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)

$T^{2}=\frac{4\pi^{2}r^{3}}{GM}$

Answer 2

$T^{2}=\frac{4\pi^{2}r^{3}}{GM}$ 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, $T=\frac{4\pi^{2}r^{3}}{G}$ $\left[T\right]=\frac{\left[L^{3}\right]}{\left[M^{-1}L^{3}T^{-2}\right]}=\left[ML^{0}T^{2}\right]$ $\therefore LHS \ne RHS$.