Using the principle of homogeneity of dimensions, which of the following is correct?

T^{2} = \frac{4 π^{2} r^{3}}{G M}

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T^{2} = 4 π^{2} r^{2}

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T^{2} = \frac{4 π^{2} r^{3}}{G}

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T = \frac{4 π^{2} r^{3}}{G}

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Solution

The correct option is D $T^{2} = \frac{4 π^{2} r^{3}}{G M}$
$T^{2} = \frac{4 π^{2} r^{3}}{G M}$
Taking dimensions on both sides, we get
$[T]^{2} = \frac{[L]^{3}}{[M^{- 1} L^{3} T^{- 2} M]} = [M^{0} L^{0} T^{2}]$
$∴ L H S = R H S$
Now, $T^{2} = 4 π^{2} r^{2}$
Taking dimensions on both sides
$[T]^{2} = [L]^{2}$ $∴ L H S \neq R H S$
Now, $T^{2} = \frac{4 π^{2} r^{3}}{G}$
Taking dimensions on both sides
$[T]^{2} = \frac{[L]^{3}}{[M^{- 1} L^{3} T^{2}]} = [M^{1} L^{0} T^{- 2}]$ $∴ L H S \neq R H S$
Now, $T = \frac{4 π^{2} r^{3}}{G}$
$[T] = \frac{[L]^{3}}{[M^{- 1} L^{3} T^{- 2}]} = [M L^{0} T^{2}]$ $∴ L H S \neq R H S$

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