Answer :
To solve this physics problem, we need to understand how a wire stretches under the influence of a force (weight in this case). This concept is explained using Young's modulus, which is a measure of the stiffness of a material.
Young's modulus [tex]E[/tex] is defined as:
[tex]E = \frac{F}{A} \cdot \frac{L_0}{\Delta L}[/tex]
where:
- [tex]F[/tex] is the force applied (which is the weight in this case),
- [tex]A[/tex] is the cross-sectional area of the wire,
- [tex]L_0[/tex] is the original length of the wire, and
- [tex]\Delta L[/tex] is the change in length (extension) of the wire.
We can rearrange this formula to find the force [tex]F[/tex]:
[tex]F = E \cdot A \cdot \frac{\Delta L}{L_0}[/tex]
Let's plug in the given values:
- The original length of the wire [tex]L_0 = 3[/tex] m.
- The extension [tex]\Delta L = 3[/tex] mm = 0.003 m.
- Diameter of the wire = 1.5 mm, so the radius [tex]r = \frac{1.5}{2} = 0.75[/tex] mm = 0.00075 m.
The cross-sectional area [tex]A[/tex] of the wire (a circle) is:
[tex]A = \pi r^2 = \pi (0.00075)^2 \approx 1.767 \times 10^{-6} \, \text{m}^2[/tex]
Substituting the values into the equation for force:
[tex]F = 9 \times 10^{10} \, \text{N/m}^2 \times 1.767 \times 10^{-6} \, \text{m}^2 \times \frac{0.003}{3}[/tex]
[tex]\approx 159.03 \, \text{N}[/tex]
The weight of the mass [tex]m[/tex] is related to the force by the equation [tex]F = mg[/tex], where [tex]g \approx 9.81 \, \text{m/s}^2[/tex] is the acceleration due to gravity. Thus:
[tex]m = \frac{F}{g} = \frac{159.03}{9.81} \approx 16.2 \, \text{kg}[/tex]
Therefore, the mass that would extend the wire by 3 mm is approximately 16.2 kg.
The correct answer is option (C) 16.2 kg.
