Question

The potential energy of a particle varies with distance $x$ from a fixed origin as $U = \dfrac{{A\sqrt x }}{{{x^2} + B}}$, where $A$ and $B$ are dimensional constants then the dimensional formula for $AB\;$is
(A) $\left[ {M{L^{\dfrac{7}{2}}}{T^{ - 2}}} \right]$
(B) $\left[ {M{L^{\dfrac{{11}}{2}}}{T^{ - 2}}} \right]$
(C) $\left[ {{M^2}{L^{\dfrac{9}{2}}}{T^{ - 2}}} \right]$
(D) $\left[ {M{L^{\dfrac{{13}}{2}}}{T^{ - 3}}} \right]$

Answer 1

Hint: The expression for the potential energy of a particle is given. It is said that the potential energy of the particle varies with the distance. In the expression for potential energy two dimensional constants $A$ and $B$ are given. Then we have to find the dimensional formula for $AB\;$.

Complete step by step solution:
Let $x$ be the distance of the particle from the fixed origin.
The potential energy of the particle is given by $U = \dfrac{{A\sqrt x }}{{{x^2} + B}}$
Here, in the denominator $B$ is added to ${x^2}$. Therefore both $B$ and $x$ are of the same dimension.
Since $x$ is the distance, the dimension of $x$ is $\left[ L \right]$.
Therefore the dimension of $B$ is $\left[ {{L^2}} \right]$.
Rearranging the potential energy of the particle, $U = \dfrac{{A\sqrt x }}{{{x^2} + B}}$
We can write $A$ as,
$A = \dfrac{{U\left( {{x^2} + B} \right)}}{{\sqrt x }}$
The formula for potential energy in general is $U = mgh$
Where $m$ stands for the mass of the particle, $g$ stands for the acceleration due to gravity, and $h$ stands for the height of the particle.
The unit of $m = Kg$
The unit of $g = m/{s^2}$
The unit of $h = m$
The unit of potential energy $mgh\;$ is $Kg{m^2}/{s^2}$
The unit of $\left( {{x^2} + B} \right) = {m^2}$
The unit of $\sqrt x = {m^{\dfrac{1}{2}}}$
Therefore, the unit of $A$ can be written as,
$A = \dfrac{{Kg{m^2}}}{{{s^2}}} \times \dfrac{{{m^2}}}{{{m^{\dfrac{1}{2}}}}} = \dfrac{{Kg{m^{\dfrac{7}{2}}}}}{{{s^2}}}$
The dimension for $Kg = \left[ M \right]$
The dimension for ${m^{\dfrac{7}{2}}} = \left[ {{L^{\dfrac{7}{2}}}} \right]$
The dimension for ${s^2} = {T^2}$
Putting all this together the dimension for $A$ can be written as,
$A = \left[ {M{L^{\dfrac{7}{2}}}{T^{ - 2}}} \right]$
The dimensional formula for $B = \left[ {{L^2}} \right]$
Therefore, the dimensional formula for $AB\;$ can be written as,
$AB = \left[ {M{L^{\dfrac{7}{2}}}{T^{ - 2}}} \right] \times \left[ {{L^2}} \right] = \left[ {M{L^{\dfrac{{11}}{2}}}{T^{ - 2}}} \right]$
Therefore,

The answer is: Option (B): $\left[ {M{L^{\dfrac{{11}}{2}}}{T^{ - 2}}} \right]$

Note:
The dimensional formula of a given physical quantity is an expression showing the dimensions of the fundamental quantities. An equation connecting the physical quantity with its dimensional formula is called the dimensional equation of that physical quantity.