Question

In brass, the velocity of longitudinal waves is $100times$ the velocity of transverse waves. If $Y=1\times {10}^{11}N/m^2,$ then the stress in the wire is

• A. ${10}^7/N/m^2$
• B. ${10}^8/N/m^2$
• C. ${10}^9/N/m^2$
• D. ${10}^{10}/N/m^2$

Answer 1

The correct option is A ${10}^7/N/m^2$

Velocity of transverse wave $V_T=\sqrt{\left(\frac{T}{m}\right)}$

$V_T=\sqrt{\left\{\frac{T}{\left(\pi r^2\rho \right)}\right\}}.............\left(1\right)$

Velocity of longitudinal wave $V_L=\sqrt{\frac{Y}{\rho }}...............\left(2\right)$

Given:

$\begin{array}{c}{V}_L=100V_T\\ \therefore \sqrt{\frac{Y}{\rho }}=100\sqrt{\left\{\frac{T}{\left(\pi r^2\rho \right)}\right\}}\\ \therefore Y={\left(100\right)}^2\times \left\{\frac{T}{\left(\pi r^2\right)}\right\}\end{array}$

As stress $=\left\{\frac{\left(force\right)}{\left(area\right)}\right\}=\left\{\frac{T}{\left(\pi r^2\right)}\right\}$

$\therefore Y={10}^4\times$stress

$\therefore$ stress $=\left(\frac{Y}{{10}^4}\right)=\left\{\frac{\left(1\times {10}^{11}\right)}{\left({10}^4\right)}\right\}=1\times {10}^{7}N/m^2$