At $23^{\circ} C$, a pipe open at both ends resonates at a frequency of $450 \,Hz$. At what frequency does the same pipe resonate on a hot day when the speed of sound is $4 \%$ higher than it would be at $23^{\circ} C$ ?
KVPY 2011, Diffcult
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(c)
For an open end pipe, Frequency, $f=\frac{v}{\lambda}=\frac{n v}{2 L}$
$\Rightarrow \quad L=\frac{n v}{2 f}$
As, length of tube remains same in both cases.
$\text { So, } L=\frac{n v_1}{2 f_1}=\frac{n v_2}{2 f_2}$
$\Rightarrow f_2=\frac{v_2}{v_1} \cdot f_1$
Here, $v_2=1.04 v_1$ and $f_1=450 \,Hz$
So, $f_2=\frac{1.04 v_1}{v_1} \times 450$
$=1.04 \times 450$
$=468 \,Hz$
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