General Physics Question 24
Question 24
- The dimensions of electrical conductivity is
(1997, 1M)
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Answer:
Correct Answer: 32. $\left[\mathrm{M}^{-1} \mathrm{~L}^{-3} \mathrm{~T}^{3} \mathrm{~A}^{2}\right]$
Solution:
- Electrical conductivity, $\sigma=\frac{J}{E}=\frac{i / A}{F / q}$
$$ =\frac{q i}{F A}=\frac{(i t)(i)}{F A}=\frac{i^{2} t}{F A} $$
$$ [\sigma]=\frac{\left[\mathrm{A}^{2}\right][\mathrm{T}]}{\left[\mathrm{MLT}^{-2}\right]\left[\mathrm{L}^{2}\right]}=\left[\mathrm{M}^{-1} \mathrm{~L}^{-3} \mathrm{~T}^{3} \mathrm{~A}^{2}\right] $$
