Physics Question: Maximum Length of Wire before Breaking
Maximum Length of Wire before Breaking
Physics Question: Maximum Length of Wire before Breaking
Question:
The breaking stress for a metal is \(7.8 \times 10^9 \, \text{N/m}^2\). Calculate the maximum length of the wire of this metal which may be suspended without breaking. The density of the metal is \(7.8 \times 10^3 \, \text{kg/m}^3\). Take \(g = 10 \, \text{N/kg}\).
Density of the metal \( \rho = 7.8 \times 10^3 \, \text{kg/m}^3 \)
Acceleration due to gravity \( g = 10 \, \text{N/kg} \)
To find the maximum length of the wire, we use the formula:
\[
\sigma = \frac{F}{A}
\]
where \( \sigma \) is the breaking stress, \( F \) is the force (weight) acting on the wire, and \( A \) is the cross-sectional area of the wire. The force acting on the wire is:
\[
F = mg
\]
where \( m \) is the mass of the wire and \( g \) is the acceleration due to gravity. The mass \( m \) is related to the density \( \rho \) and the volume \( V \) of the wire by:
\[
m = \rho \times V = \rho \times A \times L
\]
where \( L \) is the length of the wire. Substituting into the equation for force:
\[
F = \rho \times A \times L \times g
\]
Now, we substitute this expression for \( F \) into the stress formula:
\[
\sigma = \frac{\rho \times A \times L \times g}{A}
\]
Simplifying, we get:
\[
\sigma = \rho \times L \times g
\]
Solving for \( L \), we get:
\[
L = \frac{\sigma}{\rho \times g}
\]
Substituting the given values:
\[
L = \frac{7.8 \times 10^9}{7.8 \times 10^3 \times 10}
\]
Performing the calculation:
\[
L = \frac{7.8 \times 10^9}{7.8 \times 10^4} = 100,000 \, \text{m} = 100 \, \text{km}
\]
Therefore, the maximum length of the wire that may be suspended without breaking is 100,000 meters (or 100 kilometers).
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