If \( \mu_0 \) is the permeability and \( \varepsilon_0 \) is the permittivity of free space, then the velocity of light is
Answer & explanation
Correct answer: option 2
The correct answer is Option (2) → \( \displaystyle\frac{1}{\sqrt{\mu_0\varepsilon_0}} \)
Maxwell's equations predict electromagnetic waves in free space whose speed depends only on the permeability and permittivity of free space. Light is such a wave, so
\( \displaystyle c = \frac{1}{\sqrt{\mu_0\varepsilon_0}} = \frac{1}{\sqrt{4\pi\times10^{-7} \times 8.85\times10^{-12}}} \approx 3\times10^{8}\,\mathrm{m\,s^{-1}} \)
NCERT (2026-27):
Nature of Electromagnetic Waves – “… The speed of propagation of the wave is (ω/k). Using Eqs. [8.7(a) and (b)] for Ex and By and Maxwell’s equations, one finds that
\( \displaystyle \omega = ck, \ \text{where}\ c = \frac{1}{\sqrt{\mu_0\varepsilon_0}} \qquad [8.9(a)] \) ”