CLASS-12-PHYSICS -PRACTICALS -AND- READING- TABLES
CLASS XII PHYSICS PRACTICAL LIST OBSERVATION TABLES AND RESULTS
Class XII Physics Practical List
- To determine the resistivity of two/three wires by plotting a graph between potential difference and current.
- To find the resistance of a given wire/standard resistor using a meter bridge.
-
To verify the laws of combination (series) of resistances using a meter bridge.
OR
To verify the laws of combination (parallel) of resistances using a meter bridge. - To determine the resistance of a galvanometer by the half-deflection method and to find its figure of merit.
-
To convert the given galvanometer (of known resistance and figure of merit) into a voltmeter of desired range and to verify the same.
OR
To convert the given galvanometer (of known resistance and figure of merit) into an ammeter of desired range and to verify the same. - To find the frequency of AC mains using a sonometer
Least Count of Ammeter: ______ A
Least Count of Voltmeter: ______ V
(B) Observational Table (V–I Readings)| S. No. | Current, I (A) | Potential Difference, V (V) |
|---|---|---|
| 1 | 0.20 | 0.40 |
| 2 | 0.40 | 0.80 |
| 3 | 0.60 | 1.20 |
| 4 | 0.80 | 1.60 |
| 5 | 1.00 | 2.00 |
| Quantity | Symbol | Value |
|---|---|---|
| Length of the wire | L | 100 cm = 1.00 m |
| Radius of the wire | r | 0.25 mm = 2.5 × 10−4 m |
A graph is plotted between potential difference (V) on Y-axis and current (I) on X-axis. The graph is a straight line passing through the origin.
CalculationsFrom V–I graph:
Slope = V / I = Resistance (R)
Taking two points from the graph:
V1 = 2.0 V, I1 = 1.0 A
R = V / I = 2.0 / 1.0 = 2 Ω
Area of cross-section of the wire:
A = πr2
= 3.14 × (2.5 × 10−4)2
= 3.14 × 6.25 × 10−8
= 1.96 × 10−7 m2
Resistivity (ρ):
ρ = (R × A) / L
= (2 × 1.96 × 10−7) / 1.00
= 3.92 × 10−7 Ω m
The resistivity of the given wire is 3.92 × 10−7 Ω m.
Practical–2 Aim: To find the resistance of a given wire using a metre bridge. ApparatusMetre bridge, resistance wire, unknown resistance wire, standard resistance box, galvanometer, jockey, battery, key, connecting wires.
Formula UsedAt balance point of metre bridge:
R / S = l / (100 − l)
∴ R = S × l / (100 − l)
Observations (A) Least CountLeast count of metre scale: 0.1 cm
Observation Table| S. No. | Standard Resistance S (Ω) | Balancing Length l (cm) | 100 − l (cm) | Resistance R (Ω) |
|---|---|---|---|---|
| 1 | 2 | 40 | 60 | 1.33 |
| 2 | 3 | 45 | 55 | 2.45 |
| 3 | 4 | 50 | 50 | 4.00 |
For Observation 1:
S = 2 Ω, l = 40 cm
R = S × l / (100 − l)
= 2 × 40 / 60
= 1.33 Ω
For Observation 2:
S = 3 Ω, l = 45 cm
R = 3 × 45 / 55
= 2.45 Ω
For Observation 3:
S = 4 Ω, l = 50 cm
R = 4 × 50 / 50
= 4.00 Ω
Mean Resistance:
R̄ = (1.33 + 2.45 + 4.00) / 3
= 2.59 Ω (approx.)
The resistance of the given wire using a metre bridge is found to be 2.6 Ω (approximately).
Precautions- All connections should be tight and clean.
- The jockey should be pressed gently on the wire.
- The balance point should lie between 30 cm and 70 cm.
- The key should be closed only while taking readings.
Class XII Physics Practical – 3
Aim
To verify the laws of combination of resistances in series and parallel using a metre bridge.
Apparatus
Metre bridge, two given resistances R1 and R2, resistance box, galvanometer, jockey, battery, plug key and connecting wires.
Formulae Used
For series combination:
Rs = R1 + R2
For parallel combination:
1 / Rp = 1 / R1 + 1 / R2
For metre bridge:
R = S × l / (100 − l)
Observations
Least Count
Least count of metre scale = 0.1 cm
Observation Table – Series Combination
| S. No. | Standard Resistance S (Ω) | Balancing Length l (cm) | 100 − l (cm) | Rs (Ω) |
|---|---|---|---|---|
| 1 | 4 | 50 | 50 | 4.00 |
Observation Table – Parallel Combination
| S. No. | Standard Resistance S (Ω) | Balancing Length l (cm) | 100 − l (cm) | Rp (Ω) |
|---|---|---|---|---|
| 1 | 1 | 33 | 67 | 0.49 |
Calculations
Series Combination
Given: S = 4 Ω, l = 50 cm
Rs = S × l / (100 − l)
= 4 × 50 / 50
= 4 Ω
Let R1 = 2 Ω and R2 = 2 Ω
R1 + R2 = 2 + 2 = 4 Ω
Hence, resistance in series follows the law: Rs = R1 + R2
Parallel Combination
Given: S = 1 Ω, l = 33 cm
Rp = S × l / (100 − l)
= 1 × 33 / 67
= 0.49 Ω
Let R1 = 1 Ω and R2 = 1 Ω
1 / Rp = 1 / 1 + 1 / 1 = 2
Rp = 1 / 2 = 0.50 Ω
The experimental value closely matches the theoretical value.
Result
The laws of combination of resistances in series and parallel are verified using a metre bridge.
Precautions
- All connections should be tight and clean.
- The jockey should be pressed gently on the wire.
- The balance point should lie between 30 cm and 70 cm.
- The key should be closed only while taking observations.
Sources of Error
- End resistance of the metre bridge wire.
- Contact resistance of the jockey.
- Non-uniform cross-section of the bridge wire.
Class XII Physics Practical – 4
Aim
To determine the resistance of a given galvanometer by half-deflection method and to find its figure of merit.
Apparatus
Galvanometer, resistance box, high resistance box, cell, key, connecting wires.
Theory
When a current flows through a galvanometer, the deflection produced is directly proportional to the current.
If the deflection becomes half by connecting a shunt resistance in parallel with the galvanometer, the resistance of the galvanometer can be calculated using half-deflection method.
Formulae Used
Resistance of galvanometer:
G = (S × R) / (R − S)
Figure of merit:
k = I / θ
where I = current through galvanometer and θ = deflection.
Observations
Least Count
Least count of galvanometer scale = 1 division
Observation Table
| S. No. | Series Resistance R (Ω) | Shunt Resistance S (Ω) | Initial Deflection θ (divisions) | Half Deflection θ/2 (divisions) |
|---|---|---|---|---|
| 1 | 1000 | 50 | 20 | 10 |
Calculations
Calculation of Galvanometer Resistance:
Given:
R = 1000 Ω
S = 50 Ω
G = (S × R) / (R − S)
= (50 × 1000) / (1000 − 50)
= 50000 / 950
= 52.6 Ω (approx)
Calculation of Figure of Merit:
Current through galvanometer:
I = E / (R + G)
Let E = 2 V
I = 2 / (1000 + 52.6)
= 2 / 1052.6
= 1.9 × 10−3 A
Deflection θ = 20 divisions
Figure of merit:
k = I / θ
= (1.9 × 10−3) / 20
= 9.5 × 10−5 A/div
Result
The resistance of the given galvanometer is found to be 52.6 Ω.
The figure of merit of the galvanometer is 9.5 × 10−5 A per division.
Precautions
- The key should be closed only while taking observations.
- Connections should be neat and tight.
- The deflection should be noted carefully without parallax error.
- High resistance should be used to avoid large currents.
Sources of Error
- Change in cell emf during the experiment.
- Zero error in galvanometer.
- Contact resistance in wires and keys.
Class XII Physics Practical – 5
Aim
To convert the given galvanometer (of known resistance and figure of merit)
into a voltmeter of desired range and verify the same.
OR
To convert the given galvanometer (of known resistance and figure of merit)
into an ammeter of desired range and verify the same.
Apparatus
Galvanometer, resistance box, high resistance wire, standard voltmeter/ammeter, cell, key, rheostat and connecting wires.
Theory
A galvanometer can be converted into a voltmeter by connecting a high resistance in series with it. It can be converted into an ammeter by connecting a low resistance (shunt) in parallel with it.
Formulae Used
For Voltmeter Conversion:
Series resistance, R = (V / Ig) − G
For Ammeter Conversion:
Shunt resistance, S = (G × Ig) / (I − Ig)
where:
G = resistance of galvanometer
Ig = current for full-scale deflection
V = desired voltage range
I = desired current range
Given Data
Resistance of galvanometer, G = 50 Ω
Figure of merit, k = 5 × 10−5 A/div
Full-scale deflection, θ = 20 divisions
Ig = k × θ = 5 × 10−5 × 20 = 1 × 10−3 A
Calculations
(A) Conversion into Voltmeter
Desired range of voltmeter, V = 5 V
R = (V / Ig) − G
= (5 / 1 × 10−3) − 50
= 5000 − 50
= 4950 Ω
A resistance of 4950 Ω is connected in series with the galvanometer to obtain a voltmeter of 0–5 V range.
Verification Table (Voltmeter)
| S. No. | Voltage by Standard Voltmeter (V) | Voltage by Converted Voltmeter (V) |
|---|---|---|
| 1 | 1 | 1 |
| 2 | 3 | 3 |
| 3 | 5 | 5 |
(B) Conversion into Ammeter
Desired range of ammeter, I = 5 A
S = (G × Ig) / (I − Ig)
= (50 × 1 × 10−3) / (5 − 0.001)
≈ 0.050 / 4.999
= 0.01 Ω (approx)
A low resistance of about 0.01 Ω is connected in parallel with the galvanometer to obtain an ammeter of 0–5 A range.
Verification Table (Ammeter)
| S. No. | Current by Standard Ammeter (A) | Current by Converted Ammeter (A) |
|---|---|---|
| 1 | 1 | 1 |
| 2 | 3 | 3 |
| 3 | 5 | 5 |
Result
The given galvanometer is successfully converted into a voltmeter of range 0–5 V and an ammeter of range 0–5 A. The converted instruments give readings comparable to standard instruments.
Precautions
- Connections should be tight and correct.
- High resistance should be used for voltmeter conversion.
- Very low resistance wire should be used for shunt.
- The key should be closed only while taking readings.
Sources of Error
- Variation in resistance due to heating.
- Inaccurate value of galvanometer resistance.
- Zero error in standard instruments.
Class XII Physics Practical – 6
Aim
To determine the frequency of AC mains using a sonometer.
Apparatus
Sonometer, AC mains, electromagnet, weight box, hanger, pulley, meter scale, connecting wires.
Theory
The frequency of a stretched string fixed at both ends is given by:
n = (1 / 2l) × √(T / μ)
where
n = frequency of vibration of the string
l = vibrating length of the string
T = tension in the string
μ = mass per unit length of the string
When the frequency of the string is equal to the frequency of AC mains, the string vibrates in resonance.
Formulae Used
Frequency of AC mains:
n = (1 / 2l) × √(T / μ)
T = Mg
Observations
Least Count
Least count of meter scale = 0.1 cm
Determination of μ (Mass per unit length)
| Mass of wire (kg) | Length of wire (m) | μ = Mass / Length (kg m−1) |
|---|---|---|
| 0.002 | 1.0 | 0.002 |
Observation Table
| S. No. | Mass M (kg) | Tension T = Mg (N) | Resonant Length l (m) | Frequency n (Hz) |
|---|---|---|---|---|
| 1 | 0.50 | 4.9 | 0.50 | 50 |
Calculations
Mass per unit length, μ = 0.002 kg m−1
For M = 0.50 kg
T = Mg = 0.50 × 9.8 = 4.9 N
Resonant length, l = 0.50 m
Frequency:
n = (1 / 2l) × √(T / μ)
= (1 / (2 × 0.50)) × √(4.9 / 0.002)
= 1 × √2450
≈ 50 Hz
Result
The frequency of AC mains is found to be approximately 50 Hz.
Precautions
- The string should be free from kinks.
- Weights should be added gently.
- The pulley should be frictionless.
- Resonant length should be measured accurately.
Sources of Error
- Friction at the pulley.
- Air resistance.
- Non-uniformity of the wire.
Comments
Post a Comment
“✨ Have a question? Comment below or email me at abhishekupca@gmail.com” |