Question

In Kundt's tube experiment, a metal rod of length $2\text{m}$ and density $8{\text{g/cm}}^3$ clamped at middle gives dust heaps in a Kundt's tube at intervals of $10\text{cm}$. Calculate the Young's modulus of the material of rod. The speed of sound, in the gas is $360\text{m/s}$.

• A. $2.12\times {10}^{11}{\text{N/m}}^2$
• B. $2.12\times {10}^{10}{\text{N/m}}^2$
• C. $4.14\times {10}^{11}{\text{N/m}}^2$
• D. $4.14\times {10}^{10}{\text{N/m}}^2$

Answer 1

The correct option is C $4.14\times {10}^{11}{\text{N/m}}^2$

Let ${\lambda }_r$ be the wavelength of sound wave in rod and ${\lambda }_g$ be the wavelength of sound wave in gas tube.

Here
${\lambda }_r=2l_r$ and ${\lambda }_g=2\Delta l_g$
From Kundt's method

Let velocity of sound in gas be $v_g$ and in rod be $v_r$

$v_g=\frac{\Delta l_g}{l_r}{v}_r.......\left(1\right)$
Given that
$\Delta l_g=10\text{cm}=10\times {10}^{-2}\text{m}$
$l_r=2\text{m}$
Velocity of speed of sound in gas $\left(v_g\right)=360\text{m/s}$
Density $\left(\rho \right)=8{\text{g/cm}}^2=8000{\text{kg/m}}^3$
$v_r=\sqrt{\frac{Y}{\rho }}=\sqrt{\frac{Y}{8000}}$
where $Y$ is Young's modulus.
Now substituting the value in equation (1)
$\Rightarrow 360=\frac{10\times {10}^{-2}}{2}\times \sqrt{\frac{Y}{8000}}$
$\Rightarrow Y={\left(\frac{360\times 2}{{10}^{-1}}\right)}^2\times 8000=4.14\times {10}^{11}{\text{N/m}}^2$