Question

Young's modulis of brass and steel are $10\times {10}^{10}N/m$ and $2\times {10}^{11}N/m^2$, respectively. A brass wire and a steel wire of the same length are extended by $1mm$ under the same force. The radii of brass and steel wires are $R_B$ and $R_S$ respectively. Then

• A. $R_S=\sqrt{2}{R}_B$
• B. $R_S=\frac{R_B}{\sqrt{2}}$
• C. $R_S=4R_B$
• D. $R_S=\frac{R_B}{4}$

Answer 1

Answer: (B)

$R_S=\frac{R_B}{\sqrt{2}}$

We know that $Y=\frac{FL}{\pi r^{2}l}$ or $r^2=\frac{FL}{\left(Y\pi l\right)}$
$\therefore R_B^2=\frac{FL}{\left(Y_B\pi l\right)}$ and $R_S^2=\frac{FL}{\left(Y_S\pi l\right)}$
$\frac{R_B^2}{R_S^2}=\frac{Y_S}{Y_B}=\frac{2\times {10}^{10}}{{10}^{10}}=2$

$R_B^2=2R_S^2$

Thus, we get $R_S=\frac{R_B}{\sqrt{2}}$