LIST OF EXPERIMENTS PRACTICAL PHYSICS 2017-18 …cvmhssvandazhy.com/cvm hss lab manual.pdf · From...

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CVM HSS VANDAZHY PHYSICS LAB MANUAL www.cvmhssvandazhy.com

LIST OF EXPERIMENTS PRACTICAL PHYSICS 2017-18

For the Higher Secondary Practical Examination, a minimum of 22 experiments must be done and recorded. These include 10 first year experiments and 12 second year experiments and are scheduled in two cycles.

SECOND YEAR Experiments

11. Potentiometer -1 12. Potentiometer -2 13. Convex Lens 14. Concave Lens 15. Diode characteristics 16. Zener Diode 17. Ohms law -1 18. Ohms law – 2 19. Concave Mirror 20. Covex Mirror 21. Sonometer 3 22. Meter bridge

FIRST YEAR Experiments 1. Concurrent force 2. Simple Pendulum 3. Vernier Callipers 4. Sonometer - 1 5. Sonometer - 2 6. Screw Gauge 7. Helical Spring 8. Resonance column 1 9. Resonance column 2 10. Newton’s law of cooling

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OBSERVATIONS &CALCULATIONS

Scale factor 1cm =25 g wt

Body Si

No P

g wt

Q g Wt

OA cm

OB cm

OD cm

Wt. of the body OD x scale factor

Mean Weight

g wt

In Air

1 150 150 6 6 𝑊𝐴=

2 200 200 8 8

In Water

1 150 150 6 6 𝑊𝑊=

2 200 200 8 8

Weight of the body 𝑊𝐴=…………………. gm wt Mass of the body = ……………. gm = ……………….. Kg

Relative density of solid body = 𝑊𝐴

𝑊𝐴−𝑊𝑊 = ……………….

P Q O

O

B A

D

W

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CONCURRENT FORCES AIM

1. To find mass of a solid body 2. To find relative density of the solid body

APPARATUS Parallelogram apparatus, slotted weights, drawing pin, mirror piece, given body. THEORY Parallelogram law of vectors: if two forces acting at a point are represented by the adjacent sides of a parallelogram, then diagonal starting from the common point represents their resultant.

If the forces P, Q and the unknown force W are in equilibrium then Weight of the body W = OC X Scale factor

OC –Diagonal of the parallelogram with P and Q as adjacent sides

Relative density of the solid body = 𝑤𝑒𝑖𝑔 𝑕𝑡 𝑖𝑛 𝑎𝑖𝑟

𝑙𝑜𝑠𝑠 𝑜𝑓 𝑤𝑒𝑖𝑔 𝑕𝑡 𝑖𝑛 𝑤𝑎𝑡𝑒𝑟=

𝑊𝐴

𝑊𝐴−𝑊𝑊

𝑊𝐴= Weight of body in air 𝑊𝑊=Weight of body in water

PROCEDURE A sheet of paper is fixed on the board using drawing pins. Weights P = 150gm and Q=150gm are tied at the ends of two strings each passing over a pulley. The other end of the strings is connected together and from this common point, the unknown mass is suspended.

The directions of the three forces and the common knot are marked on the sheet of paper by putting dot marks. By taking suitable scale, OA and OB are marked on the paper and parallelogram is completed. From the length of diagonal OC, weight of the body is found. Now the experiment is repeated by taking P=200gm and Q=200gm. Mean weight 𝑊𝐴 is calculated

The same procedure is repeated by immersing the body in water. Thus mean weight 𝑊𝑊 is calculated. Now relative density can be found out by using equation

RESULT 1. Mass of the given body = 2. Relative density of the given body =

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Radius of bob ( r ) = 0.9 cm

Sl No

Length to the

top of bob

ℓ-r (cm)

Length of Pendulum

ℓ (cm)

Time of 20 oscillations

Period

T=𝑡

20

sec

T2

sec2

( three deci.place)

𝓵

𝑻𝟐

cm/sec2

( four deci.place)

observations Mean

(t) sec 1 2

(sec) (sec)

1 39.1 40

2 49.1 50

3 59.1 60

4 89.1 90

5 99.1 100

Mean ℓ

𝑇2 = ……… cm/s2

= ………. m/s2

Acceleration due to gravity

By calculation g = 4𝜋2 ℓ

𝑇2 = 4x3.142x………… cm/s2 = …………… m/s2

From graph g = 4𝜋2 AB

𝐵𝐶 = 4x3.142x………… cm/s2 = …………… m/s2

Time period of seconds pendulum = 2 s Length of seconds pendulum =……………...cm

= ……………...m

ℓ - T2 graph

ℓ in cm

B

𝑇2

in s

ec2

C

A

Simple pendulum

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THE SIMPLE PENDULUM

AIM 1. To draw ℓ - T2 graph of a simple pendulum 2. To determine acceleration due to gravity at the place 3. To find length of seconds pendulum

APPARATUS Simple pendulum, Meter Scale, stop clock

THEORY For small amplitudes of oscillation the time period of a pendulum is given by

T =2𝜋 ℓ

𝑔

Acceleration due to gravity at the place g = 4𝜋2 ℓ

𝑇2 , ie

ℓ

𝑇2= 𝑐𝑜𝑛𝑠𝑡𝑎𝑛𝑡

ℓ = Length of the pendulum (distance from point of suspension to the centre of the bob) T = time period (time for one oscillation)

Acceleration due to gravity g = 9.8 m/s2

The ℓ - T2 graph is a straight line and its slope gives ℓ

𝑇2

If time period of a pendulum is two, it is called as a second’s pendulum

PROCEDURE The distance from point of suspension to the top of the bob is set as 39.1 cm so that length

of pendulum ( l ) is 40 cm. Now the pendulum is allowed to oscillate with small amplitude.

The time for 20 oscillations is measured using stop clock. It is repeated and mean time (t) is found out. Hence time period T is determined. The experiment is repeated for increasing length of pendulum such that,

ℓ = 50, 60, 90, 100. In each case ℓ

𝑇2 is calculated. Then mean value of

ℓ

𝑇2 𝑎𝑛𝑑 𝑔 𝑎𝑟𝑒 𝑐𝑎𝑙𝑐𝑢𝑙𝑎𝑡𝑒𝑑

ℓ - T2 graph A graph is plotted with ℓ along the X axis and T2 along the Y axis. From the graph acceleration due to gravity and length of second’s pendulum are calculated.

RESULT 1. ℓ - T2 graph is found to be a straight line. 2. Acceleration due to gravity at the place

By calculation = …………… m/s2 From graph = ………………. m/s2

3. Length of seconds pendulum from graph =………………. m

Parallelogram Apparatus Equilibrium of three forces acing at a point

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Magnitude of one main scale division MSD =0.1 cm

Number of divisions on vernier scale n = 10

Least Count LC = 1 𝑀𝑆𝐷

𝑁 =

0.1

10 = 0.01 cm

To find dimensions of sphere

Dimension Trail

No.

MSR

cm

VSR

Divisions

VSR X LC

cm

Total reading =

MSR + (VSR X LC) Mean

Dia

met

er o

f sp

her

e

d

1

d =

……

…cm

2

3

4

5

6

7

8

Diameter of the Sphere = ………… cm = ………….. m

Radius of the sphere r = 𝒅

𝟐 = …………………………. m

Volume of the given sphere V = 𝟒

𝟑𝝅r3 = …………………m3

Mass of the sphere M = 50 gm = 50 x10-3 kg

Density of the given sphere 𝝆 = 𝑴

𝑽 = ………………. Kg/m3

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THE VERNIER CALLIPERS AIM

1. To determine dimensions of given sphere

2. To determine volume of the sphere

3. To determine density of the sphere

4. To determine dimensions of given cylinder

5. To determine volume of the cylinder

APPARATUS Vernier callipers, cylinder, sphere

THEORY The least measurement that can be taken by an instrument is called Least Count (LC)

Least Count = 𝑚𝑎𝑔𝑛𝑖𝑡𝑢𝑑𝑒 𝑜𝑓 𝑜𝑛𝑒 𝑚𝑎𝑖𝑛 𝑠𝑐𝑎𝑙𝑒 𝑑𝑖𝑣𝑖𝑠𝑖𝑜𝑛

𝑁𝑜 .𝑜𝑓 𝑑𝑖𝑣𝑖𝑠𝑖𝑜𝑛𝑠 𝑜𝑛 𝑉𝑒𝑟𝑛𝑖𝑒𝑟 𝑠𝑐𝑎𝑙𝑒 =

1 𝑀𝑆𝐷

𝑁

The measured value in vernier callipers is given by

Total reading = MSR + (VSR X LC)

MSR = Main Scale Reading (The reading of the main scale just before the zero of vernier scale is taken as MSR)

VSR = Vernier Scale Reading (The vernier scale division which coincides with any of the main scale division is taken as VSR)

Volume of cylinder 𝑉 = 𝜋𝑟2ℓ r = radius, ℓ=length of cylinder

Volume of sphere V = 4

3𝜋r3

Density of sphere 𝜌 =𝑚𝑎𝑠𝑠 𝑀

𝑣𝑜𝑙𝑢𝑚𝑒 𝑉

PROCEDURE To find volume and density of the sphere: the given sphere is gripped between the jaws of vernier callipers.MSR and VSR readings are taken. Then diameter of the sphere is found using TR = MSR + (VSR X LC). The experiment is repeated by clamping vernier callipers at various diametrically opposite points of sphere and average diameter is obtained. Then radius and hence volume of the sphere is calculated. Density can be found out by using the formulae

𝜌 =𝑚𝑎𝑠𝑠 𝑀

𝑣𝑜𝑙𝑢𝑚𝑒 𝑉

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To find dimensions of cylinder

Dimension Trail

No.

MSR

cm

VSR

Divisions

VSR X LC

cm

Total reading

MSR + (VSR X LC) Mean

Len

gth

of

cylin

der

𝓵

1

𝓵 =

……

…

c m

2

3

4

5

6

7

8

Dia

me

ter

of

cylin

der

d

1

d=

……

…

cm

2

3

4

5

6

7

8

Length of the cylinder 𝓵 = ……… c m = ………….m

Diameter of the Cylinder d = ………. cm = …………. m

Radius of the cylinder r = 𝒅

𝟐 = ……………….. m

Volume of the given cylinder 𝑽 = 𝝅𝒓𝟐𝓵 = …………………m3

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To find volume of the cylinder: The given cylinder is gripped lengthwise between jaws and length is found as before. Similarly the cylinder is gripped diameterwise between the jaws and mean diameter is determined. Then radius and volume are calculated. RESULT

1. Dimensions of sphere

a) Radius =…………….. m

2. Volume of sphere = ………………. m3

3. Density of the sphere = ……………… kg/m3

4. Dimensions of cylinder

a) Length = ……………. m

b) Radius = ……………. m

5. Volume of cylinder =………………………. m 3

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Constant mass suspended in the string = 2.5Kg

Trial No.

Frequency of Tuning fork

(𝒏) Hz

Resonating length of sonometer wire

K = 𝑛 x ℓ 1

cm 2

cm

Mean

(ℓ) cm

1 512

2 480

3 426

Mean value of K = …………Hz cm

To find unknown frequency of given tuning fork

Frequency of Tuning fork

Hz


Unknown frequency

𝑛 = K

𝐿

Hz

1 cm

2 cm

Mean

(L) cm

unknown

Sonometer

ℓ

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THE SONOMETER-1 AIM

1. To verify n x ℓ = Const. 2. To determine frequency of the given tuning fork

APPARATUS Sonometer, slotted weights, tuning fork of frequency -512 Hz, 480Hz, 426Hz & unknown frequency, rubber hammer, paper rider

THEORY The frequency ( 𝑛 ) of transverse vibrations produced in a stretched string is given by

𝑛 =1

2ℓ

𝑇

𝜇

ℓ = resonating length of sonometer wire

𝑇 = 𝑡𝑒𝑛𝑠𝑖𝑜𝑛 𝑖𝑛 𝑡𝑕𝑒 𝑠𝑡𝑟𝑖𝑛𝑔 ( T=F=mg= weight hang on wire)

𝜇 = linear density of wire ( mass per unit length = mass of sonometer wire

𝐿𝑒𝑛𝑔 𝑕𝑡 𝑜𝑓 𝑠𝑜𝑛𝑜𝑚𝑒𝑡𝑒𝑟 𝑤𝑖𝑟𝑒 )

𝐼𝑓 𝑇 𝑎𝑛𝑑 𝜇 are made as constants, then 𝑛 x ℓ = constant (K)

If L is length of wire in resonance with tuning fork of frequency 𝑛, then 𝑛 = K

𝐿

PROCEDURE The sonometer wire is stretched by suspending a constant mass = 2.5 Kg. The bridges are

kept close and paper rider is placed at the middle. Tuning fork of frequency 512 Hz is exited

and its stem is placed on the sonometer box. The bridges are moved slowly till the paper

rider is thrown off. Distance between the bridges gives the resonating length ℓ. The

experiment is repeated again and mean ℓ is determined

The same procedure is repeated by using the tuning forks of frequency 480Hz and 426Hz. 𝑛

x ℓ is found to be constant. Similarly resonating length L for the unknown frequency is

determined and hence 𝑛 is calculated.

RESULT 1. 𝑛 x ℓ is found to be constant 2. Frequency of the given tuning fork ( ) = ……………. Hz

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Frequency of tuning fork used = 426 Hz

Trial No.

Mass suspended

(m) Kg

Resonating length of sonometer wire ℓ2

cm2

K = m

ℓ2

Kg/cm2 1 cm

2 cm

Mean

(ℓ) cm

1 1.5

2 2

3 2.5

Mean value of K = ………… Kg/cm2

To find unknown mass of given body

Trial No.

Mass suspended

(M) Kg


L2

cm2

Unknown mass

M = K x L2

Kg 1

cm 2

cm

Mean

(L) cm

1 unknown

Sonometer

ℓ

m in kg

ℓ2

in c

m2

M

L2

Scale Origin (0,100) Along X axis, 1cm = 0.5 Kg

Along Y axis, 1cm = 10cm2

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THE SONOMETER-2 AIM

1. To show that 𝑚

ℓ2 =Const.

2. To Plot m - ℓ2 graph 3. To determine unknown mass of given body

APPARATUS Sonometer, slotted weights, tuning fork of frequency -426Hz, unknown mass, rubber hammer, paper rider

THEORY

The frequency of transverse vibrations produced in a stretched string is given by 𝑛 =1

2ℓ

𝑇

𝜇

ℓ = resonating length of sonometer wire 𝑇 = 𝑡𝑒𝑛𝑠𝑖𝑜𝑛 𝑖𝑛 𝑡𝑕𝑒 𝑠𝑡𝑟𝑖𝑛𝑔 𝜇 = linear density of wire (mass per unit length)

𝐼𝑓 𝑛 𝑎𝑛𝑑 𝜇 are made as constants, then 𝑇 α ℓ OR T α ℓ2

Since T = mg, Substituting m

ℓ2 = Constant (K)

If L is resonating length of wire for the unknown mass M, then M = K x L2

PROCEDURE The sonometer wire is stretched by suspending a suitable mass 1.5Kg. The bridges are kept

close and paper rider is placed at the middle. Tuning fork of frequency 426 Hz is exited and

its stem is placed on the sonometer box. The bridges are moved slowly till the paper rider is

thrown off. Distance between the bridges gives the resonating length ℓ. The experiment is

repeated again and mean ℓ is determined

The same procedure is repeated by using the same tuning fork but changing mass

suspended. m

ℓ2 is found to be constant. Similarly resonating length L for the unknown mass

is determined and hence M is calculated

A graph is plotted by taking m along X-Axis and ℓ2along Y axis.

RESULT

1. 𝑚

ℓ2 is found to be constant

2. The m- ℓ2 graph is found to be a straight line 3. Mass of the given body 1. By calculation = ……………… Kg

2. From Graph = ………………….Kg

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Reading of the pointer with dead load r0 =

Loaded mass Producing Extension

(m) in Kg

Reading of the pointer Extension

𝓍 = r - r0

cm

K = 𝑚𝑔

𝓍

N/cm

On loading

cm

On unloading

cm

Mean

r cm

m 0 +0.05

m 0 +0.1.

m 0 +0.15

m 0 +0.2.

Mean K = ………… N/cm

To find unknown mass

Loaded mass

Reading of the pointer Extension L = R-r0

cm

unknown mass

M = 𝐾 𝐿

𝑔

Kg

1 cm

2 cm

Mean R

cm

m 0 +unknown mass

Spring constant by calculation K =………… N/cm = …………..N/m

To find spring constant from graph

Spring constant K = ( 𝐴𝐵

𝐵𝐶) g =………………… N/m

Scale Origin (0, 0) Along X axis, 1cm = 0.05 Kg Along Y axis, 1cm = 1 cm

B

Mass in Kg

Exte

nsi

on

in c

m

C

A

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THE HELICAL SPRING

AIM 1. To find spring constant of the spring by load – extension method

2. To plot load - extension graph

3. To determine unknown mass of the given body

APPARATUS Helical spring, weight hanger with slotted weights, unknown mass

THEORY According to Hook’s law F =k 𝓍 where k is spring constant and 𝓍 is extension of the spring when the force is applied

Spring constant K = 𝐹𝑜𝑟𝑐𝑒 ( 𝐹)

𝐸𝑥𝑡𝑒𝑛𝑠𝑖𝑜𝑛 (𝓍) =

𝑚𝑔

𝓍

g = acceleration due to gravity If a graph is drawn with load along X axis and extension along Y axis, the graph will be a straight line. Spring constant can be determined from the graph by

K = ( 𝐴𝐵

𝐵𝐶) g

Unknown mass M = 𝐾 𝐿

𝑔 where L is the extension produced by the unknown mass.

PROCEDURE A suitable dead load m0 is suspended from the spring and the spring is brought into elastic mode. Now reading of the pointer r0 is taken.

Slotted weights 50gm are added one by one to the weight hanger and each time scale reading is taken. The slotted weights are unloaded one by one and each time reading of the pointer is noted. The mean of the readings (r) corresponding to loading and unloading is calculated. The corresponding extension 𝓍 = r - r0 for m = 50g, 100g, 150g, 200g is determined and spring constant is tabulated in each case

A graph is drawn with load along X axis and extension along Y axis, the graph will be a

straight line and spring constant is determined from the graph by the equation K = ( 𝐴𝐵

𝐵𝐶) g

The unknown mass is suspended from the weight hanger along with the dead load. The extension produced is found out and unknown mass is calculated RESULT

1. Spring constant of the helical spring

1. By calculation = ………… N/m

2. From load-extension graph = ………. N/m

2. The load-extension graph is found to be a straight line

3. Mass of the given body (by calculation) = ……………… Kg

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OBSERVATIONS &CALCULATIONS Magnitude of one pitch scale division = 1 mm

Distance moved for 4 rotations S = 4 mm

Pitch = 𝑠

4 =

4 𝑚𝑚

4 = 1 mm

Least Count = 𝑝𝑖𝑡𝑐 𝑕

𝑁𝑜 .𝑜𝑓 𝑑𝑖𝑣𝑖𝑠𝑖𝑜𝑛𝑠 𝑜𝑛 𝑡𝑕𝑒 𝑕𝑒𝑎𝑑 𝑠𝑐𝑎𝑙𝑒 =

1𝑚𝑚

100 = 0.01mm

Zero coincidence =

Zero correction Z =

To find volume of the wire

Diameter of the wire

Sl No

PSR mm

Observed HSR

Corrected HSR (HSR+Z )

Corrected HSR X LC

mm

Total Reading = PSR + (Corrected HSR x LC)

mm

1

2

3

4

5

6

7

8

Mean diameter d = ……………………. mm

Radius of the wire r = 𝒅

𝟐 = ……………… mm = ……………. m

Length of the wire 𝓵 = …………….. …..cm = …………….. m

Volume of the wire V = 𝝅𝒓𝟐𝓵 = ………………. m3

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THE SCREW GAUGE

AIM 1. To find volume of given wire

2. To determine volume of given glass plate

APPARATUS Screw gauge, glass plate, wire, reading lens

THEORY The least count of screw gauge is the minimum distance that can be measured by using it. It is equal to the distance through which tip of the screw advances for one division of rotation of the head scale

Least count L.C = 𝑃𝑖𝑡𝑐 𝑕

𝑁𝑜 .𝑜𝑓 𝑑𝑖𝑣𝑖𝑠𝑖𝑜𝑛𝑠 𝑜𝑛 𝑕𝑒𝑎𝑑 𝑠𝑐𝑎𝑙𝑒

L C of screw gauge is 0.01 mm

Pitch of the screw gauge is the distance through which tip of the screw advances for one complete rotation of the head scale The measured value using screw gauge is given by

Total reading = PSR + (Corrected HSR x LC) PSR =Pitch scale reading (observed reading on the pitch scale) HSR =Head scale reading (the division on head scale where reference line coincides)

Zero Error -If zero of head scale coincides with reference line, then there is no zero error. If zero on head scale is above the reference line, zero correction is positive, and if zero is below the reference line, zero correction is negative

Volume of the wire V = 𝜋𝑟2ℓ r = radius of the wire

ℓ=length of the wire

Volume of the glass plate V =A x t A = area of glass plate t = thickness of glass plate

PROCEDURE Pitch and LC of screw gauge is determined. To find zero correction, the head scale is completely rotated till the tips of the studs are gently touching each other. If zero of head scale coincides with reference line, then there is no zero error. If zero on head scale is above the reference line, zero correction is positive, and if zero is below the reference line, zero correction is negative.

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To find volume of glass plate

Thickness of glass plate

Si No

PSR mm

Observed HSR


Corrected HSR X LC

mm

Total Reading = PSR + (Corrected HSR x LC)

mm

1

2 3

4 5

6 7

8

Mean thickness t = ……………………. mm

Thickness of glass plate t = …………………… mm = ………………….. m

Area of the glass plate (from graph paper) A = …………… mm2 =………….. m2

Volume of the glass plate V = A x t = ………………. m3

20


To find volume of wire: The given wire is gripped between the studs of the screw gauge and PSR, HSR readings are taken

Corrected HSR = observed HSR + zero correction Diameter of wire = PSR + (Corrected HSR x LC)

The experiment is repeated for different positions of the wire. The mean diameter and radius is found out. The length of wire is determined by using a scale. Volume of the wire is calculated by using the formulae V = 𝜋𝑟2ℓ

To find volume of glass plate: the thickness of glass plate is determined by using screw gauge by the above method. Now the glass plate is placed on a graph paper and its outline is drawn. By counting number of mm squares, its area is determined. The volume is calculated by using the equation V =Area x thickness RESULT

1. Volume of the given wire = …………… m3

2. Volume of the glass plate = ……………. m3

21



Mean 𝑽𝒕 = ………………. cm/s

Velocity of sound at room temperature 𝑽𝒕 = ………………. cm/s = ………… m/s

Room temperature t = 270c

Velocity of sound in air at zero degree Celsius 𝑽𝟎 = 𝑽𝒕 − 𝟎.𝟔 𝒕 = ………… m/s

To find unknown frequency

Trial No

Frequency of tuning fork

Hz

( 𝒏 )

First resonance Length

Second resonance length

𝑉𝑡 = 2𝑛 ℓ2 − ℓ1

Velocity of Sound

cm/s

1 cm

2 cm

Mean ℓ1

cm

1 cm

2 cm

Mean ℓ2

cm

1 512

2 480

3 426


Hz

First resonance Length

Second resonance length

𝑁 =𝑉𝑡

2 𝐿2 − 𝐿1

Unknown Frequency

1 cm

2 cm

Mean 𝐿1

cm

1 cm

2 cm

Mean 𝐿2

cm Unknown

frequency

22


THE RESONANCE COLUMN-1 AIM

1. To find velocity of sound in air at room temperature

2. To find velocity of sound in air at 00 C

3. To find the unknown frequency of the given tuning fork

APPARATUS Resonance column apparatus, tuning forks, rubber hammer

THEORY If ℓ1 and ℓ2 are the first and second resonating lengths for a tuning fork of frequency n, then

velocity of sound in air at room temperature is given by

𝑉𝑡 = 2𝑛 ℓ2 − ℓ1

Resonance occurs when frequency of the tuning fork becomes equal to the frequency of the

stationary waves produced inside the resonance column. At resonance a booming sound is

heard.

Velocity of sound at 00 C is given by 𝑉0 = 𝑉𝑡 − 0.6 𝑡

Where t =room temperature in 00 C

If L1 and L2 are the first and second resonating lengths for the tuning fork of unknown

frequency N, then 𝑁 =𝑉𝑡

2 𝐿2−𝐿1

PROCEDURE

The inner tube of resonance column apparatus is kept at lowest position and an excited

tuning fork of frequency 512Hz is held at the mouth of inner tube. The tube is raised till a

booming sound is heard. The resonating length

is measured. This is repeated and mean value gives the first resonating length ℓ1.

The tube is further raised keeping the excited tuning fork above it, till another booming

sound is heard. This length is measured. This is repeated and average gives the second

resonating length ℓ2. Then velocity of sound is calculated by using the equation 𝑉𝑡 =

2𝑛 ℓ2 − ℓ1

The experiment is repeated by using tuning fork of different frequencies and average is

found out

RESULT

1. Velocity of sound at room temperature = ……………….. m/s

2. Velocity of sound in air at 00 C = …………….. m/s

3. The frequency of given tuning fork = ………….Hz

23



To find end correction

Mean end correction =

To find the ratio of frequency of two given tuning forks


in Hz

1st resonating length 2nd resonating length

𝑒 = 𝓵𝟐 − 𝟑𝓵𝟏

𝟐

End correction

1 cm

2 cm

ℓ1

Mean cm

1 cm

2 cm

ℓ2

Mean cm

512

426

Frequency of

tuning fork in Hz

1st resonating length 2nd resonating length 𝑛1

𝑛2=

𝓵′𝟐 − 𝓵′𝟏

(𝓵𝟐 − 𝓵𝟏)

1 cm

2 cm

Mean cm

1 cm

2 cm

Mean cm

512 ℓ1= ℓ2 =

426 ℓ′1 = ℓ′2=

24


THE RESONANCE COLUMN-2 AIM

1. To find the end correction

2. To find the ratio of frequency of two given tuning forks

APPARATUS Resonance column apparatus, tuning forks, rubber hammer

THEORY If ℓ1 and ℓ2 are the first and second resonating lengths for a tuning fork then end correction is given by

𝑒 = ℓ2 − 3ℓ1

2

If ℓ1 and ℓ2 are the first and second resonation lengths of a tuning fork of frequency 𝑛1 and

ℓ1′ and ℓ2

′ are that values for tuning fork of frequency 𝑛2 then ratio of frequency is given by

𝑛1

𝑛2=

ℓ′2 − ℓ′1

(ℓ2 − ℓ1)

PROCEDURE

The inner tube of resonance column apparatus is kept at lowest position and an excited

tuning fork of frequency 512Hz is held at the mouth of inner tube. The tube is raised till a

booming sound is heard. The resonating length

is measured. This is repeated and mean value gives the first resonating length ℓ1.

The tube is further raised keeping the excited tuning fork above it, till another booming

sound is heard. This length is measured. This is repeated and average gives the second

resonating length ℓ2.

The experiment is repeated by using another tuning fork of frequency 480Hz.its first and

second resonating lengths ℓ′1 and ℓ’2 are measured.

Ratio of frequency and end corrections are determined by using the above values.

RESULT

1. End correction of the resonance column apparatus =

2. Ratio of frequency of given tuning forks =

25


.

Observations and Calculation Surrounding’s temperature 𝜽𝒔 = ................ oC

Sl. No.

Time for cooling,

t (minute.)

Temperature of water in beaker

𝜽 in °𝒄

Difference of temperature,

(𝜽 − 𝜽𝒔) oC

ln(𝜽 − 𝜽𝒔)

1 0

2 1

3 2

4 3

5 4

6 5

7 6

8 7

9 8

10 9

11 10

12 12

13 14

14 16

15 18

ln(𝜽

−𝜽𝒔)

t

26


NEWTONS LAW OF COOLING

AIM 1. To study the relationship between the temperature of a hot body and time 2. To plot graph between ln(𝜽 − 𝜽𝒔) and t 3. To plot a cooling curve APPARATUS Newton’s law of cooling apparatus- a copper calorimeter, two celsius thermometers, a stop clock, a heater, liquid (water), a clamp stand. THEORY Newton’s law of cooling states that “ the rate at which a hot body loses heat is directly proportional to the difference between the temperature of the hot body and that of its surroundings and depends on the nature of material and the surface area of the body “.

−𝒅𝑸

𝒅𝒕 ∝ (𝜽 − 𝜽𝒔)

𝒅𝑸

𝒅𝒕= −𝒌(𝜽 − 𝜽𝒔) ..... (1)

Where k is the constant of proportionality For a body of mass m and specific heat s, at its initial temperature θ higher than its surrounding’s temperature 𝜽𝒔 , the rate of loss of heat

𝒅𝑸

𝒅𝒕= 𝒎𝒔

𝒅𝜽

𝒅𝒕 .......... (2)

Using Eqs. (1) and (2), the rate of fall of temperature is given by 𝑑𝜃

𝑑𝑡=

−𝑘

𝑚𝑠(𝜃 − 𝜃𝑠) or 𝑑𝜃 = 𝑘′(𝜃 − 𝜃𝑠) dt ......(3)

where k ′ = 𝑘

𝑚𝑠 is another constant, and negative sign

indicate loss of heat , On integrating eqn. 3, we get

27


Sl. No.

Time for cooling,

t (minute.)

Temperature of water in beaker

𝜽 in °𝒄

Difference of temperature,

(𝜽 − 𝜽𝒔) oC

ln(𝜽 − 𝜽𝒔)

16 20

17 22

18 24

19 26

20 28

21 30

22 34

23 38

24 42

25 46

26 50

28


𝑑𝜃

𝜃 − 𝜃𝑠= −𝑘′ 𝑑𝑡

ln(𝜽 − 𝜽𝒔) = −𝒌′𝒕 + 𝒄 ....... (4) where c is the constant of integration, taking exponential on both sides

(𝜃 − 𝜃𝑠) = 𝑒(−𝑘 ′ 𝑡+𝑐)

𝜽 = 𝜽𝒔 + 𝑪′𝒆−𝒌′𝒕 ................. (5) where C’=𝑒𝑐 , Eq (4) shows that the shape of a plot between ln(𝜽 − 𝜽𝒔) and t will be a straight line and (5) shows that a plot of 𝜽 is exponentially decreasing with t. PROCEDURE Take some liquid (water) and heat it until it boils. Using a thermometer, note down the room temperature. Then using a stand, insert the bulb into that hot water and record the temperature and time in regular intervals. Record these values in the corresponding columns in the table. Plot a graph between time t, taken along x-axis and ln(𝜽 − 𝜽𝒔) taken along y-axis. Then plot an another graph between time t, taken along x-axis and 𝜽 taken along y-axis RESULT

1. The temperature falls quickly in the beginning and then slowly as the difference of temperature goes on decreasing.

2. The graph between ln(𝜽 − 𝜽𝒔) and t is a straight line 3. The cooling curve is an exponential decay curve

29



Trial

No

External

Resistance

R

Ohm

Balancing length for

Leclanche cell Internal resistance

𝑟 = 𝑅 ℓ1−ℓ2

ℓ2

ohm

Open circuit ℓ1

cm

Closed circuit ℓ2

cm 1

2

3

4

5

6

R G

Rh

E1

E

K2

K1

A

B

30


THE POTENTIOMETER - 1 AIM

To determine the internal resistance of a Leclanche cell

APPARATUS Potentiometer, Accumulator, Leclanche cell, resistance box, rheostat, key, galvanometer

THEORY

By the principle of potentiometer, if ℓ1 is the balancing length for leclanche cell of emf E in

open circuit,then

E α ℓ1 ……………………….. (1) When the cell E is connected to an external resistance R,

𝐸 𝑅

𝑅+𝑟 α ℓ2 …………………(2)

Where ℓ2balancing length in closed circuit

Dividing (1) and (2)

The internal resistance of the cell 𝑟 = 𝑅 ℓ1−ℓ2

ℓ2

ℓ1 = Balancing length in open circuit

ℓ2 = balancing length in closed circuit

PROCEDURE

The connections are done. The primary key K1 is closed and secondary key K2 is kept open.

Now the jockey is pressed at the ends A and B of the potentiometer and rheostat is adjusted

so that deflections at the ends A and b are opposite. Keeping the key K2 open balancing

length ℓ1 is determined. Now 5 ohm resistance is introduced in the resistance box and the

key K2 is pressed. Hence the balancing length ℓ2 of closed circuit is determined. Using the

values internal resistance is determined

The experiment is repeated by increasing the value of R and in each case internal resistance

is determined

RESULT

The internal resistance of Leclanche cell varies with external resistance.

31


l no.

Mass suspended

m in kg

Resonating length ℓ 𝒍𝟐

in cm2

𝒎

𝓵𝟐

in kg/cm2 1 2 Mean

ℓ 1 0.5 2 0.6 3 0.7 4 0.8 5 0.9 6 1

mean 𝑚

ℓ2 = …………….. kg/cm2

= …………….. kg/m2

µ= 𝑴

𝑳 linear density of wire(mass per length) =1.27 × 𝟏𝟎−𝟑

kg/m

frequency of ac is given by 𝒏 = 𝒈

𝟒µ 𝒎

𝓵𝟐 = ……………. Hz

32


THE SONOMETER – 3

AIM To determine frequency of ac using a sonometer APPARATUS Sonometer, slotted weights, step down transformer (6V) ,crocodile clips ,horse shoe magnet THEORY At resonance the natural frequency of vibration n of the sonometer wire becomes equal to the applied frequency of ac and then the wire vibrates with maximum amplitude The frequency of transverse vibration produced in a stretched string is given by

𝒏 = 𝟏

𝟐𝓵 𝑻

𝝁

ℓ=resonating length of sonometer wire T= mg Tension in the string

µ= 𝑴

𝑳 linear density of wire (mass per unit length)

at resonance the frequency of ac is given by

𝒏 = 𝒈

𝟒µ 𝒎

𝓵𝟐

PROCEDURE The sonometer wire is stretched by suspending a constant mass 500 gm. The experiment is set as shown in the fig. the bridges are kept close and ac supply is switched on. The position of magnet is adjusted at the midway between the bridges. Distance between bridges A and B is adjusted till paper rider vibrates vigorously and is thrown off. Distance between the bridges gives the resonating length. The experiment is repeated once again and mean length for that corresponding mass is

calculated. Then find 𝒎

𝓵𝟐. Experiment is repeated by changing m as 600,700,800,900,1000

gm and mean 𝒎

𝓵𝟐 is calculated.

Total mass ( M ) and length of the sonometer wire ( L ) is measured and linear density can

be calculated by using the equation µ= 𝑴

𝑳

RESULT Frequency of ac mains, n = …….. Hz

33


Distant object method

Mean f =……..……cm

…………..m u - v method

u -v graph 𝟏

𝒖 –

𝟏

𝒗 graph

… ..

Trial No.

Distance from Lens to screen

f in cm 1 2

Focus object far away from ground

34


CONVEX LENS

AIM 1. To find the focal length of a convex lens by

a) Distant object method

b) u - v method

c) from u - v graph

d) from 𝟏

𝒖 –

𝟏

𝒗 graph

2. To determine the Power of lens APPARATUS Convex lens, Illuminated wire gauze, meter scale , lens stand, white screen. THEORY Distant object method: When object is placed at infinity, image is formed at focus. The distance between lens and the screen gives focal length. u - v method: If u is the object distance and v is the image distance then focal length of the convex lens is given by

From u - v method , f = 𝒖𝒗

(𝒖+𝒗)

Where u = object distance between wire gauge and mirror v = the image distance

While drawing u - v graph and 𝟏

𝒖 –

𝟏

𝒗 graph, same scale and

origin chosen from both the axis. The focal length can be find out by using the equation.

35


Observations and Calculation

Trail No

Object Distance

u in cm

Image distance

v in cm

f =𝒖𝒗

(𝒖+𝒗)

Mean f in cm

Round off to 3 decimal places

𝟏

𝒖

𝟏

𝒗

1 2 3 4 5 6 7 8

14 16 18 20 22 24 26 28

f

= …

……

…..

cm

0.071 0.062

.

.

.

.

.

.

focal length of the convex lens is given by , f = ……… cm = ………. m

Power of lens P = 𝟏

𝒇 = ……….. D

From u - v graph,

𝒇 = (𝑶𝑨+𝑶𝑩)

𝟒 = …… cm = …………. m

From 𝟏

𝒖 –

𝟏

𝒗 graph,

𝒇 = 𝟐

(𝑶𝑸+𝑶𝑹) = …….. cm = ………… m

36


From u - v graph, From 𝟏

𝒖 –

𝟏

𝒗 graph,

Power of lens, P = 𝟏

𝒇

PROCRDURE Distant object method: Lens is placed on a stand and focus an object at large distance (a tree far away from window) to form an image on screen. By varying distance between lens and screen, clear image is formed at focus. The distance between lens and the screen gives focal length. u-v method: Lens is placed at a distance (u ) from the wire gauze, as given in table and by adjusting screen clear image is formed on the screen. Now image distance (v) is measured. Using this focal length is calculated. This method is repeated for each values of (u) given in the table mean focal length is

calculated. Then u - v graph and 𝟏

𝒖 –

𝟏

𝒗 graph is plotted. Focal

length from the graph is can be calculated.

RESULT 1. Focal length of convex lens

a) Distance object method = …………. m

b) u - v method =…………. m

c) From u - v graph =…………. m

d) From 𝟏

𝒖 –

𝟏

𝒗 graph =…………. m

2. Power of lens =…………. D

𝒇 = (𝑶𝑨 + 𝑶𝑩)

𝟒 𝒇 =

𝟐

( 𝑶𝑸+𝑶𝑹 )

37



Lens used

Trail No

Object Distance

u in cm

Image distance

v in cm

Focal length

f= 𝒖𝒗

(𝒖+𝒗)

Mean focal length f in cm

Co

nve

x le

ns

1 2 3 4 5 6 7 8

14 16 18 20 22 24 26 28

f ’ =

……

……

..cm

Co

nve

x le

ns

an

d

Co

nca

ve le

ns

1 2 3 4 5 6 7 8

28 30 32 34 36 38 40 42

f =

……

……

… c

m

Focal length of the given concave lens F = 𝒇𝒇′

( 𝒇′−𝒇 ) = ………. cm

F = …………m

38


CONCAVE LENS

AIM To find focal length of concave lens by contact method APPARATUS Concave lens, convex lens, screen, illuminated wire gauge. THEORY If f is the focal length of combination of lens and f ’ is the focal length of convex lens , then the focal length of concave lens is given by

F = 𝒇𝒇′

( 𝒇′−𝒇 )

PROCRDURE

First find the focal length of convex lens ( f ’) using u-v method. Lens is placed at a distance (u ) from the wire gauze, as given in table and by adjusting screen clear image is formed. Now image distance (v) is measured. This method is repeated for each values given in the table. Now the convex and concave lens are placed in contact and stick together using insulation tap ( since concave lens can’t form real images). Now find the focal length of the combination ( f ) using u-v method as explained later. Form f and f ’ calculate f of concave lens. RESULT Focal length of the given concave lens, F = ……… m

39



Trial no Voltmeter reading V

Ammeter reading I

1 2 3 4 5 6 7 8 9

10

0 1 2

2.5 3

3.2 3.2 3.2 3.2 3.2

0 0 0

10 25 50 60 75

100 120

40


ZENER DIODE AIM To draw V-I graph APPARATUS Zener diode, voltmeter, milli ammeter, rheostat, key, battery, connecting wire. THEORY A junction diode specially designed to work only in one reverse breakdown voltage is called zener diode. In reverse bias, the potential barrier is large. Due to that reverse current through the diode is almost zero. On increasing the reverse voltage to a certain value, current increases suddenly. This voltage is called zener voltage. PROCEDURE Connections are made as shown in the figure. Using rheostat the voltage across diode is made at 0.1V and corresponding current is noted. The voltage is increased as 0.2 , 0.3 , 0.4 , 0.5...... and in each time milliammeter reading is taken. A graph is plotted with voltage along X axis and current along –ve Y axis RESULT The V-I graph is plotted.

41


42


THE DIODE CHARECTERISTICS AIM 1. To draw the V-I characteristics of a pn junction diode in forward bias. 2. To determine static resistance. 3. To determine dynamic resistance. 4. Knee voltage. APPARATUS Diode IN 4007 , milliammeter , voltmeter , battery , rheostat , key THEORY Forward bias: in forward bias, P side of diode is connected to positive terminal of battery and N side is connected to negative terminal. Forward characteristic is obtained by plotting voltage along X axis and current along Y axis.

Dynamic resistance (ac resistance) = 𝐜𝐡𝐚𝐧𝐠𝐞 𝐢𝐧 𝐬𝐦𝐚𝐥𝐥 𝐟𝐨𝐫𝐰𝐚𝐫𝐝 𝐛𝐢𝐚𝐬 𝐯𝐨𝐥𝐭𝐚𝐠𝐞

𝐜𝐡𝐚𝐧𝐠𝐞 𝐢𝐧 𝐟𝐨𝐫𝐰𝐚𝐫𝐝 𝐜𝐮𝐫𝐫𝐞𝐧𝐭

Static resistance (dc resistance) = 𝐟𝐨𝐫𝐰𝐚𝐫𝐝 𝐛𝐢𝐚𝐬 𝐯𝐨𝐥𝐭𝐚𝐠𝐞

𝐟𝐨𝐫𝐰𝐚𝐫𝐝 𝐜𝐮𝐫𝐫𝐞𝐧𝐭

Knee voltage: The forward voltage at which forward current raises sharply is called knee voltage.

43



Trial No.

Voltmeter reading V in volt

Ammeter reading

I in mA I in A

Static Resistance

R=𝑽

𝑰 in ohm

1 0.1 2 0.2 3 0.3 4 0.4 5 0.5 6 0.6 7 0.7 8 0.8 9 0.9

10 1.0 From graph

Static resistance = 𝑶𝑷

𝑶𝑸 = …………. ohm

Dynamic resistance = ∆𝑽

∆𝑰 =

𝑨𝑩

𝑩𝑪 =…………. ohm

Knee voltage =..................... volt

44


PROCEDURE The connections are made as shown in figure. Using rheostat the voltage across diode is made at 0.1V and corresponding current is noted. The voltage is increased as 0.2 , 0.3 , 0.4 , 0.5...... and in each time milliammeter reading is taken. A graph is plotted with voltage along X axis and current along Y axis. From the graph static resistance and dynamic resistance is calculated. RESULT 1. The V-I characteristics of the diode is drawn 2. The static resistance = ................... ohm 3. The Dynamic resistance = ............... ohm 4. Knee voltage =..................... volt

45


Observations and Calculation (resistor -length of wire-25 cm)

Trial no

Resistance In the box

R in ohm

Resonating length when X is in

𝒍

=(𝒍𝟏 + 𝒍𝟐)

𝟐

Mean

(𝟏𝟎𝟎 − 𝓵)

𝑿 =𝐑𝓵

(𝟏𝟎𝟎 − 𝓵)

in ohm

Left gap

𝓵1

in cm

Right gap

𝓵2

in cm

1 1

2 2

3 3

4 4

5 5

6 6

46


THE METER BRIDGE AIM 1. To determine resistance of the given wire 2. To determine resistivity of the material of given wire APPARATUS Meter bridge, given wire, resistance box, single key, connecting wire, and galvanometer THEORY The working principle of Meter Bridge is Wheatstone bridge. If l is the balancing length the bridge wire from the side of unknown resistance X and R is the known resistance, then

𝑋

𝑅=

ℓ

(100 − ℓ)

Unknown resistance 𝑿 =𝑹 𝒍

(𝟏𝟎𝟎−𝒍 )

Resistivity of the given wire, ρ = 𝝅𝒓𝟐𝑿

𝑳

Where r = radius of the wire L = length of the wire

47


To find the radius of wire using screw gauge



1𝑚𝑚

100 = 0.01mm

Zero coincidence =

Zero correction, Z =


Si No

PSR mm

Observed

HSR

Corrected HSR

= (HSR+Z )

Corrected HSR X LC

mm

Total Reading =

PSR + (Corrected HSR x LC)

mm

1

2

3

4

5

6

7

8



𝟐 = ……………… mm = ……………. m

Length of the wire 𝑳= ……………..…..cm = …………….. m

Resistivity of material of wire, ρ = 𝝅𝒓𝟐𝑿

𝑳 = …………… ohmmeter

48


PROCEDURE Connections are made as shown in figure. The unknown resistance is connected in the left gap and resistance box is introduced in the right gap. The key is closed and a suitable resistance R = 1 ohm is introduced in the box. The jockey is moved over the meter bridge wire till null deflection is obtained. The balancing length AJ = 𝑙1 is measured from the side of unknown resistance. Now X and R are interchanged and balancing length BJ = 𝑙2 is measured. The mean balancing length is obtained. Thus unknown resistance is calculated. The experiments are repeated for different values of R and mean X is calculated. Radius is determined by using a screw gauge and length by a scale. Then resistivity is calculated. RESULT 1. The resistance of the given wire, X = ............... ohm 2. The resistivity of the material wire, ρ = ........ ohm meter

49


Observations and Calculation (resistor – length of wire -50 cm)

Mean R = ………….. ohm

Resistance of the conductor R=𝑽

𝑰 = …………… ohm.

Conductance C=𝟏

𝑹 = ………….. mho

From V-I Graph

Resistance R = 𝑩𝑪

𝑨𝑩 = …………. ohm

Trial No

Ammeter Reading

I in ampere

Voltmeter reading V in volt

Resistance R=𝑽

𝑰

in ohm

1 2 3 4 5 6 7 8

50


OHM'S LAW 1 AIM 1. To plot V-I graph of the given wire 2. To determine resistance of the given wire 3. To determine conductance of the given wire 4. To determine resistivity of the given wire APPARATUS cell, key, the given wire, voltmeter, ammeter, rheostat, connecting wire THEORY Ohm's law states that at constant temperature, the potential difference across the ends of a conductor is directly proportional to current flowing through the conductor.


𝑰

from V-I Graph resistance R = 𝑩𝑪

𝑨𝑩

Conductance C=𝟏

𝑹

Resistivity of material of wire ρ = 𝝅𝒓𝟐𝑹

𝑳

where R=resistance of wire r=radius of wire L=length of wire

51


To find the radius of wire using screw gauge



1𝑚𝑚

100 = 0.01mm

Zero coincidence =

Zero correction, Z =


Si No

PSR mm

Observed HSR


Corrected HSR X LC

mm

Total Reading PSR + (Corrected HSR x LC)

mm

1 2 3 4 5 6 7 8



𝟐 = ……………… mm = ……………. m

Length of the wire 𝑳= …………….….. cm = …………….. m

Resistivity of material of wire ρ = 𝝅𝒓𝟐𝑹

𝑳 = …………… ohmmeter

52


PROCEDURE Connections are made as shown in fig. The key is pressed & rheostat is adjusted to get a current 0.8A in the ammeter. The corresponding volt meter reading is noted. The current is increased as 1A,1.2A,1.4A,1.6A...........& in each

time voltmeter reading is recorded. Now R=𝑽

𝑰 is calculated &

mean value is taken. A Voltage-current graph is plotted & slope of V-I graph gives resistance of the conductor. Measure the radius of wire using a screw gauge and length using a meter scale. Hence calculate resistivity of the conductor. RESULT

1. V-I graph of the given wire is plotted 2. Resistance of the given wire

1. By calculation=..............ohm 2. From graph=...............ohm

3. Conductance of the wire=...........mho 4. Resistivity of the wire=.............ohmmeter

53


Observations and Calculation To find the resistance of 1𝑠𝑡 wire R1 (of length 50 cm )

To find the resistance of 2𝑛𝑑 wire R2 (of length 25 cm )

Trial No

Ammeter Reading

I in ampere

Voltmeter reading

V in volt Resistance R=

𝑽

𝑰

in ohm

Mean R In ohm

1

R1

= …

……

……

…o

hm

2 3

4 5

6 7

8

Trial No

Ammeter Reading

I in ampere

Voltmeter reading


𝑽

𝑰

in ohm

Mean R In ohm

1

R2

= …

……

……

…o

hm

2

3 4

5 6

7

8

54


0HM'S LAW 2 AIM 1. Compare resistance of given two wires by ohm’s law. 2. Compare resistance of given two wires by drawing V-I graph. 3. Verify law of combination of resistance in series. 4. Verify law of combination of resistance in parallel. APPARATUS Cell, key, the given wire, voltmeter, ammeter, rheostat, connecting wire

THEORY Ohm's law states that at constant temperature, the

potential difference across the ends of a conductor is directly r to current flowing through the conductor.


𝑰

from V-I Graph resistance R = 𝑩𝑪

𝑨𝑩

Ratio of resistance of two wire = 𝐑𝟏

𝑹𝟐

When to resistance R1&R2 are connected in series the effective resistance in given by Rs= R1+R2. When to resistance R1&R2 are connected in Parallel the effective resistance in given by PROCEDURE

Connections are made as shown in fig. The key is pressed & rheostat is adjusted to get a current 0.8A in the ammeter. The corresponding volt meter reading is noted.

𝑹𝒑 =𝑹𝟏𝑹𝟐

(𝑹𝟏 + 𝑹𝟐)

55


To find the effective resistance when R1 & R2 are connected in series

Experimental value Rs= ………….ohm

To find the effective resistance when R1&R2 are connected in parallel

Experimental value Rp= ………….ohm

Ratio of resistance of two wires by ohm’s law, 𝐑𝟏

𝑹𝟐= ………….

Ratio of resistance of two wires from graph, 𝐑𝟏

𝑹𝟐= ……………

Effective resistance in series connections (Theoretical value) , Rs=R1+R2 = ……….. ohm

Effective resistance in parallel connections (Theoretical value),

𝑹𝒑 =𝑹𝟏𝑹𝟐

(𝑹𝟏+ 𝑹𝟐) = ………….ohm

Trial No

Ammeter Reading

I in ampere

Voltmeter reading


𝑽

𝑰

in ohm

Mean R In ohm

1

Rs

= …

……

……

…o

hm

2

3 4

5 6

7 8

Trial No

Ammeter Reading

I in ampere

Voltmeter reading


𝑽

𝑰

in ohm

Mean R In ohm

1

Rp

= …

……

……

…o

hm

2 3

4 5

6

7 8

56


The current is increased as ,1.2 A, 1.4 A, 1.6 A...........& in each

time voltmeter reading is recorded. Now R=𝑽

𝑰 is calculated &

mean value is taken. A V-I graph is plotted & slope of V-I graph gives resistance R1 of the conductor. Now first wire is replaced by second wire & the experiment is repeated as in the previous case. The mean value of R2 is determined. Now, R1, R2 are connected in series & parallel. The whole procedure is repeated in both cases & the effective resistance R1

& R2 are calculated. RESULT

1. Ratio of resistance of two wires by ohm’s law, 𝐑𝟏

𝑹𝟐=

2. Ratio of resistance of two wires from graph, 𝐑𝟏

𝑹𝟐=

3. Effective resistance in series connections a. Theoretical value Rs= …………..ohm b. Experimental value Rs=…………..ohm

The Theoretical value & Experimental value agrees & hence law of combination of resistance in series is verified.

4. Effective resistance in parallel connections a. Theoretical value Rp= ……………… ohm b. Experimental value Rp= ………………ohm

The Theoretical value & Experimental value agrees & hence law of combination of resistance in parallel verified.

57


Observations and Calculation Normal incidence method

focal length of a concave mirror f = 𝑹

𝟐 = ………. cm

= ………. m

Trial No.

Distance between Mirror and gauze

when image form near gauze u = v = R

in cm

Mean R in cm

1 2

58


CONCAVE MIRROR AIM To find the focal length of a concave mirror by

a) Normal incidence method

b) u - v method

c) from u - v graph APPARATUS Concave mirror, illuminated wire gauge, meter scale, mirror stand, white screen. THEORY In a concave mirror reflection takes place from inner curved surface. The distance between pole and the principle focus of the mirror is called focal length. Normal incidence method: The mirror is mounted in front of the wire gauze and adjusted till a clear image is formed by the side of the wire gauze itself. The distance between mirror and image is radius of curvature.

focal length of a concave mirror, f = 𝑹

𝟐

u - v method Focal length of the convex mirror is given by

𝟏

𝒇=

𝟏

𝒖 +

𝟏

𝒗

f = 𝒖𝒗

(𝒖+𝒗)

Where u = object distance

(distance between wire gauge and mirror) v = image distance (distance between wire image and mirror)

59


u - v method

u - v graph mean f = ……………….. m

𝒇 = (𝑶𝑨 + 𝑶𝑩)

𝟒

= ………….. cm = ………….. m

Trial No.

Object Distance

u in cm

Image Distance

v in cm

f = 𝒖𝒗

(𝒖+𝒗)

in cm

Mean f

in cm

1 36 2 38 3 40 4 42 5 44 6 46 7 48 8 50

60


From u - v graph: While drawing u - v graph same scale and same origin chosen from both the axes. The focal length can be found out by using the equation


𝟒

PROCEDURE Normal incidence method: The mirror is mounted in front of the wire gauze and adjusted till a clear image is formed by the side of the wire gauze itself. The distance between mirror and image is radius of curvature. The experiment is repeated and the mean value is calculated. u-v method: the mirror is placed 36 cm from illuminated wire gauze and adjust the screen until a clear image is formed on the screen placed near the side of wire gauze. The distance between mirror and screen is measured. The experiment is repeated for different values of u and in each case v is measured. Focal length is calculated in each case and mean value is determined. u-v graph : A graph is plotted with u along x - axis and v along y- axis as shown in figure. A bisecter to XOY is drawn ( at 45°)which meats the graph at P. the distance OA and OB is found and focal length is calculated by using the equation


𝟒

RESULT Focal length of the given concave lens

a) Normal incidence method, f =……….m

b) u - v method, f =……….m

c) from u - v graph, f =……….m

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Step 1 Step 2


Focal length of the convex mirror, f = ………….. cm =………….. m

Trial no.

Distance between Convex lens and

wire gauze

u in cm

Distance between screen and mirror

R in cm

Focal

length f= 𝑹

𝟐

in cm

Mean f

in cm

1 36 2 38 3 40 4 42 5 44 6 46 7 48 8 50

62


CONVEX MIRROR AIM To find the focal length of a convex mirror APPARATUS Convex Mirror, convex lens , Illuminated wire gauze, meter scale , lens stand, white screen. THEORY In a convex mirror, reflection take place from outer curved surface. The distance between its principle focus and pole is called focal length. IF f is the focal length and R is the radius of

curvature then f= 𝑹

𝟐 .

PROCEDURE Convex lens is placed at a distance u as given in the table. Then adjust the screen. A magnified image of the illuminated wire gauze is obtained on the screen by using a convex lens. Then the convex mirror is placed in between the screen and the lens with its reflecting surface towards the object. The position of the convex mirror is adjusted so as to get the clear image side by side of the wire gauze. The distance between the convex mirror and the screen is measured and this distance is equal to the radius of curvature. The experiment is repeated by changing distance between object and convex lens. From the value of R , focal length is calculated. And mean focal length is calculated RESULT Focal length of the convex mirror, f = ……………. m

63



Trial

No

Balancing length for Ratio of e.m.f

𝑬𝟏

𝑬𝟐 =

𝓵𝟏

𝓵𝟐

Leclanche cell 𝓵1

in cm

Daniel cell 𝓵2

in cm 1

2

3

4

5

6

64


THE POTENTIOMETER - 2 AIM To compare e.m.f of Leclanche cell and Daniel cell APPARATUS Potentiometer, Accumulator, Leclanche cell, Daniel cell, resistance box, rheostat, key, three terminal key, galvanometer THEORY By the principle of potentiometer, if ℓ1 is the balancing length for leclanche cell of emf 𝑬1 then 𝑬1 α ℓ1 …………………(1) if ℓ2 is the balancing length for Daniel cell of emf 𝑬2 then 𝑬2 α ℓ2 …………………(2) Dividing (1) and (2)

Ratio of e.m.f of the two cells 𝑬𝟏

𝑬𝟐 =

𝓵𝟏

𝓵𝟐

PROCEDURE

The connections are made as shown in the figure. The primary key K1 is closed and insert key in the three terminal keys such that 𝑬1 is connected. Now the rheostat is adjusted such that galvanometer deflects to both sides when jockey touches both ends (A and B) of potentiometer wire. Make sure that when 𝑬2 is connected , galvanometer deflect towards both sides as later. Now by connecting 𝑬1 balancing length ℓ1 is determined using the jockey. Now 𝑬2 is connected and balancing length ℓ2 is determined without any changes in rheostat. After finding one set of reading (𝓵1 & 𝓵2) , the experiment is repeated by varying the current from primary circuit by adjusting rheostat. Make sure that each time ℓ1 & ℓ2 is measured in the same setting of rheostat. Now ratio of e.m.f of the two

cells 𝑬𝟏

𝑬𝟐 is determined from the mean of

𝓵𝟏

𝓵𝟐

RESULT

Ratio of e.m.f of the two cells 𝑬𝟏

𝑬𝟐 =

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