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Question

Physics Question on Units and measurement

Dimension of resistivity is

A

ML2T−2I−1M L^{2} T ^{-2} I ^{-1}

B

ML3T−3I−2ML ^{3} T ^{-3} I ^{-2}

C

ML3T−2I−1ML ^{3} T ^{-2} I ^{-1}

D

ML2T−2I−2ML ^{2} T ^{-2} I ^{-2}

Answer

ML3T−2I−1ML ^{3} T ^{-2} I ^{-1}

Explanation

Solution

By definition R=ρlAR = \rho \frac{l}{A} ⇒ρ=RAl=VIAl=VAIl\Rightarrow\quad\rho = \frac{RA}{l} = \frac{\frac{V}{I}A}{l} = \frac{VA}{Il} ∴ [ρ]=[V][A][I][l]\therefore\,\left[\rho\right] = \frac{\left[V\right]\left[A\right]}{\left[I\right]\left[l\right]} =ML2T−2L2IL=[ML3T−2I−1]= \frac{ML^{2}T^{-2}L^{2}}{IL} = \left[ML^{3}T^{-2}I^{-1}\right]