Question

The potential energy of a particle varies with distance x with $U=\dfrac{A{{x}^{\dfrac{1}{2}}}}{{{x}^{2}}+B}$ ,where A and B are constants .the dimensional formula of $A\times B$ is:

& \text{A}\text{. }{{\text{M}}^{\text{1}}}{{\text{L}}^{\dfrac{\text{7}}{\text{2}}}}{{\text{T}}^{\text{-2}}} \\\ & \text{B}\text{. }{{\text{M}}^{\text{1}}}{{\text{L}}^{\dfrac{\text{11}}{\text{2}}}}{{\text{T}}^{\text{-2}}} \\\ & \text{C}\text{. }{{\text{M}}^{\text{1}}}{{\text{L}}^{\dfrac{\text{5}}{\text{2}}}}{{\text{T}}^{\text{-2}}} \\\ & \text{D}\text{. }{{\text{M}}^{\text{1}}}{{\text{L}}^{\dfrac{\text{9}}{\text{2}}}}{{\text{T}}^{\text{-2}}} \\\ \end{aligned}$$

Answer 1

We should have a clear idea of the principle of homogeneity and dimensions of variables. We will use the principle of homogeneity to find the dimension of A and B separately and then will solve further and find $A\times B$.

Complete step by step answer:
Firstly we will find the dimension of A and B Individually and then we will find the dimension of $A\times B$
As we know the principle of homogeneity.
So according to principle : B has the same dimension as ${{x}^{2}}$.

& \left[ {{x}^{2}} \right]=\left[ B \right] \\\ & \therefore \left[ B \right]=\left[ {{L}^{2}} \right] \\\ \end{aligned}$$ As well as $$\begin{aligned} & \left[ U \right]=\dfrac{\left[ A \right]\left[ {{x}^{\dfrac{1}{2}}} \right]}{\left[ {{x}^{2}}+B \right]} \\\ & =\left[ {{M}^{1}}{{L}^{2}}{{T}^{-2}} \right]=\dfrac{\left[ A \right]\left[ {{L}^{\dfrac{1}{2}}} \right]}{\left[ {{L}^{2}} \right]} \\\ & \therefore \left[ A \right]=\left[ {{M}^{1}}{{L}^{\dfrac{7}{2}}}{{T}^{-2}} \right] \\\ \end{aligned}$$ Now we have both the dimensions of A and B. Now we will find $$\begin{aligned} & A\times B \\\ & \left[ AB \right]=\left[ {{M}^{1}}{{L}^{\dfrac{7}{2}}}{{T}^{-2}} \right]\times \left[ {{L}^{2}} \right]=\left[ {{M}^{1}}{{L}^{\dfrac{11}{2}}}{{T}^{-2}} \right] \\\ \end{aligned}$$ Hence from all the above calculations we found that dimension of $$A\times B$$ Is $${{M}^{1}}{{L}^{\dfrac{11}{2}}}{{T}^{-2}}$$ . **Therefore the correct is (B).** **Additional Information:** Dimensional analysis is the practice of checking relative amounts of physical quantities by identifying their dimensions. It is common to be faced with a problem that uses different dimensions to express the same basic quantity. Dimensional analysis can also be used as a simple check to computations, theories and hypotheses. We know that the dimensional analysis is the practice of checking relations between physical quantities by identifying their dimensions. The dimension of any physical quantity is the combination of the essential physical dimensions that compose it. Dimensional analysis is predicated on the very fact that physical law must be independent of the units wont to measure the physical variables. It can be used to check the plausibility of derived equations, computations, and hypotheses. **Note:** The principle of homogeneity is an important and basic concept. Dimensional analysis can also be used as a simple check to computations, theories, and hypotheses. With the help of dimensional formula, we can find the unit of any variable.