If $\varepsilon_0$ is permittivity of free space, $e$ is charge of proton, $G$ is universal gravitational constant and $m_p$ is mass of a proton then the dimensional formula for $\frac{e^2}{4 \pi \varepsilon_0 G m_p{ }^2}$ is
If $\varepsilon_0$ is permittivity of free space, $e$ is charge of proton, $G$ is universal gravitational constant and $m_p$ is mass of a proton then the dimensional formula for $\frac{e^2}{4 \pi \varepsilon_0 G m_p{ }^2}$ is
- A
$\left[ M ^1 L ^1 T ^{-3} A ^{-1}\right]$
- B
$\left[ M ^0 L ^0 T ^0 A ^0\right]$
- C
$\left[ M ^1 L ^3 T ^{-3} A ^{-1}\right]$
- D
$\left[ M ^{-1} L ^{-3} T ^4 A ^2\right]$
Similar Questions
Even if a physical quantity depends upon three quantities, out of which two are dimensionally same, then the formula cannot be derived by the method of dimensions. This statement
Even if a physical quantity depends upon three quantities, out of which two are dimensionally same, then the formula cannot be derived by the method of dimensions. This statement
The dimension of stopping potential $\mathrm{V}_{0}$ in photoelectric effect in units of Planck's constant $h$, speed of light $c$, Gravitational constant $G$ and ampere $A$ is
- [JEE MAIN 2020]
If $R , X _{ L }$. and $X _{ C }$ represent resistance, inductive reactance and capacitive reactance. Then which of the following is dimensionless:
If $R , X _{ L }$. and $X _{ C }$ represent resistance, inductive reactance and capacitive reactance. Then which of the following is dimensionless:
- [JEE MAIN 2023]
The equation $\frac{{dV}}{{dt}} = At - BV$ is describing the rate of change of velocity of a body falling from rest in a resisting medium. The dimensions of $A$ and $B$ are
The equation $\frac{{dV}}{{dt}} = At - BV$ is describing the rate of change of velocity of a body falling from rest in a resisting medium. The dimensions of $A$ and $B$ are
Which of the following relation cannot be deduced using dimensional analysis? [the symbols have their usual meanings]
Which of the following relation cannot be deduced using dimensional analysis? [the symbols have their usual meanings]