
Procedure: All the procedures are same as that for the previous experiment. Instead of a
sodium lamp we use a mercury lamp in this case. The refracted spectrum consists of a number of
spectral lines of different colours (wavelengths). The prism is adjusted to be in the minimum
deviation position for each line and the corresponding angle of minimum deviation is determined
as described in the previous experiment. Using eqn.9 the dispersive powers for different pairs of
wavelengths (refractive indices) are calculated.

Fig.c

Observation and tabulation

Value of one main scale division (1 m s d) = ……………
Number of divisions on the vernier n = ……………
Value of 1 m s d
Least count (L C) = = ……………
n
[One degree = 60 minute, (1 = 60 )]
Angle of the prism A
Readings of the Ver I Ver II Mean
MSR VSR Total MSR VSR Total 2 A A
Reflected image from first face
‘a’
Reflected image from second
face ‘b’
Difference between the above readings 2 A = ab 2 A = ab
Practical-1 MCT 47

Determination of refractive indices for various colours

Reading corresponding to the Reading corresponding to the
minimum deviation position direct ray Angle of
of the refracted rays ‘y’ minimum
Colours ‘x’ deviation
of Ver 1 Ver II Ver 1 Ver II D = xy 
spectral
lines
MSR

MSR

MSR

MSR
VSR

VSR

VSR

VSR

Ver II
Mean
Total

Total

Total

Total

Ver I
Calculation
(1) Dispersive power of ………… and ……….. colours.
Refractive index of ………… colour, 1 = ………
Refractive index of ………… colour, 2 = ………
Dispersive power, … … = ………. = ………
(2) Dispersive power of ………… and ……….. colours.
Refractive index of ………… colour, 1 = ………
Refractive index of ………… colour, 2 = ………
Dispersive power, … … = ………. = ………
(3) Dispersive power of ………… and ……….. colours.
Refractive index of ………… colour, 1 = ………
Refractive index of ………… colour, 2 = ………
Dispersive power, … … = ………. = ………
Result
Dispersive power of ………… and ……….. colours = …….
Dispersive power of ………… and ……….. colours = …….
Dispersive power of ………… and ……….. colours = …….