Test: Cut Off Frequency & Wavelength

# Test: Cut Off Frequency & Wavelength

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## 12 Questions MCQ Test Electromagnetic Fields Theory (EMFT) | Test: Cut Off Frequency & Wavelength

Test: Cut Off Frequency & Wavelength for Electrical Engineering (EE) 2023 is part of Electromagnetic Fields Theory (EMFT) preparation. The Test: Cut Off Frequency & Wavelength questions and answers have been prepared according to the Electrical Engineering (EE) exam syllabus.The Test: Cut Off Frequency & Wavelength MCQs are made for Electrical Engineering (EE) 2023 Exam. Find important definitions, questions, notes, meanings, examples, exercises, MCQs and online tests for Test: Cut Off Frequency & Wavelength below.
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Test: Cut Off Frequency & Wavelength - Question 1

### The real part of the propagation constant is the

Detailed Solution for Test: Cut Off Frequency & Wavelength - Question 1

Explanation: The propagation constant is given by γ = α + jβ. Here the real part is the attenuation constant and the imaginary part is the phase constant.

Test: Cut Off Frequency & Wavelength - Question 2

### The phase constant of a wave is given by

Detailed Solution for Test: Cut Off Frequency & Wavelength - Question 2

Explanation: The phase constant of a wave in a transmission line is given by β = ω√(LC), where L and C are the specifications of the line.

Test: Cut Off Frequency & Wavelength - Question 3

### The cut off frequency of the dominant mode in a TE wave in the line having a and b as 2.5 cm and 1 cm respectively is

Detailed Solution for Test: Cut Off Frequency & Wavelength - Question 3

Explanation: The dominant mode in TE is TE10. The cut off frequency will be mc/2a, where m = 1 and a = 0.025 are given. On substituting, we get the frequency as 1 x 3 x 108/2 x 0.025 = 6 GHz.

Test: Cut Off Frequency & Wavelength - Question 4

The cut off frequency of the TE01 mode will be

Detailed Solution for Test: Cut Off Frequency & Wavelength - Question 4

Explanation: The cut off frequency consists of modes m and n. For m = 0, the dimension b will be considered. Thus the frequency is nc/2b, where c is the speed of the light.

Test: Cut Off Frequency & Wavelength - Question 5

The condition which will satisfy the dimensions of the waveguide is

Detailed Solution for Test: Cut Off Frequency & Wavelength - Question 5

Explanation: The dimensions a and b represent the broad wall and the side wall dimensions respectively. The broad wall will be greater than the side wall. Thus the condition a>b is true.

Test: Cut Off Frequency & Wavelength - Question 6

The cut off wavelength of the TE10 mode having a broad wall dimension of 5cm is

Detailed Solution for Test: Cut Off Frequency & Wavelength - Question 6

Explanation: The cut off wavelength of the waveguide is given by λc = 2a/m. on substituting for a = 0.05 and m = 1, we get λc = 2 x 0.05/1 = 0.1 units.

Test: Cut Off Frequency & Wavelength - Question 7

The broad wall dimension of a waveguide having a cut off frequency of 7.5 GHz is

Detailed Solution for Test: Cut Off Frequency & Wavelength - Question 7

Explanation: The cut off frequency and the broad wall dimension are related by fc = mc/2a. On substituting for m = 1 and fc = 7.5 GHz, we get a = 0.02 or 2 cm.

Test: Cut Off Frequency & Wavelength - Question 8

The sin θ in the waveguide refers to the ratio of the

Detailed Solution for Test: Cut Off Frequency & Wavelength - Question 8

Explanation: The ratio of the cut off frequency to the frequency at any point gives the sin θ in a waveguide.

Test: Cut Off Frequency & Wavelength - Question 9

Is the transmission of a frequency 5 GHz possible in waveguides?

Detailed Solution for Test: Cut Off Frequency & Wavelength - Question 9

Explanation: The cut off frequency for waveguide operation is 6 GHz. Thus a wave of 5 GHz is not possible for transmission in a waveguide.

Test: Cut Off Frequency & Wavelength - Question 10

The dimension for a waveguide in dominant mode with a cut off wavelength of 2 units is

Detailed Solution for Test: Cut Off Frequency & Wavelength - Question 10

Explanation: The cut off wavelength of a waveguide is given by λc = 2a/m. For the dominant mode, m = 1. Given that λc = 2, thus we get a = 4 units.

Test: Cut Off Frequency & Wavelength - Question 11

The waveguides are used in a transmission line for

Detailed Solution for Test: Cut Off Frequency & Wavelength - Question 11

Explanation: The waveguides are used to increase the transmission efficiency of the waves travelling through it.

Test: Cut Off Frequency & Wavelength - Question 12

The attenuation coefficient of the wave having a resistance of 15 ohm in a 50 ohm line is

Detailed Solution for Test: Cut Off Frequency & Wavelength - Question 12

Explanation: The attenuation coefficient of a wave with a resistance of R in a line of characteristic impedance Zo is α = R/2Zo. On substituting for R = 15 and Zo = 50, we get α = 15/(2 x 50) = 0.15 units.

## Electromagnetic Fields Theory (EMFT)

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## Electromagnetic Fields Theory (EMFT)

11 videos|46 docs|62 tests