# Solve the Following Example. 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 - Science and Technology

Sum

Solve the following example.

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.

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

(Ans: 330 m/s; 1650 m)

#### Solution

1. Time to hear first echo = 4 s
Distance travelled by sound in these 4 s = 360 × 2 m
Speed of sound = (360 xx 2)/4 = 180 m/s

2. Time taken to hear second echo = 4 s + 2 s = 6 s
Distance travelled in these 6 s = 2x
Speed of sound in air = 180 m/s
Therefore, 2x = 180 × 6
x = 540 m
Distance between the walls = 540 + 360 = 900 m

Is there an error in this question or solution?

#### APPEARS IN

Balbharati Science and Technology 9th Standard Maharashtra State Board
Chapter 12 Study of Sound
Exercise | Q 5.3 | Page 137
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