A force is represented by F = a x ^ 2 + bt(1)/(2) , where x =...

Given: $F = a{x}^{2} + bt\dfrac{1}{2}$

Dimension of \left[a\right] = \dfrac{\left[F\right]}{\left[{x}^{2}\right]} = \dfrac{\left[ML{T}^{- 2}\right]}{\left[{L}^{2}\right]} = \left[{M}^{1}{L}^{- 1}{T}^{- 2}\right]$</p><p>Dimension of \left[b\right] = \dfrac{\left[F\right]}{\left[{t}^{\dfrac{1}{2}}\right]} = \dfrac{\left[ML{T}^{- 2}\right]}{\left[{T}^{\dfrac{1}{2}}\right]} = \left[{M}^{1}{L}^{1}{T}^{- \dfrac{5}{2}}\right]$

Required value, \left[\dfrac{{b}^{2}}{a}\right] = \dfrac{\left[{M}^{2}{L}^{2}{T}^{- 5}\right]}{\left[{M}^{1}{L}^{- 1}{T}^{- 2}\right]} = \left[{M}^{1}{L}^{3}{T}^{- 3}\right]$