A brass wire 2.0 m long and 2.0 mm in diameter supports a 10.0-kg fixture. By what distance did this fixture stretch the wire? Young's modulus for brass is $10 \times 1010 \mathrm {~N} / \mathrm { m } _ { 2 }$
A) 0.62 mm
B) 0.11 mm
C) 0.33 mm
D) 0.22 mm

Question

A brass wire  2.0 m long and  2.0 mm  in diameter supports a  10.0-kg fixture. By what distance did this fixture stretch the wire? Young's modulus for brass is $10 \times 1010 \mathrm {~N} / \mathrm { m } _ { 2 }$
A)  0.62 mm
B)  0.11 mm
C)  0.33 mm
D)  0.22 mm

Quizplus · Accepted Answer

The correct answer is A.The stress on the wire is:$$\sigma = \frac{F}{A} = \frac{mg}{\pi r^2}$$where F is the force on the wire, m is the mass of the fixture, g is the acceleration due to gravity, r is the radius of the wire, and A is the cross-sectional area of the wire.The strain on the wire is:$$\epsilon = \frac{\Delta L}{L}$$where $\Delta L$ is the change in length of the wire and L is the original length of the wire.Young's modulus is defined as:$$Y = \frac{\sigma}{\epsilon}$$Substituting the expressions for stress and strain into Young's modulus, we get:$$Y = \frac{mg}{\pi r^2 L} \frac{L}{\Delta L}$$Solving for $\Delta L$, we get:$$\Delta L = \frac{mgL}{Y \pi r^2}$$Substituting the given values, we get:$$\Delta L = \frac{(10.0 \mathrm{~kg})(2.0 \mathrm{~m})}{(10 \times 10^{10} \mathrm{~N} / \mathrm{m}^2) \pi (1.0 \times 10^{-3} \mathrm{~m})^2} = 0.62 \mathrm{~mm}$$

A Brass Wire 2 10×1010 N/m210 \times 1010 \mathrm {~N} / \mathrm { m } _ { 2 }10×1010 N/m2​

A Brass Wire 2 $10 \times 1010 \mathrm {~N} / \mathrm { m } _ { 2 }$