Dimensional formula of the physical quantity, resistance is

Question

Dimensional formula of the physical quantity, resistance is (A) [M L2 T- 3 A- 2] (B) [M L- 1 T3 A- 1] (C) [M L2 T- 2 K- 1] (D) [M L- 2 T- 3 A2]

Tardigrade Admin · Accepted Answer

According to Ohm's law, V=IR ∴ R=(V/I) But V=(W/q) =(W/I t) [∵ q = I t] ∴ R=(W / I t/I) =(W/I2 t) Hence, [R]=([W]/[I2] [t])=([M L2 T- 2]/[A2 T]) =[M L2 T- 3 A- 2]

Q. Dimensional formula of the physical quantity, resistance is

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

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

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

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

According to Ohm's law,
$V = I R$
$∴ R = \frac{V}{I}$
But $V = \frac{W}{q}$
$= \frac{W}{I t}$ $[∵ q = I t]$
$∴ R = \frac{W / I t}{I}$
$= \frac{W}{I ^{2} t}$
Hence, $[R] = \frac{[ W ]}{[ I ^{2} ] [ t ]} = \frac{[ M L ^{2} T ^{- 2} ]}{[ A ^{2} T ]}$
$= [M L^{2} T^{- 3} A^{- 2}]$

Q. Dimensional formula of the physical quantity, resistance is

[ML2T−3A−2]

[ML−1T3A−1]

[ML2T−2K−1]

[ML−2T−3A2]

According to Ohm's law, V=IR ∴R=IV​ But V=qW​ =ItW​ [∵q=It] ∴R=IW/It​ =I2tW​ Hence, [R]=[I2][t][W]​=[A2T][ML2T−2]​ =[ML2T−3A−2]

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

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

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

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

According to Ohm's law,
$V = I R$
$∴ R = \frac{V}{I}$
But $V = \frac{W}{q}$
$= \frac{W}{I t}$ $[∵ q = I t]$
$∴ R = \frac{W / I t}{I}$
$= \frac{W}{I ^{2} t}$
Hence, $[R] = \frac{[ W ]}{[ I ^{2} ] [ t ]} = \frac{[ M L ^{2} T ^{- 2} ]}{[ A ^{2} T ]}$
$= [M L^{2} T^{- 3} A^{- 2}]$