KVPY Sample Paper KVPY Stream-SX Model Paper-17

In a communication system operating at wavelength 800 nm, only one percent of source frequency is available as signal bandwidth. The number of channels accommodated for transmitting TV signals of band width 6 MHz are (Take velocity of light \[c=3\times {{10}^{8}}\] m/s, \[h=6.6\times {{10}^{-34}}\] J-s)

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 Answer: C

Solution :

\[f=\frac{c}{\lambda }=\frac{3\times {{10}^{8}}}{8\times {{10}^{-\,7}}}\]\[=\frac{3}{8}\times {{10}^{15}}Hz\]

\[\therefore \] \[n=\frac{(0.01)f}{6\times {{10}^{6}}}\]

\[=\frac{\frac{3}{8}\times {{10}^{13}}}{6\times {{10}^{6}}}\]\[=\frac{1}{16}\times {{10}^{7}}\]\[=6.25\times {{10}^{5}}.\]