Question

In a resonance tube experiment the first resonance is obtained for $10cm$ of air column and the second for $32cm$ . The end correction for this apparatus is equal to
$(A)0.5cm$
$(B)1.0cm$
$(C)1.5cm$
$(D)2.0cm$

Answer 1

This question is based on the concept of end correction, so first we need to understand the basic definition of end correction. End correction is basically a short distance applied or added to the actual length of a resonance pipe which is used to calculate the precise resonance frequency of the pipe.

Complete step by step answer:
Firstly, to solve this question we need to understand the concept of resonance. This is known as resonance - when one object vibrating at the same natural frequency of a second object forces that second object into vibrational motion. The word resonance comes from Latin and means to respond - to sound out together with a loud sound.
Resonance tube works on the principle of the resonance of the air column with a tuning fork. The waves produced in the air column are transverse stationary waves. Node of the wave lies at the water level and the antinode of the wave lies on the open end of the tube.
The formula for the end correction is $\dfrac{{{l_2} - 3{l_1}}}{2}........(1)$
Now in this question we are given,
${l_1} = 10cm$
${l_2} = 32cm$
On putting the above values in equation (1), we get,
End correction $= \dfrac{{32 - (3 \times 10)}}{2}$
$= \dfrac{{32 - 30}}{2}$
$= \dfrac{2}{2}$
On further simplifying, we get,
End correction= $1cm$
So, the correct answer is $(B)1.0cm$

Note: The resonance tube experiment is used to observe the resonance phenomenon in an open ended cylindrical tube. It uses the resonance to determine the velocity of sound in air at ordinary temperatures. In this experiment the velocity of sound in air is to be found by using tuning forks of known frequency.