It is desired to increase the fundamental resonance frequency in a tube which is closed at one end. This can be achieved by
Medium
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(d) Fundamental frequency of closed pipe $n = \frac{v}{{4l}}$
where $v = \sqrt {\frac{{\gamma RT}}{M}} $ ==>$v \propto \frac{1}{{\sqrt M }}$
$\because$ ${M_{{H_2}}} < {M_{air}}\,\, \Rightarrow \,\,{v_{{H_2}}} > {v_{air}}$
Hence fundamental frequency with $H_2$ will be more as compared to air. So option $(a)$ is correct.
Also $n \propto \frac{1}{l}$, hence if $l$ decreases $n$ increases so option $(b)$ is correct.
It is well known that $(n)$ Open $= 2(n)$ Closed, hence option $(c)$ is correct.
where $v = \sqrt {\frac{{\gamma RT}}{M}} $ ==>$v \propto \frac{1}{{\sqrt M }}$
$\because$ ${M_{{H_2}}} < {M_{air}}\,\, \Rightarrow \,\,{v_{{H_2}}} > {v_{air}}$
Hence fundamental frequency with $H_2$ will be more as compared to air. So option $(a)$ is correct.
Also $n \propto \frac{1}{l}$, hence if $l$ decreases $n$ increases so option $(b)$ is correct.
It is well known that $(n)$ Open $= 2(n)$ Closed, hence option $(c)$ is correct.
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