## Solved Numericals based on SONAR Formula | Physics solved assignments

Here we are going to share some of the solved numericals based on the concept of SONAR . By practicing these examples one can know how to use SONAR formula for numericals based on this echo concept. The full form of SONAR is SOund Navigation And Ranging and this technique is used for detecting objects on or under the surface of the water. With latest advancement and technology , SONAR applications are increasing at a fast pace .For example , sonar is also used in air for robot navigation. Let us have a look at solved numericals based on sonar.

## SONAR Solved Numericals

Practicing these couple of examples with help you to perform better during exams and boot your confidence .These SONAR solved examples cover some of the tricky question also , these questions seem tricky but we have explained them in quite an easy way and you will get them in one run .Let us start with the basic SONAR formula which we are going to use for all the solved problems .

SONAR :

**Formula used : V= 2D/T where**

**V= Speed of sound**

**2D= Total distance traveled**

**T= Time taken**

### Sonar Numerical # 1

An ocean vessel is using SONAR for detecting ocean’s bottom. The speed of sound waves that travel through the water is 1450 m/s and the bottom of ocean is 1630 m from the surface .Calculate the time taken by waves to get reflected back to the surface.

**Ans :**

Given

*2D= Total distance traveled = 1630 * 2 m ( Two way distance ) *

*V= Speed of sound = 1450 m/s*

* V= 2D/T therefore T= 2D / V *

T= (1630) (2)/ 1450 = 2.25 s

### Numerical based on SONAR # 2

A ship sends a sound wave to the bottom of the sea and receives an echo after 0.3 sec .Assume speed of sound in water to be 1500 m /s . Calculate the depth of the sea .

**Ans :**

2D= Total distance traveled = to be calculated

Therefore depth of sea will be D ( half of the total )

V= Speed of sound in water = 1500 m /s

T= 0.3

V= 2D/T therefore 2D= T * V = 1500 * 0.3 = 450 m

depth of sea will be D = 225 m

**Sonar Solved Problems # 3**

There is a sonar device on a submarine and it sends out a signal .The echo is received exactly after 5 s . If the distance between the submarine and object is 3,625 m. Determine speed of sound waves in water .

**Ans. **

Given,

2D= Total distance traveled = 3,625 m* 2 m ( Two way distance )

T= 5 s

V= Speed of sound in water =

V= 2D/T = (2×3625)/5 = 1450 ms–1

### SONAR Tricky questions # 4

A guy dropped a stone from the tower which is 500 m in height .The stone falls into a pond of water at the base of the tower. After how much time the splash will be heard at the top ? The value of g is 10 ms–2 and speed of sound is 340 ms–1.

**Ans. **

Given

First we will solve the amount of time taken by stone to reach the tower base .

Distance = s = 500 m, u = 0 ms–1, g = 10 ms–2

Now use the formula s = u t + 1/2gt2

By solving , we get value of t = 10 sec

It shows that te amount of time taken by stone to reach the pond is 10 s

Second step = Use SONAR formula to calculate time taken to cover 500 m .

Give : Speed of sound = 340 ms–1

Distance = 500 m

Time taken sound to D = 500/340 = 1.5sec.

Total time taken = (10 + 1.5) s = 11.5 s

:. The splash will be heard at the top after 11.5 s

**SONAR Tricky Question # 5**

In this case we will calculate velocity instead of time .A stone is dropped into a well 44.1 m deep, and splash sound is heard after 3.13 s. (Given, g = 9.8 ms–2)

**Ans: **

** **u = 0, distance = s = 44.1 m;

g = 9.8 ms–2

Time taken by stone to reach 44.1m deep will be

Put the values in equation = s = u t + 1/2gt2

By solving , we will get t = 3 sec

The sound takes the remaining time i.e., 3.13 – 3 = 0.13 s to go to up of well through 44.1 m

Speed=d/t= 44.1 m/0.13 = = 339.23 ms–1

**SONAR Question # 6**

A ship sends out ultrasound that returns exactly after 3.42 s. Consider the speed of ultrasound in water to be 1531 m/s, calculate the distance between ship and the seabed .

**Ans. **

Given

2D= Total distance traveled = to be calculated

Therefore depth of sea will be D ( half of the total )

V= Speed of sound in water = 1531 m /s

T= 3.42 sec

V= 2D/T therefore 2D= T * V = 1531 m/s × 3.42 s = 5236 m

D= 5236 m/2 = 2618 m or 2.62 km

**Numericals based on SONAR # 7**

A submarine enters a sonar pulse, and it returns from an under water cliff in 1.02 sec. Assume speed of sound in water is 1531 m/s, determine its distance from the cliff .

**Ans : **

2D= Total distance traveled = to be calculated

Therefore depth of sea will be D ( half of the total )

V= Speed of sound in water = 1531 m /s

T= 1.02 sec

V= 2D/T therefore 2D= T * V = 2x1531x1.02 = 1561.62 m

D= 5236 m/2 = 2618 m or 2.62 km

Distance of the cliff = d/2= 1561.62/2 = 780.81 m

U guys are amazing

What about energy problems as well

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