Physics - Unit & Dimension Question with Solution | TestHub

Given, $v=\frac{A\sqrt{x}}{x+B}$....(i) Dimensions of $v=$ dimensions of potential energy $=\left[{\mathrm{ML}}^2{\mathrm{T}}^{-2}\right]$ From Eq. (i) Dimensions of $B=$ dimensions of $x=\left[{\mathrm{M}}^0{\mathrm{LT}}^0\right]$ $\therefore$ Dimensions of $A$ $\begin{array}{rl}& =\frac{\text{ dimension of }v\times \text{ dimensions of }(x+B)}{\text{ dimensions of }\sqrt{x}}\\ & =\frac{\left[{\mathrm{ML}}^2{\mathrm{T}}^{-2}\right]\left[{\mathrm{M}}^0{\mathrm{LT}}^0\right]}{\left[{\mathrm{M}}^0{\mathrm{L}}^{1/2}{\mathrm{T}}^0\right]}=\left[{\mathrm{ML}}^{5/2}{\mathrm{T}}^{-2}\right]\end{array}$ Hence, dimensions of $AB$ $=\left[{\mathrm{ML}}^{5/2}{\mathrm{T}}^{-2}\right]\left[{\mathrm{M}}^0{\mathrm{LT}}^0\right]=\left[{\mathrm{ML}}^{7/2}{\mathrm{T}}^{-2}\right]$