JEE Exam > JEE Questions > In a communication system operating at wavele...
Start Learning for Free
In a communication system operating at wavelength 800nm, only one percent of source frequency is available as signal bandwidth.The number of channels accommodated for transmitting TV signals of bandwidth 6MHz are A) 3.75*10^6 B) 4.87 * 10^5 C) 6.25 * 10^5 D) 3.86* 10^6?
Most Upvoted Answer
In a communication system operating at wavelength 800nm, only one perc...
Given:
- Wavelength of the communication system = 800 nm
- Signal bandwidth available = 1% of source frequency
- Bandwidth of TV signals = 6 MHz
To Find:
The number of channels accommodated for transmitting TV signals of bandwidth 6 MHz.
Solution:
Step 1: Calculate the source frequency
The source frequency can be calculated using the formula:
Speed of light (c) = Wavelength (λ) x Frequency (f)
Given wavelength (λ) = 800 nm = 800 x 10^(-9) m
Speed of light (c) = 3 x 10^8 m/s (approx.)
Plugging in the values, we get:
3 x 10^8 = (800 x 10^(-9)) x f
Simplifying the equation, we find:
f = (3 x 10^8) / (800 x 10^(-9))
f = 3.75 x 10^14 Hz
So, the source frequency is 3.75 x 10^14 Hz.
Step 2: Calculate the signal bandwidth
The signal bandwidth available is given as 1% of the source frequency.
Signal bandwidth = 1% of (3.75 x 10^14 Hz)
Signal bandwidth = (1/100) x (3.75 x 10^14 Hz)
Signal bandwidth = 3.75 x 10^12 Hz
Step 3: Calculate the number of channels
The number of channels that can be accommodated can be calculated using the formula:
Number of channels = Signal bandwidth / Bandwidth per channel
Given bandwidth per channel = 6 MHz = 6 x 10^6 Hz
Plugging in the values, we get:
Number of channels = (3.75 x 10^12 Hz) / (6 x 10^6 Hz)
Simplifying the equation, we find:
Number of channels = 6.25 x 10^5
Therefore, the number of channels accommodated for transmitting TV signals of bandwidth 6 MHz is 6.25 x 10^5. Hence, the correct option is C).
- Wavelength of the communication system = 800 nm
- Signal bandwidth available = 1% of source frequency
- Bandwidth of TV signals = 6 MHz
To Find:
The number of channels accommodated for transmitting TV signals of bandwidth 6 MHz.
Solution:
Step 1: Calculate the source frequency
The source frequency can be calculated using the formula:
Speed of light (c) = Wavelength (λ) x Frequency (f)
Given wavelength (λ) = 800 nm = 800 x 10^(-9) m
Speed of light (c) = 3 x 10^8 m/s (approx.)
Plugging in the values, we get:
3 x 10^8 = (800 x 10^(-9)) x f
Simplifying the equation, we find:
f = (3 x 10^8) / (800 x 10^(-9))
f = 3.75 x 10^14 Hz
So, the source frequency is 3.75 x 10^14 Hz.
Step 2: Calculate the signal bandwidth
The signal bandwidth available is given as 1% of the source frequency.
Signal bandwidth = 1% of (3.75 x 10^14 Hz)
Signal bandwidth = (1/100) x (3.75 x 10^14 Hz)
Signal bandwidth = 3.75 x 10^12 Hz
Step 3: Calculate the number of channels
The number of channels that can be accommodated can be calculated using the formula:
Number of channels = Signal bandwidth / Bandwidth per channel
Given bandwidth per channel = 6 MHz = 6 x 10^6 Hz
Plugging in the values, we get:
Number of channels = (3.75 x 10^12 Hz) / (6 x 10^6 Hz)
Simplifying the equation, we find:
Number of channels = 6.25 x 10^5
Therefore, the number of channels accommodated for transmitting TV signals of bandwidth 6 MHz is 6.25 x 10^5. Hence, the correct option is C).
Attention JEE Students!
To make sure you are not studying endlessly, EduRev has designed JEE study material, with Structured Courses, Videos, & Test Series. Plus get personalized analysis, doubt solving and improvement plans to achieve a great score in JEE.
|
Explore Courses for JEE exam
|
|
Similar JEE Doubts
Top Courses for JEEView all
In a communication system operating at wavelength 800nm, only one percent of source frequency is available as signal bandwidth.The number of channels accommodated for transmitting TV signals of bandwidth 6MHz are A) 3.75*10^6 B) 4.87 * 10^5 C) 6.25 * 10^5 D) 3.86* 10^6?
Question Description
In a communication system operating at wavelength 800nm, only one percent of source frequency is available as signal bandwidth.The number of channels accommodated for transmitting TV signals of bandwidth 6MHz are A) 3.75*10^6 B) 4.87 * 10^5 C) 6.25 * 10^5 D) 3.86* 10^6? for JEE 2024 is part of JEE preparation. The Question and answers have been prepared according to the JEE exam syllabus. Information about In a communication system operating at wavelength 800nm, only one percent of source frequency is available as signal bandwidth.The number of channels accommodated for transmitting TV signals of bandwidth 6MHz are A) 3.75*10^6 B) 4.87 * 10^5 C) 6.25 * 10^5 D) 3.86* 10^6? covers all topics & solutions for JEE 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for In a communication system operating at wavelength 800nm, only one percent of source frequency is available as signal bandwidth.The number of channels accommodated for transmitting TV signals of bandwidth 6MHz are A) 3.75*10^6 B) 4.87 * 10^5 C) 6.25 * 10^5 D) 3.86* 10^6?.
In a communication system operating at wavelength 800nm, only one percent of source frequency is available as signal bandwidth.The number of channels accommodated for transmitting TV signals of bandwidth 6MHz are A) 3.75*10^6 B) 4.87 * 10^5 C) 6.25 * 10^5 D) 3.86* 10^6? for JEE 2024 is part of JEE preparation. The Question and answers have been prepared according to the JEE exam syllabus. Information about In a communication system operating at wavelength 800nm, only one percent of source frequency is available as signal bandwidth.The number of channels accommodated for transmitting TV signals of bandwidth 6MHz are A) 3.75*10^6 B) 4.87 * 10^5 C) 6.25 * 10^5 D) 3.86* 10^6? covers all topics & solutions for JEE 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for In a communication system operating at wavelength 800nm, only one percent of source frequency is available as signal bandwidth.The number of channels accommodated for transmitting TV signals of bandwidth 6MHz are A) 3.75*10^6 B) 4.87 * 10^5 C) 6.25 * 10^5 D) 3.86* 10^6?.
Solutions for In a communication system operating at wavelength 800nm, only one percent of source frequency is available as signal bandwidth.The number of channels accommodated for transmitting TV signals of bandwidth 6MHz are A) 3.75*10^6 B) 4.87 * 10^5 C) 6.25 * 10^5 D) 3.86* 10^6? in English & in Hindi are available as part of our courses for JEE.
Download more important topics, notes, lectures and mock test series for JEE Exam by signing up for free.
Here you can find the meaning of In a communication system operating at wavelength 800nm, only one percent of source frequency is available as signal bandwidth.The number of channels accommodated for transmitting TV signals of bandwidth 6MHz are A) 3.75*10^6 B) 4.87 * 10^5 C) 6.25 * 10^5 D) 3.86* 10^6? defined & explained in the simplest way possible. Besides giving the explanation of
In a communication system operating at wavelength 800nm, only one percent of source frequency is available as signal bandwidth.The number of channels accommodated for transmitting TV signals of bandwidth 6MHz are A) 3.75*10^6 B) 4.87 * 10^5 C) 6.25 * 10^5 D) 3.86* 10^6?, a detailed solution for In a communication system operating at wavelength 800nm, only one percent of source frequency is available as signal bandwidth.The number of channels accommodated for transmitting TV signals of bandwidth 6MHz are A) 3.75*10^6 B) 4.87 * 10^5 C) 6.25 * 10^5 D) 3.86* 10^6? has been provided alongside types of In a communication system operating at wavelength 800nm, only one percent of source frequency is available as signal bandwidth.The number of channels accommodated for transmitting TV signals of bandwidth 6MHz are A) 3.75*10^6 B) 4.87 * 10^5 C) 6.25 * 10^5 D) 3.86* 10^6? theory, EduRev gives you an
ample number of questions to practice In a communication system operating at wavelength 800nm, only one percent of source frequency is available as signal bandwidth.The number of channels accommodated for transmitting TV signals of bandwidth 6MHz are A) 3.75*10^6 B) 4.87 * 10^5 C) 6.25 * 10^5 D) 3.86* 10^6? tests, examples and also practice JEE tests.
|
|
Explore Courses for JEE exam
|
|
Suggested Free Tests
Signup for Free!
Signup to see your scores go up within 7 days! Learn & Practice with 1000+ FREE Notes, Videos & Tests.
|
© EduRev
|
Education Revolution
|
Follow Us
|
Signup to see your scores
go up within 7 days!
Access 1000+ FREE Docs, Videos and Tests
Takes less than 10 seconds to signup