Q153 / 160 · TS EAMCET 2023 (13 May Shift 2) · Physics
The speed of electromagnetic waves in a medium is $1.5 \times 10^8 \mathrm{~ms}^{-1}$. If relative permittivity of that medium is 2 , then its magnetic susceptibility is (speed of light in vacuum is $3 \times 10^8 \mathrm{~ms}^{-1}$ ).
- Correct answerA2
- B3
- C1
- D-1.5
Solution
Speed of electro magnetic wave, $\mathrm{v}=1.5 \times 10^8 \mathrm{~m} / \mathrm{s}$ Relative permittivity, $\epsilon_{\mathrm{r}}=2$
$\begin{aligned}<br />& v=\frac{1}{\sqrt{\epsilon_0 \epsilon_{\mathrm{r}}} \mu_0 \mu_{\mathrm{r}}}=\frac{\mathrm{c}}{\sqrt{\mu_{\mathrm{r}} \in_{\mathrm{r}}}}\left(\because \mathrm{C}^2=\frac{1}{\mu_0 \varepsilon_0}\right) \\<br />& \Rightarrow 1.5 \times 10^8=\frac{3 \times 10^8}{\sqrt{\mu_{\mathrm{r}} \times 2}}<br />\end{aligned}$
Magnetic susceptibility $\mu_{\mathrm{r}}=2$
$\begin{aligned}<br />& v=\frac{1}{\sqrt{\epsilon_0 \epsilon_{\mathrm{r}}} \mu_0 \mu_{\mathrm{r}}}=\frac{\mathrm{c}}{\sqrt{\mu_{\mathrm{r}} \in_{\mathrm{r}}}}\left(\because \mathrm{C}^2=\frac{1}{\mu_0 \varepsilon_0}\right) \\<br />& \Rightarrow 1.5 \times 10^8=\frac{3 \times 10^8}{\sqrt{\mu_{\mathrm{r}} \times 2}}<br />\end{aligned}$
Magnetic susceptibility $\mu_{\mathrm{r}}=2$
Hence, the correct answer is option: 'A'.