1

Hooke’s Law (القانون ھوك)

Room No: 1A31

Objective- To determine the spring constant زنبرك ثابت (k) by Hooke’s law.

Formula used-

Part-1 (Spring Constant by Hooke’s law)

Total force on the spring after putting some weight (for table 1)

' 0F F

0mg k L

mg k L

.g

L mk

Compare with y m x , we get

gSlope

k

So, for table1

Where g = 9.8 m/s2

Observation table-1

m (kg) x 10-3

(x-axis)ΔL increase (m) x 10-2 ΔL decrease (m) x 10-2 ΔL average (m) x 10-2

(y-axis)50100150200

Part-2 Spring Constant by time oscillation

When the weight oscillates into the spring, the equation is defined as

0

22

0

2

4( )

m mT

k

T m mk

gk

Slope

1

2

222 0

2

20

0 0

44

Compare with

4Slope =

4x

mT m

k k

y m x C

k

m CC C Slope m m

k Slope

Where C is Y-intercept

for table2

Observation table-2

m (kg) x10-3

(x-axis)

t1(Sec) t2 (Sec) t3 (Sec) t (Sec) T=t/10(Sec)

T2 (Sec2) x10-2

(y-axis)50100150200

Graph-1

0 50 100 150 200 2500

2

4

6

8

10

L

(m) x

10-2

m (kg) x 10-3

22 1

32 1

( ) x 10

( ) x 10

y ySlope

x x

gk

Slope

24k

Slope

3

Graph-2

0 50 100 150 2000

5

10

15

20

25

30T2 (S

ec2 ) x

10-2

m (kg) x 10-3

Result-19.8

( / )(.......)

gk N m

Slope

Result-224 39.5

( / )(.......)

k N mSlope

24k

Slope

22 1

32 1

( ) 10.

( ) 10

y ySlope

x x

1

Boyle’s Law (القانون بویل)

PHY-103

Room No: 1A20

Objective- To determine the atmospheric pressure ( الجويالضغط ) (Pa) by Boyle’s law.

Formula used-

constantPV

Where constant ( ).( . )ah P l a

So, ( ).( . )aPV h P l a

1( ).a

ah P

l PV

Saya

DPV

1( ).ah P D

l

0D So, ( ) 0ah P

Boyle’s experiment

aP h

2

2

Observation table

y x

A (cm) x (cm) y (cm) ( )h y x (cm. Hg)(x-axis)

( )l A x (cm)

11( )cm

l 1 21

( ).10cml

(y-axis)606060606060

Graph

x-axis: 1 cm= 2 (cm.Hg)

y-axis: 1 cm= 1 x 10-2 (cm-1)

2 22 1

2 1

2

( )x 10 (......) x 10

( )

(.......) x 10

/ (.......)

( (......)) (........) ( . )

y ya slope

x x

b Intercept

h b a

Pa h cm Hg

0 2 4 6 8 100

1

2

3

4

5

6

7

8

1/l x

10-2

h (cm.Hg)

Result: Atmospheric pressure is found to be Pa =.......... (cm.Hg)

1

Free fall ( )

Room No: 1A24

Objective- To determine the acceleration due to gravity الأرضیةالجاذبیة تسارع (g) by free fall.

Formula used-

Distance travelled by any object in particular time t is given by

21 2 'S ut g t

At start 0u 21 2 'S g t

2 2.

't S

g

Compare with y m x , we get

2

'Slope

g

So,

where,

g = 9.8 m/s2 (theoretical value of g)

g’ = experimental value of g

Observation table

S x 10-2(m)(x-axis)

t1 x10-3

(sec)t2 x10-3

(sec)t3x10-3

(sec)t x10-3

(sec)2 2

4

t (sec )

10. 10-6. 104 t2 (sec2) x 10-2

(y-axis)

8070605040

2'g

Slope

3

2

Graph

X-axis: 1cm=10 x 10-2 (m)

Y-axis: 1cm=2 x 10-2 (sec2)

0 10 20 30 40 50 60 70 800

2

4

6

8

10

12

14

16

18

t2 (sec

2 ) x 1

0-2

S (m) x 10-2

Result:

2'g

Slope = 22

....... ( / sec )(.....)

m

' 9.8 ......% x 100 x 100 ...... %

9.8

g gerror

g

22 1

22 1

( ) 10.

( ) 10

y ySlope

x x

1

Viscosity ( اللزوجةمعامل )

Room No: 1A26

Objective- To determine the coefficient of viscosity (η) for pure glycerine by Stokes’ law.

Formula used-

Force on the steel balls (downward)

31

4

3 sF mg V g r g

Force exerted on ball due to liquid/fluid (upward)

32

4

3 LF r g

Drag force by Stokes’ law exerted on the spherical objects in a continuous viscous fluid(upward)

3 6 tF v r

So, balance the force

1 2 3

3 3

3

2

F = F + F

4 46

3 34

( ). . 63

( )2. .

9

s L t

s L t

s Lt

r g r g v r

g r v r

v g r

Compare with y m x , we get

( )2.

9s Lslope g

So, finally

where, g = 9.8 m/s2

s = density of solid ball = 7800 kg/m3

L = density of liquid glycerine = 1260 kg/m3

radius of balls2

Dr

s = distance from 140 – 70 = 70 cm (fixed)vt = (s / t) x 10-2 (m/s)

2 1. ( ).

9 s Lgslope

4

2

Observation table

D(m) x10-3

r = D/2(m) x10-3

r2 (m2) x10-6

(x-axis)

t1 (sec) t2 (sec) t3 (sec) t (sec) vt = (s/t) x 10-2

(m/s)(y-axis)

6.344.763.973.17

Graph

x-axis: 1 cm= 1 x 10-6 (m2)

y-axis: 1 cm= 1 x 10-2 (m/s)

0 1 2 3 4 5 6 7 8 9 10 110

1

2

3

4

5

6

7

8

v t x 1

0-2 (

m/s

)

r2 x 10-6 (m2)

Result:

2 1' . ( ).

9 s Lgslope

2 1

x 9.8 x (7800 1260)x9 (.........)

= ........... (Pa.sec)

η = 0.934 Pa.Sec (theoretical value) at 25°C

- ' 0.934 -.......% error = x100 x100 .......%

0.934

22 1

62 1

( ) x10

( ) x10

y ySlope

x x

1

Force table ( )

Room No: 1A30

Objective- To compare the resultant angle (θR) & resultant force (R) by practical, calculationand graphical methods.

Methods

(i) By practical

F1 =........ gwt (say) & θ1 = ......˚ (say)

F2 =........ gwt (say) & θ2 = ......˚ (say)

Put the weight F1 & F2 and on the other hand put the weight R until it balanced(Note down R). Adjust the ring at the centre that it did not touch anywhere.

R = ......... gwt (balance & write)

Calculate 0' 180 ......... 180 ........R R

(ii) By calculation

2 21 2 1 2 2 1

1 01 1 2 2

2 cos ( - ) .......gwt

cos coscos ........R

R F F F F

F F

R

θR’

F2

F1

R

θR

θ1

θ2

R’

5

2

(iii) By graphical

1 cm = 20 gwt => 1 gwt = 1/20 cm

F1 = ...... gwt = ....../20 =..... cm at θ1 = ....˚F2 = ....... gwt = ....../20 =..... cm at θ2 = ....˚

θR

Calculate R = 20 x .......= ....... gwt

θR = (........)0

Results

(Comparison table)

Method θR (degree) R (gwt)PracticalCalculationGraphical

F2 =..... cm

R =..... cm x 20 =......gwt

Note θR

Θ1 = .....0

Θ2 =.....0F1=.....cm

1

Simple Pendulum ( )

Room No: 1A42

Objective: To determine the acceleration due to gravity (g) by the Simple Pendulum.

Formula Used:

For small amplitudes the period of a simple pendulum depends only on its length and the valueof the acceleration due to gravity

2T L g

Where

T – Time periodL – Length of threadg – Acceleration due to gravity

Making Square

22 4

.T Lg

Compare with y = m x we get Slope =24

g

So,

Observation table

L x 10-2

(m) X-axist1(sec)

for 10 Osc.t2 (Sec)for 10 Osc.

t3 (Sec)for 10 Osc.

t (Sec)T =

10

t

(Sec)

T2 (sec2)Y-axis

4050607080

24g

Slope

2

Graph

X-axis: 1cm=10 (m)Y-axis: 1cm=1 (sec2)

0 10 20 30 40 50 60 70 800

1

2

3

4

5

6

7

8

9

10

11

12

T2

(sec

2 )

L x 10-2(m)

2 12

2 1

( )

( ) 10

y ySlope

x x

= ………

Result

So, the experimental value of g will be

24'g

Slope

24

.......

= …….. (m/s2)

g = 9.8 (m/s2) Theoretical value

Percentage Error

' 9.8 .......% 100 x 100 .........%

9.8

g gerror

g

1

Planck’s constant ( )

Room No: 1A44Objective: To determine the value of Planck’s constant (h).

Formula Used:

Total energy of photons

0. .E K E W

0sh f eV W

Cut off voltage/Stopping voltage

Wheref = c/λ = frequency of light (Hz) or sec-1

c – Speed of light = 3 x 108 m/se – Charge of electron = 1.6 x 10-19 Cλ – Wavelength of different light

Compare with y m x c

Slope = h/e

So,

Observation table

Color λ x 10-9

(m)14( ) 10f c

(Hz)X-axis

Vs (V)(n=1)

Vs (V)(n=2)

sV (Volts)Y-axis

Yellow 578.00Green 546.07Blue 435.84Purple1 404.66Purple2 365.48

h e Slope

0.s

WhV f

e e

2

Note:(1) Use dark room.(2) Use filters only for Yellow and Green light.(3) Put the multimeter at 2V DC & switch on the h/e apparatus.

Graph

X-axis: 1cm=0.5 x 1014 (Hz)Y-axis: 1cm=0.2 (Volts)

0.0 0.5 1.0 1.5 2.0 2.5 3.0 3.5 4.0 4.5 5.0 5.5 6.0 6.5 7.0 7.5 8.0 8.50.0

0.2

0.4

0.6

0.8

1.0

1.2

1.4

1.6

1.8

2.0

Vs (

Vol

ts)

14( ) 10c

f Hz

W0 {

Result

Experimental value of Planck’s constant

19' 1.6 x 10 x ................ ..................( . )h e Slope J s

Theoretical value of Planck’s constant (h) is 6.626 x 10-34 J.s

Percentage Error

' 6.626 .........% 100 100 ...........%

6.626

h herror

h

142 114

2 1

( )(.........)x10

( ) 10

y ySlope

x x

1

Ohm’s Law (قانون أوم)

Room No:1A45

Objective: To verify the Ohm’s law for series and parallel connections.

1. Ohm’s Law

[For R1=2 Ω] [For R2=5 Ω]

I (Amp)x-axis

V (Volts)y-axis

R1= V/I (Ω)

0.10.20.30.40.5

11

....................

5 5

RR 2

2

..................

5 5

RR

0.0 0.1 0.2 0.3 0.4 0.50.00.10.20.30.40.50.60.70.80.91.01.1

2 11

2 1

( )......

( )

y yR Slope

x x

V(V

olts

)

I (Amp)

0.0 0.1 0.2 0.3 0.4 0.50.00.20.40.60.81.01.21.41.61.82.02.22.4

V(V

olts

)

I (Amp)

2 12

2 1

( )......

( )

y yR Slope

x x

Note: Plot graph only for this section.

I (Amp)x-axis

V (Volts)y-axis

R2= V/I (Ω)

0.10.20.30.40.5

2

2. Resistance in series connection

I (Amp) V (Volts) RS= V/I (Ω)0.10.20.3

From experiment .......3

SS

RR

In series connection 1 2 ........ .......... ..........SR R R

3. Resistance in parallel connection

I (Amp) V (Volts) Rp= V/I (Ω)0.10.20.3

From experiment ...........3

pp

RR

In parallel connection 1 2

1 2

......................

...........p

R RR

R R

Circuit Diagrams

In Series (Tawali) In Parallel (Tawazi)

Result: Ohm’s law verified.

1

Absorption Coefficient of γ- rays ( )

Room No:1A46

Objective: To determine the absorption coefficient of γ- rays (μ).

Formula Used:

When a gamma ray passes through matter, the probability for absorption is proportional to thethickness of the layer, the density of the material, and the absorption cross section of thematerial. The total absorption shows an exponential decrease of intensity with distance from theincident surface:

xc o cI I e

ln oc

c

Ix

I

Compare with y m x

where, x is the distance from the incident surfaceμ = nσ is the absorption coefficient, measured in cm−1

Steps

(1) Background intensity [Put the Al sheet in the 2nd last row]

1 2 3 ... ... ......... / min

3 3BG BG BG

BG

I I II C

(2) Original intensity [Put Pb source in the last row]

1 2 3 ...... ..... ............ / min

3 3o o o

o

I I II C

Source intensity without background

....... ...... ....... / minoc o BGI I I C

(3) Put Cobalt black slice 2 at a time above the Pb & Al sheet and note down I1, I2 & I3 forobservation table.

Slope

2

Observation table

No ofSlice

x (cm)X- axis

I1

(C/min)I2

(C/min)I3

(C/min)I(C/min)

c BGI I I oc

c

I

Iln oc

c

II

Y-axis2 0.44 0.86 1.28 1.610 2.0

Note:(1) Set the Time 60 Sec.(2) H.V. = 500 V(3) Press H.V.(4) Press count.(5) Each Cobalt sheet has a thickness 0.2 cm.

Graph

X-axis: 1cm=0.2 (cm)Y-axis: 1cm=0.2

0.0 0.2 0.4 0.6 0.8 1.0 1.2 1.4 1.6 1.8 2.0 2.20.0

0.2

0.4

0.6

0.8

1.0

1.2

ln (I

oc/Ic

)

x (cm)

-12 1

2 1

( )........ cm

( )

y ySlope

x x

Result

Apply this formula for absorption coefficient of Gamma-rays:

1........ ( )Slope cm

Note: Press H.V. down to zero then switch off the machine. Don’t switch off directly.

1

Capacitors (المكثفات)

Room No: 1B8

Objective: To determine the time constant (t) by charging of a capacitor.

Formula Used:

Equation for charging of a capacitor

0 (1 )t RCV V e

where

t – Time constantR – Resistance = 1 MΩC – Capacitance = 100 μF

When t = RC

10 (1 )V V e

The theoretical value of time constant

t = RC = 1 MΩ x 100 μF = 1 x 106 x 100 x 10-6 = 100 Sec

Observation table

t (Sec)x-axis

V (Volts)y1-axis(Graph-1)

0

0

lnV

V V

y2-axis(Graph-2)

0306090120150180210

00.63V V

2

240270300330360390420450480510540570600 ….. = V0

Graph

X-axis: 1cm=60 (sec)Y-axis: 1cm=1 (Volts)

0 60 120 180 240 300 360 420 480 540 6000

1

2

3

4

5

6

7

8

9

10

V(Vo

lts)

t (sec)

t ' = ...... Sec

V = 0.63 x .V0..... = ....... Volt

3

X-axis: 1cm=60 (sec)Y-axis: 1cm=1

0 60 120 180 240 300 360 420 480 540 6000

1

2

3

4

5

ln(V

0/(V

0-V

))

t (Sec)

ln(V0/(V

0-V)) = t /RC

RC=1/slope (sec)

slope = ......

t' = RC=1/..... =...... sec

Circuit diagram

Note:(1) First discharge the capacitor, if showing any voltage.(2) Switch on green button & Voltmeter power button simultaneously (without stopping note

down the t and V).(3) Put on the voltmeter dc at 10 V.(4) For Y1-axis, divide the Multi-meter voltage by 10.(5) 1 small block means 6 sec.

Result: We have found the time constant (t') = ……Sec for graph1 & (t') = ……Sec for graph2.