GATE 2026 EC Q.33: solution and answer

GATE 2026 · ECQ.33 · 1 mark · NAT

The cutoff frequency (in GHz) for the dominant $TE_{10}$ mode of an air-filled rectangular waveguide of inner dimension $0.28$ inch $\times$ $0.14$ inch is ____. (rounded off to two decimal places).

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Answer and solution

ANSWER

The cut-off frequency of the dominant $TE_{10}$ mode is $f_{c, 10} = c /2 a = (3 \times 1 0^{8}) / (2 \times 7.112 \times 1 0^{- 3} m) \approx 21.09 GHz$ .

Official answer: 20.50 to 21.50Source: Official answer key, GATE 2026 EC (IIT Guwahati)

For a rectangular waveguide with $a > b$ , the dominant mode is $TE_{10}$ . Its cut-off depends only on the broad dimension: $f_{c, 10} = \frac{c}{2 a}$ , where $c = \frac{1}{μ _{0} ϵ _{0}} = 3 \times 1 0^{8} m/s$ for air ( $μ_{0} = 4 π \times 1 0^{- 7} H/m$ , $ϵ_{0} = 8.854 \times 1 0^{- 12} F/m$ ).
Take the broad wall $a = 0.28 inch$ . The narrow wall $b = 0.14 inch$ plays no part in $TE_{10}$ .
Convert to SI units with $1 inch = 2.54 cm$ : $a = 0.28 \times 2.54 cm = 0.7112 cm = 7.112 \times 1 0^{- 3} m$ .
Cut-off wavelength: $λ_{c, 10} = 2 a = 1.4224 \times 1 0^{- 2} m$ .
$f_{c, 10} = \frac{c}{2 a} = \frac{3 \times 1 0 ^{8}}{1.4224 \times 1 0 ^{- 2}} Hz \approx 2.1091 \times 1 0^{10} Hz = 21.09 GHz$
Check: the next modes are $TE_{20}$ and $TE_{01}$ , both cut off at $c / a = c /2 b \approx 42.18 GHz$ (here $b = a /2$ ). $TE_{10}$ has the lowest cut-off, so it is indeed the dominant mode. Answer: $21.09 GHz$ .

KEY IDEA

For $TE_{mn}$ in an air-filled rectangular guide, $f_{c, mn} = \frac{c}{2} (\frac{m}{a})^{2} + (\frac{n}{b})^{2}$ . With $m = 1$ , $n = 0$ this becomes $f_{c, 10} = c /2 a$ , which depends only on the broad dimension $a$ . The whole problem comes down to converting $a$ from inches to metres correctly.

WATCH OUT FOR

Using the narrow dimension $b = 0.14$ inch, or writing $f_{c} = c / a$ without the factor 2. Both give about $42.18 GHz$ , which is the $TE_{01}$ / $TE_{20}$ cut-off, not $TE_{10}$ . Another slip is forgetting to convert inches to metres, which is off by a factor of about 39. (Taking $c = 2.998 \times 1 0^{8} m/s$ gives $21.08 GHz$ , which is still inside the key range.)

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GATE 2026 EC Q.33: solution and answer

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