Dimensional Analysis
A physicist examines the coefficient of viscosity that appears when describing the slow flow of honey through a narrow tube. What is the dimensional formula of this coefficient?
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About This Question
- Subject
- physics
- Chapter
- physics and measurement
- Topic
- dimensional analysis
- Difficulty
- Hard
- Year
- 2025
Solution
Correct Answer:
[M1L−1T−1]
As outlined in NCERT Class 11, Chapter 2 (Units and Measurements), the dimensions of a coefficient follow from the defining relation in which it appears. The viscous force is given by F = η A (dv/dx), where η is the coefficient of viscosity, A is area and dv/dx is the velocity gradient. Rearranging gives η = F / (A × dv/dx). The velocity gradient dv/dx has dimensions [L^{1}T^{-1}] / [L^{1}] = [T^{-1}]. Substituting, η = [M^{1}L^{1}T^{-2}] / ([L^{2}][T^{-1}]) = [M^{1}L^{-1}T^{-1}]. The option [M^{1}L^{1}T^{-2}] is wrong because it equals force, not viscosity. The option [M^{1}L^{-2}T^{-1}] is wrong as the length power is miscalculated. The option [M^{1}L^{-1}T^{-2}] is wrong since it matches pressure. A final check confirms the SI unit of viscosity, the pascal-second, reduces to kg per metre per second, which matches [M^{1}L^{-1}T^{-1}].
This hard difficulty physics question is from the chapter physics and measurement, covering the topic of dimensional analysis. It appeared in the 2025 exam.
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