A) \[[{{E}^{-2}}{{V}^{1}}{{T}^{-2}}]\]
B) \[[{{E}^{-2}}{{V}^{-2}}{{T}^{1}}]\]
C) \[[{{E}^{1}}{{V}^{-2}}{{T}^{-2}}]\]
D) \[[{{E}^{-2}}{{V}^{-2}}{{T}^{-2}}]\]
Correct Answer: C
Solution :
Let surface tension, \[S\propto {{E}^{a}}{{v}^{b}}{{T}^{c}}\] or \[S=K{{E}^{a}}{{v}^{b}}{{T}^{c}}\] Writing the dimensions on both sides \[[{{M}^{1}}{{L}^{0}}{{T}^{-2}}]=K{{[{{M}^{1}}{{L}^{2}}{{T}^{-2}}]}^{a}}{{[{{M}^{0}}L{{T}^{-1}}]}^{b}}{{[{{M}^{0}}{{L}^{0}}T]}^{c}}\]\[[{{M}^{1}}{{L}^{0}}{{T}^{-2}}]=K[{{M}^{a}}{{L}^{2a+b}}{{T}^{-2a-b+c}}]\] Comparing the power on both sides, we have \[a=1\] \[2a+b=0\] \[\Rightarrow \] \[b=-2a\] \[\Rightarrow \] \[b=-2\times 1=-2\] and \[-2a-b+c=-2\] \[\Rightarrow \] \[-2xl-(-2)+c=-2\] \[\Rightarrow \] \[-2+2+c=-2\] \[\Rightarrow \] \[c=-2\] Hence, dimensions of surface tension are \[[{{E}^{1}}{{v}^{-1}}{{T}^{-2}}]\].
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