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(D) Let the relation be $m = K \rho^a v^b g^c$,where $K$ is a dimensionless constant.Dimensional formulas are:$m = [M^1 L^0 T^0]$$\rho = [M^1 L^{-3} T

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$m \propto \frac{\rho v^6}{g^3}$

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(D) Let the relation be $m = K \rho^a v^b g^c$,where $K$ is a dimensionless constant.Dimensional formulas are:$m = [M^1 L^0 T^0]$$\rho = [M^1 L^{-3} T^0]$$v = [M^0 L^1 T^{-1}]$$g = [M^0 L^1 T^{-2}]$Substituting these into the equation:$[M^1 L^0 T^0] = [M^1 L^{-3} T^0]^a [M^0 L^1 T^{-1}]^b [M^0 L^1 T^{-2}]^c$$[M^1 L^0 T^0] = [M^a L^{-3a+b+c} T^{-b-2c}]$Comparing powers of $M, L,$ and $T$:For $M$: $a = 1$For $T$: $-b - 2c = 0 \Rightarrow b = -2c$For $L$: $-3a + b + c = 0$Substituting $a=1$ and $b=-2c$ into the $L$ equation:$-3(1) + (-2c) + c = 0$$-3 - c = 0 \Rightarrow c = -3$Now,$b = -2(-3) = 6$Thus,$m \propto \rho^1 v^6 g^{-3} = \frac{\rho v^6}{g^3}$.

$A$ body of mass $m$ is moved by a flowing river. The mass $m$ depends on the velocity of the river $v$,the density of water $\rho$,and the acceleration due to gravity $g$. Then $m \propto$ ?

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