Question

A 5 m aluminium wire $(Y = 7\times10^{10} Nm^{-2})$ of diameter 3 mm supports a 40 kg mass. In order to have the same elongation in a copper wire $(T =12\times10^{10}Nm^{-2})$ of the same length under the same weight, the diameter (in mm) should be

• A. 1.75
• B. 2.0
• C. 2.3
• D. 5.0

Answer 1

Answer: (C)

Young's modulus $Y=\frac{stress}{strain}=\frac{F/A}{l/L}=\frac{F . L}{\pi r^2 l}$
Given $Y_1=7\times10^{10} Nm^{-2}$
$Y_2=12\times10^{10} Nm^{-2}$
$r_1=\frac{D_1}{2}=\frac{3}{2}mm,r_2=\frac {D_3}{2}$
= Taking ratio of Young's modulus and putting $r=\frac{D}{2}, we get$
$\frac{Y_2}{Y_1}=\left(\frac{D_1}{D_2}\right)^2$
$\frac{12\times10^{10}}{7\times10^{10}}=\left(\frac{3}{D_2}\right)^2$
$\Rightarrow \frac{3}{D_2}= \sqrt{\frac{12}{7}}$
$\Rightarrow D_2=3\sqrt{\frac{7}{12}}\approx 2.3 mm$