This question was previously asked in

PTCUL AE E&M 2017 Official Paper (Set B)

Option 4 : 15.8 GHz

__Concept__:

Minimum operating frequency or the cut off frequency for a rectangular waveguide is given by:

\({f_c} = \frac{c}{2}\sqrt {\frac{{{m^2}}}{{{a^2}}} + \frac{{{n^2}}}{{{b^2}}}} \)

a = length of the waveguide

b = height of the waveguide

m,n = modes of operation

__Calculation__:

Given, a = 3 cm

b = 1 cm

The minimum frequency in TE11 is nothing but the cut-off frequency calculated as:

\({f_c} = \frac{{3 \times {{10}^8}}}{2}\sqrt {\frac{1}{{{{\left( {3 \times {{10}^{ - 2}}} \right)}^2}}} + \frac{1}{{{{\left( {1 \times {{10}^{ - 2}}} \right)}^2}}}} \)

\({f_c} = 150 \times {10^6}\sqrt {11111.11} \;\;\)

\({f_c} = 150 \times {10^6} \times 105.41\)

_{fC} = 15.81 GHz