Question

A resonance tube is resonated with tuning fork of frequency 256 Hz. If the length of first and second resonating air columns are 32 cm and 100 cm, then end correction will be

• A. 1 cm
• B. 2 cm
• C. 4 cm
• D. 6 cm

Answer 1

The correct option is B 2 cm

First resonating length:

$\frac{\lambda }{4}=L_1+e$

Here,

$\lambda$ is wavelength

$L_1$ is length of first resonating column

$e$ is the end correction

$\frac{\lambda }{4}=32+e$ ...........(1)

Second resonating length:

$\frac{3\lambda }{4}=L_2+e$

Here,

$\lambda$ is wavelength

$L_2$ is length of second resonating column

$e$ is the end correction

$\frac{3\lambda }{4}=100+e$ ...........(2)

Solving the equation (1) and (2),

(2) - (1) gives,

$\frac{\lambda }{2}=100-32$

$\frac{\lambda }{2}=68$

$\lambda =136cm$

Hence,

$\frac{\lambda }{4}=32+e$

$e=\frac{136}{4}-32$

$=34-32=2cm$

Therefore the end correction is equal to 2 cm