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(A) The dimensions of the variables are: $[t] = T^1$, $[\rho] = ML^{-3}$, $[r] = L^1$, $[\eta] = ML^{-1}T^{-1}$, and $[\sigma] = ML^{-3}$.Given the re

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(A) The dimensions of the variables are: $[t] = T^1$, $[\rho] = ML^{-3}$, $[r] = L^1$, $[\eta] = ML^{-1}T^{-1}$, and $[\sigma] = ML^{-3}$.Given the relation $t = A \rho^{a} r^{b} \eta^{c} \sigma^{d}$, we equate the dimensions on both sides:$T^1 = (ML^{-3})^a (L)^b (ML^{-1}T^{-1})^c (ML^{-3})^d$$T^1 = M^{a+c+d} L^{-3a+b-c-3d} T^{-c}$Comparing the powers of $M, L,$ and $T$ on both sides:For $M$: $a+c+d = 0$For $L$: $-3a+b-c-3d = 0$For $T$: $-c = 1 \implies c = -1$Substituting $c = -1$ into the $M$ equation: $a+d-1 = 0 \implies a+d = 1$.Substituting $c = -1$ into the $L$ equation: $-3a+b+1-3d = 0 \implies b - 3(a+d) + 1 = 0$.Since $a+d = 1$, we get $b - 3(1) + 1 = 0 \implies b - 2 = 0 \implies b = 2$.Finally, calculating the value: $\frac{b+c}{a+d} = \frac{2 + (-1)}{1} = \frac{1}{1} = 1$.

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