SATHEE: General Physics Question 24

General Physics Question 24

Question 24

The dimensions of electrical conductivity is

(1997, 1M)

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Answer:

Correct Answer: 32. $[M^{- 1} L^{- 3} T^{3} A^{2}]$

Solution:

Electrical conductivity, $σ = \frac{J}{E} = \frac{i / A}{F / q}$

$= \frac{q i}{F A} = \frac{(i t) (i)}{F A} = \frac{i^{2} t}{F A}$

$[σ] = \frac{[A^{2}] [T]}{[{MLT}^{- 2}] [L^{2}]} = [M^{- 1} L^{- 3} T^{3} A^{2}]$

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General Physics Question 24

Question 24

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