The potential energy of a particle varies with distance $x$ from a fixed origin as $U\, = \,\frac{{A\sqrt x }}{{{x^2} + B}}$ Where $A$ and $B$ are dimensional constants then find the dimensional formula for $A/B$

A physical quantity $\vec{S}$ is defined as $\vec{S}=(\vec{E} \times \vec{B}) / \mu_0$, where $\vec{E}$ is electric field, $\vec{B}$ is magnetic field and $\mu_0$ is the permeability of free space. The dimensions of $\vec{S}$ are the same as the dimensions of which of the following quantity (ies)?

$(A)$ $\frac{\text { Energy }}{\text { charge } \times \text { current }}$

$(B)$ $\frac{\text { Force }}{\text { Length } \times \text { Time }}$

$(C)$ $\frac{\text { Energy }}{\text { Volume }}$

$(D)$ $\frac{\text { Power }}{\text { Area }}$

If momentum $[ P ]$, area $[ A ]$ and time $[ T ]$ are taken as fundamental quantities, then the dimensional formula for coefficient of viscosity is :

The equation of a circle is given by $x^2+y^2=a^2$, where $a$ is the radius. If the equation is modified to change the origin other than $(0,0)$, then find out the correct dimensions of $A$ and $B$ in a new equation: $(x-A t)^2+\left(y-\frac{t}{B}\right)^2=a^2$.The dimensions of $t$ is given as $\left[ T ^{-1}\right]$.

Consider two physical quantities A and B related to each other as $E=\frac{B-x^2}{A t}$ where $E, x$ and $t$ have dimensions of energy, length and time respectively. The dimension of $A B$ is