MCQ
When the observer moves towards the stationary source with velocity, $V_1$, the apparent frequency of emitted note is $F_1$. When the observer moves away from the source with velocity $V_1$, the apparent frequency is $F_2$. If $V$ is the velocity of sound in air and $\frac{F_1}{F_2}=2$ then $\frac{V}{V_1}=$ ?
  • A
    2
  • 3
  • C
    4
  • D
    5

Answer

Correct option: B.
3
(b) :When the observer moves towards the stationary source, apparent frequency of emitted note,
$
F_1=\left(\frac{V+V_1}{V}\right) n ...(i)
$
When the observer moves away from the stationary source, apparent frequency of emitted note
$
F_2=\left(\frac{V-V_1}{V}\right) n
$ ...(ii)
Dividing eqn. (i) by eqn. (ii), we get
$
\begin{aligned}
& \frac{F_1}{F_2}=\frac{V+V_1}{V-V_1} \\
& \because \quad \frac{F_1}{F_2}=2 \text { (given) } \therefore \quad \frac{V+V_1}{V-V_1}=2 \\
& \Rightarrow \quad V+V_1=2 V-2 V_1 \Rightarrow V=3 V_1 \text { or } \frac{V}{V_1}=3
\end{aligned}
$

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