4{{l}_{1}}}\] and \[{{n}_{2}}=\frac{v}{4{{l}_{2}}}\] Þ \[\frac{{{n}_{1}}}{{{n}_{2}}}=\frac{{{l}_{2}}}{{{l}_{1}}}=\frac{26}{25}\] ?..(ii) After solving these equation \[{{n}_{1}}=260Hz\],\[{{n}_{2}}=250\]\[Hz\]

"> 4{{l}_{1}}}\] and \[{{n}_{2}}=\frac{v}{4{{l}_{2}}}\] Þ \[\frac{{{n}_{1}}}{{{n}_{2}}}=\frac{{{l}_{2}}}{{{l}_{1}}}=\frac{26}{25}\] ?..(ii) After solving these equation \[{{n}_{1}}=260Hz\],\[{{n}_{2}}=250\]\[Hz\]

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Two closed pipe produce 10 beats per second when emitting their fundamental nodes. If their length are in ratio of 25 : 26. Then their fundamental frequency in Hz, are [MH CET 2002]

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

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\[{{n}_{1}}-{{n}_{2}}=10\] ..?(i) Using \[{{n}_{1}},=\frac{v}{4{{l}_{1}}}\] and \[{{n}_{2}}=\frac{v}{4{{l}_{2}}}\] Þ \[\frac{{{n}_{1}}}{{{n}_{2}}}=\frac{{{l}_{2}}}{{{l}_{1}}}=\frac{26}{25}\] ?..(ii) After solving these equation \[{{n}_{1}}=260Hz\],\[{{n}_{2}}=250\]\[Hz\]

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