The dimensional formula for resistivity is

Question

The dimensional formula for resistivity is (A)  [ML3T-3A-2]  (B) [ M-1L2T-3A-2 ] (C) [ M3L-2T-1 ] (D) [ ML3T-4A-3 ]

Tardigrade Admin · Accepted Answer

As we know that resistance, R=ρ (l/A) where, ρ = resistivity l= length of conductor wire and A = area of cross- section of conductor wire. or ρ =(RA/l) ⇒ Dimensions, ρ =([ML2T-3A-2][L2]/[L]) =[M L3T-3A-2]

Q. The dimensional formula for resistivity is

$[M L^{3} T^{- 3} A^{- 2}]$

[ $M^{- 1} L^{2} T^{- 3} A^{- 2}$ ]

[ $M^{3} L^{- 2} T^{- 1}$ ]

[ $M L^{3} T^{- 4} A^{- 3}$ ]

As we know that resistance, $R = ρ \frac{l}{A}$ where, $ρ =$ resistivity $l =$ length of conductor wire and A = area of cross- section of conductor wire. or $ρ = \frac{R A}{l}$ $\Rightarrow$ Dimensions, $ρ = \frac{[ M L ^{2} T ^{- 3} A ^{- 2} ] [ L ^{2} ]}{[ L ]}$ $= [M L^{3} T^{- 3} A^{- 2}]$

Q. The dimensional formula for resistivity is

[ML3T−3A−2]

[ M−1L2T−3A−2 ]

[ M3L−2T−1 ]

[ ML3T−4A−3 ]

As we know that resistance, R=ρAl​ where, ρ= resistivity l= length of conductor wire and A = area of cross- section of conductor wire. or ρ=lRA​ ⇒ Dimensions, ρ=[L][ML2T−3A−2][L2]​ =[ML3T−3A−2]

$[M L^{3} T^{- 3} A^{- 2}]$

[ $M^{- 1} L^{2} T^{- 3} A^{- 2}$ ]

[ $M^{3} L^{- 2} T^{- 1}$ ]

[ $M L^{3} T^{- 4} A^{- 3}$ ]

As we know that resistance, $R = ρ \frac{l}{A}$ where, $ρ =$ resistivity $l =$ length of conductor wire and A = area of cross- section of conductor wire. or $ρ = \frac{R A}{l}$ $\Rightarrow$ Dimensions, $ρ = \frac{[ M L ^{2} T ^{- 3} A ^{- 2} ] [ L ^{2} ]}{[ L ]}$ $= [M L^{3} T^{- 3} A^{- 2}]$