completed physics manual

Physics Laboratory Manual

Fig.1 Vernier Caliper

(a). Zero Error

(b)Positive error (c) Negative error

Adithya Institute of Technology

Physics Laboratory ManualExpt. No.Date

1. VERNIER CALIPER

Aim:

To find the breadth of a given object using vernier caliper.

Apparatus Required:

Vernier caliper and the given object.

Procedure,

A vernier attached to a movable jaw sliding on a metal scale with a fixed jaw at the

zero ends is called the vernier caliper. The upper fixed jaw and movable jaw are attached

so as the distance between their outer ends equals the distance between the inner ends of

the lower jaws. There is a stud to fix the vernier. The pointed strip at the farther end is

intended for depth measure.

(i) Determination of Least Count

Least count of a measuring instrument is the smallest length that can be

accurately measured with it. In the vernier caliper it is the difference between one main

scale division (MSD) and one vernier division (VSD). After ascertaining the value of

one main scale division, the zero of the vernier is fixed to a definite main scale division

and the distance between the first and the last vernier scale division (VSD) is found in

terms of the main scale divisions. The value of the vernier division in terms of MSD is

calculated. Then the least count is determined by the following formula:



To find the least count:

Least count = 1 MSD – 1 VSD

10 VSD coincides with 9 MSD

Therefore,

10 VSD = 9 MS

Observation

Least Count = 0.01 x10-2 m Zero Error = div

Zero Correction = x 10-2m

S.N

O

MSR

X 10-2m

VSC

Div

VSR=VSC X

L.C

X 10-2m

Observed reading

OR= MSR + VSR

X 10-2m

Correct Reading

CR=OR ± ZC)

X 10-2m

1

2

3

4

5

Mean value of breath



1 VSD = 9/10 MSD

1 MSD = 1 mm

Therefore,

LC = 1 MSD – 1 VSD

= 1 mm – 9/10 mm

= 1/10 mm

= 0.1mm or .01 cm

Usually the readings are noted in cms only.

(ii) Determination of Zero error

(a) Zero error:

When the fixed and movable jaws are brought into contact, if the zero of the

vernier coincides exactly with the zero of the main scale, then there is no zero error.

(b) Positive error:

When the two jaws are in contact, if the zero of the vernier lies right to the zero

of the main scale (i.e. in the positive direction of measurement) the error is positive. The

coinciding division of the vernier with any MSD is implied by the least count to get the

value of the error.

(1) If the yth division of vernier coincides with any one of the main scale division ,

then

Zero error = +y × LC

Zero correction = -y × LC

For example, if the fifth division of the vernier scale exactly coincides with any one of the

main scale division,



CALCULATION



If the zero of the vernier lies left to the zero of the main scale (i.e. in the

negative direction of measurement) the error is negative. The coinciding division of the

vernier with any MSD is mulplied by the least count to get the val;ue of the error.

(2) If the yth division of the vernier coincides with any one main scale division,

then , Zero error = -y × LC

Zero correction = +y × LC

For example, if the third division of the vernier scale exactly coincides with any

one of the main scale division, then Zero error = +5 × LC

Zero correction = -5 × LC

= +3 ×0.01cm

= +0.03 cm

(iii) Determination of the breadth of the given object

The given object is held between the two lower jaws of vernier caliper as in fig 1.

The main scale reading which is just preceding or exactly coinciding with the zero of vernier

scale is noted as the MSR. The vernier scale division that coincides exactly with any of the

main scale division is noted as VSD. Then, the breadth of the given object is determined by

the given formula.

Breadth of the object = [MSR + VSR]

Where VSR = VSC × LC ± ZC

The object is held in different positions and the experiment is repeated.

Result: The breadth of the given object = ……………. × 10-2m.



Fig.1. Screw Gauge


Physics Laboratory ManualExpt. No:

Date:

2. SCREW GAUGE

Aim:

To find the thickness of the given object using screw gauge.

Apparatus Required:

Screw gauge and the given object.

Procedure:

A uniformly threaded screw works in a hollow cylindrical nut, whose left end is

attached to a U-shaped frame. A scale with its division corresponding to the screw

threads engraved on the cylindrical nut is called pitch scale. A plane stud is opposite to

the plane tip of the screw. Its head is mild and to avoid over screwing a ratchet

arrangement is fixed. The beveled edge of the metal sleeve attached to the head of the

screw is divided into a definite number of divisions known as head scale.

(i) Determination of least count :

Least count is the smallest length that can be accurately measured by the

instrument. The least count is calculated by dividing the pitch by the number of divisions

on the head scale,

One Pitch scale division or Pitch

Total number of divisions on the head scale

Distance moved by the head scale on the pitch scale

No of rotations given to the head scale

One Pitch scale division = 1x10-3m

Then, Least count = 1/100 x10-2m , usually here the readings are noted in mm only.


Least Count (LC) =

Pitch =


Observation

Least Count = 0.01x10-3 m Zero Error = div

Zero Correction = x10-3m

S.N

O

PSR

X 10-3m

HSC

Div

HSR=HSC X LC

X 10-3m

Observed

reading

OR= PSR +

HSR

X 10-3m

Correct

Reading

CR=OR ± ZC

X 10-3m

1

2

3

4

5


Mean


(ii) Determination of Error:

(a) Zero error:

When the top of the screw is made to contact the fixed stud, if the zero of the

head scale coincides exactly with the zero of the pitch scale on the reference line, then there

is no zero error.

(b) Positive error

If the head scale zero has not crossed the pitch scale reference line, the error is

positive. If yth division on head scale coincides with pitch scale reference line, the error is +y

divisions and zero correction is (-y ×LC).

For example, as in fig., when 5th division on head scale coincides with pitch scale

reference line, the zero error is + 5 divisions and the zero correction is -5 × LC

i.e. Zero correction = -5 × 0.01x10-3m

= -0.05x10-3m

(c) Negative error:

If the head scale zero has crossed the pitch scale reference line, error is

negative. If yth division head scale division coincides with pitch scale reference line, the

error is – (100-y) division and zero correction is + (100-y) × LC.

For example, as in fig.b, when 95th division on head scale coincides with pitch scale

reference line, the zero error is –(100-95) divisions. i.e. -5 divisions and the zero correction

is +5 × LC



i.e. Zero correction = +5 × 0.01x10-3m

= +0.05x10-3m

(iii) Determination of the thickness of a wire

The given wire is gently gripped between a fixed stud and the screw tip. The

pitch scale division just in front of the head scale and the coinciding division of the head

scale with the reference line are noted.

Head scale reading (HSR) = HSC×LC

Observed reading (OR) = PSR + HSR

Correct Reading = OR ± ZC

Calculation,



Result

Thickness of a given wire = ………………….. ×10-3m.



Fig.1.Overview of Travelling Microscope


1 VSD = cm


Date:

3. TRAVELLING MICROSCOPE

Aim

To find the radius of the given capillary tube.

Apparatus Required

Travelling microscope, a reading lens, capillary tube and a stand.

Procedure

The Travelling microscope consists of a compound microscope sliding along a

graduated vertical pillar. The vertical pillar is fixed to a horizontal base resting on screws.

Two screws at the base of the microscope are used for its leveling. The main scale

divisions are marked on the base and on the pillar. Vernier scale is attached to the base of

the pillar.

(i) Determination of Least count

The value of one main scale division (MSD) is 0.05 cm. The vernier scale is

divided into 50 divisions which is equivalent to 49 main scale divisions.

L.C = 1 MSD - 1 VSD

20 MSD = 1 cm

1 MSD = 1/20 cm = 0.05 x10-2m


Here 50 VSD = 49 MSD

1 49 49 1

50 50 20

1 VSD =0.049 x10-2m


MSD or


Observation

Least Count = 0.001x10-2m

Cross wire PositionMicroscope Readings Diameter

A ~ B

X 10-2m

MSR

X 10-2m

VSC

Div

TR= MSR+ (VSCx

LC) X 10-2mVertical crosswire

Left (A)

Right (B)

Horizontal

crosswire

Bottom (B)

Mean

Mean radius of the capillary tube = ………………. ×10-2 m

LC = 1 MSR - 1 VSR

= 0.05-0.049

LC = 0.001 x10-2m



(ii) Determination of radius of the Capillary tube

First the capillary tube is placed on the separate stand and the Travelling

microscope is adjusted to view the capillary tube. The microscope is focused in the inner

bore of capillary tube. The vertical cross wire is made to coincide with any one end(say left)

and by moving the head slightly while viewing, if the cross wire shifts with respect to the

focused point, then by pulling the eye piece slightly out or pushing in, adjustment is made to

avoid parallax error. Main scale and vernier scale readings (R1) are taken (a reading lens is

used to observe the vernier coincidence, carefully). Now moving the microscope in

horizontal direction, vertical cross wire is made to coincide with other end (say right) of the

inner bore of capillary tube. Main scale and vernier scale readings (R2) are taken. The

difference between these two readings gives the diameter and hence the radius of the inner

bore of the capillary tube. Then the microscope is moved vertically so that the horizontal

cross wore is made to coincide with the top end of inner bore. Corresponding main scale and

vernier scale readings (R3) are taken. Now moving the microscope in vertical direction,

horizontal cross wire is made to coincide with other end (say bottom) of the inner bore of

capillary tube. Main scale and vernier scale readings (R4) are taken. The difference between

these two readings gives the diameter and hence the radius of the inner bore of the capillary

tube.

Result:

Radius of the capillary tube = …………………… ×10-2 m



Fig.1. Spectrometer

Expt. No:



Date:

4. SPECTROMETER

Aim:

To determine the difference between vernier A and vernier B in a spectrometer.

Apparatus Required :

Spectrometer, a reading lens and a spirit level.

Description of the Instrument:

Spectrometer consists of collimator, telescope, vernier table and prism table.

Collimator consists of a brass tube with collimating lenses at one end and a vertical slit of

adjustable width at the other end. To obtain the parallel beam of light, the distance between

the slit and the lens is adjusted with the help of a side screw attached to the collimator. The

telescope is an ordinary refracting telescope with an objective lens near the collimator and an

eye piece at the other end. The telescope is fitted on one arm of the spectrometer and can be

rotated about the central axis. The prism table is made up of two circular platforms one

above the other connected together by three leveling screws. The vernier table has two

verniers, VA and VB each having main scale and vernier scale. Here one main scale division

is equal to half a degree. Each vernier scale has 30 divisions, which is equal to 29 main scale

divisions.

Preliminary Adjustments to be followed before taking measurements:(i) Eyepiece

The telescope is turned towards and focused at the white wall the the eyepiece is moved

inwards or outwards until the cross wires are clearly seen.

(ii) Telescope

The telescope is turned towards a distant object where it is adjusted for parallel rays.

The distance between the eyepiece and the objective is adjusted till a clear imagethe crowire.



Observation

LC = 1' Total reading, TR = MSR + (VSC ×LC) degrees

S. No

Position

of

telescope

Vernier A Vernier B VA ~

VB

degrees

MSR

Degrees

VSC

Div

TR

Degrees

MSR

Degrees

VSC

Div

TR

Degrees

1. Left

2. Right



(iii) Slit

The slit is made narrow with the help of the screw, provided aside of the

slit. This controls the amount of light passing through the collimator.

(iv) Collimator

The telescope is brought in line with the collimator. The slit is

illuminated by a source of light. If the image of the slit appears blurred,

then the screw of the collimator is adjusted until a clear image is seen

when viewed through the telescope. Now the rays of light emerging from

the collimator will be rendered parallel.

(v) Prism table

By releasing the clamping screw, the prism table is raised so that the table

top is in line with the base of the collimator. The prism table is made

horizontal by using the center spirit level. The spirit level is placed

perpendicular to the line joining the two screws and the bubble brought to

the center by adjusting the third screw. By keeping the spirit level at

different places the prism table is made horizontal.


1VSD =

Physics Laboratory ManualDetermination of Least count

The main scale is graduated in half degree. There are 30 divisions in the vernier.

Least count = 1MSD - 1VSD.

1 MSR = ½ degree = 30'

Total no of divisions on vernier scale = 30


30 VSD = 29 MSD

2930

1VSD = 29/30 × 30' = 29'

LC = 1MSD - 1VSD

= 30'- 29'

Least count (L.C) =

Measurement

Before doing any experiment using spectrometer, the above mentioned initial

adjustments have to be made. While performing the experiment, the main scale and vernier

scale readings are noted from both the verniers in VA and VB in degrees and minutes.

Therefore,

Total reading = MSR + (VSC ×LC) degrees


MSD


Calculation,

Result

The readings are noted and the difference between vernier A and vernier B is found to

be …………………….. Degrees.



Fig.1. Experimental set up for Particle size determination


2 1 23 34 4

Sin θm

mN

dm

D

Physics Laboratory Manual 1, 2, 3, 4 order of spectrum

Expt. No:

Date:

1. LASER EXPERIMENTS

Aim:

To determine

(i) the particle size of the lycopodium powder using semiconductor laser

(ii) the wavelength of the given laser source using grating

Apparatus Required:

Semiconductor laser, lycopodium powder coated glass slide, Slide holder Screen,

scale, grating,

Formula,

(i) The size of the particle D = 1.22mλ/Sinθ m

θ = tan-1(r/d)

where

D - Size of the particle (× 10-2m)

λ - Wavelength of the laser source (× 10-10m)

d - Distance between the particle slide and the screen. (× 10-2m)

m - Order of diffraction (No unit)

r - Radius of the nth order ring (× 10-2m)

θ - Angle of diffraction ( degrees)

(ii) Wavelength of the laser source, λ = m

where θm = tan-1 degrees


Physics Laboratory ManualObservation

(i) To determine size of the particle

Distance

between screen

and glass plate,

d,X 10-2m

Order of

diffraction

(m) (No

unit)

Radius of

mth order

ring, r

X 10-2m

Angle of

diffraction

θ=[tan -1(r / d)]

degrees

Particle size

D = 1.22mλ/Sinθ

X 10-6m

1.

2.

3.

4.

5.

Mean



Fig.2. Experimental set up for laser- wavelength determination

λ - Wavelength of the laser source (× 10-10m)

N - Number of lines per meter in the grating (lines/m)


θ - Angle of diffraction (degrees)

dm – distance of the mth order spectrum from zeroth order(× 10-2m

D - Distance between grating and the screen (× 10-2m)



Fig.3 Determination of wavelength of the laser source


Physics Laboratory ManualProcedure

(i) Determination of particle size

The lycopodium powder is sprinkled on the glass plate and placed in between a laser

beam and a screen as shown in fig 1. A diffraction pattern is obtained as shown in fig.2. The

distance between the glass plate and the screen is adjusted to get the clear and more number

of orders of the fringes. By measuring the radius(r) of the first order circle and the distance

between the screen and slide (d), the particle size (D) can be calculated using the given

equation relating to the radius of the circle and the angle of diffraction corresponding to the

mth ring.

(ii) Determination of wavelength of the laser source

The grating is kept between laser light and the screen as shown in fig 2. Laser

beam gets diffracted by the grating. Diffraction pattern of several orders is formed on the

screen. The distance between the screen and the grating D is measured. The distance (d)

between the zero order and first order is measured. Experiment is repeated for different

values of D and d. using the formula λ is measured. Readings are tabulated.

S.No Distance between screen

and grat-ing, D×10-2m

Order of diffraction

m

(No unit)

Distance between zeroth

order and mth order (d)×10-2m

Angle of dif-fraction

ө=tan-1(d/D)

degrees

Wavelengthλ = [sin θ/

Nm]m

×10-10mLeft Right Mean



Mean λ

(ii) DETERMINATION OF WAVELENGTH OF THE LASER SOURCE:

Calaulation,



Result

1. The grain size of the lycopodium powder =……………….. m

2. The wavelength of the given laser source =……………….. (×10-10m)



Fig.1. Air Wedge Arrangement


λl2

Physics Laboratory Manual Fig.2. Interference- fringe pattern

Expt. No:

Date:

2. AIR WEDGE-THICKNESS OF A THIN WIRE

Aim:

To find the thickness of a thin wire by forming interference fringes using air-wedge

arrangement.

Apparatus Required:

Two optically plane rectangular glass plates, thin wire, Travelling microscope,

Sodium vapour lamp etc.,

Formula,

Thickness of the thin wire ,

t = m

Where ,

- Wavelength of Sodium vapour lamp (5893 ×10-10m)

- Bandwidth (distance between any two dark or bright frings) (×10-2m)

l - Distance between the edge of contact and the wire( length of the air wedge)

(×10-2m)



Determination of band width () using Traveling Microscope:

S.NoOrder of the

fringesMicroscope Reading Bandwidth of

10 fringesBandwidth of one fringe x 10-2 m

MSRx 10-2 m

VSC DIV

TRx 10-2 m

Mean bandwidth = x 10-2 m

Calculation,



Principle

When a normally reflected monochromatic light falls on a wedge shaped air film and

if viewed through a microscope, alternate dark and bright interference bands of equal

thickness are observed.

Procedure

Two optically plane glass plates are tied together at one end .At the other end a thin

wire is introduced with its length perpendicular to the length of the plate. The plates are tied

together at this end also. A thin air wedge of steadily increasing thickness is formed between

the plates. Light from Sodium vapour lamp is rendered parallel by a convex lens and made

to fall on the glass plate kept inclined at angle of 45 to the horizontal. The partially

reflected light travels vertically downwards and it is incident normally on the air wedge. An

interference pattern consisting of a number of alternate dark and bright bands will be

obtained. This is viewed by the Travelling microscope arranged above the glass plates. The

first well-defined dark band near the left end is taken as the n th Band. The microscope is

adjusted such that its vertical cross wire is coinciding with the n th band. The reading on the

horizontal scale of the microscope is noted. Cross wire is made to coincide with successive

fifth fringes and the corresponding microscope readings are noted. The readings are to be

taken up to fifty fringes. From these readings, the mean fringe width () is found. The

distance ‘l’ is measured with the help of the Travelling microscope and by using the given

formula thickness of the thin wire is calculated.

Result


Physics Laboratory Manual Thickness of the given thin wire =…………………………m.



Fig.1.Ultrasonic interferometer

Expt. No:

Date:

3. ULTRASONIC INTERFEROMETER –VELOCITY OF

ULTRASONIC WAVES AND COMPRESSIBILITY OF LIQUID


2d x

1 V2

ρ


Aim:

(i) To determine the velocity of ultrasonic waves in the given liquid using

Ultrasonic interferometer.

(ii) To determine the compressibility of the given liquid.

Apparatus required:

Ultrasonic interferometer, Measuring cell, Frequency generator, liquid, etc,

Formula,

(i) Velocity of Ultrasonic waves in the given liquid v = nλ ms-1

Where λ = m

(ii)Compressibility of the given liquid k = ρ m2N-1

λ - Wavelength of Ultrasonics (m)

n - Frequency of the generator which excites the crystal (Hertz)

d - Distance moved by the micrometer screw (m)

x - Number of oscillations (No unit)

ρ – Density of the given liquid (kgm-3)

(i) To determine the velocity of ultrasonic waves Frequency= Hz

S.NoMicrometer Readings Distance

for maximum Current(d)

𝜆 = 2d/ x

X 10-3mPSR

X 10-3mHSCDiv

TR= PSR+(HSCx LC) X 10-3m

V= nλ


Physics Laboratory Manual X 10-3m

Mean ( 𝜆) = X 10-3m

Theory

An Ultrasonic Interferometer is a simple and direct device to determine the velocity of

ultrasonic waves in liquid with a high degree of accuracy. Here the high frequency generator

generates variable frequency, which excites quartz crystal placed at the bottom of the

measuring cell. The excited quartz crystal generates ultrasonic waves in the experimental


Physics Laboratory Manualliquid. The liquid will now serve as an acoustical grating element. Hence when ultrasonic

waves pass through the rulings of grating, successive maxima and minima occur, satisfying

the condition for diffraction. In high frequency generator two knobs are provided for initial

adjustments. One is marked with "Adj" (set) and the other with "Gain" (sensitivity). With

knob marked "Adj" the position of the needle on the ammeter is adjusted and with the knob

marked "Gain", the sensitivity of the instrument can be increased for greater deflection, if

desired.

Procedure

The electrodes are connected to the output terminal of the frequency generator

through a shielded cable. The cell is filled with the experimental liquid before switching ON

the generator. Now, when the frequency generator is switched ON, the Ultrasonic waves

move normal from the Quartz crystal till they are reflected back by the movable reflector

plate. Hence, standing waves are formed in the liquid in between the reflector plate and the

Quartz crystal. The distance between the reflector and crystal is varied using the micrometer

screw such that the anode current of the generator increases to a maximum and then

decreases to a minimum and again increases to a maximum. The distance of separation

between two successive maximum or two successive minimum in the anode current is equal

to half the wavelength of the Ultrasonic waves in the liquid. Therefore, by noting the initial

and final position of the micrometer screw for one complete oscillation (maxima- minima-

maxima) the distance moved by the reflector can be determined.

Calculation,



To minimize the error, the distance (d) moved by the micrometer screw is noted for 'x'

number of oscillations (successive maxima), by noting the initial and final reading in the

micrometer screw and is tabulated. From the total distance (d) moved by the micrometer

screw and the number of oscillations (x), the wavelength of ultrasonic waves can be


Physics Laboratory Manualdetermined using the formula f... = 2d / x. From the value of f... and by noting the frequency

of the generator (n), the velocity of the Ultrasonic waves can be calculated using the given

formula. (1) After determining the velocity of the Ultrasonic waves in liquid, the

compressibility of the liquid is calculated using the given formula. (2)

Result

(i) The velocity of Ultrasonic waves in the given liquid = ms-1

(ii) Compressibility of the given liquid = m2N-1



SPECTROMETER

Expt. No:


sinө mλ

sinө mN

Physics Laboratory ManualDate:

4. SPECTROMETER – WAVELENGTH OF MERCURY SOURCE USING

GRATING

Aim:

(i) To determine the number of lines per meter on the grating

(ii) To find the wavelength of the prominent spectral lines in the mercury source.

Apparatus Required:

Spectrometer, Plane transmission grating, Mercury vapour lamp, Reading lens etc.,

Formula,

(i) Number of lines per meter on the grating,

N = lines/m

(iii) Wavelength of a spectral line, λ = Ǻ

where

- Angle of diffraction (degree)


Procedure

(i) Adjustment of the grating for Normal Incidence

Preliminary adjustments of the spectrometer are made. The grating is mounted

on the grating table with its ruled surface facing the collimator. The slit is illuminated by a

source of light (either sodium or mercury vapour lamp) and is made to coincide with the

vertical cross wire. The vernier scales are adjusted to read 0 deg and 180 deg for the direct

ray (fig .1). The telescope is rotated through an angle of 90 deg and is fixed (fig 2). The

Observation


sinө mλ

Physics Laboratory Manual(i) To determine the number of lines per meter of the grating

λ =

S.

No

Telescope readings

(degrees)Difference2θ (deg) Mean

2θ

(deg)

Meanangle

θ

N=

×105

lines /mLeft Right

A1 B2 A1 B2 A(A1-A2)

B(B1-B2)

grating table is adjusted until the image coincides with the vertical cross wire. Both the


sinө mλ

Physics Laboratory Manualgrating table and the telescope is fixed at this position (fig.3). Now rotate the vernier table

through 45 deg in the same direction in which the telescope has been previously rotated. The

light from the collimator incidents normally( perpendicularly) on the grating. The telescope

is released and is brought in line with the direct image of the slit. Now the grating is said to

be in Normal incidence position (fig. 4)

(ii) Determination of Number of lines per metre of the grating:

The slit is illuminated by mercury green of known wavelength. The telescope is

released to catch the diffracted image of the first order on the left side of the central direct

image. The readings in the two verniers are noted. It is then rotated to the right side to catch

the diffracted image of the first order, the corresponding readings are noted. The difference

between the positions of the right and left sides gives twice the angle of diffraction 2.

For a first order spectrum m = 1 and for green line , λ = 5461Ǻ,

The number of lines per metre on the grating (N) is calculated using the formula

N = lines/m

(iii) Determination of wavelength of prominent spectral lines of the mercury

spectrum

The grating for normal incidence is not disturbed. The telescope is brought round on

both sides of the direct image to view the first order spectrum. The vertical cross wire of the

telescope is made to coincide successively with each one of the prominent lines and the

readings are taken. Similarly the corresponding readings of the same prominent lines for the

CALCULATION



first order on the other side are taken. The observations are tabulated. The angle of


sinө mN

Physics Laboratory Manualdiffraction for each prominent line is determined as before.

The wavelength of the lines of the mercury source is calculated using the formula.,

λ = Ǻ

Result

(i) The number of lines per meter on the grating = ………………. lines /m.

(ii) The wavelengths of the different spectral lines of the mercury spectrum are

λV = ………… Ǻ λY1 = ……………. Ǻ

λB = ………… Ǻ λY2 = ……………. Ǻ

λBG = ………….Ǻ λR1 = ……………. Ǻ

λG = ………… Ǻ λR2 = ……………. Ǻ


Physics Laboratory ManualFig. 1 Lee's Disc Apparatus


MSRd(r+2h)

πr2(2r+2h)(θ1 - θ2 )

dθ

dt


Date:

5.LEES DISC-CO-EFFICIENT OF THERMAL CONDUCTIVITY OF

A BAD CONDUCTOR

Aim:

To determine the thermal conductivity of a bad conductor such as cardboard or

glass using lee’s disc apparatus.

Apparatus required:

Lee’s disc apparatus, card board, thermometer, Stop watch, Vernier caliper,

Screw gauge, Steam boiler, etc,

Formula,

The co-efficient of Thermal conductivity

K = W/mK

M- Mass of the brass disc (×10-3 kg)

S- specific heat of the material of the disc( J /Kg/K)

R = - Rate of cooling at steady state temperature θ2 (°C/sec) (from graph)

dθ – Change in temperature (°C)

dt – Change in time (sec)

d- Thickness of the bad conductor (×10-3 m)

r- Radius of the metallic disc or cardboard (×10-2 m)

h- Thickness of the metallic disc (×10-3 m)

θ1-Temperature of the steam in ºC

θ2- Steady temperature of the metallic disc in ºC

(i) To find the thickness of the brass disc (h)


Mean

Physics Laboratory ManualZE = div

LC = 0.01X10-3m ZC = x10-3m

S.NOPSR

X 10-3m

HSC

Div

HSR=HSC X LC

X 10-3m

Observed reading

OR= PSR + HSR

X 10-3m

Correct Reading

CR=OR ± ZC

X 10-3m

1

2

3

4

5

Principle


Physics Laboratory Manual The Principle involved in the Lee’s disc is conduction of heat. In steady state the

temperature of the lower metal plate becomes constant and heat is lost by the metal disc and

heat is gained by the bad conductor.

Procedure: Lee’s disc consists of thick brass disc A suspended horizontally by strings

from a metal ring attached to a retort stand. The cardboard disc of the same diameter is

placed on the brass disc. A steam chamber B of the same cross section as the brass disc is

placed on the cardboard. The thermometers T1 and T2 are inserted into the both discs A and

B to record the temperature on the two sides of the cardboard.

The main thickness of the cardboard disc is determined using screw gauge. The

radius r of the lower metallic disc A is found using vernier calipers. The mass M of the

lower metallic disc is also found with the help of weighing balance. Steam is allowed to pass

through the steam chamber. Heat is conducted through the cardboard to the metal disc A.

The thermometers indicate rise of temperature.

(a) Static part: Determination of steady state temperature θ1 and θ2

Steam is passed through the steam chamber for a sufficiently longtime until

thermometers T1 and T2 indicate steady temperatures θ1 and θ2 respectively. When the

temperatures of thermometers T1 and T2 remain steady it means that the whole arrangement

has reached steady state. In this state heat is conducted into the lower slab through the

cardboard disc, is just equal to the heat radiated by the flat bottom of the lower slab.

(b) Dynamic part: Determination of Rate of cooling (R) of the brass disc

After noting the steady temperature θ2°C, The cardboard disc is now removed

and A is heated in direct contact with the steam chamber B until its temperature rises by

above 10º above the steady temperature θ2. Now the slab A is suspended separately and

allowed to cool. When its temperature reaches (θ2+5)º a stop clock is started and the time is

(ii) To find the thickness of the card board (d)

ZE = div


dθ dt

ABBC

Physics Laboratory Manual LC = 0.01x10-3m ZC = x10-3m

S.N

O

PSR

X 10-3m

HSC

Div

HSR=HSC X LC

X 10-3m

Observed reading

OR= PSR + HSR

X 10-3m

Correct Reading

CR=OR ± ZC

X 10-3m

1

2

3

4

5

Model graph

R- By graphical method

R = ,

dθ = (θ2 +0.5)-( θ2-0.5) = 1

dt = BC

recorded for every 1 degree fall of temperature until its temperature reaches (θ2-5)º. A graph


Mean

Physics Laboratory Manualis drawn taking time along the x axis and temperature along the y axis. By drawing a tangent

to the graph at the temperature θ2, the corresponding rate of fall of temperature R of the brass

disc A can be found.

(iv) To find rate of cooling (R) by experiment

Calculations

Mass of the brass disc A M = 850× 10-3 Kg

Radius of the brass (card board) disc r = ................... × 10-2 m

Thickness of the brass disc h =................. × 10-3 m

Thickness of the bad conductord = ............... × 10-3 m

Specific heat capacity of brass disc S = 370 J/Kg/K

Steady temperature of the steam chamber,θ1 = ................ °C

Steady temperature of brass disc θ2= ................ °C

Rate of fall of temperature R = .................. °C/ sec

Variation of temperature with te : (Rate of cooling)


Physics Laboratory ManualTemperature

°C Time

Second

Calculation,



Result

Thermal conductivity of the given bad conductor K = .........................W/mK



Fig. 1. Hysteresis circuit diagram


DATE:

EX:

(i tan θ).x.y.n.Be(d2 + L2)2

r2LD


6. HYSTERESIS LOSS IN A FERROMAGNETIC MATERIAL

Aim:

To find the hysteresis loss or energy loss per unit volume per cycle of magnetization

of a ferromagnetic substance.

Apparatus Required :

Ferromagnetic uniform rod, Batteries, Ammeter, Key, Commutator , Rheostat,

Solenoid, Compensating coil, Deflection magnetometer, Screw gauge.

Formula:

Intensity of magnetization is given by

I = J/m2/cycle

Where,

i tan θ- Area of the hysteresis loop(m3)

x and y – Scale factors on x and y axes

n - Number of turns per metre in the solenoid

Be- Horizontal component of the earth’s magnetic field (40×104 Tesla)

θ – Deflection in the magnetometer (degrees)

r – Radius of the ferromagnetic rod (×10-3 m)

d – Distance between the rod and centre of compass box (×10-2 m)

L – Length of the ferromagnetic rod (×10-2 m)



Fig.2.Hysteresis loop

Principle.

From the magnetic effects of currents, we can also prove that, when a current (I) ampere


Physics Laboratory Manualpasses through the solenoid, the magnetising field (H) will be given by

H = μ0 n I

where

μo - Permeability of vacuum ( 4 π x 10-7 henry/metre. )

n - Number of turns/ metre in the solenoid

The energy loss per cycle of magnetization per unit volume of the specimen is given by the

area of I - H curve.

Procedure

Now the ferromagnetic rod is placed inside the solenoid along its axis such that

the two ends of the rod are equidistant from the two ends of the solenoid and bottom end is

in the plane of the magnetic needle of the compass box. Now the circuit is switched on and

rheostat adjusted for maximum current (say 5 amps). There will be deflection in the compass

box. (This need not to be recorded.) Then, slowly the The hysteresis apparatus containing

solenoid and deflection magnetometer is first set in Tan A position. The compass box is

adjusted such that the aluminium pointer reads 0-0. Then the circuit connections are made as

shown in figure without the ferromagnetic rod inside the solenoid. A deflection will be

observed in the compass box, when a maximum current of 5 to 6 ampere flows in the circuit.

Now the position of the compensating coil is adjusted such that the aluminium pointer reads

0- O. It will be seen that, after this adjustment for any current in the circuit (below the

maximum current), the aluminium pointer will always read a - O. The current is brought to

minimum and then the circuit is switched off.

current is reduced in steps of 0.5 amps. When the current is 0.5 amps, the circuit switched off and the commutator reversed. Then

Observation

(i) To find tan θ



S.No.Current (I)

Ampere

Deflections in the Compass box Tan θ

degreesθ1 (deg) θ2 (deg) Mean

θ ( deg)

13

22.5

32

41.5

51

60.5

70

8-0.5

9-1.0

10-1.5

11-2.0

12-2.5

13

14-3.0

15


Physics Laboratory Manual-3.0

16-1.5

17-1.0

18-0.5

190

200.5

211.0

221.5

232.0

242.5

253.0

(i) To find the diameter of the rod

ZE = div

LC = 0.01 X 10-3m ZC = X 10-3m


Mean


S.NoPSR

X 10-3m

HSC

Div

HSR=HSC X

LC

X 10-3m

Observed reading

OR= PSR + HSR

X 10-3m

Correct Reading

CR=OR ± ZC

X 10-3m

1

2

3

4

5

Mean diameter of the rod = ………………………… X 10-3m

Radius of the rod, r = …………………………. X 10-3m

Calculation.

be - 0.5 amp. Then the current is increased to its maximum value in steps of 0.5 amp up to –

5 amp. Again it is decreased to - 0.5 in steps of 0.5 amp. The circuit is switched off and


Physics Laboratory Manualcommutator reversed. Then the circuit is switched on and ammeter will read 0.5 amp. This is

considered to be + 0.5 amp. Again the current IS increased upto + 5 amp in steps of 0.5 amp.

This is called one cycle. Like this, by passing the current in cyclic order, the ferromagnetic

rod is magnetised and demagnetised alternatively.

When the operation is completed 20 times, the readings are recorded during the

21st and 22nd cycles. If these two sets of readings are not one and the same repeat 10 more

times before taking readings. All readings are recorded in table.

The distance between the compass box and rod d, the number of turns per metre

length of the solenoid, n, the radius of the rod r are determined. Using the readings, a graph

is drawn between the current (I in amp) in X axis and tan e in Y axis. This graph will be as

shown in fig.9.2. The area of the graph is also found out. Then the energy loss per unit

volume per cycle of magnetization is found out using the given formula.



Result

Hysteresis loss of the material of the given sample = …………………J/cycle/volume.



Expt. No:


Mgl3

4bd3s

Physics Laboratory ManualDate:

7. YOUNG’S MODULUS – NON UNIFORM BENDING

Aim:

To determine the young’s modulus of the material of a given rod by non- uniform

bending.

Apparatus Required:

A uniform rectangular beam made of wood or iron, two equal knife edges, a weight

hanger with slotted weights, vernier calipers. Screw gauge, Travelling microscope, pin, etc.,

Formula,

Young’s modulus of the material,

Y = Nm-2

Where

M- Load producing the depression ‘s’ (× 10-3 kg)

g – Acceleration due to gravity (9.8m/s2)

l - Length of the beam between two knife edges (× 10-2m)

b – Breadth of the beam(× 10-2m)

d – Thickness of the beam (× 10-3m)

s – Depression produced for a load ‘M’ (× 10-2m)

youngs modulus table



Procedure


Physics Laboratory ManualThe given beam is symmetrically placed over the knife edges. AB is the length of the

bam. At the center of the beam C a weight hanger is suspended. A pin is fixed vertically at C

using some wax. A Travelling microscope is focused on the tip of the pin. The beam is

brought to elastic mode by periodical loading and unloading. The reading in the vertical

scale of the microscope is noted. Weights are added in equal steps of M kg and the

corresponding readings are noted. The readings are noted while unloading also. The length

of the beam/(AB) between the knife edges is measured. The breadth (b) and thickness (d) of

the beam are measured with a vernier caliper and screw gauge respectively. The experiment

is repeated by changing the distance between the knife edges.

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


Mean


E = div

LC = 0.01x10-2m ZC = x10-2m

S.N

O

MSR

×10-2m

VSC

Div

VSR=VSC X

L.C

×10-2m

Observed reading

OR= MSR + VSR

X 10-2m

Correct Reading

CR=OR ± ZC

X 10-2m

1

2

3

4

5



Calculations

M = ……………….. ×kg

g = 9.8ms-2

l = …………………×10-2m

b = …………………×10-2m

d = …………………×10-3m

s = …………………×10-2m



(iii) To find the thickness of the beam (d)

ZE = div

LC = 0.01 X 10-3m ZC = X 10-3m

S.NoPSR

X 10-3m

HSC

Div

HSR=HSC X

LC

X 10-3m

Observed reading

OR= PSR + HSR

X 10-3m

Correct Reading

CR=OR ± ZC

X 10-3m



Result

Young’s modulus of the given material of the rod = ………………….. … Nm-2.



Fig.1 Band gap of a Semiconductor

Model graph


Expt. No:

Date:

∆ lnR ∆


8. BAND GAP DETERMINATION OF A SEMICONDUCTORAim:

To determine the band gap energy of a semiconductor by studying the variation of the

resistance of the thermistor with varying temperature.

Apparatus require:

Thermistor, Ammeter, Voltmeter, Oven, Thermometer, Water bath etc.,

Formula,

Band gap Eg = 2 × kB × slope eV

Where Slope =

Rs – Resistance (ohm).

T – Absolute temperature (K)

kB – Boltzmann’s constant = 1.380662 × 10-23 J/K

Principle

A Semiconductor has the energy band structure of an insulator with the difference that

the forbidden energy gap is less than 2 eV. For germanium the energy band is 0.7 eV. And

for silicon it is 1.1 eV at 0 K. Energy of this magnitude cannot be imparted to an electron by

an applied electric field. Hence the valence band reminds full and conduction band empty.

These substances are insulators at room temperature. If the temperature is increased some of

the valence electrons may gain thermal energy greater than Eg. Consequently they move into

conduction band. When an electron moves from valence band to conduction band a

Observation: Determination of current for various temperatures

Power supply = ……………………… Volts


1T

V Is


Procedure

The circuit is as shown in figure (1). Themistor and the thermometer is immersed in a

water (or) oil bath, in such a way that the thermometer is kept near by the diode. The power


S.NoTemperatur

°C

Temperatur

K

1

T(K-1)

I

(× 10-6

amp)

Rs =

(Ohm)

ln Rs

1.

2.

3.

4.

5.

6.

7.

8.

9.

10.

Physics Laboratory Manualsupply is kept constant (say 3 volts). The heating mantle is switched on and the oil bath is

heated up to 70°C. Now the heating mantle is switched off and the oil bath is allowed to cool

slowly. For every one degree, fall of temperature the micro ammeter reading (I) is noted.

A graph is plotted taking 1/T along x-axis and lnR along Y axis. A straight line is

obtained as shown in model graph. By finding the slope of the straight line, the band gap

energy can be calculated using the given formula. The same procedure can be repeated for

various constant power supply (4 volts, 5 volts).

Calculations

Slope =

kB = 1.380662× 10-23 J/K



Result

The band gap energy of the given thermistor, Eg = ……………………….. eV



Fig.1.Carey Foster's Bridge


R (l1-l2)

πr2X L


Expt. No:

Date:

9. CAREY FOSTER'S BRIDGE- SPECIFIC RESISTANCE OF A WIRE

Aim:

To determine the resistance of a given coil of wire and hence determine its specific

resistance using Carey Foster's bridge.

Apparatus Required:

A Carey Foster's bridge, Two equal (10 ohm) resistances, standard resistance box, 2V

DC regulated power supply, standard resistance box, Galvanometer, High Resistance , metal

and alloy wires and screw gauge.

Formula:

Resistance of the unknown coil X = R+ (l2- l1) x

Resistance per meter of the bridge wire,x = ohm/m

Specific resistance of a coil,

S = ohm.m

where

R – The resistance in the standard resistance box (Ohm)

x – Resistance per meter of the Carey foster bridge wire (ohm/m)

l1 – Balancing length when X is on the left gap and R is in the right gap (× 10-2m)

l2- Balancing length when X and R are interchanged (× 10-2m)

r – Radius of the resistance wire (× 10-2m)

L – Length of the resistance wire (× 10-2m)


Physics Laboratory ManualS – Specific resistance of the wire (ohm.m)

Observation

(i) Determination of resistance per meter (x) of the bridge:


S.No

Resistance

(R)

Ohms

Balancing length

x = R/( l1- l2)

ohm/m

l1

×10-2 m

l2

× 10-2 m

1

2

3

4

5

6.

Physics Laboratory ManualProcedure:

(i) Determination of the resistance per meter(x) of the bridge wire AB:

The connections are made as shown in the fig. P and Q is equal resistances, a

thick copper strip of 0 resistance is connected in the place of x and a standard low

resistance (0.1 ohm)in the place of R. The jockey is pressed near A and moved towards B

touching every point on the wire AB. The balancing point J1 is found out at which the

deflection in the galvanometer Zero. If J1 is the position of the jockey for null deflection, the

balancing length AJ1 (l1) is noted. The copper strip and the standard resistance interchanged.

Now the balancing length AJ2 is (l2) measured. The resistance per meter of the bridge wire

(r) is calculated by using the formula,

x= R/ (l2-l1) ohm/m

(ii) Determination of the unknown resistance (X) of the given wire:

Equal resistance P and Q are connected in the inner gaps. The resistance box R

is connected in the right gap and the given coil of unknown resistance X in the left gap. A 2

volt power supply is connected to the middle strip and the other terminal to the jockey

through a high resistance. A suitable resistance says 5 ohms is introduced in the resistance

box and the circuit is closed. The jockey is pressed near A and then near B. If the deflections

are in opposite directions the connections are correct. The jockey now pressed at the middle

of the bridge wire and the resistance in R is adjusted until the deflection in the galvanometer

(ii) Determination of unknown resistance (X) of the given coil



is almost zero. The value of resistance to be included in the resistance box R should be

rounded this value. If J1 is the balancing point, the balancing length AJ1 is measured. The

coil and R are interchanged. The new balancing length AJ2 is measured. The resistance of

the coil X is calculated by using the formula,


S.No Resistance

(R)

Ohm

Balancing length

X = R+( l2- l1)x

ohm

l1

×10-2 m

l2

× 10-2 m

1

2

3

4

5

6.

Physics Laboratory ManualX = R+(l1-l2)x ohm

(iii) Determination of the specific resistance of the given wire:

The resistance X of the wire is found as before. The length of the wire (L) is measured

after stretching it without any kink. The radius r of the wire is found with a screw gauge by

measuring its diameter at various points along the wire. The specific resistance of the wire is

calculated by using the formula (3).

(iii) Determination of diameter of the given coil

Least Count = 0.01 X 10-3m Zero Error = div

Zero Correction = X 10-3m



S.N

O

PSR

X 10-3m

HSC

Div

HSR=HSC X LC

X 10-3m

Observed reading

OR= PSR + HSR

X 10-3m

Correct Reading

CR=OR ± ZC

X 10-3m

1

2

3

4

5

Mean

Radius of the given coil (r) = ……………………… ×10-3m

Calculations,



Result

1. The resistance of the wire = ………………………… ohm.

2. Specific resistance of the

given wire. =………………………….. ohm-meter.



Fig.1. Viscosity of a liquid – Poiseuille’s flow


gr4 ht 8lV


Date:

10. CO-EFFICIENT OF VISCOSITY – POISEUILLE’S FLOW METHOD

Aim:

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

method.

Apparatus Required:

Burette, Capillary tube, Travelling microscope, Beaker, Stop clock, Metre scale,

Rubber tube etc.,

Formula;

The coefficient of viscosity of the liquid = Nsm-2

Where ;

g – Acceleration due to gravity (9.8 ms-2)

- Density of water (1000 kg m-3)

r - Radius of the capillary tube (× 10-2 m)

l – Length of the capillary tube (× 10-2 m)

V – Volume of the flow of the liquid in cubic meter (10-6m3).

h - Mean pressure head (× 10-2 m)

t - Time taken for liquid flow



table,


Physics Laboratory ManualProcedure,

A clean graduated burette is vertically mounted. A capillary tube of length about 20

centimeters is connected to the bottom of the burette using a pinch cock provided at the

rubber tubing. The capillary tube is adjusted to be horizontal. The burette is filled with given

liquid. The pinch cock is completely opened and the liquid is allowed to flow. A beaker is

used to collect the out flowing liquid. When the liquid level crosses 0cc mark a stop clock is

started and the time is noted for the liquid level to reach 5,10,15….40cc.The readings are

tabulated then the vertical heights of the marks 0,5,10……40cc are measured from the

access of the capillary tube. From these measurements the mean pressure head (h) can be

calculated.


gr4 ht 8lV

Physics Laboratory Manual (ii) To determine diameter of the capillary tube

Least Count = 0.001x10-2m

Cross wire PositionMicroscope Readings Diameter

A ~ B

X 10-2m

MSR

X 10-2m

VSC

Div

TR= MSR+ (VSCx

LC) X 10-2mVertical crosswire

Left (A)

Right (B)

Horizontal

crosswire

Bottom (B)

Mean

Diameter of the capillary tube, d = ……………….×10-2 m

Radius of the capillary tube r = ………………. ×10-2 m

Length of the capillary tube, l = ………………. ×10-2 m

Calculations


Physics Laboratory Manual = Nsm-2

Density of the liquid, ρ = 1000kg/m3.

Acceleration due to gravity g = 9.8ms-2.

Radius of the capillary tube r = ……………….×10-2 m

Length of the capillary tube l = ……………….×10-2 m

Volume of the liquid V = ……………….×10-6 m3.

ht =……………….×10-2 m.sec

Calculation,



Result:

The coefficient of viscosity of the given liquid,

sm-2



Fig .1. Position of

spectrometer- Angle of the prism

Fig .2. Position of spectrometer- Angle of minimum deviation


(µv - µr ) ( µy -1)


Date:

11.SPECTROMETER – DISPERSIVE POWER OF THE PRISM

Aim

To determine the refractive index of the material of the prism for various colors in the

mercury light spectrum and hence to calculate its dispersive power.

Apparatus Required

Spectrometer, prism, mercury vapour lamp, prism holder and reading lens.

Formula

Refractive index, µ = (No unit)

Dispersive power of a prism, ω = (No unit)

Where,

A – Angle of the prism (degrees)

D – Angle of minimum deviation (degrees)

µv - Refractive index of the prism for violet line (No unit)

µr - Refractive index of the prism for red line (No unit)

µy - Refractive index of the prism for red line(No unit

Procedure:

(i) Determination of angle of prism (A)

The given prism is mounted vertically at the centre of the prism table with its refracting edge

facing the collimator. Now the parallel rays of light emerging out from the collimator falls

alm0st equally on the two faces of the prism ABC as shown in Fig.1. The telescope is turned


Sin

Sin

A+D 2

A2

Physics Laboratory ManualTABLE


Physics Laboratory Manualto catch the reflected image from one face of' the prism and fixed in that position. The tan-

gential screw is adjusted until the vertical cross-wire coincides with the fixed edge of the im-

age of the slit. The readings on both the verniers are noted. Similarly the readings corre-

sponding to the reflected image of the slit on the other face are also taken. The difference be-

tween the two readings of the same vernier gives twice the angle of the prism. Hence, the an-

gle of the prism 'A' is determined.

(ii) Determination of angle of minimum deviation (D)

The prism table is rotated so that the beam of light from the collimator is incident on

one face of the prism and emerges out from the other face. The telescope is rotated to catch

the refracted image of the yellow slit. The prism table is rotated in such a direction so that

the refracted image moves towards the direct beam. The telescope is rotated carefully to

have the image in the held of view. At one stage, the image stops momentarily and turns

back. This is the position of the minimum deviation. The telescope is rotated and made to

coincide with the violet slit. The telescope is fixed in this position and refracted ray reading

of the telescope is noted. The experiment is repeated for red slit. The prism is removed and

the direct reading of the slit is taken. The difference between the direct reading and the re-

fracted ray reading corresponding to the minimum deviation gives the angle of minimum de-

viation 'D'. The dispersive power is calculated using the given formula.


Physics Laboratory ManualTable


µ2 + µ1

2(µ1 ~ µ2) (µ - 1)


(iii) To find dispersive power of the prism

S.Noµ1

(No unit)

µ1

(No unit)

µ1 ~ µ2

(No unit)

µ =

(No unit)

ω =

(No unit)



Calculation,



Result

1. Angle of the prism A = …………………degrees.

2. Angle of the minimum deviation D = ………………...degrees

3. Refractive index of the material of the prism µ = ………………… (No unit)

4. Dispersive power of the given prism = …………………(No unit)


3MDgl2

2bd3s


Date:

12. YOUNG’S MODULUS – UNIFORM BENDINGAim :

To determine the young’s modulus of the material of a given rod by uniform bending.

Apparatus Required;

A uniform rectangular beam made of wood or iron, two equal knife edges, two weight

hangers with slotted weights, vernier caliper, Screw gauge, Travelling microscope, pin, etc.,

Formula:

Young’s modulus of the material,

Y = Nm

Where,

M- Load producing the depression‘s’ (× 10-3 kg)

D – Distance between the weight hanger and any one of the adjacent knife edge (× 10-2m)

g – Acceleration due to gravity (9.8m/s2)

l - Length of the beam between two knife edges (× 10-2m)

b – Breadth of the beam (× 10-2m)

d – Thickness of the beam (× 10-2m)

s – Elevation produced for a load ‘M’ (× 10-2m)


Physics Laboratory ManualTabulation,

S.NoLoad10-3

Kg

Microscope Reading 10-2 m Elevation for M x10-3Kg

Y x 10-2m

Loading Unloading Mean10 -2 m

MSR10-2 m

VSCdiv

TR10 -2 m

MSR10-2m

VSC

div

TR

10-2m

1.

2.

3.

4.

5.

6.


Physics Laboratory ManualProcedure

The given beam is symmetrically placed over the knife edges. AB is the length of the

beam. Two weight hangers are suspended, one each on either side of the knife edge at equal

distance from the knife edge. A pin is fixed vertically exactly at the center of the beam i.e. at

C using some wax. A Travelling microscope is focused on the tip of the pin. The beam is

brought to elastic mode by periodical loading and unloading of weights on both the weight

hangers simultaneously. The reading in the vertical scale of the microscope is noted.

Weights are added in equal steps of M kg on both the weight hangers and the corresponding

readings are noted. The readings are noted while unloading also. The length of the beam the

weight hanger and any one of the adjacent knife edge (D) is measured. The breadth (b) and

thickness (d) of the beam are measured with a vernier caliper and screw gauge respectively.

The experiment is repeated by changing the distance between the knife edge and the weight

hangers.


Mean

Physics Laboratory Manual(ii) To find the breadth of the beam using vernier caliper (b)

ZE = divLC = 0.01cm ZC = cm

S.N

O

MSR

× 10-2m

VSC

Div

VSR=VSC X

L.C

× 10-2m

Observed reading

OR= MSR + VSR

× 10-2m

Correct Reading

CR=OR ± ZC)

× 10-2m

1

2

3

4

5

Breadth of the beam (b) = ………….×10-2m


Physics Laboratory ManualCalculations

M = ……………….. kg

g = 9.8ms-2

D = ………………..×10-2m

l = …………………×10-2m

b = …………………×10-2m

d = …………………×10-2m

s = …………………×10-2m



(iii) To find the thickness of the beam (d)

ZE = div

LC = 0.01 X 10-3m ZC = X 10-3m

S.N

O

PSR

X 10-3m

HSC

Div

HSR=HSC X LC

X 10-3m

Observed reading

OR= PSR + HSR

X 10-3m

Correct Reading

CR=OR ± ZC

X 10-3m

1

2

3

4

5

Thickness of the beam (d) = …………………. ×10-3m


Mean


Result

Young’s modulus of the given material of the rod = ………………….. … Nm-2.



Fig .1 Torsional pendulum

Expt. No:

Date:


2m (d22 – d 1

2 ) To 2 ( T2

2-T12)

8πI l r4 To

2


13. TORSIONAL PENDULUM- RIGIDITY MODULUS

Aim:

To determine

(i) the moment of inertia of a disc and

(ii) rigidity modulus of the material of a wire by Torsional oscillations.

Apparatus required;

Torsion pendulum, two equal cylindrical masses, Stop clock , Screw gauge, Meter

scale etc.,

Formula:

Moment of inertia of the disc I = Kgm2

Rigidity modulus of the material of the wire n = Nm-2. Wherem- Mass of each of the two symmetrical weights placed on the disc (×10-3kg)

d1 – Closest distance between suspension wire and the centre of mass of a cylinder (×10-2m)

d2 - Farthest distance between suspension wire and the centre of mass of a cylinder ×10-2m

To – Time period of oscillation without any weights (sec.)

T1 – Time period when two equal masses placed on the disc at a distance d1(sec.)

T2 – Time period when two equal masses placed on the disc at a distance d2(sec.)

l- Length of the suspension wire(×10-2m)

r- Radius of the wire(×10-3m)


Physics Laboratory Manual(iii) To find period of oscillation (T):

Procedure

One end of a long, uniform wire whose rigidity modulus is to be determined is

clamped by a vertical chuck. To the lower end, a heavy uniform circular disc is attached by

another chuck. The length of the suspension 'l' is fixed to a particular value (say 60 cm or 70

cm). The suspended disc is slightly twisted so that it executes torsional oscillations. Care is

taken to see that the disc oscillates without wobbling. The first few oscillations are omitted.

By using the pointer, (a mark made in the disc) the time taken for 10 complete oscillations

are noted.Two trials are taken. The meantime period T0 (time for one oscillation) is found.

Two equal cylindrical masses are placed on the disc symmetrically on either side, close to

the suspension wire (at the minimum distance). The closest distance 'd l' from the centre of

the mass of the cylinder and the centre of the suspension wire is found. The disc with masses

at distance 'dl' is made to executive torsional oscillations by twisting the disc. The time taken

for 10 oscillations is noted. Two trials are taken. Then the mean time period 'T l' is


Position of masses

Time period for 10 oscillationsTime period for one oscillation

seconds

Trial – 1seconds

Trial - 2seconds

Meanseconds

Without any masses

To =

With masses at closest distance d1 = cm T1 =

With masses at closest distance d1 = cm T2 =

Physics Laboratory Manualdetermined.

Two equal masses are now moved to the extreme ends so that the edges of masses

coincide with the edge of the disc and the centres are equi-distant. The distance 'd 2' from the

centre of the mass of the cylinder and the centre of the suspension wire is noted. The disc

with masses at distance 'd2' is allowed to execute tensional oscillations by twisting the disc.

The time taken for 10 oscillations is noted and time period 'T2' is calculated. The diameter of

the wire is accurately measured at various places along its length with screw gauge. From

this, the radius of the wire is calculated. The moment of inertia of the disc and the rigidity

modulus of the wire are calculated using the given formulae.

Calculations

Mean radius of the wire r = ……………………….×10-3m

Length of the wire l = …………………………×10-2m

Mass of one of the symmetrical weights m = ………………………..×10-2m

Distance at which masses are placed closest d1 = ………………………..×10-2m

Distance at which masses are placed farthest d2 = ……………………….×10-2m

Period of oscillations without any weights T0 = ……………………… secs

Period of oscillations with masses at d1 T1 = ……………………….secs

Period of oscillations with masses at d2 T2 = ……………………….secs



(ii) Determination of the diameter of a wire:

ZE = div LC = 0.01mm ZC = mm

S.N

O

PSR

X 10-3m

HSC

Div

HSR=HSC X LC

X 10-3m

Observed reading

OR= PSR + HSR

X 10-3m

Correct Reading

CR=OR ± ZC

X 10-3m

1

2

3

4

5



S.N

O

PSR

X 10-3m

HSC

Div

HSR=HSC X LC

X 10-3m

Observed reading

OR= PSR + HSR

X 10-3m

Correct Reading

CR=OR ± ZC

X 10-3m

Diameter of the suspension wire = …………………. ×10-3m


Mean


Result

Moment of inertia of the disc (I) =…………………………….. kgm2

Rigidity modulus .of the material of the wire n = ……………………………. Nm - 2


completed physics manual

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