Home
Class 9
PHYSICS
A sonar signal from a ship is emitted un...

A sonar signal from a ship is emitted underwater towards the sea bed. It takes 0.7 s for the signal to bounce back from the sea bed. If sound travels at 1500 ` m s ^(-1)` in water, how deep is the sea?

A

525 m

B

1050 m

C

1071m

D

2143 m

Text Solution

AI Generated Solution

The correct Answer is:
To find the depth of the sea using the sonar signal information provided, we can follow these steps: ### Step 1: Understand the problem A sonar signal is emitted from a ship, travels to the sea bed, and then reflects back to the ship. The total time taken for the signal to travel to the sea bed and back is given as 0.7 seconds. ### Step 2: Identify the known values - Speed of sound in water (v) = 1500 m/s - Total time taken for the signal to return (t) = 0.7 s ### Step 3: Calculate the total distance traveled by the sound The total distance traveled by the sound wave is twice the depth of the sea (h), since it goes down to the sea bed and comes back up: \[ \text{Total distance} (s) = 2h \] ### Step 4: Use the formula for speed The formula for speed is given by: \[ v = \frac{s}{t} \] Rearranging this formula to find the distance (s): \[ s = v \cdot t \] ### Step 5: Substitute the known values into the equation Substituting the values of speed and time into the equation: \[ s = 1500 \, \text{m/s} \times 0.7 \, \text{s} \] \[ s = 1050 \, \text{m} \] ### Step 6: Relate the total distance to the depth Since the total distance \( s \) is equal to \( 2h \): \[ 2h = 1050 \, \text{m} \] ### Step 7: Solve for the depth (h) To find the depth \( h \): \[ h = \frac{1050 \, \text{m}}{2} \] \[ h = 525 \, \text{m} \] ### Final Answer The depth of the sea is **525 meters**. ---

To find the depth of the sea using the sonar signal information provided, we can follow these steps: ### Step 1: Understand the problem A sonar signal is emitted from a ship, travels to the sea bed, and then reflects back to the ship. The total time taken for the signal to travel to the sea bed and back is given as 0.7 seconds. ### Step 2: Identify the known values - Speed of sound in water (v) = 1500 m/s - Total time taken for the signal to return (t) = 0.7 s ...
Promotional Banner

Topper's Solved these Questions

  • FOOTSTEPS TOWARDS CBSE BOARD

    MTG IIT JEE FOUNDATION|Exercise Section - D |14 Videos
  • FORCE AND LAWS OF MOTION

    MTG IIT JEE FOUNDATION|Exercise OLYMPIAD /HOTS CORNER |20 Videos

Similar Questions

Explore conceptually related problems

Sound signal of frequency 40 kHz is sent vertically into sea water. The signal gets reflected from the ocean bed and returns to the surface 0.60 s after it was emitted. The speed of sound in sea water is 1500 ms^(-1) . What is the depth of the sea and wavelength of this signal in water.

An ultrasound signal of frequency 50 kHz is sent vertically into sea water. The signal gets reflected from the ocean bed and returns to the surface 0.80 s after it was emitted. The speed of sound in sea water is 1500ms^-1 . a. Find the depth of the sea. b. What is the wavelength of this signal in water.

A man is swimming at a depth d in a sea at a distance L (gt gt d) from a ship (S). An explosion occurs in the ship and after hearing the sound the man immediately moves to the surface. It takes 0.8 s for the man to rise to the surface after he hears the sound of explosion. 0.2 s after reaching the surface he once again hears a sound of explosion. Calculate L. Give: speed of sound in air =340 ms^(-1) , Bulk modulus of water =2xx10^(9) Pa

An echo-sounder in a trawler (fishing boat) receives an echo from a shoal of fish 0.4s after it we sent. If the speed of sound in water is 1500 m/s, how deep is the shoal?

MTG IIT JEE FOUNDATION-FOOTSTEPS TOWARDS NEET-Multiple Choice Question
  1. A stone is released from the top of a tower of height 19.6m. Calculate...

    Text Solution

    |

  2. A mountaineer of weight 300 N climbs up a rock face of vertical height...

    Text Solution

    |

  3. A sonar signal from a ship is emitted underwater towards the sea bed. ...

    Text Solution

    |

  4. An object moving with uniform circular motion shows

    Text Solution

    |

  5. When a bicycle is motion the force of friction exerted by the ground o...

    Text Solution

    |

  6. The distance between two bodies becomes 6 times more than the usual di...

    Text Solution

    |

  7. A body possess potential energy of 460 J whose mass is 20 kg and is ra...

    Text Solution

    |

  8. A key of a mechanical piano is struck gently and then struck again but...

    Text Solution

    |

  9. A particle moves 3 m north , then 4 m east and finally 6m south . Cal...

    Text Solution

    |

  10. When we stops pedalling a bicycle we are riding, the bicycle beigns to...

    Text Solution

    |

  11. A planet has density p, radius R and acceleration due to gravity as g....

    Text Solution

    |

  12. A car completes its journey in a straight line in three equal parts wi...

    Text Solution

    |

  13. A particle starts its motion from rest under the action of a constant ...

    Text Solution

    |

  14. If a planet existed whose mass was twice that of Earth and whose radiu...

    Text Solution

    |

  15. Best relationship between momentum and kinetic energy possesses by an ...

    Text Solution

    |

  16. A wave in slinky travelled to and fro in 5 s. The length of the slinky...

    Text Solution

    |

  17. The diagram shows the displacement-time graph for a particle moving in...

    Text Solution

    |

  18. Pick the fundamental law of motion

    Text Solution

    |

  19. A solid of density D is floating in a liquid of density d. If V is the...

    Text Solution

    |

  20. A body of mass 20 kg falls through a distance of 50 cm. Then the loss ...

    Text Solution

    |