Correct Answer - Option 1 : 2.94 GHz

__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 × 10^{10} cm/s

__Calculation__:

For TE_{10} mode, m = 1, and n = 0.

The cut-off frequency will be:

\({{f}_{c\left( mn \right)}}=\frac{c}{2}\sqrt{{{\left( \frac{1}{a} \right)}^{2}}+{{\left( \frac{0}{b} \right)}^{2}}}\)

\({{f}_{c\left( mn \right)}}=\frac{c}{2a}\)

With a = 5.1 cm, the cut-off frequency will be:

\({{f}_{c\left( mn \right)}}_{TE_{(1,0)}}=\frac{3\times 10^{10}}{2\times 5.1}\)

**f**_{c(1,0)} = 2.94 GHz