A dimensionless quantity is constructed in te | vedclass

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Question

(A) For the quantity to be dimensionless,$e^\alpha \varepsilon_0^\beta h^\gamma c^\delta = M^0 L^0 T^0 A^0$.Substituting the dimensions:$[e] = AT$$[\v

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$(2n, -n, -n, -n)$

Vedclass · Answer

(A) For the quantity to be dimensionless,$e^\alpha \varepsilon_0^\beta h^\gamma c^\delta = M^0 L^0 T^0 A^0$.Substituting the dimensions:$[e] = AT$$[\varepsilon_0] = M^{-1} L^{-3} T^4 A^2$$[h] = M L^2 T^{-1}$$[c] = L T^{-1}$$(AT)^\alpha (M^{-1} L^{-3} T^4 A^2)^\beta (M L^2 T^{-1})^\gamma (L T^{-1})^\delta = M^0 L^0 T^0 A^0$Comparing powers of $M, L, T, A$:$M: -\beta + \gamma = 0 \Rightarrow \gamma = \beta$$A: \alpha + 2\beta = 0 \Rightarrow \alpha = -2\beta$$L: -3\beta + 2\gamma + \delta = 0 \Rightarrow -3\beta + 2\beta + \delta = 0 \Rightarrow \delta = \beta$$T: \alpha + 4\beta - \gamma - \delta = 0 \Rightarrow -2\beta + 4\beta - \beta - \beta = 0$ (Consistent)Let $\beta = -n$,then $\alpha = 2n, \gamma = -n, \delta = -n$.Thus,$(\alpha, \beta, \gamma, \delta) = (2n, -n, -n, -n)$.

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