JEE Advanced · 2018 · 20 May 2018 · Paper 2 · Waves & Sound

Successive resonances differ by half a wavelength, so $\frac{\lambda }{2}=83.9-50.7=33.2$ cm and $\lambda =66.4$ cm. The speed of sound is $v=f\lambda =500\times 0.664=332$ m/s. Because $\frac{\lambda }{4}=16.6$ cm is far less than $50.7$ cm, the $50.7$ cm resonance is not the fundamental but the next harmonic at $\frac{3\lambda }{4}$; writing $50.7=\frac{3\lambda }{4}-e$ (and $83.9=\frac{5\lambda }{4}-e$) gives an end correction $e=-0.9$ cm. So the wavelength $66.4$ cm and the speed $332$ m/s are correct, while the end correction is not $+0.9$ cm and $50.7$ cm is not the fundamental harmonic.