Question

If velocity $\left(v\right)$, time $\left(t\right)$, and mass $\left(m\right)$ are taken as fundamental quantities, then the dimensional formula for gravitational potential would be___

Answer 1

Step 1: Dimensional Formula

1. The dimensional formula is the formula derived using the fundamental quantities.
2. These fundamental quantities cannot be derived from any other relation.
3. There are seven such fundamental quantities:

1. mass
2. length
3. time
4. current
5. temperature
6. amount of substance
7. luminosity

Step 2: Gravitational Potential

Gravitational potential can be expressed as

$P=mgh$

where $m$ is the mass, $g$ is the acceleration due to gravity, and $h$ is the height of the body.

Upon substituting their units we get,

$P=kg\times m{s}^{-2}\times m$

$=kg{m}^2{s}^{-2}$

Therefore, the dimensional formula is $\left[M^1{L}^2{T}^{-2}\right]$.

Step 3: Dimensional formula in terms of $\left[vmt\right]$

Let $P=\left[v^x{m}^y{t}^z\right]$

$\Rightarrow$$\left[M^1{L}^2{T}^{-2}\right]=\left[{\left(m{s}^{-1}\right)}^{x}k{g}^y{s}^z\right]$

$\Rightarrow$$\left[M^1{L}^2{T}^{-2}\right]=\left[m^x{s}^{-x}k{g}^y{s}^z\right]$

$\Rightarrow$$\left[M^1{L}^2{T}^{-2}\right]=\left[m^x{s}^{-x+z}k{g}^y\right]$

$\Rightarrow$$\left[M^1{L}^2{T}^{-2}\right]=\left[M^y{L}^x{T}^{-x+z}\right]$

$\therefore y=1,x=2,-x+z=-2$

$\Rightarrow z=-2+2=0$

$\therefore P=\left[v^2{m}^1{t}^0\right]$

Hence, if velocity $\left(v\right)$, time $\left(t\right)$, and mass $\left(m\right)$ are taken as fundamental quantities, then the dimensional formula for gravitational potential would be $\left[v^2{m}^1{t}^0\right]$.