Two closed organ pipes of length \(100 \mathrm{~cm}\) and \(101 \mathrm{~cm}\) produces 16 beats in \(20…

1. \(303 \mathrm{~ms}^{-1}\)
2. \(332 \mathrm{~ms}^{-1}\)
3. \(323.2 \mathrm{~ms}^{-1}\)
4. \(300 \mathrm{~ms}^{-1}\).

Solution

Frequency of fundamental mode of vibrations of $1^{\text {st }}$ closed organ pipe, From question, $v_1-v_2=\frac{16}{20}$ $\begin{aligned} & \Rightarrow \frac{v}{4 \times 100}-\frac{v}{4 \times 101}=\frac{16}{20} \\ & \Rightarrow \frac{v}{4} \cdot\left[\frac{1}{100}-\frac{1}{101}\right]=\frac{16}{20} \\ & \Rightarrow \frac{v}{4}\left[\frac{101-100}{101 \times 100}\right]=\frac{16}{20} \Rightarrow \frac{v}{4} \times \frac{1}{101 \times 100}=\frac{16}{20} \\ & \Rightarrow v=\frac{16 \times 101 \times 4 \times 100}{20}=32320 \mathrm{~cm} / \mathrm{s} . \\ & =323.2 \mathrm{~m} / \mathrm{s} .\end{aligned}$

Asked in: NEET 2002

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