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Test: Cut Off Frequency & Wavelength - Electrical Engineering (EE) MCQ


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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) 2024 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) 2024 Exam. Find important definitions, questions, notes, meanings, examples, exercises, MCQs and online tests for Test: Cut Off Frequency & Wavelength below.
Solutions of Test: Cut Off Frequency & Wavelength questions in English are available as part of our Electromagnetic Fields Theory (EMFT) for Electrical Engineering (EE) & Test: Cut Off Frequency & Wavelength solutions in Hindi for Electromagnetic Fields Theory (EMFT) course. Download more important topics, notes, lectures and mock test series for Electrical Engineering (EE) Exam by signing up for free. Attempt Test: Cut Off Frequency & Wavelength | 12 questions in 10 minutes | Mock test for Electrical Engineering (EE) preparation | Free important questions MCQ to study Electromagnetic Fields Theory (EMFT) for Electrical Engineering (EE) Exam | Download free PDF with solutions
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

Answer: a
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

Answer: a
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.

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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

Answer: d
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

Answer: d
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

Answer: b
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

Answer: a
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

Answer: b
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

Answer: c
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

Answer: a
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

Answer: b
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

Answer: a
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

Answer: c
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.

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