The potential energy of a particle moving along $x$-direction varies as $V=\frac{A x^2}{\sqrt{x}+B}$

Question

The potential energy of a particle moving along $x$-direction varies as $V=\frac{A x^2}{\sqrt{x}+B}$. The dimensions of $\frac{A^2}{B}$ are

Professor PC Thomas Classes · Answer

(c) $: V=\frac{A x^2}{\sqrt{x}+B}$ Dimensions of $V=\left[\mathrm{M}^1 \mathrm{~L}^2 \mathrm{~T}^{-2}ight]$ Dimensions of $x^2=\left[\mathrm{L}^2ight]$ Dimension of $\sqrt{x}=\left[\mathrm{L}^{1 / 2}ight]$ From the principle of homogeneity $B=\left[L^{1 / 2}ight]$ $ \begin{aligned} & {\left[\mathrm{M}^1 \mathrm{~L}^2 \mathrm{~T}^{-2}ight]=\frac{A\left[\mathrm{~L}^2ight]}{\left[\mathrm{L}^{1 / 2}ight]}} \ & \Rightarrow A=\left[\mathrm{M}^1 \mathrm{~L}^{1 / 2} \mathrm{~T}^{-2}ight] \Rightarrow A^2=\left[\mathrm{M}^2 \mathrm{~L}^1 \mathrm{~T}^{-4}ight] \ & 	herefore \quad \frac{A^2}{B}=\frac{\left[\mathrm{M}^2 \mathrm{~L}^1 \mathrm{~T}^{-4}ight]}{\left[\mathrm{L}^{1 / 2}ight]}=\left[\mathrm{M}^2 \mathrm{~L}^{1 / 2} \mathrm{~T}^{-4}ight] \end{aligned} $