Question

The breaking stress for a metal is $7.8 \times 10^{9} Nm ^{-2}$. The density of the metal is $7800\, kg\, m ^{-3}$. If $g=10\, N\, kg ^{-1}$, Find the maximum length of the wire (in $km$ ) made of this metal which may be suspended without breaking.

Answer 1

If $L$ is the maximum length of the wire that can be suspended.
Then, $F=m g$, where $m=\rho L A$ (where $A=$ cross-sectional area)

$\therefore$ Maximum stress $=\frac{m g}{A}=g \cdot \frac{\rho L A}{A}=\rho L g$
Given, breaking stress $\rho_{b}=7.8 \times 10^{9} Nm ^{-2}$
$\therefore \rho L g=7.8 \times 10^{9}$
$\Rightarrow L=7.8 \times 10^{9} \times \frac{1}{7800 \times 10}$
or $L=10^{5} m =100\, km$