PHYSICS FOR CHEMISTRY PRACTICALS

(PH 1102)

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Index

S.No Date Name of Experiment Page No.

Staff signature

Marks

1.

2.

3.

4.

5.

6.

7.

8.

Total Marks

2

Date: Dept. No:

Experiment No:

YOUNG’S MODULUS BY STRETCHING

Aim:To determine the Young’s modulus of the material of a given wire

by horizontal stretching.

Apparatus Required:

Vernier Microscope, wire, digital screw gauge, a hanger with slotted weights, etc...

Formula:

E =mgLπ r2 l Nm-2

Where,

E – Young’s Modulus of Elasticity (Nm-2)m –mass for which extension is determined (kg).L – Length of the wire (m)r – Radius of the given wire (m)g – 9.8 ms-2.l - Mean extension of the wire for the mass ‘m’ (m).

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Tabular Column:

1. To determine the extension in the wire:Least Count = 0.001 cmT.R = M.S.R + (V.C×L.C)

Load×10-3 kg

Readings of the Vernier Microscope(cm) Mean×10-2 m

Extension for a load of 4m×10-2 m

Mean extension (l) for m

Loading Unloading

M.S.R VC T.R M.S.R VC T.R

W

W+500

W+1000

W+1500

W+2000

W+2500

W+3000

W+3500

Mean (l) for m=

4

2. To determine the radius of the given wire using digital Screw gauge:

S.No Diameter of the given wire (d) ×10-3 m

1.2.3.4.5.

Mean diameter (d) = ---------------------------×10-3 m.

r =d2

r = ----------------×10-3m

L = Length of the wire = ------------------×10-2m

Calculation:

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Result:

The Young’s Modulus of the material of the given wire (E) = ----------------Nm-2

Staff Signature

Date:

6

Date: Dept. No:

Experiment No:

RIGIDITY MODULUS – TORSIONAL PENDULUM

Aim:

To determine the rigidity modulus of the material of the given wire using torsional pendulum.

Apparatus Required:

A circular metal disc, suspension wire, meter scale, stop-watch, digital vernier calipers and digital screw gauge.

Formula:

G = 8 πIr4 ( l

T2 ) Nm-2

Where,G - Rigidity modulus of the material of the wire (Nm-2)π - 3.14 (no unit)l - Length of the torsion wire (m)r - Radius of the given wire (m)T- Time Period of oscillation (in seconds)I - Moment of inertia of the circular disc about the axis of rotation,

I = M R2

2 kg m2

M - Mass of the circular disc (kg)R -Radius of the disc. (m)

Tabular Column:

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1. To determine the Time period of oscillations :

S.No Length of the

pendulum

l

(×10-2m)

Time taken for 10 oscillations (sec)

Time Period of

oscillations

T (sec)

lT2 ms-2

Trial 1 Trial 2 Mean

1.

2.

3.

4.

5.

Mean (l

T2 ) = --------------ms-2

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2. To determine the radius of the given wire using digital Screw gauge

S.No Diameter of the given wire d ×10-3m

1.2.3.4.5.

Mean diameter (d) = ---------------×10-3mRadius (r) = --------------×10-3m

Observations:

Mass of the disc found using digital balance M= ------------------×10-3kgCircumference of the disc, 2πR = -----------×10-2m

R = −−−−−¿2π

¿×10-2m.

Calculation:

I = M R2

2 kg m2

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Result:

1. The moment of inertia of the disc = ------------------(Kg m2)

2. Rigidity Modulus of the material of the given wire = ---------------(Nm-2)

Staff Signature

Date:

Date: Dept. No:10

Experiment No:

SURFACE TENSION AND INTERFACIAL SURFACE TENSION BY DROP WEIGHT METHOD

Aim: (i)To determine the surface tension of water and(ii)To determine the interfacial tension between water and kerosene by the

method of drops.

Apparatus:A glass funnel, a glass tube of suitable diameter, a small rubber tubing,

pinch clip, water and kerosene, digital balance, Hare’s apparatus and digital screw gauge.

Formula:

1. Surface tension of water by the method of drops is given by

σ = mg3.8 r Nm-1

Where,σ – Surface Tension (Nm-1)m - Mass of one drop (kg)g - Acceleration due to gravity (9.8 ms-2)r - Radius of the glass tube. (m)

2. Interfacial surface tension between kerosene and water is given by

σ = mg3.8 r [1-

ρ2

ρ1] Nm-1

Where,σ – Surface Tension (Nm-1)m - Mass of one dropg - Acceleration due to gravity (9.8 ms-2)r - Radius of the glass tube. (m)ρ1 – Density of water (kg m-3)ρ2 – Density of liquid (kg m-3)

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Tabular Column:

1. To determine the surface tension of water by method of drops:

Mass of the beaker (W) = ---------------×10-3kg

2. To determine the interfacial surface tension of water by method of drops:

Mass of the beaker+liquid+drops (W) ---------------×10-3kg

S.No Mass of the beaker+liquid+drops

Mass of one drop

Mean×10-3kg

1. W+30

2. W+60

3. W+90

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S.No Mass of the beaker +drops

Mass of one drop

Mean×10-3kg

1. W+302. W+60

3. W+90

3. Hare’s apparatus:

S.No Height of the water column

(h1)

Height of the liquid column

(h2)

ρ2

ρ1=

h1

h2

1.23.4.5.

Meanρ2

ρ1 = ---------

4. To determine the radius of the tube using digital Screw gauge

S.No Diameter of the given wire d ×10-3m

1.2.3.4.5.

Mean diameter (d) = ---------------×10-3m

Radius (r) = --------------×10-3m

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Calculation:

Result:

1. The Surface tension of water = -------------------------- Nm-1.2. The interfacial Surface tension

between water and the given Liquid = ------------------ Nm-1.

Staff Signature

Date:

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Date: Dept. No:Experiment No:

SURFACE TENSION OF A LIQUID BY CAPILLARY RISE

Aim : To determine the surface tension of a liquid by capillary rise method.

Apparatus: Capillary tube, travelling microscope, a beaker containing the given liquid, etc.

Formula :

Surface tension of the liquid = h rρg2 Nm1.

Where r Radius of the capillary tube (m)

h Capillary rise (m)

Density of the liquid (kg/m3)

g Acceleration due to gravity = 9.8 ms2

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Observations:

Table1: To find the capillary rise

Correct reading (CR) = M.S.R. + V.S.C. x L.C. L.C. = 0.001 cm.

S.No.

Reading of the meniscus in the capillary tube (h1)

Reading of the meniscus in the steel needle (h2)

Capillary rise h = (h1 h2)

102mMSR VSC CR MSR VSC CR

Mean height (h) = ............. 102 m

Table2:

To find the radius of the capillary tube

L.C. = 0.001 cm.

S.No. Position of the cross-wire

Microscope reading Diameter of the capillary tube R1 R2

102 m

MSR VSC MSR + (VSC LC)

1 Left R1

2 Right R2

3 Top R1

4 Bottom R2

Mean diameter, d = …………………… 102 m

Mean radius r = d2 = …………………… 102 m

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Calculation:

= hrρg2 Nm1.

r

h

g

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Result:

The surface tension of the given liquid was found to be = ---------------N/m.

Staff Signature

Date:

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Date: Dept. No:

Experiment No:

COEFFICIENT OF VISCOSITY OF THE GIVEN LIQUID – POISEUILLE’S FLOW METHOD

Aim:

To determine the coefficient of viscosity of the given liquid (water) by capillary flow method.

Apparatus required:

Graduated burette, beaker, Capillary tube, and stop clock, meter scale, stand, liquid

Formula:

η = πρg r4

8 lV ( h t ) Nsm-2

Where,

η – Coefficient of viscosity of the given liquid (Nsm-2)r- Radius of the capillary tube (m)h- Pressure head (m)ρ- Density of the liquid (Kg m3)g- Acceleration due to gravity (ms-2)l- Length of the capillary tube (m)V -volume of the liquid (cc) (1 cc = 10-6 m3)T-time taken for flow of 5cc liquid(s)

r=√ mπρL ×10-2m

Where,

r- Radius of the capillary tubem - Mass of the mercury pellet (Kg)

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π – 3.14ρ- Density of Mercury (13,600 Kg m-3)L - Length of the mercury pellet in the capillary tube (m)

Tabular column:

1. To determine ‘ht’

BuretteLevel

Height (h)

×10-2m

Time noted while crossing

level (s)

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Range(cc)

Time for flow of 5cc liquid ‘t’ (s)

Height to initial reading

h1

(m)

Height to final reading

h2

(m)

Pressure head

h =h1+h2

2 – h0

(m)

ht

ms

2. To determine the radius of the capillary tube:

1. Length of the mercury pellet in the capillary tube = ----------------×10-2m

2. Mass of the empty dish (m1) = ----------------×10-3kg

3. Mass of the empty dish + mercury pellet (m2) = ----------------×10-3kg

4. Mass of mercury (m = m2-m1) = ----------------×10-3kg

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r=√ mπρL ×10-2m

Calculation:

Result:

The Coefficient of viscosity of a given liquid by capillary tube method =

----------------------------- Nsm-2

Staff Signature

Date:

Date: Dept. No:

22

Experiment No:

SPECIFIC HEAT OF LIQUID – ELECTRICAL HEATINGAim:

To determine the specific heat capacity of a liquid using method of electrical heating

Apparatus Required:Calorimeter, Ammeter, Voltmeter, Rheostat, Battery

Formula

Heat lost = Heat gained

VIt=[ w1+(w2−w1) s2 ](θ2−θ1)

s2=1

w 2−w 1 [ VItθ2−θ1

−w1 s1] J/Kg/K

WhereV is the voltmeter reading in voltsI is the ammeter reading in amperest is the time of flow of current in secondsw1 is the mass of calorimeter with stirrer in kgw2 is the mass of calorimeter with stirrer and water in kgs1 is the specific heat of calorimeter (J/Kg/K)s2 is the specific heat of liquid (J/Kg/K)θ1 is room temperature (°C)θ2 is final corrected temperature (°C)

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OBSERVATION:

Ammeter reading (I) = _________ A

Voltmeter reading V = __________ Volts

Time of flow of current (t) = ________ Seconds

Mass of calorimeter with stirrer (w1) = __________ g

Mass of calorimeter with stirrer and water (w2) = __________ g

Room temperature θ1=¿ _________ ℃

Final corrected temperature θ2=¿ _________ ℃

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Half-time correction:

Time Temperature

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Calculation:

Result: Specific heat capacity of a given liquid is determined as ________.

Staff Signature

Date:

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Date: Dept. No:

Experiment No:

SONOMETER – VERIFICATION OF LAWSAim:

To verify the first and second law of transverse vibrations in stretched strings using a sonometer.

Apparatus Required:Sonometer, tuning forks, weight hanger with slotted weights,

Rubber hammer, and Bridges and Paper rider.

Laws of transverse vibrations in stretched strings:

First Law

The frequency (n) of tranverse vibrations of a stretched string is inversely proportional to the length (l) of the vibrating segment provided the tension (T) and mass per unit length (m) of the wire remains constant.

nl = constantSecond Law

The frequency (n) of transverse vibrations of a stretched string is directly proportional to the square root of the tension (√T), if the length (l) of the vibrating segment and mass per unit length (m) of the wire are maintained a constant.

(√T )l = constant

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Tabular Column:

1. To prove first law:

Mass suspended in the string (M) = ------------------------Kg.

S.No Frequency of tuning fork ‘n’Hz

Length of vibrating segment ‘l’

×10-2 m

nlHz - m

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2. To prove second law:

Frequency of tuning fork = ------------------Hz

S.No Mass suspended in string ‘M’ ×10-3 kg

Tension in string T= Mg

(N)

Tension √T Length of the vibrating segment’l’

×10-2 m

(√T )l

Calculation:

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Result: The first and second laws of transverse vibrations in stretched strings are verified using a sonometer.

Staff Signature

Date:

30

Date: Dept. No:

Experiment No:

COMPOUND BAR PENDULUM

Aim:

To determine the periods of oscillation of the compound pendulum about various points of suspension on both sides of the centre of gravity, to draw a suitable graph and use it to determine the radius of gyration about the centre of gravity of the bar and the value of acceleration due to gravity at the place.

Apparatus Required:

Compound pendulum, knife – edge, metre scale and stop watch.

Formula:

g = 4π2¿) ms-2

Where,

g- acceleration due to gravity(ms-2)π- 3.14 (no unit)l- Length of the equivalent simple pendulum(x10-2 m)T –Time period of Oscillation (s)

Tabular Column:

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1. To determine the time period ( T)

Hole No

Distance of knide

edge from the fixed end

Al

(×10-2)

Time taken for 10 oscillations (s)

Time periodT (s)Trial 1 Trial 2 Mean

Model graph:

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2. To determine g

Time period (T) from the

graph(s)

Length of the equivalent pendulum ( x10-2 m)

g=4π2¿)ms-2

PR QS l =PR+QS2

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3. To determine the radius of gyration (k):

S.No h1= PS2 h2=

QR2

k = √h1 h2

Calculation:

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Result:

1. The acceleration due to gravity at the place (g) =------------ (ms-2)2. The radius of gyration of the compound bar pendulum about its centre of

gravity (k) = ------------------------- (m).

Staff signature

Date:

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