Q. From the dimensional consideration, which of the following equation is correct
A) $T = 2\pi r\sqrt{\begin{array}{r} \frac{R^{3}}{GM} \\ \ \end{array}}$
B) $T = 2\pi\sqrt{\begin{array}{r} \frac{GM}{R^{3}} \\ \ \end{array}}$
C) $T = 2\pi\sqrt{\begin{array}{r} \frac{GM}{GR^{2}} \\ \ \end{array}}$
D) $T = 2\pi\sqrt{\begin{array}{r} \frac{R^{2}}{GM} \\ \ \end{array}}$
Explanation
By substituting the dimensions in $T = 2\pi\sqrt{\begin{array}{r} \frac{R^{3}}{GM} \\ \ \end{array}}$ we get $\sqrt{\begin{array}{r} \frac{L^{3}}{M^{- 1}L^{3}T^{- 2} \times M} \\ \ \end{array}} = T$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.