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  1. Problems
  2. /A wire 2.6 m long and 1.1 mm across carries 75 N and becomes 1.0 mm longer…
EasyPhysicsHigh School (11–12)Mechanics › Elasticity and Mechanical Properties of SolidsNumericalBoard ExamA-Level

A wire 2.6 m long and 1.1 mm across carries 75 N and becomes 1.0 mm longer…

StemPy LabContributed by StemPy Lab

Try it on paper first. A hint, the key idea and the worked solution are here when you need them.

A wire 2.6 m2.6\ \text{m}2.6 m long and 1.1 mm1.1\ \text{mm}1.1 mm across carries 75 N75\ \text{N}75 N and becomes 1.0 mm1.0\ \text{mm}1.0 mm longer. Find the stress, the strain and Young's modulus. Which metal could it be?

Check your answer

Work it out on paper, then type your number. The answer itself stays hidden.

Pa
Pa
Need a reminder? The key ideaShowHide

Young's modulus: Y=F/AΔL/L=FLA ΔLY = \dfrac{F/A}{\Delta L/L} = \dfrac{FL}{A\,\Delta L}Y=ΔL/LF/A​=AΔLFL​. A wire stretches ΔL=FLAY\Delta L = \dfrac{FL}{AY}ΔL=AYFL​ and acts as a spring with k=YALk = \dfrac{YA}{L}k=LYA​. A wire breaks when its stress reaches the breaking stress, whatever its length.

The worked solution

Have you tried it?

Work it out on paper first. It is fine to get stuck or make a mistake: that is how we learn. When you are ready, go through the solution one step at a time, and check each step against your own.

Step 1. Area. A=π(0.55×10−3 m)2=9.50×10−7 m2A = \pi(0.55 \times 10^{-3}\ \text{m})^2 = 9.50 \times 10^{-7}\ \text{m}^2A=π(0.55×10−3 m)2=9.50×10−7 m2.

Step 2. Stress. 75 N9.50×10−7 m2=7.89×107 Pa\dfrac{75\ \text{N}}{9.50 \times 10^{-7}\ \text{m}^2} = 7.89 \times 10^{7}\ \text{Pa}9.50×10−7 m275 N​=7.89×107 Pa.

Step 3. Strain. 1.0×10−3 m2.6 m=3.85×10−4\dfrac{1.0 \times 10^{-3}\ \text{m}}{2.6\ \text{m}} = 3.85 \times 10^{-4}2.6 m1.0×10−3 m​=3.85×10−4.

Step 4. Modulus. Y=7.89×107 Pa3.85×10−4=2.05×1011 PaY = \dfrac{7.89 \times 10^{7}\ \text{Pa}}{3.85 \times 10^{-4}} = 2.05 \times 10^{11}\ \text{Pa}Y=3.85×10−47.89×107 Pa​=2.05×1011 Pa.

Answer: 7.89×107 Pa7.89 \times 10^{7}\ \text{Pa}7.89×107 Pa, 3.85×10−4 Pa3.85 \times 10^{-4}\ \text{Pa}3.85×10−4 Pa, 2.05×1011 Pa2.05 \times 10^{11}\ \text{Pa}2.05×1011 Pa: steel.

Watch out. Change millimetres to metres in both the diameter and the stretch.

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