Physical World Units and Measurements - Result Question 29
31. The dimensional formula for permeability $\mu$ is given by
(a) $[MLT^{-2} A^{-2}]$
[1991]
(b) $[M^{0} L^{1} T]$
(c) $[M^{0} L^{2} T^{-1} A^{2}]$
(d) None of the above
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Answer:
Correct Answer: 31. (a)
Solution:
- (a) Permeability of free space,
$ \mu_0=\frac{2 \pi \times \text{ force } \times \text{ distance }}{(\text{ current })^{2} \times \text{ length }} $
So, dimensional formula
$ \mu_0=\frac{[MLT^{-2}][L]}{[A^{2}][L]}=[MLT^{-2} A^{-2}] $
Also find the dimensional formula by using the relation,
Speed of light, $c=\frac{1}{\sqrt{\mu_0 \varepsilon_0}}$
