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
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
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
Solution
Virtual height of F2-layer is 350 km and critical frequency is 6 MHz
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
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
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
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
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
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
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
Solution
Modulating and demodulating device
Modulating and demodulating device
Q7.Ozone layer blocks the radiations of wavelength
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 â„« to 4000 â„«. Thus, it blocks the radiations of wavelength less than 3×10-7 m (or 3000 â„«).
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 â„« to 4000 â„«. Thus, it blocks the radiations of wavelength less than 3×10-7 m (or 3000 â„«).
Q8.Optical fibre works on the principle of
Solution
Total internal reflection of light
Total internal reflection of light
Q9.The maximum range, dmax of radar is
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
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
Solution
High frequency carrier wave provides a larger band width.
High frequency carrier wave provides a larger band width.
