A train is moving on a straight track with speed 20 ms-1 .It is blowing its whistle at the frequency of 1000 Hz. The percentage change in the frequency heard by a person standing near the track as the train passes him is (speed of sound = 320 ms-1 ) close to :

  • 6%

  • 12%

  • 18%

  • 18%


B.

12%

Apparent frequency heard by the person before crossing the train.

straight f subscript 1 space equals space open parentheses fraction numerator straight c over denominator straight c minus straight v subscript straight s end fraction close parentheses space straight f subscript straight o space equals space open parentheses fraction numerator 320 over denominator 320 minus 20 end fraction close parentheses space 1000
Similarly, apparent frequency heard, after crossing the trainsstraight f subscript 2 space equals space open parentheses fraction numerator straight c over denominator space straight c plus straight V subscript straight s end fraction close parentheses straight f subscript straight o space equals space open parentheses fraction numerator 320 over denominator 320 plus 20 end fraction close parentheses 1000
left square bracket straight c space equals speed space of space sound right square bracket
increment straight f space equals space straight f subscript 1 minus straight f subscript 2 space equals space open parentheses fraction numerator 2 cv over denominator straight c squared minus straight v subscript straight s superscript 2 end fraction close parentheses straight x space straight f subscript straight o
or space fraction numerator increment straight f over denominator straight f subscript straight o end fraction straight x 100 space equals space open parentheses fraction numerator 2 cv over denominator straight c squared minus straight v subscript straight s superscript 2 end fraction close parentheses straight x 100
space equals space fraction numerator 2 space straight x space 320 space straight x space 20 over denominator 300 space straight x space 340 end fraction straight x 100
equals space fraction numerator 2 space space straight x space 32 space straight x space 20 over denominator 3 space straight x 34 end fraction space equals space 12.54 space equals space 12 percent sign

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A diode detector is used to detect an amplitude modulated wave of 60% modulation by using a condenser of capacity 250 pico farad in parallel with a load resistance 100 kilo ohm. Find the maximum modulated frequency which could be detected by it. 

  • 10.62 MHz

  • 10.62 kHz

  • 5.31 mHz

  • 5.31 mHz


B.

10.62 kHz

The frequency is given as

μ =0.6
R = 100 k = 100 x1000 Ω
c = 250 pico farad = 250 x 10-12 F
So,

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Arrange the following electromagnetic radiations per quantum in the order of increasing energy:
A: Blue light
B: Yellow light
C: X-ray
D: Radiowave

  • D, B, A, C

  • A, B, D, C

  • C, A, B, D

  • C, A, B, D


A.

D, B, A, C

As, we know energy liberated, E = hc/λ

i.e straight E space proportional to space 1 over straight lambda
So, lesser the wavelength greater will be energy liberated by electromagnetic radiations per quantum.

As, order of wavelength is given by

X- rays, VIBGYOR, Radio waves
therefore, the order of electromagnetic radiations per quantum.

D<B<A<C

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Choose the correct statement:

  • In amplitude modulation, the amplitude of the high-frequency carrier wave is made to vary in proportion to the amplitude of the audio signal.

  • In amplitude modulation, the frequency of the high-frequency carrier wave is made to vary in proportion to the amplitude of the audio signal.

  • In frequency modulation, the amplitude of the high-frequency carrier wave is made to vary in proportion to the amplitude of the audio signal.

  • In frequency modulation, the amplitude of the high-frequency carrier wave is made to vary in proportion to the amplitude of the audio signal.


B.

In amplitude modulation, the frequency of the high-frequency carrier wave is made to vary in proportion to the amplitude of the audio signal.

As, we know, an amplitude modulated wave, the bandwidth is twice the frequency of modulating the signal. Therefore, amplitude modulation (AM), the frequency of the high-frequency carrier wave is made to vary in proportion to the amplitude of the audio signal.

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A signal of 5 kHz frequency is amplitude modulated on a carrier wave of frequency 2 MHz. The frequencies of the resultant signal is/are:

  • 2005 kHz, and 1995 kHz

  • 2005 kHz, 2000 kHz and 1995 kHz

  • 2000 kHz and 1995 kHz

  • 2000 kHz and 1995 kHz


C.

2000 kHz and 1995 kHz

Frequency associated with AM are
fc - fm, f, fc + fm
according to the question
fc = 2 MHz = 2000 kHz

fm = 5 kHz
Thus, frequency of the resultant signal is are carrier frequency fc = 2000 kHz, LSB frequency  fc-fm = 2000 kHz-5kHz = 1995 kHz and USB frequency fc+fm = 2005 kHz
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