Correct Answer - Option 1 : 7.5 GHz

__Concept:__

The cutoff frequency of the waveguide is given by:

\({f_c} = \frac{c}{{2\sqrt {{\mu _b}{\epsilon_b}} }}\sqrt {{{\left( {\frac{m}{a}} \right)}^2} + {{\left( {\frac{n}{b}} \right)}^2}} \)

where c = 3 × 108 m/s

The lowest propagating mode is TE10 cutoff frequency

\({f_c} = \frac{c}{{2a}}\)

__Calculation__:

Given:

a = 2 cm = 0.02m

\({f_c} = \frac{3\times10^8}{{2\times0.02}}\)

f_{c} = 7.5 GHz