Correct Answer - Option 1 : 15 mm

__Concept__:

The dominant mode in a particular waveguide is the mode having the lowest cut-off frequency.

The cut-off frequency for a rectangular waveguide with dimension ‘a (length)’ and ‘b (width)’ is given as:

\({{f}_{c\left( mn \right)}}=\frac{c}{2}\sqrt{{{\left( \frac{m}{a} \right)}^{2}}+{{\left( \frac{n}{b} \right)}^{2}}}\)

'm' and 'n' represents the possible modes.

c = speed of light = 3 × 1010 cm/s

**Calculation:**

TE_{10} mode means m = 1, n = 0

The cut - off frequency of the dominant mode TE10 of the rectangular waveguide is:

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

Where a is the dimension of the inner broad wall

\(\begin{array}{l} a = 3 \times \frac{{{{10}^{10}}}}{{2 \times 10 \times {{10}^9}}}\\ a = 1.5\;cm \end{array}\)

= 15 mm