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  1. Class 9
  2. Chapter 12 Class 9 - Sound

Transcript

NCERT Question 8 Two children are at opposite ends of an aluminum rod. One strikes the end of the rod with a stone. Find the ratio of times taken by the sound wave in air and in aluminum to reach the second child. Taking Speed of sound in air and aluminum from Table 12.1 NCERT Speed of sound in air = 346 m/s Speed of sound in aluminum = 6420 m/s Sound travels for the same distance in both cases. Let the distance travelled by sound = d We know that, Speed = ๐ท๐‘–๐‘ ๐‘ก๐‘Ž๐‘›๐‘๐‘’/๐‘‡๐‘–๐‘š๐‘’ โˆด Time = ๐ท๐‘–๐‘ ๐‘ก๐‘Ž๐‘›๐‘๐‘’/๐‘†๐‘๐‘’๐‘’๐‘‘ Air Time in air = ๐ท๐‘–๐‘ ๐‘ก๐‘Ž๐‘›๐‘๐‘’/(๐‘†๐‘๐‘’๐‘’๐‘‘ ๐‘–๐‘› ๐‘Ž๐‘–๐‘Ÿ) = ๐‘‘/346 Aluminum Time in aluminum = ๐ท๐‘–๐‘ ๐‘ก๐‘Ž๐‘›๐‘๐‘’/(๐‘†๐‘๐‘’๐‘’๐‘‘ ๐‘–๐‘› ๐‘Ž๐‘™๐‘ข๐‘š๐‘–๐‘›๐‘–๐‘ข๐‘š) = ๐‘‘/6420 Finding ratio (๐‘‡๐‘–๐‘š๐‘’ ๐‘–๐‘› ๐‘Ž๐‘–๐‘Ÿ)/(๐‘‡๐‘–๐‘š๐‘’ ๐‘–๐‘› ๐‘Ž๐‘™๐‘ข๐‘š๐‘–๐‘›๐‘–๐‘ข๐‘š) = ๐‘‘/346 รท๐‘‘/6420 (๐‘‡๐‘–๐‘š๐‘’ ๐‘–๐‘› ๐‘Ž๐‘–๐‘Ÿ)/(๐‘‡๐‘–๐‘š๐‘’ ๐‘–๐‘› ๐‘Ž๐‘™๐‘ข๐‘š๐‘–๐‘›๐‘–๐‘ข๐‘š) = ๐‘‘/346 ร— 6420/๐‘‘ (๐‘‡๐‘–๐‘š๐‘’ ๐‘–๐‘› ๐‘Ž๐‘–๐‘Ÿ)/(๐‘‡๐‘–๐‘š๐‘’ ๐‘–๐‘› ๐‘Ž๐‘™๐‘ข๐‘š๐‘–๐‘›๐‘–๐‘ข๐‘š) = 6420/346 (๐‘‡๐‘–๐‘š๐‘’ ๐‘–๐‘› ๐‘Ž๐‘–๐‘Ÿ)/(๐‘‡๐‘–๐‘š๐‘’ ๐‘–๐‘› ๐‘Ž๐‘™๐‘ข๐‘š๐‘–๐‘›๐‘–๐‘ข๐‘š) = 3210/173 (๐‘‡๐‘–๐‘š๐‘’ ๐‘–๐‘› ๐‘Ž๐‘–๐‘Ÿ)/(๐‘‡๐‘–๐‘š๐‘’ ๐‘–๐‘› ๐‘Ž๐‘™๐‘ข๐‘š๐‘–๐‘›๐‘–๐‘ข๐‘š) = 18.55 Time in air = 18.55 ร— Time in aluminum Hence, ratio is 18.55

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CA Maninder Singh
CA Maninder Singh is a Chartered Accountant for the past 9 years and a teacher from the past 16 years. He teaches Spoken English, Written English, Grammar and Vocabulary at Englishtan.