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  4. A closed organ pipe of length L (L/4) is in resonance with a tuning fork. If a hole is made in the pipe at a distance L/4 from closed-end, it will be in resonance again, when :

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Q. A closed organ pipe of length $L$ $\frac{L}{4}$ is in resonance with a tuning fork. If a hole is made in the pipe at a distance L/4 from closed-end, it will be in resonance again, when :

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Solution:

Solution
Standing wave cannot be formed. It becomes combination of two pipes one closed and other open pipe. They will not satisfy the condition of standing wave simultaneously.
$\frac{n_{1} V}{2 \left(\right. \frac{3 L}{4} \left.\right)} = \frac{n_{2} V}{4 \left(\right. \frac{L}{4} \left.\right)}$
$\frac{n_{1} 2 V}{3 L} = \frac{n_{2} V}{L} = \frac{V}{4 L}$
$n_{1} = \frac{3}{8} \, \, n_{2} = \frac{1}{4}$ Which are not integer so not possible