Question

A tuning fork vibrating at frequency $1000\text{Hz}$ produces resonance in a resonance column tube. The upper end is open and the lower end is closed by the water whose height can be varied. The successive resonances are observed at lengths $10\text{cm}$ and $27\text{cm}$. Then, the speed of sound in air is [neglect end corrections]

• A. $340\text{m/s}$
• B. $330\text{m/s}$
• C. $343\text{m/s}$
• D. $353\text{m/s}$

Answer 1

The correct option is A $340\text{m/s}$

Given that,
Frequency of tuning fork $\left(f\right)=1000\text{Hz}$




The column filled with water behaves as a closed organ pipe.

Modes of vibration of an air column in a closed organ pipe are given by

$f_c=(2n+1)\frac{v}{4l}$ where $n=0,1,\mathrm{2...}$
or
$l=\frac{(2n+1)v}{4f_c}$

For first resonance
$l_1=\frac{v}{4f_c}...(n=0)$

For second resonance
$l_2=\frac{3v}{4f_c}...(n=1)$

$\Rightarrow l_2-l_1=\frac{v}{2f_c}$

From the data given in the question,

$(27-10)\times {10}^{-2}=\frac{v}{2\times 1000}$

[$f_c=f=1000\text{Hz}$]

$\Rightarrow v=17\times {10}^{-2}\times 2000\Rightarrow v=340\text{m/s}$

Thus, option (a) is the correct answer.