Question

2. A source emits radio signals of 100 MHz and
moves towards an observer with a speed of 9Kms.
The apparent change in the frequency of the signals is
1) 3KHz 2) 30KHz 3) 3MHz 4) 30MHz​

Answer 1

Answer:

b

Explanation:

because apparent is directly proportional to it

Answer 2

ANSWER:

The apparent change in frequency will be 3 KHz.

EXPLANATION:

   The apparent change in the frequency signal when the source is moving towards an observer while emitting a frequency of 100 MHz can occur only due to Doppler effect. According to Doppler effect, the frequency of the source emitter will vary depending on the position of source with respect to the observer. So in this case, the "source is moving towards" the observer, so the Doppler frequency can be determined using the below equation.

$f=\left(\frac{c}{c-v_{s}}\right) f_{0}$

   Here, c is the speed of wave in the medium, vs is the speed of the source which is moving towards the observer, the minus sign is used before the speed of source as the direction of source emitter is moving towards the observer and f0 is the source emitter frequency.  

As the radio signals are emitted in air, the speed of radio signals are 3 * 105 Km/s and the speed of the source emitter is given as 9 km/s and the source frequency is given as 100 MHz. So the Doppler frequency will be

$f=\left(\frac{c}{c-v_{s}}\right) f_{0}$

$f=\left(\frac{3 * 10^{5}}{3 * 10^{5}-9}\right) * 100 * 10^{6}$

So the apparent change in the frequency will be the difference between the "frequency emitted by" the source and the "frequency received by" observer due to Doppler effect.

Thus

$\text { Apparent change in frequency }=f-f_{0}$

Thus the "apparent change in frequency" will be 3 KHz.