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Question

(A) Given,power radiated $P$ is $P = \mu_{0}^{x} m^{y} \omega^{z} c^{u}$.Substituting the dimensions of the physical quantities:$[P] = [ML^{2}T^{-3}]$

Vedclass · Accepted Answer

$x=1, y=2, z=4$ and $u=-3$

Vedclass · Answer

(A) Given,power radiated $P$ is $P = \mu_{0}^{x} m^{y} \omega^{z} c^{u}$.Substituting the dimensions of the physical quantities:$[P] = [ML^{2}T^{-3}]$$[\mu_{0}] = [MLT^{-2}A^{-2}]$$[m] = [L^{2}A]$$[\omega] = [T^{-1}]$$[c] = [LT^{-1}]$Equating dimensions: $[ML^{2}T^{-3}] = [MLT^{-2}A^{-2}]^{x} [L^{2}A]^{y} [T^{-1}]^{z} [LT^{-1}]^{u}$.Comparing powers of $M, L, T, A$:$M: x = 1$$A: -2x + y = 0 \implies y = 2x = 2$$L: x + 2y + u = 2 \implies 1 + 2(2) + u = 2 \implies u = -3$$T: -2x - z - u = -3 \implies -2(1) - z - (-3) = -3 \implies -2 - z + 3 = -3 \implies z = 4$.Thus,$x=1, y=2, z=4, u=-3$.

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