Question

Let n1 harmonic is corresponding to 50.7 cm & n₂ harmonic is corresponding 83.9 cm. since both one consecutive harmonics.
$\therefore their difference=\frac{\lambda }{2}$
$\therefore \frac{\lambda }{2}=(83.9-50.7)cm$
$$\begin{gathered} \frac{\lambda }{2}=33.2cm \\ \lambda =66.4 cm \end{gathered}$$
$\therefore \frac{\lambda }{4}=16.6 cm$
length corresponding to fundamental mode must be close to  $\frac{\lambda }{4}$& 50.7 cm must be an odd multiple
of this length 16.6 × 3 = 49.8 cm. therefore 50.7 is 3^|rd harmonic If end correction is e, then
$e+50.7=\frac{3\lambda }{4}$
$e=49.8-50.7=-0.9 cm$
speed of sound, v = f$\lambda$
$\therefore v=500\times 66.4 cm/sec=332.000m/s$