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  • If momentum P, area A, and time T are taken to be fundamental quantities, then the dimensional formula for energy is:- Option: 1 <img alt="P A^{1 / 2} T^{-1}" src="https://learn.careers360.com/latex-image/?P%20A%5E%7B1%20/%202%7

If momentum P, area A, and time T are taken to be fundamental quantities, then the dimensional formula for energy is:-
Option: 1 P A^{1 / 2} T^{-1}
Option: 2 P^2 A T^{-2}
Option: 3 P A^{-1} T^{-2}
Option: 4 P^{\frac{1}{2}} A T^{-1}

Answers (1)

best_answer

\\ \text{Let energy,}\\ U \propto P^{a} A^{b} T^{c} \\ U=k P^{a} A^{b} T^{c} \ldots \ldots(i)\\ where \ \mathrm{k} \ \text{is dimensionless constant of proportionality}. \\ From (i)\\ \left[M L^{2} T^{-2}\right]=\left(M L T^{-1}\right)^{a} (L^2)^b T^{c}=M^{a} L^{a+2 b} T^{-a+c} \\ \text{Applying the principle of homogeneity of dimensions}, \\ we \ get\\ a=1, \\ a+2 b=2\Rightarrow b=\frac{2-a}{2}=\frac{2-1}{2}=\frac{1}{2}\\ -\mathrm{a}+\mathrm{c}=-2\Rightarrow \mathrm{c}=-2+\mathrm{a}=-2+1=-1 \\ \therefore from \ (i), U=k\left[P^{1} A^{1 / 2} T^{-1}\right]

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