Physics - Unit & Dimension Question with Solution | TestHub

From the dimensional checking of the equation, we can say the dimensions of $\left[B\right]=\left[x^2\right]=\left[L^2\right]$

And dimensions of $\left[A\right]=\frac{\left[x^2\right]}{\left[tE\right]}=\frac{\left[L^2\right]}{\left[T\right]\left[M{L}^2{T}^{-2}\right]}=\frac{1}{\left[M{T}^{-1}\right]}$

Hence, $\left[A\right]=\left[M^{-1}T\right]$

Required value, $\left[AB\right]=\left[L^2{M}^{-1}{T}^1\right]$