Question

Three metal rods are located relative to each other as shown in the figure, where $L_1+L_2=L_3$. Values of density and Youngs's modulus for three materials are:
${\rho }_1=2.7\times {10}^{3}kg/m^3,Y_1=7\times {10}^{10}pa$
${\rho }_1=11.3\times {10}^{3}kg/m^3,Y_2=1.6\times {10}^{10}pa$
${\rho }_1=8.8\times {10}^{3}kg/m^3,Y_3=11\times {10}^{10}pa$
If $L_3=1.5m$ what must be the ratio $\frac{L_1}{L_2}$ be if a sound wave is to travel the length of rods $1$ and $2$ in the same time as required to travel the lengthy of rod $3$.

• A. $6.8$
• B. $3.4$
• C. $4$
• D. $1.2$

Answer 1

The correct option is B $3.4$

$\begin{array}{c}{\rho }_1=2.7\times {10}^{3}Kg/m^3,Y_1=7\times {10}^{10}Pa\\ {\rho }_2=11.3\times {10}^{3}Kg/m^3,Y_2=1.6\times {10}^{10}Pa\\ L_3=1.5m\end{array}$

$\begin{array}{c}\text{know,}\\ \text{speed of sound through material}=\sqrt{\frac{Y}{\rho }}\\ \text{Given}\\ t_1+t_2=t_3\end{array}$

$\frac{L_1}{\sqrt{\frac{Y_1}{{\rho }_1}}}+\frac{L_2}{\sqrt{\frac{Y_2}{{\rho }_2}}}=\frac{L_3}{\sqrt{\frac{Y_3}{{\rho }_3}}}$

$\frac{L_1}{L_2}\left(\sqrt{\frac{{\rho }_1}{Y_1}}\right)+\sqrt{\frac{{\rho }_2}{Y_2}}=1.5\sqrt{\frac{{\rho }_3}{Y_3}}$

$\begin{array}{c}\frac{L_1}{L_2}\times \sqrt{\frac{{\rho }_1}{Y_1}}=1.5\sqrt{\frac{{\rho }_3}{Y_3}}\sqrt{\frac{{\rho }_2}{Y_2}}\therefore L_3=1.5\\ \frac{L_1}{L_2}=\frac{1.5\sqrt{\frac{{\rho }_3}{Y_3}}\sqrt{\frac{{\rho }_2}{y_2}}}{\sqrt{\frac{1_1}{y_1}}}=3.4\end{array}$