Q 11-14-020NEETJEE MainMedium
In a resonance tube experiment, the first and second resonances occur at air-column lengths of $16$ cm and $49$ cm. The end correction of the tube is
Answer: (D) $0.5$ cm
$e = \dfrac{l_2 - 3l_1}{2} = \dfrac{49 - 48}{2} = 0.5$ cm.
Solution by Sreeraj P, M.Sc Physics