Question

If $F$ is given by $F$ = ($\alpha /\beta$) ($1-e^{-\beta t^2/m}$), where $F$ is force and $m$ & $t$ are mass and time, then dimension of $\alpha$ will be :

Answer 1

Since, $F=\frac{\alpha }{\beta }(1-e^{-\frac{\beta t^2}{m}})$

Since exponential is an dimensionless quantity

Hence,

$-\frac{\beta t^2}{m}$ is a dimensionless quantity

$\therefore$ $\left[\frac{\beta t^2}{m}\right]=\left[M^0{L}^0{T}^0\right]$

$\Rightarrow \beta \left[\frac{T^2}{M}\right]=\left[M^0{L}^0{T}^0\right]$

Hence, $\beta =\frac{\left[M^0{L}^0{T}^0\right]\left[M\right]}{\left[T^2\right]}$

and the two quantities cannot subtracted added until they don't have same dimension

$\left[F\right]=\frac{\alpha }{\beta }$

since, $\left[F\right]=\left[M^1{L}^1{T}^{-2}\right]$

$\left[M^1{L}^1{T}^{-2}\right]=\frac{\left[\alpha \right]}{\left[M^1{L}^0{T}^{-2}\right]}$

$\therefore \left[\alpha \right]=\left[M^1{L}^1{T}^{-2}\right]\left[M^1{L}^0{T}^{-2}\right]$

$\left[\alpha \right]=\left[M^2{L}^1{T}^{-4}\right]$