Question

A steel wire of $1.5m$ long of diameter $0.25cm$ and a brass wire of $1.0m$ long of diameter $0.25m$ are loaded as shown. Calculate, the elongation of the brass wire.

• A. $1.1\times 10^{-4}m$
• B. $1.3\times 10^{-4}m$
• C. $1.5\times 10^{-4}m$
• D. $1.7\times 10^{-4}m$

Answer 1

Answer: (B)

If a wire of length $L$ and radius $r$ increase in length by $\Delta L$ under
a suspended load $Mg$,
then the Youngs modulus of the material of the wire is
$Y=\frac{MgL}{\left(\pi r^2\right)\times \Delta L}$
or $\Delta L=\frac{MgL}{\left(\pi r^2\right)Y}$
For the brass wire, the suspended load is
$6.0kg\times 9.8N/kg$
$=58.8N$
$\therefore$ Elongation of the brass wire is given by
$\Delta L_{\text{Brass }}=\frac{(Mg{)}_{Br}\times L_{Br}}{\left(\pi r^2\right)\times Y_{Br}}$
$=\frac{58.8\times 1.0}{3.14\times {\left(0.125\times 10^{-2}\right)}^2\times 0.91\times 10^{11}}$
$=13\times 10^{-4}m$