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The potential energy of a particle moving along x-direction varies as $V=\dfrac{Ax^2}{\sqrt{x+B}}$. The dimensions of $\dfrac{A^2}{B}$ are:
A
$[M^{3/2}L^{1/2}T^{-3}]$
B
$[M^{1/2}LT^{-3}]$
C
$[M^2L^{1/2}T^{-4}]$
D
$[ML^2T^{-4}]$
Detailed Solution
By homogeneity, $B=[L]$. Since $V=[ML^2T^{-2}]=\dfrac{A[L^2]}{[L^{1/2}]}$, $A=[ML^{1/2}T^{-2}]$. So $\dfrac{A^2}{B}=\dfrac{M^2LT^{-4}}{L^{1/2}}=[M^2L^{1/2}T^{-4}]$.