Question

At room temperature, if the relative permittivity of water be $80$ and the relative permeability be $0.0222$, then the velocity of light in water is ________ $m/s^{-1}$

• A. $3\times {10}^8$
• B. $2.25\times {10}^8$
• C. $2.5\times {10}^8$
• D. $3.5\times {10}^8$

Answer 1

Answer: (B)

$2.25\times {10}^8$

Formula

$V=\frac{1}{\sqrt{{\mu }_0{ϵ}_0}}{V}^1=\frac{1}{{\mu }_r{ϵ}_r}\Rightarrow \frac{V}{V^1}=\sqrt{\frac{{\mu }_r{ϵ}_r}{{\mu }_0{ϵ}_0}}$

$\Rightarrow \frac{V^1}{V}=\sqrt{\frac{{\mu }_0}{}{\mu }_r\times \frac{{ϵ}_0}{{ϵ}_r}}=\sqrt{\frac{1}{80}\times \frac{1}{0.0222}}\Rightarrow V^1=\sqrt{\frac{1}{80}\times \frac{1}{0.0222}}\times V=\sqrt{\frac{1}{80}\times \frac{1}{0.022}}\times 3\times {10}^8=2.249\times {10}^8\approx 2.22\times {10}^{8}m/s$