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The length of air column in a resonance tube for fundamental mode is 16 cm and that for second resonance is 50.25 cm. Find the end correction. |
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Answer» Data: l1 = 16 cm, l2 = 50.25 cm End correction, e = \(\frac{I_2-3I_1}2\) = \(\frac{50.25-3\times16}2\) = \(\frac{2.25}2\) = 1.125 cm |
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