Question

A body emitting sound of frequency $350Hz$ is dropped from a balloon rising vertically upwards with constant velocity $5m/\sec$ . The frequency of sound as felt by the observer in the balloon $2\sec$ after the release is: (Velocity of sound in air is $335m/s$ ; acceleration due to gravity is $10m{s^{ - 2}}$ )
A. $330Hz$
B. $235Hz$
C. $340Hz$
D. $355Hz$

Answer 1

Hint : To solve this question, first we will rewrite the given facts related to this question, and then we will conclude whether the acceleration is applying upwards or downwards. And finally, we will apply the formula to find the frequency according to the given datas.

Complete Step By Step Answer:
As per the question, there are given information about the question:
Frequency of the sound that is emitted by the body, $f = 350Hz$ ,
And, the Velocity of the observer, ${v_o} = 5\,m/s$ ,
As the frequency of sound felt by the observer, in the balloon $2\sec$ after the release, the acceleration of sound due to gravity is constant i.e.. $g = 10m{s^{ - 2}}$ .
So, the velocity of sound, ${v_s} = g \times t = 10 \times 2 = 20m.{s^{ - 1}}$ ;
here, $g$ is the acceleration due to gravity and $t$ is time taken while releasing.
Therefore, the frequency of sound as felt by the observer in the balloon $2\sec$ after the release is:
Frequency, ${f’} = (\dfrac{{v - {v_s}}}{{v + {v_o}}}).f$
Here, $v$ is the velocity of sound in air which is given.
${v_s}$ is the velocity of sound.
${v_o}$ is the velocity of the observer.
and $f$ is the velocity of sound emitted by the body.
$\begin{gathered} \Rightarrow f = (\dfrac{{335 - 20}}{{335 + 5}}) \times 350 \\\ \,\,\,\,\,\,\,\,\ = 330Hz \\\ \end{gathered}$
Hence, the correct option is (A.) $330Hz$ .

Note :
Let's examine the physics of a hot air balloon using a sample calculation. The heated air inside the envelope is at roughly the same pressure as the outside air. The heated air inside the envelope is at roughly the same pressure as the outside air.