# Question 1 Page 172 - Chapter 12 Class 9 - Sound

Last updated at June 26, 2019 by Teachoo

Last updated at June 26, 2019 by Teachoo

Transcript

Question 1 Page 172 A submarine emits a sonar pulse, which returns from an underwater cliff in 1.02 s. If the speed of sound in salt water is 1531 m/s, how far away is the cliff? Time taken by the sound to return = t = 1.02s Speed of sound in water = v = 1531 m/s Let the distance between the submarine and the cliff = d Hence, Total distance covered by the sound = 2d Now, Speed of sound = (๐ท๐๐ ๐ก๐๐๐๐ ๐ก๐๐๐ฃ๐๐๐๐๐ ๐๐ฆ ๐ ๐๐ข๐๐)/(๐๐๐๐ ๐๐๐๐๐) v = 2๐/๐ก v ร t = 2d 2d = v ร t d = (๐ฃ ร ๐ก)/2 d = (1531 ร 1.02)/2 d = (1531 ร 102)/(2 ร 100) d = (1531 ร 51)/100 d = 78081/100 d = 780.81 m Distance between submarine and cliff is 780.81 m.

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Question 1 Page 172 You are here

Class 9

Chapter 12 Class 9 - Sound

About the Author

Davneet Singh

Davneet Singh is a graduate from Indian Institute of Technology, Kanpur. He has been teaching from the past 9 years. He provides courses for Maths and Science at Teachoo.