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

Correct Answer: 600

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

Breaking stress (\( \sigma \)) is defined as:

\[ \sigma = \frac{T}{A} = \frac{mg}{A} \]

Here, \( T \) is tension, \( m \) is mass, and \( A \) is cross-sectional area.
Given that \( m = \rho A \ell \) (where \( \ell \) is the wire length), the equation becomes:

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

Solving for \( \ell \):

\[ \ell = \frac{\sigma}{\rho g'} \]

Substituting the provided values:

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