ISRO EC 2023 Q.40: solution and answer

Official answer: BSource: ISRO's revised and final answer key of 16 February 2024

1. A TEM wave has no field component along the direction of propagation, so $E_z=H_z=0$.

2. For any guided mode the cut-off wavenumber satisfies $k_c^2={\omega }^2\mu ϵ-{\beta }^2$. For a TEM wave the phase constant is that of a plane wave in the same medium, $\beta =\omega \sqrt{\mu ϵ}$, so $k_c=0$.

3. The cut-off frequency is $f_c=\frac{k_c}{2\pi \sqrt{\mu ϵ}}=0$, that is, DC. A TEM wave can propagate at every frequency down to zero, as on a coaxial line or a two-wire line.

4. Infinite cut-off would mean the wave never propagates, the opposite of the TEM case, so (A) is wrong.

5. The expression $2c/\lambda$ is not a cut-off frequency of a TEM wave. For a rectangular waveguide the TE and TM modes have $f_c=\frac{c}{2}\sqrt{{\left(\frac{m}{a}\right)}^2+{\left(\frac{n}{b}\right)}^2}$, which is non-zero, so (C) is wrong.

6. Since the cut-off frequency of a TEM wave is zero (DC), option (B) is correct and (D) is not.