Two tuning forks $A$ and $B$ sounded together give $6$ beats per second. With an air resonance tube closed at one end, the two forks give resonance when the two air columns are $24 cm$ and $25 cm$ respectively. Calculate the frequencies of forks.
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(c)
Let the frequency of the first fork be $f_1$ and that of second be $f_2$.
We then have, $f_1=\frac{v}{4 \times 24}$ and
$f _2=\frac{ v }{4 \times 25}$
We also see that $f _1 > f _2$
$\therefore f _1- f _2=6$
$\text { and } \frac{ f _1}{ f _2}=\frac{24}{25}$
Solving $(i)$ and $(ii)$, we get $f _1=150\,Hz$ and $f _2=144\,Hz$
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