Physics And Measurement Question 196

Question: The potential energy of a particle varies with distance $ \chi $ as $ U=\frac{Ax _{{}}^{1/2}}{x _{{}}^{2}+B} $ where A and B are constants. The dimensional formula for A x B is

Options:

A) $ M _{{}}^{1}L _{{}}^{7/2}T _{{}}^{-2} $

B) $ M _{{}}^{1}L _{{}}^{1/2}T _{{}}^{-2} $

C) $ M _{{}}^{1}L _{{}}^{5/2}T _{{}}^{-2} $

D) $ M _{{}}^{1}L _{{}}^{9/2}T _{{}}^{-2} $

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Answer:

Correct Answer: B

Solution:

[b] Here $ x^{2} $ has the dimensions of $ L^{2},B=[L^{2}] $ .

Also $ ML^{2}{{T}^{-2}}=\frac{A{{L}^{1/2}}}{L^{2}} $ or $ A=M{{L}^{7/2}}{{T}^{-2}} $

$ \therefore A\times B=M{{L}^{11/2}}{{T}^{-2}} $



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