Question

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

• A. $T^2=\frac{4{\pi }^2{r}^3}{GM}$
• B. $T^2=4{\pi }^2{r}^2$
• C. $T^2=\frac{4{\pi }^2{r}^3}{G}$
• D. $T=\frac{4{\pi }^2{r}^3}{G}$

Answer 1

Answer: (A)

$T^2=\frac{4{\pi }^2{r}^3}{GM}$
Taking dimensions on both sides, we get
${\left[T\right]}^2=\frac{{\left[L\right]}^3}{\left[M^{-1}{L}^3{T}^{-2}M\right]}=\left[M^0{L}^0{T}^2\right]$
$\therefore LHS=RHS$
Now, $T^2=4{\pi }^2{r}^2$
Taking dimensions on both sides
${\left[T\right]}^2=\left[L^2\right]$
$\therefore LHS\ne RHS$
Now, $T^2=\frac{4{\pi }^2{r}^3}{G}$
Taking dimensions on both sides
${\left[T\right]}^2=\frac{{\left[L\right]}^3}{\left[M^{-1}{L}^3{T}^{-2}\right]}=\left[M^1{L}^0{T}^2\right]$
$\therefore LHS\ne RHS$
Now, $T=\frac{4{\pi }^2{r}^3}{G}$
$\left[T\right]=\frac{\left[L^3\right]}{\left[M^{-1}{L}^3{T}^{-2}\right]}=\left[M{L}^0{T}^2\right]$
$\therefore LHS\ne RHS$.