question_answer

A parachutist jumps from the top of a very high tower with a siren of frequency 800 Hz on his back. Assume his initial velocity to be zero. After falling freely for 12s, he observes that the frequency of sound heard by him reflected from level ground below him is differing by 700Hz w.r.t. the original frequency. What was the height of tower. Velocity of sound in air is 330 m/s, and\[g=10m/{{s}^{2}}\].

A)  511.5 m.

B)  1057.5 m.

C) 757.5 m.

D) 1215.5 m.            

Correct Answer: B

Solution :

[b] Let the sound observed by the parachutist at \[{{t}_{0}}=12s\] be produced at \[{{t}_{1}}s\]. Velocity of source at the instant of sound\[=g{{t}_{1}}\]and velocity of observer at the instant of observing same sound\[=g{{t}_{0}}\]. Hence the relation between apparent frequency\[f'\]and original frequency\[f\]will be \[f'=f\left( \frac{v+g{{t}_{0}}}{v-g{{t}_{1}}} \right)\] Here \[f=800\,Hz,\]  \[g=10\,\,m/{{s}^{2}}\]   \[v=330\,m/s\],     \[{{t}_{0}}=12s\] and \[f'=800+700=1500\,Hz\] Putting these, we get, \[{{t}_{1}}=9s\] Now the distance travelled by sound in \[({{t}_{0}}-{{t}_{1}})\,\sec \] is \[v({{t}_{0}}-{{t}_{1}})=\left( h-\frac{1}{2}gt_{0}^{2} \right)+\left( h-\frac{1}{2}gt_{1}^{2} \right)\] Putting the values, we get, h = 1057.5m.