question_answer

If energy E, velocity V and time T were taken as fundamental units, the dimensions for surface tension in these units are.

A) \[{{E}^{\,-2}}\,V{{T}^{\,-2}}\]                      

B) \[{{E}^{\,-2}}\,{{V}^{\,-2}}T\]

C) \[E{{V}^{\,-2}}{{T}^{\,-2}}\,\]          

D) \[{{E}^{-2}}{{V}^{\,-2}}{{T}^{\,-2}}\,\]

Correct Answer: C

Solution :

[c] Surface tension \[\frac{F}{L}=\frac{ML{{T}^{-2}}}{L}=M{{T}^{-2}}\] \[M{{L}^{-2}}\propto {{\left[ M\,({{L}^{2}}{{T}^{-2}}) \right]}^{a}}{{(L{{T}^{-1}})}^{b}}\,{{(T)}^{c}}\] \[M{{L}^{-2}}\propto \,{{M}^{a}}\,{{L}^{2a+b}}{{T}^{-2a-b+c}}\] \[a=1\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,2a+b=0-2a-b+c=-2\] \[\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,c=-2\] \[T={{E}^{1}}{{V}^{-2}}{{T}^{-2}}\]