Dimensional Analysis Application
Mediumphysics
Using dimensional analysis, determine which of the following cannot be the formula for the time period T of a simple pendulum (where l is length, g is acceleration due to gravity, and m is mass)?
Select the correct option:
Solution
Incorrect! Answer:
T = 2π√(gl)
- Goal: Check for dimensional homogeneity. Dimension of T must be [T1].
- Term Analysis:
- l/g=[LT−2][L]=[T2]=[T1]. This is consistent with T.
- gl=[LT−2][L]=[L2T−2]=[LT−1]. This is the dimension of Velocity, not time.
- Dimensionless constants: 2π, 1/2π, and (m/m) are dimensionless and do not affect the validity relative to units.
- Conclusion: Formula 2πgl is dimensionally incorrect.
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About This Question
- Subject
- physics
- Chapter
- units and measurements
- Topic
- dimensional analysis application
- Difficulty
- Medium
- Year
- 2025
Solution
Correct Answer:
T = 2π√(gl)
- Goal: Check for dimensional homogeneity. Dimension of T must be [T1].
- Term Analysis:
- l/g=[LT−2][L]=[T2]=[T1]. This is consistent with T.
- gl=[LT−2][L]=[L2T−2]=[LT−1]. This is the dimension of Velocity, not time.
- Dimensionless constants: 2π, 1/2π, and (m/m) are dimensionless and do not affect the validity relative to units.
- Conclusion: Formula 2πgl is dimensionally incorrect.
This medium difficulty physics question is from the chapter units and measurements, covering the topic of dimensional analysis application. It appeared in the 2025 exam.
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