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An engine approaches a high vertical wall with constant speed of 90 m/sec.  When it is at a distance of 1050 m from the wall, it blows a whistle, whose echo is heard after 5 second.  Calculate the speed of sound in air.
Asked by Topperlearning User | 04 Jun, 2014, 01:23: PM

Let V be the velocity of sound and u be the velocity of source and echo is heard after time t.  Let the engine whistle when it is at P and hear the echo at Q.

Therefore, Distance travelled by engine in time t.

PQ = ut .........................(i)

and the distance travelled sound in time t

PR + RQ = Vt .........................(ii)

Add (i) and (ii), we get PR + RQ + PQ = 2PR = (V + u)t

Here u = 90 m/s = (V + 90) x 5

5V = 1650 or V = 330 m/s

Answered by | 04 Jun, 2014, 03:23: PM
CBSE 11-science - Physics
Asked by Topperlearning User | 04 Jun, 2014, 01:23: PM
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Asked by Topperlearning User | 04 Jun, 2014, 01:23: PM
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Asked by Topperlearning User | 04 Jun, 2014, 01:23: PM
CBSE 11-science - Physics
Asked by Topperlearning User | 04 Jun, 2014, 01:23: PM