At a temperature of $27^{\circ} \mathrm{C}$, two identical organ pipes produce notes of frequency $140…
At a temperature of $27^{\circ} \mathrm{C}$, two identical organ pipes produce notes of frequency $140 \mathrm{~Hz}$. If the temperature of one pipe is raised to $57.75^{\circ} \mathrm{C}$, then the number of beats produced per second is
7
5
3
9
Solution
Here, $n_0=140 \mathrm{~Hz}, T_0=27^{\circ} \mathrm{C}=300 \mathrm{~K}$ and
$T_2=57.75^{\circ} \mathrm{C}=330.75 \mathrm{~K}$
The variation of speed of sound with temperature,
$\frac{v_t}{v_0}=\sqrt{\frac{T}{T_0}}$
Let $v_0=$ speed of sound at the room temperature.
$v_t=v_0 \sqrt{\frac{330.75}{300}}=1.05 v_0$
Hence, the beats produced in oragan pipe.
Number of beats, $n=n_t-n_0=140 \times 1.05-140=7$
$\left(\because n_t=\frac{v_t}{4 L} \text { and } n_0=\frac{v_0}{4 L}\right)$
So, the correct option is (a).