Question

The elastic constant and density of sea water are respectively 4.312 x 10⁹ N/m² and 2.2 x 10³ kg/m³. In a sonar experiment it was found that the acoustic echo from the sea bed is heard after 3 second. The depth of the sea is​

Answer 1

The depth of the sea bed is 2.1 km

Therefore, option (2) is correct.

Explanation:

Given:

The elastic constant as 4.312 x 10⁹ N/m²

The density of sea water as 2.2 x 10³ kg/m³

The echo is heard after 3 seconds

To find out:

The depth of the sea

Solution:

Let the depth of the sea is d

Given that

E = 4.312 x 10⁹ N/m²

ρ = 2.2 x 10³ kg/m³

Velocity of the sound in water

$v=\sqrt{\frac{E}{\rho}}$

$\implies v=\sqrt{\frac{4.312\times 10^9}{2.2\times 10^3}}$

$\implies v=\sqrt{1.96\times 10^6}$

$\implies v=1.4\times 10^3$ m/s

Since the echo is heard after 3 seconds i.e. in 3 seconds sound travels a distance 2d

Therefore, the depth of the sea

$2d=\text{speed}\times\text{time}$

$\implies 2d=1.4\times 10^3\times 3$

$\implies d=2.1\times 10^3$ m

$\implies d=2.1$ km

Hope this answer is helpful.

Know More:

Q: The velocity of sound waves 'v' through a medium may be assumed to depend on:

(i) The density of the medium 'd' and

(ii) The modulus of elasticity 'E'

Deduce by the method of dimensional formula for the velocity of sound. (k=1)

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