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Physics Question on physical world

If energy (E)energy\,(E), velocity (V)velocity\,(V) and time (T)time\,(T) are chosen as the fundamental quantities, the dimensional formula of surface tension will be

A

[EV−2T−2][EV^{-2}T^{-2}]

B

[E−2V−1T−3][E^{-2}V^{-1}T^{-3}]

C

[EV−2T−1][EV^{-2}T^{-1}]

D

[EV−1T−2][EV^{-1}T^{-2}]

Answer

[EV−2T−2][EV^{-2}T^{-2}]

Explanation

Solution

Let S=kEaVbTcS=kE^aV^bT^c where kk is a dimensionless constant. Writing the dimensions on both sides, we get [M1L0T−2]=[ML2T−2]a[LT−1]b[T]c[M^1L^0T^{-2}]=[ML^{2}T^{-2}]^a[LT^{-1}]^b[T]^c =[MaL2a+bT−2a−b+c]=[M^aL^{2a+b}T^{-2a-b+c}] Applying principle of homogeneity of dimensions, we get, a=1…(i)a = 1\ldots (i) 2a+b=0…(ii)2a+b=0\ldots(ii) −2a−b+c=−2…(iii)-2a-b+c=-2\ldots (iii) Adding (ii)(ii) and (iii)(iii), we get c=−2c=-2 From (ii)(ii), b=−2a=−2b = -2a = -2 ∴S=kEV−2 T−2\therefore S=kEV^{-2}\,T^{-2} or [S]=[EV−2T−2][S]=[EV^{-2}T^{-2}]