Question

A student is performing the experiment of a resonance column. The diameter of the column tube is 4cm. The frequency of the tuning fork is 512Hz. The air temperature is ${{38}^{0}}C$in which the speed of sound is $336m{{s}^{-1}}$. 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) 14.0cm
(B) 15.2cm
(C) 16.4cm
(D) 17.6cm

Answer 1

Hint: The reading of the water level in the column is the sum of actual reading and end correction. Since, it has been given that first resonance occurs then, it means that this reading is equal to the distance between the node and the antinode formed which is equal to wavelength upon four. Using these above conditions, we shall proceed ahead to calculate the actual reading of the water column.

Complete step-by-step answer:
Let us first assign some terms that we are going to use in our solution.
Let the actual reading of the water level in the column be equal to L.
And, the end correction is given by ‘e’.
Now, the wavelength at first resonance (say $\lambda $) can be calculated as:
$\Rightarrow \lambda =\dfrac{v}{f}$
Where,
v is the speed of air at the given temperature. And,
f is the frequency of the tuning fork.
Putting their respective values in the above equation, we get:
$\Rightarrow \lambda =\dfrac{336}{512}m$
$\begin{align}
  & \Rightarrow \lambda =\dfrac{336\times 100}{512}cm \\
 & \therefore \lambda =65.6cm \\
\end{align}$
Now, we will calculate the end correction of the water column. It is given by:
$\Rightarrow e=0.3D$
Where, D is the diameter and is equal to 4cm.
$\begin{align}
  & \Rightarrow e=0.3\times 4cm \\
 & \therefore e=1.2cm \\
\end{align}$
Now, we can write the condition for first resonance as:
$\Rightarrow L+e=\dfrac{\lambda }{4}$
Putting the values of all the known terms and calculating for L, we get:
$\begin{align}
  & \Rightarrow L=\dfrac{\lambda }{4}-e \\
 & \Rightarrow L=\dfrac{65.6}{4}-1.2 \\
 & \Rightarrow L=(16.4-1.2)cm \\
 & \therefore L=15.2cm \\
\end{align}$
Hence, at first resonance, the absolute reading of the water column is equal to 15.2cm .

So, the correct answer is “Option B”.

Note: In the above problem, one should always take care of the end correction. If we ignore this end correction, then we concur a small error in our answer and this could be given as an option in the question. So, we should always be sure to include all the possible measurement errors in our solution before reaching a final answer.