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}}] $