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(D) The potential energy $V$ has dimensions of work or energy,which is $[M L^2 T^{-2}]$.In the denominator $(x + B)$,since $x$ is a distance,the dimen

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$M^1 L^{7/2} T^{-2}$

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(D) The potential energy $V$ has dimensions of work or energy,which is $[M L^2 T^{-2}]$.In the denominator $(x + B)$,since $x$ is a distance,the dimension of $B$ must be the same as the dimension of $x$. Therefore,$[B] = [L]$.The expression is $V = \frac{A\sqrt{x}}{x + B}$. Rearranging for $A$,we get $A = \frac{V(x + B)}{\sqrt{x}}$.Substituting the dimensions: $[A] = \frac{[M L^2 T^{-2}] [L]}{[L^{1/2}]} = [M L^2 T^{-2}] [L^{1/2}] = [M L^{5/2} T^{-2}]$.Now,the dimension of $AB$ is $[A][B] = [M L^{5/2} T^{-2}] [L] = [M L^{7/2} T^{-2}]$.

List-$I$	List-$II$
$A$. Coefficient of viscosity	$I$. $[ML^{-1}T^{-2}]$
$B$. Surface tension	$II$. $[ML^2T^{-2}]$
$C$. Pressure	$III$. $[ML^0T^{-2}]$
$D$. Surface energy	$IV$. $[ML^{-1}T^{-1}]$

The potential energy of a particle varies with distance $x$ from a fixed origin as $V = \frac{A\sqrt{x}}{x + B}$,where $A$ and $B$ are constants. The dimensions of $AB$ are

Similar Questions

$A$ dimensionless quantity

Match List-$I$ with List-$II$.
List-$I$ List-$II$
$A$. Coefficient of viscosity $I$. $[ML^{-1}T^{-2}]$
$B$. Surface tension $II$. $[ML^2T^{-2}]$
$C$. Pressure $III$. $[ML^0T^{-2}]$
$D$. Surface energy $IV$. $[ML^{-1}T^{-1}]$

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