From the equation tan theta = frac{rg}{v^2},o | vedclass

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(C) The dimension of $	an 	heta$ is $[M^0 L^0 T^0]$ as it is a dimensionless quantity.For the right-hand side,the dimensions are: $[r] = [L]$,$[g] =

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Dimensionally correct only

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(C) The dimension of $	an 	heta$ is $[M^0 L^0 T^0]$ as it is a dimensionless quantity.For the right-hand side,the dimensions are: $[r] = [L]$,$[g] = [L T^{-2}]$,and $[v^2] = [L^2 T^{-2}]$.Thus,the dimension of $\frac{rg}{v^2}$ is $\frac{[L] [L T^{-2}]}{[L^2 T^{-2}]} = [M^0 L^0 T^0]$.Since both sides have the same dimensions,the equation is dimensionally correct.However,the correct physical formula for the angle of banking is $	an 	heta = \frac{v^2}{rg}$.Therefore,the given equation is numerically incorrect.

From the equation $\tan \theta = \frac{rg}{v^2}$,one can obtain the angle of banking $\theta$ for a cyclist taking a curve (the symbols have their usual meanings). Then say,it is

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