QUIZ-14-COMMUNICATION-SYSTEM

COMMUNICATION SYSTEM Quiz-14

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 one of the following statements is wrong?

•  Radio waves in the frequency range 30 MHz to 60 MHz are called sky waves
•  Radio horizon of the transmitting antenna for space wave is d_T=√( 2Rh_T )  (R=radius of earth,h_T=height of transmitting antenna)
•  Within the skip distance neither the ground waves nor the sky waves are received
•  The principle of fibre optical communication is total internal reflection

Solution
Within the skip distance neither the ground waves nor the sky waves are received

Q2.A step index fibre has a relative refractive index of 0.88%. What is the critical angle at the corecladding interface

•  60°
•  75°
•  45°
•  None of these

Solution
Here (μ₁-μ₂)/μ₁ =0.88/100⇒μ₂/μ₁ =0.9912 

∴ Critical angle θ_c=sin^-1⁡(μ₂/μ₁ )=sin^-1(0.9912)=82°24'

Q3. Venus looks brighter than other stars, due to

•   Atomic fusion takes place on its surface
•  It is closer to the earth than other stars
•  It has higher density than other stars
•  It is heavier than other stars

Solution
Venus looks brighter than other stars because it is closer to earth than other stars.

Q4. The SKIP ZONE in Radio Wave Transmission is that range where

•  There is no reception of either ground wave or sky wave
•  The reception of ground wave is maximum but that of sky wave is minimum
•  The reception of ground wave is minimum, but that of sky wave is maximum
•  The reception of both ground and sky wave is maximum

Solution
There is no reception of either ground wave or sky wave

Q5.An earth satellite emits a radio signal of frequency 10⁸ Hz. An observer on the ground detects beats between the received signal and a local signal also of frequency 10⁸ Hz. At a particular moment, the beat frequency is 2400 Hz. What is the component of satellite velocity directed towards earth at this moment?

•  1080 ms^-1
•  1800 ms^-1
•  3600 ms^-1
•  7200 ms^-1

Solution
Here, v-v^'=∆v=2400 Hz 

As ∆v=vv/c or v=c∆v/v=(3×10⁸×2400)/10⁸ =7200 ms^-1  

Q6. A diode AM detector with the output circuit consisting of R=1 kW andC=1 μF would be more suitable for detecting a carrier signal of

•  0.1 kHz
•  0.5 kHz
• 1 kHz
•  10 kHz

Solution
Given R=1 kΩ Or R=1×10³ C=1μF C=1×10^-6 F 

In this condition frequency of carrier signal 1/R_c <<f_c 

 1/1×10³×10^-6 <<f_c ∴f_c>>1 kHz Because frequency greater than 1 So f_c=10 kHz

Q7.If sky wave with frequency of 50 MHz is incident on D region at an angle of 30°, then angle of refraction is

•  15°
•  30°
•  60°
•  45°

Solution
For D-region, N=10⁹ m^-3 μ=√(1-(81.45 N)/v² )=√(1-(81.45×10⁹)/(50×10⁶ )² )=1 μ=sin⁡i/sin⁡r =1 Or sin⁡r=sin⁡i or r=i=30°

Q8. What is the refractive index of the fiber core?

•  1.556
•  1.453
•  1.425
•  1.626

Solution
Taking refraction from air to core of optical fiber. Then, μ₁/μ₀ =sin⁡ 10°/sin⁡ 7° =0.1736/0.1219=1.424 or μ₁=1.424 (asμ₂=1)

Q9.A laser beam of pulse power 10¹² W is focused on an object of area 10^-4 cm². the energy flux in Wcm² at the point of focus is

•  10²⁰
•  10¹⁶
•  10⁸
•  10⁴

Solution
Electric flux ϕ=(pulse power)/area =10¹²/10^-4 =10¹⁶ Wcm^-1

Q10. When the modulation percentage is 75, an AM transmitter produces 10 kW. How much of this is carrier power?

•  10 kW
•  13.33 kW
•  7.5 kW
• 7.81 kW

Solution
P_t=P_c (1+m²/2) P_c=P_t/1+m²/2 =10/1+1/2 (3/4)² =10×32/41 P_c=7.81 kw