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 :
We know that Surface tension \[\text{(S)=}\frac{\text{Force }\!\![\!\!\text{ F }\!\!]\!\!\text{ }}{\text{Lenght }\!\![\!\!\text{ L }\!\!]\!\!\text{ }}\] So, \[[S]=\frac{[MK{{T}^{-2}}]}{[L]}=[M{{L}^{0}}{{T}^{-2}}]\] Energy (E) = Force \[\times \] displacement \[\Rightarrow \,\,\,\,\,[E]=[M{{L}^{2}}{{T}^{2}}]\] Velocity (v) \[\text{=}\frac{\text{displacement}}{\text{time}}\] \[\Rightarrow \,\,\,\,\,\,\,\,\,\,\,[v]=[L{{T}^{-1}}]\] A.s, \[S\propto {{E}^{a}}{{v}^{b}}{{T}^{c}}\] where, a, b, c are constants. From the principle of homogeneity, [LHS] = [RHS] \[\Rightarrow \,\,\,\,\,\,\,\,\,[M{{L}^{0}}{{T}^{-2}}]={{[M{{L}^{2}}{{T}^{-2}}]}^{a}}{{[L{{T}^{-1}}]}^{b}}{{[T]}^{c}}\] \[\Rightarrow \,\,\,\,\,\,\,\,\,[M{{L}^{0}}{{T}^{-2}}]=[{{M}^{a}}{{L}^{2a+b}}{{T}^{-2a-b+c}}]\] Equating the power on both sides, we get \[a=1,2a+b=0,b=-2\] \[\Rightarrow \,\,\,\,\,\,\,\,\,\,-2a-b+c=-2\] \[\Rightarrow \,\,\,\,\,\,\,\,\,\,\,c=(2a+b)-2=0-2=-2\] So \[[S]\,=[E{{v}^{-2}}{{T}^{-2}}]\,=[E{{v}^{-2}}{{T}^{-2}}]\]
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