Q. When sound is produced in an aeroplane moving with a velocity of $200\ ms^{- 1}$horizontal its echo is heard after $10\sqrt{5}$s. if velocity of sound in air is $300\ ms^{- 1}$the elevation of aircraft is
A) 250 m
B) $250\ \sqrt{5}m$
C) 12.50 m
D) 2500 m
Explanation
(d) From the figure sound travel ABC
[image:15_image91.emf]
Given, time for the echo=$10\sqrt{5}s$
Velocity of the plane=$200ms^{- 1}$
Hence,
$OC = 200 \times 5\sqrt{5} = 2236\ m$
$BC =$velocity of sound$\ \times 5\sqrt{5}$
$\Longrightarrow BC = 300 \times 5\sqrt{5} = 3354\ m$
$\therefore OB = \sqrt{BC^{2} - OC^{2}}$
OB=2500m
The plane is 2500 m above the ground.
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.