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

Answer:

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}}v=

ρ

E

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

2.2×10

3

4.312×10

9

\implies v=\sqrt{1.96\times 10^6}⟹v=

1.96×10

6

\implies v=1.4\times 10^3⟹v=1.4×10

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}2d=speed×time

\implies 2d=1.4\times 10^3\times 3⟹2d=1.4×10

3

×3

\implies d=2.1\times 10^3⟹d=2.1×103

³m

=2.1⟹d=2.1 km