Physics And Measurement Question 199
Question: Which of the following product of $ e,h,\mu ,G $ (where $ \mu $ is the permeability) be taken so that the dimensions of the product are same as that of the speed of light?
Options:
A) $ he _{{}}^{-2}\mu _{{}}^{-1}G _{{}}^{0} $
B) $ h _{{}}^{2}eG _{{}}^{0}\mu $
C) $ h _{{}}^{0}e _{{}}^{2}G _{{}}^{-1}\mu $
D) $ hGe _{{}}^{-2}\mu _{{}}^{0} $
Show Answer
Answer:
Correct Answer: A
Solution:
[a] Here $ v=e^{a}h^{b}{{\mu }^{c}}G^{d} $ .
Taking the dimensions, $ M^{0}L{{T}^{-1}}A^{0}={{[AT^{1}]}^{a}}{{[ML^{2}{{T}^{-1}}]}^{b}}{{[ML{{T}^{-2}}{{A}^{-2}}]}^{c}}[{{M}^{-1}}L^{3}{{T}^{-2}}] $
There will be four simultaneous equations by equating the dimensions of M, L, T, and A.
These are a-2c=0, a-b-2c solving for a, b, c, and d, we get a=-2, b=1, c=-1, d=0
Thus, $ v={{e}^{-2}}h{{\mu }^{-1}}G^{0} $
BETA
