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(D) The fundamental frequency of a closed pipe is given by $n = \frac{v}{4l}$,where $v$ is the speed of sound and $l$ is the length of the tube.$1$. R

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All of the above

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(D) The fundamental frequency of a closed pipe is given by $n = \frac{v}{4l}$,where $v$ is the speed of sound and $l$ is the length of the tube.$1$. Replacing air with hydrogen gas: Since $v = \sqrt{\frac{\gamma RT}{M}}$,the speed of sound is inversely proportional to the square root of the molar mass $(v \propto \frac{1}{\sqrt{M}})$. Because the molar mass of hydrogen $(M_{H_2})$ is less than that of air $(M_{air})$,the speed of sound in hydrogen is higher,which increases the frequency $n$. Thus,option $(A)$ is correct.$2$. Decreasing the length of the tube: Since $n \propto \frac{1}{l}$,decreasing the length $l$ increases the fundamental frequency $n$. Thus,option $(B)$ is correct.$3$. Opening the closed end of the tube: An open pipe has a fundamental frequency $n_{open} = \frac{v}{2l}$,which is twice the fundamental frequency of a closed pipe $(n_{closed} = \frac{v}{4l})$. Thus,opening the end increases the frequency. Option $(C)$ is correct.Since all three methods increase the fundamental frequency,the correct answer is $(D)$.

It is desired to increase the fundamental resonance frequency in a tube which is closed at one end. This can be achieved by

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