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Physics Question on Surface tension

If energy E, velocity v and time T are taken as fundamental quantities, the dimensional formula for surface tension is

A

[Ev−2T−2][E{{v}^{-2}}{{T}^{-2}}]

B

[E2vT−2][{{E}^{2}}v{{T}^{-2}}]

C

[Ev−2T−1][E{{v}^{-2}}{{T}^{-1}}]

D

[E−2v2T−1][{{E}^{-2}}{{v}^{2}}{{T}^{-1}}]

Answer

[Ev−2T−2][E{{v}^{-2}}{{T}^{-2}}]

Explanation

Solution

The dimensions of given fundamental quantities are given below: Energy =[ML2T−2]=[M{{L}^{2}}{{T}^{-2}}] Velocity (v)=(LT−1)(v)=(L{{T}^{-1}}) Time (T)=[T](T)=[T] and surface tension [S]=[MT−2][S]=[M{{T}^{-2}}] Let [S]∝[E]a×[v]b×[T]c[S]\propto {{[E]}^{a}}\times {{[v]}^{b}}\times {{[T]}^{c}} [S]=k[E]a×[v]b×[T]c[S]=k{{[E]}^{a}}\times {{[v]}^{b}}\times {{[T]}^{c}} Putting the dimensions of both sides, we have [MT−2]=[ML2T−2]a×[LT−1]b×[T]c[M{{T}^{-2}}]={{[M{{L}^{2}}{{T}^{-2}}]}^{a}}\times {{[L{{T}^{-1}}]}^{b}}\times {{[T]}^{c}} [ML0T−2]=[M]a×[L]2a+b×[T]−2a−b+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=1a=1 2a+b=02a+b=0 −2a−b+c=−2-2a-b+c=-2 So, we get a=1, b = - 2 and c = - 2 Hence the dimensional formula for surface tension is, T=[Ev−2T−2]T=[E{{v}^{-2}}{{T}^{-2}}]