Question

At room temperature, if the relative permittivity of water is $80$ and the relative permeability be $0.0222$ then the velocity of light in water is ____________$\mathrm{m}/\mathrm{s}$

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

Answer 1

The correct option is B

$2.25\times {10}^8$

Answer(B)

Explanation for correct option:

Step 1:

Velocity of light

$\mathrm{V}=1/\sqrt{{\mathrm{\mu }}_0}{\mathrm{\epsilon }}_0$

Velocity of light in water

$\mathrm{V}\text{'}=1/\sqrt{{\mathrm{\mu }}_{\mathrm{r}}}{\mathrm{\epsilon }}_{\mathrm{r}}$

Step 2:

Relative premittivity of water $=80$

Relative permeability $=0.0222$

Step 3:

From the formulae:

$\frac{\mathrm{V}\text{'}}{\mathrm{V}}=\sqrt{\frac{{\mathrm{\mu }}_0}{{\mathrm{\mu }}_{\mathrm{r}}}}\times \sqrt{\frac{{\mathrm{\epsilon }}_0}{{\mathrm{\epsilon }}_{\mathrm{r}}}}$

Step 4:

Putting all given values ina bove formulae:

$$\begin{gathered} \frac{\mathrm{V}\text{'}}{\mathrm{V}}=\sqrt{\frac{1}{80}}\times \sqrt{\frac{1}{0.0222}} \\ \mathrm{V}\text{'}=\sqrt{\frac{1}{80}}\times \sqrt{\frac{1}{0.0222}}\times \mathrm{V} \\ \mathrm{V}\text{'}=\sqrt{\frac{1}{80}\times }\sqrt{\frac{1}{0.0222}}\times \mathrm{V} \\ \mathrm{V}\text{'}=\sqrt{\frac{1}{80}\times }\sqrt{\frac{1}{0.0222}}\times 3\times {10}^8 \\ \mathbf{V}\mathbf{\text{'}}\mathbf{=}\mathbf{2}\mathbf{.}\mathbf{25}\mathbf{\times }{\mathbf{10}}^{\mathbf{8}}\mathbf{m}\mathbf{/}\mathbf{s} \end{gathered}$$

Hence option (B) is correct option.