Question

Two wires of length $l$, radius $r$ and length $2l$, radius $2r$ having same Young's modulus $Y$ are hung with weight $mg$. Net elongation is

• A. $3mgl/2\pi r^{2}Y$
• B. $1mgl/2\pi r^{2}Y$
• C. $2mgl/3\pi r^{2}Y$
• D. $3mgl/3\pi r^{2}Y$

Answer 1

The correct option is A $3mgl/2\pi r^{2}Y$

$\begin{array}{cc}\text{Net elongation}& =\Delta L_1+\Delta L_2\\ \Delta L_1& =\text{elongation is}{1}^{\text{st}}\text{wire}\\ \Delta L_2& =\text{elongation is}{2}^{\text{nd}}\text{wire}\end{array}$

$\begin{array}{c}\text{For}{1}^{\text{st}}\text{wire,}\\ \begin{array}{c}Y=\frac{MgL}{\pi r^2\Delta L_1}\\ \Delta L_1=\frac{MgL}{\pi r^{2}Y}\end{array}\end{array}$

$\begin{array}{c}\text{For}{2}^{\text{nd}}\text{wire}\\ Y=\frac{\mathrm{mg}2L}{\pi (2r{)}^2\Delta L_2}\\ \Delta L_2=\frac{2mgL}{4\pi r^{2}Y}\end{array}$

$\begin{array}{c}\Delta L_{\text{net}}=\frac{mgL}{\pi r^{2}Y}+\frac{mgL}{2\pi r^{2}Y}\\ \Delta L_{\text{net}}=\frac{3}{2}\frac{mgL}{\pi r^{2}Y}\end{array}$

Hence option A is correct