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Sunil is standing between two walls. The wall closest to him is at a distance of 360 m. If he shouts, he hears the first echo after 4 s and another after another 2 seconds. - Science and Technology

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Question

Sunil is standing between two walls. The wall closest to him is at a distance of 660 m. If he shouts, he hears the first echo after 4 s and another after another 2 seconds.

  1. What is the velocity of sound in air?
  2. What is the distance between the two walls? 
Sum
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Solution

Given - a distance of the nearest wall from Sunil = 660 m

Time after which Sunil hears first echo, (t1) = `(4 sec)/(2 sec)` = 2 s

Time after which Sunil hears second echo (t2) = `(6 sec)/(2 sec)` = 3 s

Velocity = `"distance"/"time"`

Let d1 be the distance of closest wall.

Let d2 be the distance of the other closest wall.

V = `"d"_1/"t"_1 = 660/2` = 330 m/s

Velocity = 330 m/s

Velocity = `"d"_2/"t"_2`

330 = `"d"_2/3`

330 × 3 = `"d"_2`

`"d"_2 = 990`

Distance between the two walls -

d+ d= 660 + 990 = 1650 m

Distance between the two walls - 1650 m

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Chapter 12: Study of Sound - Exercise [Page 137]

APPEARS IN

Balbharati Science and Technology [English] 9 Standard Maharashtra State Board
Chapter 12 Study of Sound
Exercise | Q 5. c. | Page 137

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