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(A) Using the principle of dimensional homogeneity,we assume $F = k V^a \rho^b g^c$,where $k$ is a dimensionless constant.The dimensions of the quanti

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$V \rho g$

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(A) Using the principle of dimensional homogeneity,we assume $F = k V^a \rho^b g^c$,where $k$ is a dimensionless constant.The dimensions of the quantities are:$[F] = [M L T^{-2}]$$[V] = [L^3]$$[\rho] = [M L^{-3}]$$[g] = [L T^{-2}]$Substituting these into the equation:$[M L T^{-2}] = [L^3]^a [M L^{-3}]^b [L T^{-2}]^c$$[M L T^{-2}] = [M^b L^{3a - 3b + c} T^{-2c}]$Comparing the powers of $M, L,$ and $T$ on both sides:For $M: b = 1$For $T: -2c = -2 \Rightarrow c = 1$For $L: 3a - 3b + c = 1$Substituting $b=1$ and $c=1$ into the equation for $L$:$3a - 3(1) + 1 = 1$$3a - 2 = 1$$3a = 3 \Rightarrow a = 1$Thus,the expression is $F = k V^1 \rho^1 g^1$. Taking $k=1$,we get $F = V \rho g$.

If the buoyant force $F$ acting on an object depends on its volume $V$ immersed in a liquid,the density $\rho$ of the liquid,and the acceleration due to gravity $g$,then the correct expression for $F$ can be:

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