Question

The potential energy of a particle varies with distance x from a fixed origin as $V\, = \,\dfrac{{A\sqrt x }}{{x + \,B}}$ where $A$ and $B$ are constants. The dimensions of $AB$are
(a). $\left[ {{M^1}{L^{\dfrac{5}{2}}}{T^{ - 2}}} \right]$
(b). $\left[ {{M^1}{L^2}{T^{ - 2}}} \right]$
(c). $\left[ {{M^{\dfrac{3}{2}}}{L^{\dfrac{5}{2}}}{T^{ - 2}}} \right]$
(d). $\left[ {{M^1}{L^{\dfrac{7}{2}}}{T^{ - 2}}} \right]$

Answer 1

- Hint: In this question you need to first understand that they are asking about the dimension of AB, so we need to find A and B individually and then multiply them in order to get the dimensional formula for AB. The dimensions are based on the parameters used in calculating the term.

Complete step-by-step solution -

First of all, let's discuss some terms.
Potential Energy-
An object can store energy as a result of its position and this stored energy of position is known as Potential energy. In simple words, Potential energy can be defined as the form of energy that an object gains when it is lifted.
Some examples of Potential Energy are-
Gravitational Potential Energy (associated with Gravitational Force)
Elastic Potential Energy (based on elasticity)
Chemical Potential Energy (based on structural arrangement of atoms/molecules)
Electrical Potential Energy (based on electrical charges)
Magnetic Potential Energy (based on magnetic properties)
Now we will solve the query.
Given that Potential energy of a particle varies with a distance from fixed origin as $V\, = \,\dfrac{{A\sqrt x }}{{x\, + \,B}}$
Also, we can see that $x$ and $B$ are added together. Hence it means that their dimensions are the same.
i.e. dimension of $x$ = dimension of $B$ ( i.e. = $L$)
Also, we know that dimension of potential energy $V = \,\left[ {M{L^2}{T^{ - 2}}} \right]$ - (1)
Now, dimension of $V$= dimension of \left\\{ {A\sqrt x } \right\\}/ dimension of \left\\{ {x\, + \,B} \right\\}
$\Rightarrow \,\,V\, = \,\dfrac{{\left[ A \right]\left[ {\sqrt x } \right]}}{{\left[ {x\, + \,B} \right]}}$
$\Rightarrow \,\,\left[ A \right]\, = \,\dfrac{{\left[ V \right]\,\left[ {x\, + \,B} \right]}}{{\left[ {\sqrt x } \right]}}$
$\Rightarrow \,\,\left[ A \right]\,\, = \,\dfrac{{\left[ {M{L^2}{T^{ - 2}}} \right]\left[ L \right]}}{{\left[ {{L^{\dfrac{1}{2}}}} \right]}}$ (from equation 1)
$\Rightarrow \,\,\left[ A \right]\,\, = \,\left[ {M{L^{\dfrac{5}{2}}}{T^{ - 2}}} \right]$
Now, we have found the dimension of $A$, and we already have the dimension of $B$, now we can find the dimension of $AB$
Hence,
$\Rightarrow \,\,\left[ {AB} \right]\,\, = \,\,\left[ A \right]\left[ B \right]\,\, = \,\,\left[ {M{L^{\dfrac{5}{2}}}{T^{ - 2}}} \right]\,\,\left[ L \right] \\\ \Rightarrow \,\,\left[ {M{L^{\dfrac{7}{2}}}{T^{ - 2}}} \right] \\\$
Hence, the correct option is (D) $\left[ {M{L^{\dfrac{7}{2}}}{T^{ - 2}}} \right]$

Note- An object can store energy as a result of its position and this stored energy of position is known as Potential energy. In simple words, Potential energy can be defined as the form of energy that an object gains when it is lifted.