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A whistling engine is approaching a stationary observer with velocity 110 m/s. The velocity of sound is 330 m/s. The ratio of frequencies as heard by observer at the time of approaching and passing of engine is : |
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Answer» Solution :`v_(1) ((V)/(V - U_(s))) v and v_(2) = ((V)/(V + U_(s)))` V `therefore (v_(1))/(v_(2)) = (V + U_(s))/(V - U_(s)) = (330 + 110)/(330 - 110) = (v_(1))/(v_(2))= 2 ` HENCE the CORRECT CHOICE is (d). |
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