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
Concept. With end correction $e$: $l_1 + e = \dfrac{\lambda}{4}$ and $l_2 + e = \dfrac{3\lambda}{4}$, so $l_2 + e = 3(l_1 + e)$.
Step 1. Solve for $e$.
$$e = \frac{l_2 - 3l_1}{2} = \frac{49 - 48}{2} = 0.5\ \text{cm}$$
Answer. $0.5$ cm (Option D).
Solution by Sreeraj P, M.Sc Physics