Communication Systems
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Two sources of sound placed close to each other, are emitting progressive waves given by
y1 = 4sin 600π t and y2 = 5 sin 608πt
An observer located near these two sources of sound will hear
-
4 beats per second with intensity ratio 25: 16 between waxing and waning
-
8 beats per second with intensity ratio 25:16 between waxing and waning
-
8 beats per second with intensity ratio 81:1 between waxing and waning
-
4 beats per second with intensity ratio 81:1 between waxing and waning
D.
4 beats per second with intensity ratio 81:1 between waxing and waning
Given, y1 = 4sin 600π t and y2 = 5 sin 608πt
Comparing with general equation
y = a sin 2πft
we get, f2 = 300 Hz and f2 304 Hz
Number of beats = f2-f1 = 4s-1
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A hemisphere is positively charged uniformly. The electric field at a point on a diameter away from the centre is directed
perpendicular to the diameter
parallel to the diameter
at an angle tilted towards the diameter
at an angle tilted away from the diameter
A.
perpendicular to the diameter
When the point is on the diameter and away from the centre of hemisphere which is charged uniformly and positively, the component of electric field intensity parallel to the diameter cancel out. So the electric field is perpendicular to the diameter.
A sound absorber attenuates the sound level by 20 dB. The intensity decreases by a factor of
1000
10000
10
100
D.
100
Let intensity of sound be I and I'. Loudness of sound initially
β2 - β1 = 20
∴ I' = 100 I
A source emits electromagnetic waves of wavelength 3m. One beam reaches the observer directly and other after reflection from a water surface, travelling 1.5 m extra distance and with an intensity reduced to 1/4 as compared to intensity due to the direct beam alone. The resultant intensity will be.
(1/4) fold
(3/4) fold
(5/4) fold
(9/4) fold
D.
(9/4) fold
We know that a phase change of π occurs when the reflection takes place at the boundary of the denser medium. This is equivalent to a path difference of λ/2.
Therefore, Total phase difference = π -π = 0
Thus, the two waves superimpose in phase.
Resultant,
What maximum frequency can be reflected from ionosphere?
5 MHz
6 GHz
5 KHz
500 MHz
A.
5 MHz
The ionosphere extends from a height of 80 km to 300 km. The refractive index of ionosphere is less than its free space value. That is, it behaves as a rarer medium. As we go deep into the ionosphere, the refractive index keeps on decreasing. The bending of beam (away from the normal) will continue till it reaches critical angle after which it will be reflected back. The different points on earth receive signals reflected from different depths of the ionosphere. There is a critical frequency f, (S to 100 MHz) beyond which the waves cross the ionosphere and do not return back to earth.
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Mock Test Series
Mock Test Series



