A) \[[E{{V}^{-\,2}}\,{{T}^{-\,1}}]\]
B) \[[E{{V}^{-\,1}}\,{{T}^{-\,2}}]\]
C) \[[E{{V}^{-\,2}}\,{{T}^{-\,2}}]\]
D) \[[{{E}^{-\,2}}\,{{V}^{-\,1}}\,{{T}^{-\,3}}]\]
Correct Answer: C
Solution :
Let surface tension \[[s]\,\,=\,\,\left[ {{E}^{x}}{{V}^{y}}{{T}^{z}} \right]\] We have \[[s]=\left[ M{{T}^{-2}} \right]\] \[\left[ E \right] = \left[ M{{L}^{2}}{{T}^{-}}^{2} \right] ,\,\,\left[ V \right] = \left[ L{{T}^{-}}^{1} \right]\,\,and\,\,\left[ T \right] = \left[ T \right]\] \[=\,\,{{\left[ M{{L}^{2}}{{T}^{-2}} \right]}^{x}}{{\left[ L{{T}^{-1}} \right]}^{y}}\,{{\left[ T \right]}^{z}}\] \[\left[ {{M}^{1}}{{L}^{0}}{{T}^{-2}} \right]\,\,=\,\,\left[ {{M}^{x}}{{L}^{2x+y}}\,{{T}^{-2x-y+z}} \right]\] \[x=1\] \[2x+y=0\,\,\,\,\Rightarrow \,\,\,y=-2\] \[-2x-y+z=-2\] \[\Rightarrow \,\,\, -2\times 1-\left( -2 \right)+z=-\,2\] \[\Rightarrow \,\,z=-\,2\] Hence \[\left[ s \right]\,\,\,=\,\,\,\left[ E{{V}^{-}}^{2}{{T}^{-}}^{2} \right]\]
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