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} $
Show Answer
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}} $
