MINIMUM AUDIBLE frequency = 20 \[HZ\]. \[\Rightarrow \] \[\frac{v}{4l}=20\Rightarrow l=\frac{336}{4\times 20}=4.2\]\[m\] "> MINIMUM AUDIBLE frequency = 20 \[HZ\]. \[\Rightarrow \] \[\frac{v}{4l}=20\Rightarrow l=\frac{336}{4\times 20}=4.2\]\[m\] ">
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If the velocity of sound in air is 336 m/s. The maximum length of a closed pipe that would produce a just audible sound will be [KCET 2001]

Joint Entrance Exam - Main (JEE Main) Physics in Joint Entrance Exam - Main (JEE Main) . 8 months ago

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MINIMUM AUDIBLE frequency = 20 \[HZ\]. \[\Rightarrow \] \[\frac{v}{4l}=20\Rightarrow l=\frac{336}{4\times 20}=4.2\]\[m\]
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