Question

The density and breaking stress of a wire are $6 \times 10^4 \, \text{kg/m}^3$ and $1.2 \times 10^8 \, \text{N/m}^2$ respectively. The wire is suspended from a rigid support on a planet where the acceleration due to gravity is $\frac{1}{3}$ of the value on the surface of Earth. The maximum length of the wire without breaking is ________ m (take $g = 10 \, \text{m/s}^2$).

Answer 1

Given:
Density of wire, $\rho = 6 \times 10^4 \, \mathrm{kg/m^3}$
Breaking stress, $\sigma = 1.2 \times 10^8 \, \mathrm{N/m^2}$
Acceleration due to gravity on the planet, $g' = \frac{g}{3} = \frac{10}{3} \, \mathrm{m/s^2}$

The breaking stress ($\sigma$) is given by:

$\sigma = \frac{T}{A} = \frac{mg}{A}$

Where $T$ is the tension, $m$ is the mass, and $A$ is the cross-sectional area.
Since $m = \rho A \ell$ (where $\ell$ is the length of the wire), we have:

$\sigma = \frac{(\rho A \ell) g'}{A} = \rho \ell g'$

Rearranging for $\ell$:

$\ell = \frac{\sigma}{\rho g'}$

Substituting the given values:

$\ell = \frac{1.2 \times 10^8}{6 \times 10^4 \times \frac{10}{3}} = 600 \, \mathrm{m}$