Applying the principle of homogeneity of dimensions, determine which one is correct. where T is time period, G is gravitational constant, M is mass, r is radius of orbit.

(A) \mathrm{T}^2=\frac{4 \pi^2 \mathrm{r}}{\mathrm{GM}^2}

(B) \mathrm{T}^2=4 \pi^2 \mathrm{r}^3

(D) \mathrm{T}^2=\frac{4 \pi^2 \mathrm{r}^2}{G M}

JEE Mains · Physics · STD 11 - Units,dimensions and measurement

Applying the principle of homogeneity of dimensions, determine which one is correct. where \(\mathrm{T}\) is time period, \(\mathrm{G}\) is gravitational constant, \(M\) is mass, \(r\) is radius of orbit.

Pick an option to check your answer.

Verified Solution

Correct Answer

(C) \(\mathrm{T}^2=\frac{4 \pi^2 \mathrm{r}^3}{G M}\)

Detailed explanation

According to principle of homogeneity dimension of \(LHS\) should be equal to dimensions of RHS so option \((3)\) is correct.

\(\mathrm{T}^2=\frac{4 \pi^2 \mathrm{r}^3}{\mathrm{GM}}\)

\({\left[\mathrm{T}^2\right]=\frac{\left[\mathrm{L}^3\right]}{\left[\mathrm{M}^{-1} \mathrm{~L}^3 \mathrm{~T}^{-2}\right][\mathrm{M}]}}\)

(Dimension of G is \(\left[\mathrm{M}^{-1} \mathrm{~L}^3 \mathrm{~T}^{-2}\right]\) )

\(\left[\mathrm{T}^2\right]=\frac{\left[\mathrm{L}^3\right]}{\left[\mathrm{L}^3 \mathrm{~T}^{-2}\right]}=\left[\mathrm{T}^2\right]\)

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