ISRO EC 2023 Q.35: solution and answer

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

1. A hollow metallic waveguide supports a propagating mode only for $f>f_c$, where $f_c$ is the cutoff frequency of that mode.

2. For such a mode the phase constant along the guide is ${\beta }_g={\beta }_0\sqrt{1-{\left(\frac{f_c}{f}\right)}^2}$, where ${\beta }_0=\frac{2\pi }{{\lambda }_0}$ is the free-space phase constant.

3. The guide wavelength is $${\lambda }_g=\frac{2\pi }{{\beta }_g}=\frac{{\lambda }_0}{\sqrt{1-{\left(\frac{f_c}{f}\right)}^2}}.$$

4. For $f>f_c$ we have $0<\sqrt{1-(f_c/f{)}^2}<1$, so the denominator is less than 1 and ${\lambda }_g>{\lambda }_0$ at every propagating frequency.

5. Hence (a) is true. Option (b) would need $f_c=0$, which is a TEM wave, and a hollow single-conductor guide cannot carry one. Option (c) would need the square-root factor to exceed 1, which is impossible. Option (d) is wrong because (a) holds.