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Question: If energy E, velocity (V) and time (T) are chosen as the fundamental quantities, then the dimensions...

If energy E, velocity (V) and time (T) are chosen as the fundamental quantities, then the dimensions of surface tension will be –

A

EV-2T -1\text{E}\text{V}^{\text{-2}}\text{T}\text{ }^{\text{-1}}

B

EV-1T -2\text{E}\text{V}^{\text{-1}}\text{T}\text{ }^{\text{-2}}

C

EV-2T -2\text{E}\text{V}^{\text{-2}}\text{T}\text{ }^{\text{-2}}

D

E-2V-1T -3E^{\text{-2}}V^{\text{-1}}\text{T}\text{ }^{\text{-3}}

Answer

EV-2T -2\text{E}\text{V}^{\text{-2}}\text{T}\text{ }^{\text{-2}}

Explanation

Solution

Surface Tension = EaVbTc= \text{ }\text{E}^{a}V^{b}T^{c}

[M1 L0 T−2]=[M1 L2 T−2]a[M0 L1 T−2]b[T1]c\left[ \mathrm { M } ^ { 1 } \mathrm {~L} ^ { 0 } \mathrm {~T} ^ { - 2 } \right] = \left[ \mathrm { M } ^ { 1 } \mathrm {~L} ^ { 2 } \mathrm {~T} ^ { - 2 } \right] ^ { \mathrm { a } } \left[ \mathrm { M } ^ { 0 } \mathrm {~L} ^ { 1 } \mathrm {~T} ^ { - 2 } \right] ^ { \mathrm { b } } \left[ \mathrm { T } ^ { 1 } \right] ^ { \mathrm { c } } M1L0T -2 = Ma L2a + b T -2a - 2b + ca = 1M^{1}L^{0}\text{T}\text{ }^{\text{-2}}\ = \text{ }\text{M}^{a}\text{ }\text{L}^{\text{2a } + \text{ b}}\text{ }\text{T}^{\text{ -2a - 2b } + \text{ c}}\text{a } = \text{ 1}

b = -2\text{b } = \text{ -2}

c = -2\text{c } = \text{ -2}