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Q.
In a resonance tube experiment, the first two resonances are observed at length $10.5\, cm$ and $29.5\, cm$. The third resonance is observed at the length_________(cm).
Waves
Solution:
$\frac{\lambda}{2}=l_{2}-l_{1}=29.5-10.5=19\, cm$
Third resonance will be observed at
$l_{3}=l_{2}+\frac{\lambda}{2}$
$=29.5+19=48.5\, cm$