## COMMUNICATION SYSTEM Quiz-17

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. The closed structure of co-axial cable prevents inner copper wire or core or from radiating signal power. The statement is
•  True
•  False
•  Neither true nor false
•  Partly true and partly false
Solution
The statement is perfectly true

Q2.Sinusoidal carrier voltage of frequency 1.5 MHz and amplitude 50 V is amplitude modulated by sinusoidal voltage of frequency 10 kHz producing 50% modulation. The lower and upper side-band frequencies in kHz are
•  1490, 1510
•  1510, 1490
•  1/1490,1/1510
•  1/1510,1/1490
Solution
Here, fc=1.5 MHz=1500 kHz,fm=10 kHz ∴ Low side band frequency =fc-fm=1500 kHz-10 kHz=1490 kHz Upper side band frequency =fc+fm=1500 kHz+10 kHz=1510 kHz

Q3. Of the following which is preferred modulation scheme for digital communication
•   Pulse Code Modulation (PCM)
•  Pulse Amplitude Modulation (PAM)
•  Pulse Position Modulation (PPM)
•  Pulse Width Modulation (PWM)
Solution
Pulse Code Modulation (PCM)

Q4. The velocity factor of a transmission line is x. If dielectric constant of the medium is 2.6, the value of x is
•  0.26
•  0.62
•  2.6
•  6.2
Solution
v.f.=1/√k=1/√2.6=0.62

Q5.In ruby laser, the stimulated emission is due to transition from
•  Metastable state to any lower state
•  Any higher state to lower state
•  Metastable state to ground state
•  Any higher state to ground state
Solution
For a ruby laser, a crystal of ruby is formed into a cylinder. A fully reflecting mirror is placed on one end and a partially reflecting mirror on the other. A high-intensity lamp is spiraled around the ruby cylinder to provide a flash of white light that triggers the laser action. The green and blue wavelengths in the flash excite electrons in the chromium atoms to a higher energy level. Upon returning to their normal state, the electrons emit their characteristic ruby-red light. The mirrors reflect some of this light back and forth inside the ruby crystal, stimulating other excited chromium atoms to produce more red light, until the light pulse builds up to high power and drains the energy stored in the crystal. In ruby laser stimulated emission is due to transition from metastable state to ground state.

Q6. A laser beam is used for locating distant objects because
•  It is monochromatic
•  It is coherent
• It is not absorbed
•  It has small angular speed
Solution
A LASER (Light Amplification by Stimulated Emission of Radiation) is an optical source that emits photons in a coherent beam. The property of coherent beam is used I many military applications to enhance the locating of distant objects. Since, a laser beam typically has low divergence, the laser light appears as a small spot even at long distances, the user simply places the spot on the desired target and barrel of the gun is aligned.

Q7.An amplitude modulated wave is modulated to 50%. What is the saving in power if carrier as well as one of the side bands are suppressed
•  70%
•  65.4%
•  94.4%
•  25.5%
Solution
Psb=Pc (ma/2)2=Pc (0.5)2/4=0.0625 Pc Also P=Pc (1+(ma2)/2)=Pc (1+(0.5)2/2)=1.125Pc ∴% saving =(1.125Pc-0.0625Pc)/(1.125Pc )×100=94.4%

Q8.The relation between the maximum electron density Nmax and the critical frequency fc for the ionosphere can be given as
•  fc=√9 Nmax
•  fc=√9  Nmax
•  fc=9√Nmax
•  None of these
Solution
If maximum electron density of the ionosphere in Nmax per m3, then fc=9(Nmax )1/2. Above fc, a wave will penetrate the ionosphere and is not reflected by it.

Q9.The distance of coverage of a transmitting antenna is 12.8 km. Then, the height of the antenna is (Given that radius of earth = 6400 km)
•  6.4 m
•  12.8 m
•  3.2 m
•  16 m
Solution
Given d=12.8 km,R=6400 km We have d=√2hR h=d2/2R=(12.8)2/(2×6400)=12.8 m

Q10. A telephone link operating at a central frequency of 10 GHz is established. If 1% of this available then how many telephone channels can be simultaneously given when each telephone covering a band width of 5 kHz
•  2×104
•  2×106
•  5×104
•  5×106
Solution
1% of 10 GHz=10×109×1/100=108 Hz Number of channels =108/(5×103 )=2×104

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