COMMUNICATION SYSTEM Quiz-18
Dear Readers,
As per analysis for previous years, it has been observed that students preparing for NEET find Physics out of all the sections to be complex to handle and the majority of them are not able to comprehend the reason behind it. This problem arises especially because these aspirants appearing for the examination are more inclined to have a keen interest in Biology due to their medical background.
Furthermore, sections such as Physics are dominantly based on theories, laws, numerical in comparison to a section of Biology which is more of fact-based, life sciences, and includes substantial explanations. By using the table given below, you easily and directly access to the topics and respective links of MCQs. Moreover, to make learning smooth and efficient, all the questions come with their supportive solutions to make utilization of time even more productive. Students will be covered for all their studies as the topics are available from basics to even the most advanced. .
Q1. Arrange the following communication frequency bands in the increasing order of frequencies
1. AM broadcast
2. Cellular mobile radio
3. F.M. broadcast
4. Television UHF
5. Satellite communication
Solution
Q2.The principle used in the transmission of signal through an optical fibre is
Solution
The optical fibres are used to transmit light signals from one place to another without any practical loss in the intensity of light signal. It works on the principle of total internal reflection.
The optical fibres are used to transmit light signals from one place to another without any practical loss in the intensity of light signal. It works on the principle of total internal reflection.
Q3. Electromagnetic waves with frequencies greater than the critical frequency of ionosphere cannot be used for communication using sky wave propagation, because
Solution
The sky waves are the radiowaves of frequency range between 2 MHz to 30 MHz. The ionosphere reflects these radiowaves back to earth during their propagation through atmosphere.
The sky waves are the radiowaves of frequency range between 2 MHz to 30 MHz. The ionosphere reflects these radiowaves back to earth during their propagation through atmosphere.
The refractive index of ionosphere is less than its free space value. That is, it behaves as a rarer medium. Therefore, the wave will turn away from the normal when it enters the ionosphere. As we go deep into the ionosphere (electron density N is large), the refractive index keeps on decreasing. The refraction or bending of the beam will continue till it reaches critical angle after which it will be reflected back. If the frequency is too high, then after a certain value, the electron density N may never be so high as to produce enough bending for attainment of critical angle or condition of reflection. This is called critical frequency (fc ).
For frequencies higher than this value, the refractive index of the ionosphere becomes very high, so they cross the ionosphere and do not return back to the earth.
Q4. Which fibers are less expensive and simple to construct?
Solution
Multi-mode step index fibers are less expensive and easy to construct
Multi-mode step index fibers are less expensive and easy to construct
Q5. Intelsat satellite is used for
Solution
Intelsat satellite is used for intercontinental communication
Intelsat satellite is used for intercontinental communication
Q6. For good demodulation of AM signal of carrier frequency f, the value of RC should be
Solution
For good demodulation, 1/f<<RC or RC>>1/f
For good demodulation, 1/f<<RC or RC
Q7.In hydrogen spectrum, the wavelength of Hα line is 656 nm, whereas in the spectrum of a distant galaxy, Hα wavelength is 706 nm. Estimated speed of the galaxy with respect to earth is
Solution
The wavelength of light emitted by a moving object is shifted. This effecd is called the Doppler’s shift given by λ=λ0 (1+v/c) …(i) Where λis perceived wavelength, v the velocity and c the speed of light. From Eq. (i) v=c((λ-λ0)/λ0 ) Given, λ0=656 nm,λ=706 nm, c=3×108 ms-1 v=3×108 ((706-656)/656) v=3×108×50/656≈2×107 ms-1
The wavelength of light emitted by a moving object is shifted. This effecd is called the Doppler’s shift given by λ=λ0 (1+v/c) …(i) Where λis perceived wavelength, v the velocity and c the speed of light. From Eq. (i) v=c((λ-λ0)/λ0 ) Given, λ0=656 nm,λ=706 nm, c=3×108 ms-1 v=3×108 ((706-656)/656) v=3×108×50/656≈2×107 ms-1
Q8.The wavelength of red light from He-Ne laser is 633 nm in air but 474 nm in the aqueous humor inside the eye ball. Then the speed of red light through the aqueous humor is
Solution
μ∝ 1/v∝1/λ ∴V∝λ V1/V2 =λ1/λ2 ⇒(3×108)/V2 =633/474⇒V2=3×108×474/633 =2.25×108 m/s
μ∝ 1/v∝1/λ ∴V∝λ V1/V2 =λ1/λ2 ⇒(3×108)/V2 =633/474⇒V2=3×108×474/633 =2.25×108 m/s
Q9.In the communication systems, AM is used for broadcasting because
Solution
In the communication system, amplitude modulation is used for broadcasting to avoide receiver complexity.
In the communication system, amplitude modulation is used for broadcasting to avoide receiver complexity.
Q10. In an FM system a 7 kHz signal modulate 108 MHz carrier so that frequency deviation is 50 kHz. The carrier swing is
Solution
Carrier swing =(Frequency deviation)/(Modulation frequency)=50/7=7.143
Carrier swing =(Frequency deviation)/(Modulation frequency)=50/7=7.143
