Question

In a resonance pipe, the first and second resonances are obtained at depths $22.7cm$ and $70.2cm$ respectively. What will be the end correction?
A) $1.05cm$
B) $115.5cm$
C) $92.5cm$
D) $113.5cm$

Answer 1

Hint: Recall that for the first resonance, distance between the node and antinode is $\lambda /4$ and for second resonance, the distance between first node and final antinode is $3\lambda /4$ . Use the proportionality between depths at which resonance is obtained and distances between node and antinodes and simplify to get the answer.

Formula Used:
Depth at which resonance is obtained is directly proportional to the distance between the node and antinode. i.e. $l \propto \lambda $.

Complete step by step solution:
The first resonance is obtained at a depth of $22.7cm$ . Let this be ${l_1}$ . The distance between node and antinode for this will value is $\lambda /4$
The second resonance is obtained at a depth of $70.2cm$ . Let this be ${l_2}$ . The distance between node and antinode for this will value is $3\lambda /4$
Now, we know that the depth of resonance is directly proportional to the distance between node and antinode. Hence, for this question, we conclude that
${l_1} \propto \lambda /4$ and ${l_2} \propto 3\lambda /4$
Now, if we divide the second equation by the first equation, the proportionality will change into equality. Hence, $\dfrac{{{l_2}}}{{{l_1}}} = \dfrac{{3\lambda /4}}{{\lambda /4}}$ ……(A)
Let the end correction be $x$ . This end correction must be added both to recorded values of depth of resonance to get the final value of the experiment. Hence, the actual values of resonance will be ${l_1} + x$ and ${l_2} + x$ . Therefore, equation (A) transforms to
$\dfrac{{{l_2} + x}}{{{l_1} + x}} = \dfrac{{3\lambda /4}}{{\lambda /4}}$
On solving this, we get $\dfrac{{{l_2} + x}}{{{l_1} + x}} = 3$
$ \Rightarrow {l_2} + x = 3{l_1} + 3x$ or, ${l_2} - 3{l_1} = 2x$
Putting the values given in question, we get
$70.2 - (3 \times 22.7) = 2x$
$70.2 - 68.1 = 2x$
$ \Rightarrow x = 1.05cm$

Hence, option A is the correct answer.

Note: While performing this experiment, the distance between node and antinode for the first resonance is always $\lambda /4$ and for the second resonance, the distance between node and antinode is always $3\lambda /4$ . This remains constant for every experiment performed.