If $G$ is the universal gravitational constant and $g$ is the acceleration due to gravity,then the dimensions of $\frac{G}{g}$ will be ...................
- A$[M^{-1} L^2]$
- B$[M^{-1} L]$
- C$[M^{-2} L]$
- D$[M^{-1} L^{-2}]$
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Similar Questions
Which of the following is not a dimensionless quantity?
The dimensional formula for angular momentum is
MediumIIT 1983
View SolutionWhich of the following physical quantities has no dimensions?
Easy
View SolutionIf $e$ is the electronic charge,$c$ is the speed of light in free space,and $h$ is Planck's constant,the quantity $\frac{1}{4 \pi \varepsilon_{0}} \frac{| e |^{2}}{h c}$ has dimensions of .......
DifficultJEE MAIN 2021
View SolutionMatch List-$I$ with List-$II$:
List-$I$ List-$II$ $(a)$ $h$ (Planck's constant) $(i)$ $[M L T^{-1}]$ $(b)$ $E$ (kinetic energy) $(ii)$ $[M L^2 T^{-1}]$ $(c)$ $V$ (electric potential) $(iii)$ $[M L^2 T^{-2}]$ $(d)$ $P$ (linear momentum) $(iv)$ $[M L^2 I^{-1} T^{-3}]$
Choose the correct answer from the options given below:
| List-$I$ | List-$II$ |
| $(a)$ $h$ (Planck's constant) | $(i)$ $[M L T^{-1}]$ |
| $(b)$ $E$ (kinetic energy) | $(ii)$ $[M L^2 T^{-1}]$ |
| $(c)$ $V$ (electric potential) | $(iii)$ $[M L^2 T^{-2}]$ |
| $(d)$ $P$ (linear momentum) | $(iv)$ $[M L^2 I^{-1} T^{-3}]$ |
Choose the correct answer from the options given below:
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