Q. A student is performing the experiment of Resonance Column. The diameter of the column tube is $4cm.$ The frequency of the tuning fork is $512Hz.$ The air temperature is $38℃$ in which the speed of sound is $336m/s.$ The zero of the meter scale coincides with the top end of the Resonance column tube. When the first resonance occurs, the reading of the water level in the column is
A) $14.0\ cm$
B) $15.2\ cm$
C) $16.4\ cm$
D) $17.6\ cm$
Explanation
(b) $\frac{v}{4(\mathcal{l +}e)} = f$
$\mathcal{\Rightarrow l +}e = \frac{V}{4f}$
$\mathcal{\Rightarrow l =}\frac{V}{4f} - e$
Here $e = (0.6)r = (0.6)(2) = 1.2\ cm$
So $\mathcal{l =}\frac{336 \times 10^{2}}{4 \times 512} - 1.2 = 15.2\ cm$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.