Q 11-01-012NEETNEET 2021Top questionEasy
If E and G respectively denote energy and gravitational constant, then $\dfrac{E}{G}$ has the dimensions of :
Answer: (B) $[\mathrm{M^2}][\mathrm{L^{-1}}][\mathrm{T^0}]$
$[E] = [\mathrm{ML^2T^{-2}}]$.
From $F = \dfrac{Gm_1m_2}{r^2}$: $[G] = \dfrac{[\mathrm{MLT^{-2}}][\mathrm{L^2}]}{[\mathrm{M^2}]} = [\mathrm{M^{-1}L^3T^{-2}}]$.
$$\left[\frac{E}{G}\right] = \frac{[\mathrm{ML^2T^{-2}}]}{[\mathrm{M^{-1}L^3T^{-2}}]} = [\mathrm{M^2L^{-1}T^0}]$$
Solution by Sreeraj P, M.Sc Physics