Question

Using the principle of homogeneity of dimensions, which of the following is correct?
$\text{A}\text{. }{{T}^{2}}=\dfrac{4{{\pi }^{2}}{{r}^{3}}}{GM}$
$\text{B}\text{. }{{T}^{2}}=4{{\pi }^{2}}{{r}^{2}}$
$\text{C}\text{. }{{T}^{2}}=\dfrac{4{{\pi }^{2}}{{r}^{3}}}{G}$
$\text{D}\text{. }T=\dfrac{4{{\pi }^{2}}{{r}^{3}}}{G}$

Answer 1

Hint: According to the principle of homogeneity of dimensions, in an equation of physical quantities, the dimensional formula of the expressions on both the sides of the expression must be the same. Go through each option and check which one obeys the principle of homogeneity.

Complete step by step answer:

According to the principle of homogeneity of dimensions, in an equation of physical quantities, the dimensional formula of the expressions on both sides of the expression must be the same. For example, $\text{speed = }\dfrac{\text{distance}}{\text{time}}$. Here, the quantity speed will have the dimensional formula equal to the ratio of the dimensional formulas of distance and time respectively.
To give a correct option to this question, we have to go through each option and check which one obeys the principle of homogeneity.
Let us first write down the dimensional formulas for T, G, M and r so that we can directly substitute.
The dimension of M is mass. Therefore, the dimensional formula of M is [M].
(Dimensional formulas are enclosed in square brackets).
Dimension of r is length (since it is radius). Hence, its dimensional formula is [L].
T has the dimension of time and its dimensional formula is [T].
You may know that G is a universal constant used in the equation of gravitational force between two bodies.
Its dimensional formula is $\left[ G \right]=\left[ {{M}^{-1}}{{L}^{3}}{{T}^{-2}} \right]$.
Numbers do not have any dimensions. They are dimensional quantities. Therefore, they are not included in the equation of dimensional formula.
Let us now begin to analyse each option.
Option A:
The given equation is ${{T}^{2}}=\dfrac{4{{\pi }^{2}}{{r}^{3}}}{GM}$.
The dimensional formula of ${{T}^{2}}$ is $\left[ {{T}^{2}} \right]$. ………(i)
Dimensional formula of $\dfrac{4{{\pi }^{2}}{{r}^{3}}}{GM}$ is $\left[ \dfrac{4{{\pi }^{2}}{{r}^{3}}}{GM} \right]=\left[ \dfrac{{{L}^{3}}}{{{M}^{-1}}{{L}^{3}}{{T}^{-2}}.M} \right]=\left[ {{M}^{0}}{{L}^{0}}{{T}^{2}} \right]=\left[ {{T}^{2}} \right]$. ………(ii).
If we compare (i) and (ii), we get that the dimensional formula ${{T}^{2}}$ and $\dfrac{4{{\pi }^{2}}{{r}^{3}}}{GM}$ is same i.e. $\left[ {{T}^{2}} \right]$.
Therefore, the dimensional formula on each side of the equation is the same. Therefore, the principal of homogeneity holds true.
Hence, the correct option A.

Note: One may argue that more than one option may be correct since we have not checked the other options. The argument is valid. However, it is not required in this case.
Consider the equation in option A i.e. ${{T}^{2}}=\dfrac{4{{\pi }^{2}}{{r}^{3}}}{GM}$. We got the dimensional formula of $\dfrac{4{{\pi }^{2}}{{r}^{3}}}{GM}$ as $\left[ {{T}^{2}} \right]$. Now if we change anything like remove or add any quantity (other than numbers) involved in the given equation the dimensional formula will change and we do not want that to happen.