When an air column at $15\,^oC$ and a tunning fork are sounded together then $4$ beats per second are produced, the frequency of the fork is less then that of air column. When the temperature falls to $10\,^oC$ , then the beat frequency decreases by one. The frequency of the fork will be ..... $Hz$ $[V_{sound}$ at $0\,^oC = 332\,m/s]$
Diffcult
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$\frac{{{{\rm{V}}_{15}}}}{{2l}} - {\rm{n}} = 4$
$\frac{{{{\rm{V}}_{10}}}}{{2l}} - {\rm{n}} = 3$
$\frac{V_{15}}{V_{10}}=\frac{n+4}{n+3}$ but $V_{t}=V_{0}+0.6 t$
$\frac{332+9}{332+6}=\frac{n+4}{n+3}$
$\mathrm{n}=110 \mathrm{\,Hz}$
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