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:
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:

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

Answer 2

Step 1: Identify the limiting condition
For a vertical wire hanging under its own weight, the maximum stress occurs at the top. To avoid breaking, this maximum stress must not exceed the breaking stress \( \sigma_b \). The tensile stress at the top due to the wire’s own weight is \[ \sigma = \rho\, g'\, L, \] where \( \rho \) is density, \( g' \) is the local acceleration due to gravity, and \( L \) is the wire length.

Step 2: Use the given data
\(\rho = 6 \times 10^4\ \text{kg/m}^3,\quad \sigma_b = 1.2 \times 10^8\ \text{N/m}^2,\quad g' = \frac{g}{3} = \frac{10}{3}\ \text{m/s}^2.\)

Step 3: Apply the no-break condition
Set \( \sigma = \sigma_b \): \[ \rho\, g'\, L = \sigma_b \quad \Longrightarrow \quad L = \frac{\sigma_b}{\rho\, g'}. \] Compute: \[ \rho\, g' = \left(6 \times 10^4\right)\left(\frac{10}{3}\right) = \frac{6 \times 10^5}{3} = 2 \times 10^5, \] \[ L = \frac{1.2 \times 10^8}{2 \times 10^5} = 0.6 \times 10^{3} = 600\ \text{m}. \]

Final answer

600