Question

A sonar emits pulses on the surface of water which are detected after reflection from its bottom at a depth of 1531 m.If the time interval between the imision and detection of pulse is 2sec. find the speed of sound in water.​

Answer 1

Answer:-

• Depth of sea=d=1531m
• Time interval=2s
• Speed of sound=v=?

We know that

$\boxed{\sf 2d=vt}$

$\\ \sf\longmapsto v=\dfrac{2d}{t}$

$\\ \sf\longmapsto v=\dfrac{2\times 1531}{2}$

$\\ \sf\longmapsto v=\dfrac{3062}{2}$

$\\ \sf\longmapsto v=1531m/s$

Answer 2

Given Information:

• Depth of water = 1531 m
• Time interval between the imision and detection of pulse = 2 seconds

Need To Calculate:

• Speed of sound in water

Using Concept:

• The echo method can be used to determine the speed of sound in air. For this, sound is produced from a place at a known distance let us say is at d at least 50 m from the reflecting surface. The time interval t in which the echo reaches the place from where the sound was produced, is noted by a stop watch having the least count 0.001 s. Then the speed of sound is calculated by using this formula:

Formula:

• $\hookrightarrow \blue{ \boxed{ \bf{V \: = \: \dfrac{total \: distance \: which \: had \: been \: travelled}{time \: interval} }}}$

In short:

• $\hookrightarrow \: \blue{\boxed{\bf{ \dfrac{2d}{t} \: ms {}^{ - 1} }}}$

Calculations:

Using the formula of determination of sound by echo method where V is the speed of sound:

• $\hookrightarrow \bf{V \: = \: \dfrac{2d}{t} }$

Substituting values:

• $\hookrightarrow \bf{V \: = \: \dfrac{2 \: \times \: 1531}{2} }$

Cancelling terms:

• $\hookrightarrow \bf{V \: = \: \dfrac{ \cancel2 \: \times \: 1531}{ \cancel2} }$

Hence:

• $\hookrightarrow \bf{V \: = \: {1531} \: ms {}^{ - 1} }$

Answer:

• Speed of sound in water is 1531 ms-¹