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COMMUNICATION SYSTEM Quiz-13

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. In the given detector circuit, the suitable value of carrier frequency is
  •  <<109 Hz
  •  <<105 Hz
  •  >>109 Hz
  •  None of these
Solution
Using 1/fcarrier   <<RC We get time constant, RC=1000×10-12=10-9 s Now v=1/T=1/10-9 =109 Hz Thus the value of carrier frequency should be much less than 109 Hz, say 100 kHz

Q2.In earth's atmosphere, for F2-layer, the virtual height and critical frequency in night time are
  •  210 km and 5 MHz
  •  250 km and 6 MHz
  •  280 km and 7 MHz
  •  350 km and 6 MHz
Solution
Virtual height of F2-layer is 350 km and critical frequency is 6 MHz

Q3. The bit rate for a signal, which has a sampling rate of 8 kHz and where 16 quantisation levels have been
used is
  •   32000 bits/s
  •   16000 bits/s
  •   64000 bits/s
  •   72000 bits/s
Solution
If n is the number of bits per sample, then number of quantisation levels =2n Since the number of quantisation levels is 16 ⇒2n=16⇒n=4 ∴ bit rate = sampling rate × no. of bits per sample =8000×4=32,000 bits/s

Q4. A broken ligament is being ‘welded’ back in place using 20 ms pulses from a 0.5 W laser operating at a wavelength of 632 nm. The number of photons in 5 pulses of laser are
  •  1.59×10-18
  •  3.18×10-17
  •  1.59×1023
  •  3.18×10-16
Solution
The power of laser is 0.5 W, let n photon/s are incident by laser pulse on the broken ligament, then n ×hv=0.5 W ⇒nhc/λ=0.5 ⇒n=0.5×λ/hc =0.5×632×10-9/6.626×10-34×3×108 =1.59×1018photon/s So, number of photons contained in 5 pulses are, n×5×(20×103 )=1.59×1023

Q5.Small vales of numerical aperture (NA) decrease the pulse dispersion but increase losses due to
  •  Scattering
  •  Absorption
  •  Bending
  •  Microbending
Solution
Pulse dispersion ∝ numerical aperture of fiber. When numerical aperture of an optical fibre is small, then the energy losses will increase due to micro-bending  

Q6. A modem is a
  •  Modulating device only
  •  Demodulating device only
  • Modulating and demodulating device
  •  Receiving device
Solution
Modulating and demodulating device

Q7.Ozone layer blocks the radiations of wavelength
  •  Less than 3×10-7 m
  •  Equal to 3×10-7 m
  •  More than 3×10-7 m
  •  All of these
Solution
Ozone layer extends from 30 km to nearly 50 km above the earth’s surface in ozone sphere. This layer absorbs the major part of ultraviolet radiations coming from the sun and does not allow them to reach the earth’s surface. The range of ultraviolet radiations is 100 Å to 4000 Å. Thus, it blocks the radiations of wavelength less than 3×10-7 m (or 3000 Å).

Q8.Optical fibre works on the principle of
  •  Interference of light
  •  TTotal internal reflection of light
  •  Reflection of light
  •  Diffraction of light
Solution
Total internal reflection of light

Q9.The maximum range, dmax of radar is
  •  Proportional to the cube root of the peak transmitted power
  •  Proportional to the fourth root of the peak transmitted power
  •  Proportional to the square root of the peak transmitted power
  •  Not related to the peak transmitted power at all
Solution
Maximum Range of the radar is given by Rmax=( Pt A2 S/4πλ2 Pmin )1/4 Where Pt: peak value of transmitted power A: capture area of the receiving antenna S: radar cross-sectional area λ: wavelength of RADAR wave Pmin: minimum receivable power of the receiver

Q10. Large band width for higher data rate is achieved by using
  •  High frequency carrier wave
  •  Low frequency carrier wave
  •  High frequency audio wave
  • Low frequency audio wave
Solution
High frequency carrier wave provides a larger band width.

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