Question

Which of the following is dimensionless?

• A. $\frac{{\mathrm{v}}^2}{\mathrm{rg}}$
• B. $\frac{{\mathrm{v}}^2\mathrm{g}}{\mathrm{r}}$
• C. $\frac{v^{2}g}{\mathrm{r}}$
• D. ${\mathrm{v}}^2\mathrm{rg}$

Answer 1

The correct option is A

$\frac{{\mathrm{v}}^2}{\mathrm{rg}}$

Explanation for the correct option:

A: $\frac{{\mathrm{v}}^2}{\mathrm{rg}}$

1. Given $\mathrm{v}$ is the speed,$\mathrm{r}$is the radius, and $\mathrm{g}$ is the gravitational acceleration.
2. Check it out by putting $\left[\mathrm{v}\right]=\left[{\mathrm{LT}}^{-1}\right],\left[\mathrm{r}\right]=\left[\mathrm{L}\right],\left[\mathrm{g}\right]=\left[{\mathrm{LT}}^{-2}\right]$

$\frac{{\mathrm{v}}^2}{\mathrm{rg}}=\frac{{\left({\mathrm{LT}}^{-1}\right)}^2}{{\mathrm{L}}^1\left({\mathrm{L}}^1{\mathrm{T}}^{-2}\right)}={\mathrm{M}}^0{\mathrm{L}}^0{\mathrm{T}}^0$

Explanation for the incorrect option(s):

B: $\frac{{\mathrm{v}}^2\mathrm{g}}{\mathrm{r}}$

$\frac{{\mathrm{v}}^2\mathrm{r}}{\mathrm{g}}=\frac{{\left({\mathrm{L}}^1{\mathrm{T}}^{-1}\right)}^2{\mathrm{L}}^1}{{\mathrm{L}}^{-2}}={\mathrm{L}}^2$

C: $\frac{{\mathrm{v}}^2\mathrm{g}}{\mathrm{r}}$

$\frac{{\mathrm{v}}^2\mathrm{g}}{\mathrm{r}}=\frac{{\left({\mathrm{LT}}^{-1}\right)}^2{\mathrm{LT}}^{-2}}{\mathrm{L}}={\mathrm{L}}^2{\mathrm{T}}^{-4}$

D: ${\mathrm{v}}^2\mathrm{rg}$

${\mathrm{v}}^2\mathrm{rg}={\left({\mathrm{LT}}^{-1}\right)}^2\left({\mathrm{L}}^1\right)\left({\mathrm{LT}}^{-1}\right)={\mathrm{L}}^4{\mathrm{T}}^{-3}$

Hence, option (A) is the correct answer.