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.You need to login to perform this action.
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