Basic Maths and Units and DimensionsMedium
Question
The potential energy of a particle changes with distance $x$ from a fixed origin as $V = \frac{A\sqrt{x}}{x + B}$, where A and B are constant with appropriate dimensions. The dimension of AB are $\_\_\_\_$ .
Options
A.$\left\lbrack M^{1}{\text{ }L}^{5/2}{\text{ }T}^{- 2} \right\rbrack$
B.$\left\lbrack M^{3/2}{\text{ }L}^{5/2}{\text{ }T}^{- 2} \right\rbrack$
C.$\left\lbrack M^{1}{\text{ }L}^{2}{\text{ }T}^{- 2} \right\rbrack$
D.$\left\lbrack M^{1}{\text{ }L}^{7/2}{\text{ }T}^{- 2} \right\rbrack$
Solution
Sol. [Key Concept: Dimensionally $\lbrack B\rbrack = \lbrack x\rbrack = L$; from $\lbrack U\rbrack = \lbrack\frac{A\sqrt{x}}{x + B}\rbrack$ get $\lbrack A\rbrack = {ML}^{5/2}T^{- 2}$, hence $\lbrack AB\rbrack = {ML}^{7/2}T^{- 2}$.]
$${\lbrack B\rbrack = \lbrack x\rbrack = L }{\lbrack U\rbrack = \left\lbrack \frac{A\sqrt{x}}{B + x} \right\rbrack \Rightarrow ML^{2}{\text{ }T}^{- 2} = \frac{\lbrack A\rbrack L^{1/2}}{\text{ }L} }{\lbrack A\rbrack = {ML}^{5/2}{\text{ }T}^{- 2} }$$
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