Physics NCERT Exemplar Solutions Class 12th Chapter Fifteen Communication Systems Class 12th Physics Physics NCERT Exemplar Solutions Class 12th Chapter Five

Only 2% of the optical source frequency is the available channel bandwidth for an optical communicating system operating at 1000 nm. If an signal requires a bandwith of 8 kHz, how many channels can be accommodated for transmission:

Option 1 - <p>375 × 10<sup>7</sup></p>

Option 2 - <p>75 × 10<sup>7</sup></p>

Option 3 - <p>375 × 10<sup>8</sup></p>

Option 4 - <p>75 × 10<sup>9</sup></p>

3 Views|Posted 8 months ago

Asked by Shiksha User

2 Answers

Sort by:

A

Answered by

Alok Kumar Singh | Contributor-Level 10

8 months ago

Correct Option - 2

Detailed Solution:

$v = f λ$

$f = \frac{v}{λ} = \frac{3 * 1 0^{8}}{1 0^{3} * 1 0^{- 9}} = 3 * 1 0^{14} H_{z}$

Channels = $\frac{2 % o f 3 * 1 0^{14} H_{z}}{8 * 1 0^{3}}$

$\frac{2 % o f 3 * 1 0^{14} H_{z}}{8 * 1 0^{3}}$

$\frac{2 % o f 3 * 1 0^{14} H_{z}}{8 * 1 0^{3}}$

$\frac{= \frac{2 x}{100} 3 * 1 0^{14}}{8 * 1 0^{3}} = 75 * 1 0^{7}$

Upvote

Thumbs Down Icon

A

Answered by

Alok Kumar Singh | Contributor-Level 10

8 months ago

$v = f ?$

$f = \frac{v}{?} = \frac{3 * 1 0^{8}}{1 0^{3} * 1 0^{? 9}} = 3 * 1 0^{14} H_{z}$

Channels = $\frac{2 ? ? % ? ? o f ? ? 3 * 1 0^{14} H_{z}}{8 * 1 0^{3}}$

$\frac{2 ? ? % ? ? o f ? ? 3 * 1 0^{14} H_{z}}{8 * 1 0^{3}}$

$\frac{2 ? ? % ? ? o f ? ? 3 * 1 0^{14} H_{z}}{8 * 1 0^{3}}$

$\frac{= \frac{2 x}{100} 3 * 1 0^{14}}{8 * 1 0^{3}} = 75 * 1 0^{7}$

Upvote

Thumbs Down Icon

Similar Questions for you

The TV transmission tower at a particular station has a height of 125 m. For doubling the coverage of its range, the height of the tower should be increased by

According to question, we can write

$d = \sqrt[]{2 h R} \Rightarrow \frac{d_{2}}{d_{1}} = \sqrt[]{\frac{h_{2}}{h_{1}}} \Rightarrow h_{2} = {(\frac{d_{2}}{d_{1}})}^{2} h_{1} = {(\frac{2}{1})}^{2} \times 125 = 500 m$

Increment in height of tower = h₂ – h₁ = 500 – 125 = 375 m

A signal containing multiple frequencies is shown in figure A. It is passed through 2 different filters and then outputs are shown in figure B & C.

Low pass filter will allow low frequency signal to pass while high pass filter allow high frequency to pass through

An antenna is mounted on a 400m tall building. What will be the wavelength of signal that can be radiated effectively by the transmission tower upto a range of 44kg?

$λ > h \Rightarrow λ > 400 m$

A wave of frequency of 10 kHz and amplitude 20 volts is being used to modulate amplitude of carrier wave of frequency 1000 kHz and peak voltage 40 volts. Choose correct option regarding carrier modulated wave.


µ = A? /A? = 0.5 {A? = 20 volt, A? = 40 volt}
m (t) = A? sin ω? t {ω? = 2π×10? }
c (t) = A? sin ω? t {ω? = 2π×10×10³}
C? (t) = (A? + A? sin ω? t) sin ω? t ⇒ A? {1+ µsin ω? t} sin ω? t

For VHF signal broadcasting........ km² of maximum service area will be covered by an antenna tower of height 30m, if the receiving antenna is placed at ground. Let radius of the earth be 6400km. (Round off to the Nearest Integer) (Take π as 3.14)

Kindly consider the following figure

Taking an Exam? Selecting a College?

Get authentic answers from experts, students and alumni that you won't find anywhere else.

On Shiksha, get access to

66K

Colleges

|

1.2K

Exams

|

6.9L

Reviews

|

1.8M

Answers

Learn more about...

Physics NCERT Exemplar Solutions Class 12th Chapter Fifteen 2025

Physics NCERT Exemplar Solutions Class 12th Chapter Fifteen 2025

View Exam Details

Most viewed information

Share Your College Life Experience

Didn't find the answer you were looking for?

Search from Shiksha's 1 lakh+ Topics

or

Ask Current Students, Alumni & our Experts

Quick Actions

Community GuidelinesUser Points System

QuestionsDiscussionsTags

Have a question related to your career & education?

or

See what others like you are asking & answering