A) \[[E{{v}^{-2}}{{T}^{-2}}]\]
B) \[[{{E}^{2}}v{{T}^{-2}}]\]
C) \[[E{{v}^{-2}}{{T}^{-1}}]\]
D) \[[{{E}^{-2}}{{v}^{2}}{{T}^{-1}}]\]
Correct Answer: A
Solution :
The dimensions of given fundamental quantities are given below: Energy \[=[M{{L}^{2}}{{T}^{-2}}]\] Velocity \[(v)=(L{{T}^{-1}})\] Time \[(T)=[T]\] and surface tension \[[S]=[M{{T}^{-2}}]\] Let \[[S]\propto {{[E]}^{a}}\times {{[v]}^{b}}\times {{[T]}^{c}}\] \[[S]=k{{[E]}^{a}}\times {{[v]}^{b}}\times {{[T]}^{c}}\] Putting the dimensions of both sides, we have \[[M{{T}^{-2}}]={{[M{{L}^{2}}{{T}^{-2}}]}^{a}}\times {{[L{{T}^{-1}}]}^{b}}\times {{[T]}^{c}}\] \[[M{{L}^{0}}{{T}^{-2}}]={{[M]}^{a}}\times {{[L]}^{2a+b}}\times {{[T]}^{-2a-b+c}}\] Comparing the powers of both sides, we have \[a=1\] \[2a+b=0\] \[-2a-b+c=-2\] So, we get a=1, b = - 2 and c = - 2 Hence the dimensional formula for surface tension is, \[T=[E{{v}^{-2}}{{T}^{-2}}]\]
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