### Waves - Result Question 77

#### 79. A source and an observer move away from each other, with a velocity of $10 m / s$ with respect to ground. If the observer finds the frequency of sound coming from the source as $1950 Hz$, then original frequency of source is (velocity of sound in air $=340 m / s$ )

(a) $1950 Hz$

(b) $2068 Hz$

(c) $2132 Hz$

(d) $2486 Hz$

[2001]

## Show Answer

#### Answer:

Correct Answer: 79. (b)

Solution:

- (b) According to Doppler’s effect

$ n^{\prime}=(\frac{v-v_0}{v-v_s}) n=(\frac{340-10}{340+10}) n=\frac{330}{350} \times 1950 $

$ ={ }^{`} 068 Hz $