JEE Advanced · Physics · 16. Waves & Sound
A student is performing the experiment of resonance column. The diameter of the column tube is \(4 \mathrm{~cm}\). The frequency of the tuning fork is \(512 \mathrm{~Hz}\). The air temperature is \(38^{\circ} \mathrm{C}\) in which the speed of sound is \(336 \mathrm{~m} / \mathrm{s}\). 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 \(14.0 \mathrm{~cm}\)
- B \(15.2 \mathrm{~cm}\)
- C \(16.4 \mathrm{~cm}\)
- D \(17.6 \mathrm{~cm}\)
Answer & Solution
Correct Answer
(B) \(15.2 \mathrm{~cm}\)
Step-by-step Solution
Detailed explanation
Considering the end correction \([e=0.3 D\) where \(\mathrm{D}=\) diameter \(]\)
\(f=n\left[\frac{v}{4(l+e)}\right]\) For first resonance, \(n=1\)
\(\therefore f=\frac{v}{4(l+0.3 D)} \Rightarrow l=\frac{v}{4 f}-0.3 D\)
or, \(l=\left(\frac{336 \times 100}{4 \times 512}\right)-0.3 \times 4=15.2 \mathrm{~cm}\)
\(f=n\left[\frac{v}{4(l+e)}\right]\) For first resonance, \(n=1\)
\(\therefore f=\frac{v}{4(l+0.3 D)} \Rightarrow l=\frac{v}{4 f}-0.3 D\)
or, \(l=\left(\frac{336 \times 100}{4 \times 512}\right)-0.3 \times 4=15.2 \mathrm{~cm}\)
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