High School

A 3-meter long wire with a diameter of 1.5 mm has a Young's modulus of \(9 \times 10^{10}\) N/m². What mass hung from the wire would extend it by 3 mm?

Group of answer choices:

A. 159 kg
B. 5.20 kg
C. 16.2 kg
D. 64.8 kg
E. None of the above

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.