Question

The potential energy of a particle varies with distance x from a fixed origin as $V=\left(\frac{A\sqrt{x}}{x+B}\right);$ where A and B are constants. The dimensions of A B are:

• A. $[M{L}^{5/2}{T}^{-2}]$
• B. $[M{L}^2{T}^{-2}]$
• C. $[M^{3/2}{L}^{3/2}{T}^{-2}]$
• D. $[M{L}^{7/2}{T}^{-2}]$

Answer 1

Answer: (D)

Given, $v=\frac{A\sqrt{x}}{x+B}$ ?(i) Dimensions of v= dimensions of potentialenergy $=[M{L}^2{T}^{-2}]$ From Eq. (i), Dimensions ofB= dimensions of $x=[M^{0}L{T}^o]$ $\therefore$ Dimensions of A $=\frac{\text{dimensions}\text{of}v\times \text{dimensitons}\text{of}(x+B)}{\text{dimensions}\text{of}\sqrt{x}}$ $=\frac{[M{L}^2{T}^{-2}][M^{0}L{T}^0]}{[M^0{L}^{1/2}{T}^0]}$ $=[M{L}^{5/2}{T}^{-2}]$ Hence, dimensions ofAB $=[M{L}^{5/2}{T}^{-2}][M^{0}L{T}^0]$ $=[M{L}^{7/2}{T}^{-2}]$