1) Aim
2) Apparatus
3) Diagram
4) Formula with explanation/law
5) Theory
6) Observations
7) Observation Table
8) Calculation
9) Result/Conclusion
10) Precautions
11) Questions
12) Graph
आता आपण Experiment चे Solutions पाहण्यास सुरुवात करुया :-
Class 12 Physics Practical No 5 : Sonometer Law of Tension Solutions Maharashtra Board
[Page No 43]
Aim :
To study the relationship between the length of a given wire and tension for constant frequency.
Apparatus :
Sonometer, slotted weights with hanger, a tuning fork of unknown frequency, rubber pad, light paper rider, meter scale.
Diagram :
Formula :
n = 1/2l*√T/m
Where,
n = frequency of vibration of the wire.
T = tension applied to the wire.
m = mass per unit length of the wire
l = vibrating length of the wire.
T = 4n²l²m
i.e. T*α*l² when n and m are kept constant
Theory :
The frequency of stationary transverse vibrations of a stretched string is directly proportional to the square root of the tension in the wire, when length of vibrating string and mass per unit length of the wire are kept constant.
∴ n*α*√T
∴ n/√T = constant
For verification l*α√T or √T/l = constant
[Page No 44]
Observation :
g = 980 cm/s² Frequency of tuning fork n = 512 Hz.
| Obs. No. |
Mass attached to the wire including hange Mgm |
Tension T=mg dyne |
Vibrating length of wire | l2 cm2 |
T/l2 dyne/cm2 |
||
|---|---|---|---|---|---|---|---|
| l1 cm | l2 cm | l3 cm | |||||
| 1 | 1000 | 98*104 | 9.9 | 9.8 | 9.85 | 97.01 | 1.010*104 |
| 2 | 1500 | 147*104 | 11.2 | 11.1 | 11.15 | 124.3 | 1.182*104 |
| 3 | 2000 | 196*104 | 13.5 | 13.6 | 13.55 | 183.5 | 1.068*104 |
| 4 | 2500 | 245*104 | 14.9 | 15.0 | 14.95 | 223.6 | 1.095*104 |
| 5 | Unknown 3000 | Tx295*104 | 16.0 | 16.1 | lx16.05 | 257.5 | 1.146*104 |
Calculations :
| Obs.No. | 1 | 2 | 3 | 4 | 5 |
|---|---|---|---|---|---|
| log T=a | 1.9912 | 2.1673 | 2.2923 | 2.3892 | 2.4698 |
| log l2=b | 1.9868 | 2.0944 | 2.2637 | 2.3496 | 2.4698 |
| a-b=c | 0.0044 | 0.0729 | 0.0286 | 0.0396 | 0.0590 |
| Antilog c=T/l2 | 1.010*104 | 1.182*104 | 1.068*104 | 1.095*104 | 1.146*104 |
[Page No 47]
Result :
1) As T/l² = constant, when n and m are kept constant. The law of tension is verified.
2) The graph of l² against T is straight line ∴ T*α*l² Hence law of tension is verified.
3) unknown mass by calculation = 2995.6 g
4) unknown mass by graph = 3346.0 g
Precautions :
i) Wire should be free form kinks and of uniform cross section.
ii) Keep the paper rider at the mid point of knife - edges.
iii) The stem of tuning fork should rest on the sonometer box.
iv) Strike the tuning fork on the rubber pad gently.
Questions :
1) State the second law of vibrating string.
Ans : The second law of vibrating strings states that the frequency of vibration is inversely proportional to the length of the string, given the tension and linear mass density remain constant, expressed as f ∝ 1/L, where f is the frequency and L is the length of the string.
2) Why do we keep frequency constant instead of keeping vibrating length constant to verify second law of vibrating string?
Ans : We keep the frequency constant instead of the vibrating length because it is easier to control and maintain a constant frequency using a vibrating source, ensuring accurate measurements and a consistent reference for comparison, whereas maintaining a constant vibrating length is more challenging and may introduce additional variables affecting the experiment's accuracy.
3) How many nodes and antinodes are formed when the wire vibrates in fundamental mode?
Ans : In the fundamental mode of vibration, the wire forms one node at each fixed end and one antinode at the midpoint, resulting in a total of two nodes and one antinode along the vibrating wire.
Graphs :
A graph of l² against T
अशा प्रकारे मित्रांनो आपण इयत्ता बारावी Physics Practical Number 5 - Sonometer Law of Tension याचे Solutions पहिले.
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