JEE Mains · Physics · STD 11 - 1. units,dimensions and measurement
If Surface tension \((S)\), Moment of Inertia \((I)\) and Planck’s constant \((h)\), were to be taken as the fundamental units, the dimensional formula for linear momentum would be
- A \(S^{1 /2} I^{1 /2} h^0\)
- B \(S^{1 /2} I^{3 /2} h^{-1}\)
- C \(S^{3 /2} I^{1 /2} h^0\)
- D \(S^{1 /2} I^{1 /2} h^{-1}\)
Answer & Solution
Correct Answer
(A) \(S^{1 /2} I^{1 /2} h^0\)
Step-by-step Solution
Detailed explanation
\begin{array}{l} P = k\,{s^a}{i^b}{h^c}\\ Where\,k\,is\,{\rm{dimensionless}}\,{\rm{constant}}\\ ML{T^{ - 1}} = {\left( {M{T^{ - 2}}} \right)^a}\left( {M{L^2}} \right){\left( {M{L^2}{T^{ - 1}}} \right)^c}\\ a + b + c = 1\\ 2b + 2c = 1\\ - 2a - c = - 1\\ a =…
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