A sonar emits pulses on the surface of a pond which are detected after reflection from the bottom. If the time interval between the emission and detection of the pulse is $2 s$ , find the depth of the pond. Take the velocity of sound in water as $1531 m s^{- 1}$

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Solution

Let, $d$ is the depth of the pond.
Distance traveled by the sound between emission and detection will be $2 d$ .
So, $2 d = v t$ , here $v = velocity of sound in water = 1531 m / s$ and $t = total time = 2 s$
$∴ d = \frac{v t}{2} = \frac{1531 \times 2}{2} = 1531 m$

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A sonar emits pulses on the surface of a pond which are detected after reflection from the bottom. If the time interval between the emission and detection of the pulse is 2 s, find the depth of the pond. Take the velocity of sound in water as 1531 ms−1

Let, d is the depth of the pond.Distance traveled by the sound between emission and detection will be 2d.So, 2d=vt, here v=velocity of sound in water=1531 m/s and t=total time=2 s ∴d=vt2=1531×22=1531m

A sonar emits pulses on the surface of a pond which are detected after reflection from the bottom. If the time interval between the emission and detection of the pulse is $2 s$ , find the depth of the pond. Take the velocity of sound in water as $1531 m s^{- 1}$

Let, $d$ is the depth of the pond.
Distance traveled by the sound between emission and detection will be $2 d$ .
So, $2 d = v t$ , here $v = velocity of sound in water = 1531 m / s$ and $t = total time = 2 s$
$∴ d = \frac{v t}{2} = \frac{1531 \times 2}{2} = 1531 m$