QUIZ-11-COMMUNICATION-SYSTEM

COMMUNICATION SYSTEM Quiz-11

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. Which of the following is the disadvantage of FM over AM

•  Larger band width requirement
•  Large noise
•  Higher modulation power
•  Low efficiency

Solution
Frequency modulation requires much wider channel (7 to 15 times) as compared to AM

Q2.According o Hubble’s law the red-shift (Z) of a receding galaxy and its distance r from earth are related as

•  Z∝r
•  Z∝ 1/r
•  Z∝1/r²
•  Z∝r^{(3/2 )}

Solution
Hubble’s law is the statement is physical cosmology that the red-shift (Z) in light coming from different galaxies is proportional to their distance (r). ie,Z∝r It is considered the first observational basis for the expanding space and today serves as one of the most often cited pieces of evidence in support of the Big Bang.

Q3. The electron density of E,F₁,F₂ layers of ionosphere is 2×10¹¹,5×10¹¹ and 8×10¹¹ m^(-3) respectively. What is the ratio of critical frequency for reflection of radiowaves

•  2∶4∶3
•  4:3:2
•  2:3:4
•  3:2:4

Solution
f_c∝(N)^(1/2)⇒(f_c )_E:(f_c )_(F_1 ):(f_c )_(F_2 )=(2×10¹¹ )^(1/2):(5×10¹¹ )^(1/2):(8×10¹¹ )^(1/2)=2∶3∶4

Q4. A laser beam of pulse power 10¹² W is focused on an object of area 10^(-4) cm². The energy flux in Watt cm^(-2). The energy flux in Watt cm^(-2) at the point of focus is

•  10²⁰
•  10¹⁶
•  10⁸
•  10⁴

Solution
Electric flux ϕ= (pulse power)/area=10¹²/10^(-4) =10⁶ Wcm^(-2)

Q5.Which of the following frequencies will be suitable for beyond the horizon communication

•  10 kHz
•  10 MHz
•  1 GHz
•  1000 GHz

Solution
10 MHz 

Q6. An antenna is of height 500 m. What will be its range (Radius of earth is 6400 km)?

•  800 km
•  100 km
• 50 km
•  80 km

Solution
Suppose height of antenna is h. Due to curvature of earth the TV signals will be received on earth in a circle of radius d. Reception range is given by d=√2hR Here, h=500 m,R=6400 km ∴d=√(2×500×6400×10³) =80000 m=80 km

Q7.Advantage of optical fibre is

•  High band width and EM interference
•  Low band width and EM interference
•  High band width low transmission capacity and no EM interference
•  High band width, high data transmission capacity and no EM interference

Solution
Few advantages of optical fibres are that the number of signals carried by optical fibres is much more than that carried by the Cu wire or radio waves. Optical fibres are practically free from electromagnetic interference and problem of cross talks whereas ordinary cables and microwave links suffer a lot from it.

Q8. Basically, a communication system is a

•  Transmitter
•  TReceiver
•  Messenger
•  None of these

Solution
A communication system acts like a messenger

Q9.When electromagnetic waves enter the ionised layer of ionosphere, then the relative permittivity ie,dielectric constant dielectric constant of the ionised layer

•  Does not change
•  Appears to increase
•  Appears to decrease
•  Sometimes appears to increase and sometimes to decrease

Solution
When electromagnetic waves enter the ionised layer of ionosphere, then the relative permittivity of the ionised layer appears to decrease

Q10. In co-axial cables, the repeater spacing is of the order of

•  20 km
•  2 km
•  200 km
• 2000 km

Solution
20 km