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Published on: 27/05/2021
CBSE 12th Standard Physics Subject Communication Systems HOT Questions 1 Mark Questions With Solution 2021
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1.
You are given three semiconductors A,B,C with respective band gaps of 3eV, 2eV and 1eV for use in a photodetector to detect \(\lambda \) = 1400nm . Select the suitable semiconductor. Give reasons
2.
A microwave telephone link operating at the cenral frequency of 10 GHz has been established .If 2 % of this is available for microwave communication channel, then how many telephones channels can be simultaneously granted if each telephone is allotted a band width of 8 KHz
3.
A TV tower has a height of 110m. How much population is covered by the TV broadcast if the average population density around the tower is 1000 km-2? Given that radius of Earth = 6.4 x 106m
4.
The TV transmission tower at a particular place has a height of 160m. What is its coverage range? By how much should the height be increased to double its coverage range? Given that radius of earth = 6400 km
5.
A schematic arrangement for transmitting a message signal (20 Hz to 20kHz) is given below:
Give two drawbacks from which this arrangement suffers. Describe briefly with the help of a block diagram the alternative arrangement for the transmission and reception of the message signal
1.
( )
Energy corresponding to \(\lambda \) = 1400nm = 1400 x 10-9 m is
E = \(\frac { hc }{ \lambda } =\frac { 1.42\times { 10 }^{ -19 } }{ 1.6\times { 10 }^{ -19 } } eV=1eV\)
For detection E must be equal to greater then Eg. Hence only suitable semiconductor is C.
2.
( )
Microwave communication channel width = \(\frac { 2 }{ 100 } \times 10GHz\)
= 0.2 Ghz
band width of channel = 8 KHz
= 2.5 x 104
3.
( )
Radius of the area covered by TV broadcast is
d = (2Rh)1/2
= 37500m = 37.5 km
= 4.4 x 106
4.
( )
d = (2 x 6400 x 103 x 160)1/2 = 45255m
Coverage range , d = (2Rh)1/2
h2 - 4h1 = 4 x 160 = 640 m
5.
( )
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