Question

$\frac{\alpha }{t^2}=Fv+\frac{\beta }{x^2}$

Find the dimensional formula for $\left[\alpha \right]$ and $\left[\beta \right]$ using the given equation

(here t = time, F = force, v = velocity, x = distance)

Answer 1

Since dimension of $Fv=\left[Fv\right]=\left[M^1{L}^1{T}^{-2}\right]\left[L^1{T}^{-1}\right]=\left[M^1{L}^2{T}^{-3}\right]$
so $\left[\frac{\beta }{x^2}\right]$ should also be $M^1{L}^2{T}^{-3}$
$\frac{\left[\beta \right]}{\left[x^2\right]}=M^1{L}^2{T}^{-3}$ $\left[\beta \right]=M^1{L}^4{T}^{-3}$
and $[Fv+\frac{\beta }{x^2}]$ will also have dimension $M^1{L}^2{T}^{-3}$ so L.H.S. should also have the same dimension $M^1{L}^2{T}^{-3}$
so $\frac{\left[\alpha \right]}{\left[t^2\right]}=M^1{L}^2{T}^{-3}=\left[\alpha \right]=M^1{L}^2{T}^{-1}$