Question

In a resonance tube experiment, the tuning fork of frequency $\;256$ Hz, pipe being $30$ cm out. Also, there is resonance between tuning fork frequencies $480$ Hz with the same pipe being $15$ cm out. The diameter of the pipe approximate is?
$\left( {\text{A}} \right){\text{ 10cm}}$
$\left( {\text{B}} \right){\text{ 7}}{\text{.1cm}}$
$\left( {\text{C}} \right){\text{ 9}}{\text{.2cm}}$
$\left( {\text{D}} \right){\text{ 3}}{\text{.55cm}}$

Answer 1

We will use the formula for velocity of sound in air as a function of frequency, length of air column and radius of the pipe.
$v = 4f(l + 0.6r)....\left( 1 \right)$ Where, $v =$ velocity of sound in air, $f =$ resonance frequency in Hz, $l =$ length of air column and $r =$ radius of pipe.

Complete step by step answer:
Given that for the first case,
${f_1} = 256\;{\text{Hz}},{{\;}}{{\text{l}}_1} = 30\;{\text{cm}}$. So, we have from $\left( 1 \right)$
$\Rightarrow v = 4 \times 256 \times (30 + 0.6r)....\left( 2 \right)$
For the second case, we have
${f_2} = 480\;{\text{Hz}},{{\;}}{{\text{l}}_2} = 15\;{\text{cm}}$, so, we have from $\left( 1 \right)$
$\Rightarrow v = 4 \times 480 \times (15 + 0.6r).....\left( 3 \right)$
As the speed of sound will be the same in air at a given temperature, we can say that the LHS and hence RHS of equations $\left( 2 \right)$ and $\left( 3 \right)$ will be the same. On equating, we have
$\Rightarrow 4 \times 256 \times (30 + 0.6r) = 4 \times 480 \times (15 + 0.6r)$
Cancelling the equating term we get,
$\Rightarrow 256 \times 30 + 256 \times 0.6r = 480 \times 15 + 480 \times 0.6r$
Taking variable as RHS and remaining terms as LHS, we get
$\Rightarrow 256 \times 30 - 480 \times 15 = 256 \times 0.6r + 480 \times 0.6r$
Let us multiply the terms we get,
$\Rightarrow 7680 - 7200 = 153.6r + 288r$
On subtracting the terms we get,
$\Rightarrow 480 = 134.4r$
Let us take $r$ as RHS and remaining as terms are in divide we get
$\Rightarrow r = \dfrac{{480}}{{134.4}}$
$\Rightarrow r = 3.571429cm$
Also,$d = 2r$
Substitute the value of $r$ we get.
$= 2 \times 3.571429$
On multiplying the terms we get,
$= 7.142857cm$

Therefore, the diameter of the pipe is ${\text{7}}{\text{.1cm}}$. So, Option $\left( {\text{B}} \right)$ is correct.

Note:
We have to keep in mind that the pipe being out means the length of the air column and not the length underwater. Also, it is asked in the question to find the diameter of the pipe and not the radius. As the formula involves radius, we need to be careful while calculating the final answer because $3.55{\text{ }}cm$ is also in the option.