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 $\left[ T ^{-1}\right]$.
$A =\left[ L ^{-1} T \right], B =\left[ LT ^{-1}\right]$
$A =[ LT ], B =\left[ L ^{-1} T ^{-1}\right]$
$A =\left[ L ^{-1} T ^{-1}\right], B =\left[ LT ^{-1}\right]$
$A =\left[ L ^{-1} T ^{-1}\right], B =[ LT ]$
If velocity $v$, acceleration $A$ and force $F$ are chosen as fundamental quantities, then the dimensional formula of angular momentum in terms of $v,\,A$ and $F$ would be
If Surface tension $(S)$, Moment of Inertia $(I)$ and Planck’s constant $(h)$, were to be taken as the fundamental units, the dimensional formula for linear momentum would be
If energy $(E),$ velocity $(V)$ and time $(T)$ are chosen as the fundamental quantities, the dimensional formula of surface tension will be
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 dimensional formula for $A B$ is