JEE Mains · Physics · STD 11 - Units,dimensions and measurement
- JEE Mains 2019
- Hard
- MCQ
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
Pick an option to check your answer.
Answer & Solution
Correct Answer
(A) \(S^{1 /2} I^{1 /2} h^0\)
Step-by-step Solution
Detailed explanation
\(\begin{array}{l}
P = k\,{s^a}{i^b}{h^c}\\
Where\,k\,is\,{\rm{dimensionless}}\,{\rm{constant}}\\
ML{T^{ - 1}} = {\left( {M{T^{ - 2}}} \right)^a}\left( {M{L^2}} \right){\left( {M{L^2}{T^{ - 1}}} \right)^c}\\
a + b + c = 1\\
2b + 2c = 1\\
- 2a - c = - 1\\
a = \frac{1}{2}\,\,\,\,\,\,\,b = \frac{1}{2}\,\,\,\,\,\,c = 0\\
{S^{1/2}}{i^{1/2}}{h^0}
\end{array}\)
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 = \frac{1}{2}\,\,\,\,\,\,\,b = \frac{1}{2}\,\,\,\,\,\,c = 0\\
{S^{1/2}}{i^{1/2}}{h^0}
\end{array}\)
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