The velocity v of a particle depends upon its position x according...

According to the principle of homogeneity of dimensions, each term in an equation must have the same dimensional formula.

Given equation: $v = A\sqrt{x} + \dfrac{B}{x^2}$

The dimensions of velocity $v$ are $[L T^{-1}]$ and distance $x$ are $[L]$.

Equating dimensions of the first term to $v$:
$[A\sqrt{x}] = [v]$
$[A] [L^{1/2}] = [L T^{-1}]$
$[A] = \dfrac{[L T^{-1}]}{[L^{1/2}]} = [L^{1/2} T^{-1}]$

Equating dimensions of the second term to $v$:
\left[\dfrac{B}{x^2}\right] = [v]$<br>$\dfrac{[B]}{[L^2]} = [L T^{-1}]$<br>$[B] = [L T^{-1}] \times [L^2] = [L^3 T^{-1}]$<br><br>Now, finding the dimensional formula for the product$A \times B$:<br>$[A \times B] = [L^{1/2} T^{-1}] \times [L^3 T^{-1}] = [L^{7/2} T^{-2}]$<br><br>Including mass, the dimensional formula is$[M^0 L^{7/2} T^{-2}]$.<br><br>Answer:$[M^0 L^{7/2} T^{-2}]