Q140 / 160 · AP EAMCET 2019 (21 Apr Shift 1) · Physics

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

  • Correct answer
    A
    7
  • B
    5
  • C
    3
  • D
    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).

Hence, the correct answer is option: 'A'.

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