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(B) The dimensions of the given quantities are:$[v] = [L T^{-1}]$$[r] = [L]$$[g] = [L T^{-2}]$Now,check the dimensions of each option:Option $A$: $[v^

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$\frac{v^2}{rg}$

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(B) The dimensions of the given quantities are:$[v] = [L T^{-1}]$$[r] = [L]$$[g] = [L T^{-2}]$Now,check the dimensions of each option:Option $A$: $[v^2 r / g] = \frac{[L^2 T^{-2}] [L]}{[L T^{-2}]} = [L^2]$Option $B$: $[v^2 / (rg)] = \frac{[L^2 T^{-2}]}{[L] [L T^{-2}]} = \frac{[L^2 T^{-2}]}{[L^2 T^{-2}]} = [M^0 L^0 T^0] = 1$ (Dimensionless)Option $C$: $[v^2 / (g/r)] = [v^2 r / g] = [L^2]$Option $D$: $[v^2 r g] = [L^2 T^{-2}] [L] [L T^{-2}] = [L^4 T^{-4}]$Therefore,the expression $\frac{v^2}{rg}$ is dimensionless.

Given that $v$ is the speed,$r$ is the radius,and $g$ is the acceleration due to gravity. Which of the following is dimensionless?

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