JEE Physics 2025 Question with Solution

Given: Relative permeability $\mu_r = \frac{10}{\pi}$ and relative permittivity $\epsilon_r = \frac{1}{0.0885}$.

Find: By how many times the speed of light in vacuum is greater than that in the medium.

The speed of light in a medium is related to the relative permeability and relative permittivity by

$$\frac{c}{v} = \sqrt{\mu_r \epsilon_r}$$

Substitute the given values:

$$\frac{c}{v} = \sqrt{\frac{10}{\pi} \cdot \frac{1}{0.0885}}$$

Using $\pi \approx 3.1416$,

$$\frac{10}{\pi} \approx 3.183$$

and

$$\mu_r \epsilon_r \approx \frac{3.183}{0.0885} \approx 35.97$$

Therefore,

$$\frac{c}{v} = \sqrt{35.97} \approx 6$$

So, the velocity of light in vacuum is greater than that in the medium by $6$ times.

The solution contains arithmetic inconsistencies in one approach, but both the stated correct answer and the correct relation $\frac{c}{v} = \sqrt{\mu_r\epsilon_r}$ lead to the final value $6$.

Given:

• $\mu = \mu_0 \mu_r$
• $\epsilon = \epsilon_0 \epsilon_r$
• $v = \frac{1}{\sqrt{\mu \epsilon}}$
• $c = \frac{1}{\sqrt{\mu_0 \epsilon_0}}$

Find: The factor by which $c$ exceeds $v$.

Write the ratio:

$$\frac{c}{v} = \frac{1/\sqrt{\mu_0\epsilon_0}}{1/\sqrt{\mu\epsilon}} = \sqrt{\frac{\mu\epsilon}{\mu_0\epsilon_0}} = \sqrt{\mu_r\epsilon_r}$$

Now substitute:

$$\frac{c}{v} = \sqrt{\frac{10}{\pi} \cdot \frac{1}{0.0885}} \approx \sqrt{35.97} \approx 5.99$$

Hence,

$$\frac{c}{v} \approx 6$$

Therefore, the required numerical value is $6$.