Question

The Young’s modulus of brass and steel are $10\times {10}^{10}N{m}^{-2}and2\times {10}^{11}N{m}^{-2}$ respectively. A brass wire and a steel wire of the same length are extended by 1 mm under the same force. The radii of the brass and steel wires are $R_B$ and $R_s$ respectively. Then

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

Answer 1

The correct option is B $R_S=\frac{R_B}{\sqrt{2}}$

$Y=\frac{F}{\pi R^2}\times \frac{1}{△l}$
F, l and $△l$ are constants
Therefore $R^2\propto \frac{1}{Y}$

$\frac{R_S^2}{R_B^2}=\frac{Y_B}{Y_S}=\frac{{10}^{11}}{2\times {10}^{11}}=\frac{1}{2}$

$or\frac{R_S}{R_B}=\frac{1}{\sqrt{2}}or{R}_s=\frac{R_B}{\sqrt{2}}$