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A ship in a sea sends SONAR waves straight down into the seawater from the bottom of the ship. The signal reflects from the deep bottom bed rock and returns to the ship after 3.5 s. After the ship moves to 100km it sends another signal which returns back after 2s. Calculate the depth of the sea in each case and also compute the difference in height between two cases.​

Answer 1

Explanation:

Since speed of sound in water isn't given, I'm gonna assume it to be 1500 m/sec

In each case sound is going to travel twice the distance in specified time.

For first case:

2d = 1500*3.5 = 5250

d = 2625m or 2.625km

For second case

2d` = 1500*2

d` = 1500m or 1.5km

Difference in distances = 2.625 - 1.5 = 1.125km

Answer 2

SOLUTION :-

We know that,

Speed of SONAR waves in water,

$\sf \blue{c = 1500 {ms}^{ - 1}}$

[NOTE :- t is taken two times in each case because the waves will travel two times that is, from ship to sea bed and vice versa.]

Time taken to reflect from the bottom of the sea,

$\sf 2t = 3.5 \: sec$

$\sf t = \frac{3.5}{2}$

$\sf \therefore \fbox{t = 1.75 \: sec}$

Case I :-

Distance covered in forward and reflected backward, $\sf (d_1)$ $\sf \pink{= c \times t}$

$\sf \green{d_1 = 1500 \times 1.75 = 2625 \: m}$

Now,

After ship move in a distance = $\sf 100 \: km$

Time taken to reflect by the waves,

$\sf 2t = 2 \: sec$

$\sf t = \frac{2}{2} \: sec$

$\sf \therefore \fbox{t = 1 \: sec}$

Case II :-

Distance covered by the waves, $\sf (d_2)$ $\sf \pink{= c \times t}$

$\sf \green{d_2 = 1500 \times 1= 1500}$

The different between the height of two cases $\sf \red{ = (d_1) - (d_2)}$

$\sf = 2625 - 1500$

$\sf = \fbox{1125 \: m}$

Hence,

$\sf \fbox \orange{h_{difference} = 1125 \: metres}$