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JEE MAIN 2019
09-01-19 S2
Question
In a communication system operating at wavelength 800 nm , only one percent of source frequency is available as signal bandwidth. The number of channels accomodated for transmitting TV signals of band width 6 MHz are (take velocity of light $\mathrm{c}=3 \times 10^8 \mathrm{~m} / \mathrm{s}, \mathrm{h}=6.6 \times 10^{-34} \mathrm{~J}-\mathrm{s}$ )
Select the correct option:
A
$3.75 \times 10^6$
B
$3.86 \times 10^6$
C
$6.25 \times 10^5$
D
$4.87 \times 10^5$
✓ Correct! Well done.
✗ Incorrect. Try again or view the solution.
Solution
$\begin{aligned} & v=\frac{3 \times 10^8}{800} \times 10^9=\frac{3 \times 10^{15}}{8} \\ & \text { Singal bandwidth }=\frac{3 \times 10^{15}}{8} \times 0.01 \\ & \text { No. of Channels }=\frac{3 \times 10^{13}}{8 \times 6 \times 10^6}=6.25 \times 10^5\end{aligned}$
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