A closed organ pipe of radius $r_{1}$ and an open organ pipe of radius $r_{2}$ and having same length $L$ ' resonate when excited with a given tuning fork. Closed organ pipe resonates in its fundamental mode where as open organ pipe resonates in its first overtone, then:-

r_{2} - r_{1} = L

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r_{2} = r_{1} = L / 2

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r_{2} - 2 r_{1} = 2.5 L

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2 r_{2} - r_{1} = 2.5 L

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Solution

The correct option is C $r_{2} - 2 r_{1} = 2.5 L$
For closed organ pipe with end correction
$V_{n}^{1} = \frac{(2 n - 1) V}{4 (L + 0.6 r_{1})}$ where, $0.6 r_{1}$ is end correction
For fundamental node i.e., n=1
$V_{2}^{1} = \frac{V}{4 (L + 0.6 r_{1})}$
For open organ pipe with end correction
$V_{m} = \frac{m V}{2 (L + 2 \times {0.6}_{2})}$ where, $0.6 r_{2}$ is end correction
For first overtone i.e, m=2
$V_{2} = \frac{2 V}{2 (L + 1 - 2 r_{2})}$
But we know, $V_{2}^{1} = V_{2}$
$\frac{V}{4 (L + 0.6 r_{1})} = \frac{2 V}{2 (L + 1 - 2 r_{2})}$
$L + 1 - 2 r_{2} = 4 L + 2.4 r_{1}$
$r_{2} - 2 r_{1} = 2.5 L$

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