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Physics-

General

Easy

Question

A sound source is falling under gravity. At some time t=0, the detector lies vertically hallow sources at a depth $H blank$ as shown in figure. If v is the velocity of sound and $f subscript 0 end subscript$ is frequency recorded after t=2s is

$f_{0}$
$\frac{f_{0} (v + 2 g)}{v}$
$\frac{f_{0} (v - 2 g)}{v}$
$f_{0} (\frac{v}{v - 2 g})$

The correct answer is: $f subscript 0 end subscript open parentheses fraction numerator v over denominator v minus 2 g end fraction close parentheses$

For maxima
Path Difference = (Even multiple) $fraction numerator lambda over denominator 2 end fraction rightwards double arrow capital delta x equals open parentheses 2 n close parentheses fraction numerator lambda over denominator 2 end fraction$
For 5 maximum responses
$rightwards double arrow 2 equals 2 open parentheses 5 close parentheses fraction numerator lambda over denominator 2 end fraction open curly brackets because increment x equals open parentheses 2 n close parentheses fraction numerator lambda over denominator 2 end fraction close curly brackets rightwards double arrow lambda equals fraction numerator 2 over denominator 5 end fraction equals 0.4 m$
$v subscript s end subscript equals 0 plus g cross times 2 equals 2 g$
As source is moving towards the observer.
$therefore blank f equals fraction numerator v over denominator v minus v subscript s end subscript end fraction f subscript 0 end subscript equals fraction numerator v f subscript 0 end subscript over denominator v minus 2 g end fraction$

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