JAMB 2023 Physics · Question 5

A wire of radius 0.2 mm is extended by 0.5% of its length when supported by a load of 1.5 kg. Determine the…

Question

A wire of radius 0.2 mm is extended by 0.5% of its length when supported by a load of 1.5 kg. Determine the Young's modulus for the material of the wire. [Take g = 10 ms−2^{-2}]

  1. A.2.4×1010(Nm−2)2.4×10^{10}(Nm^{-2})
  2. B.1.5×1010(Nm−2)1.5×10^{10}(Nm^{-2})
  3. C.2.4×109(Nm−2)2.4×10^9(Nm^{-2})
  4. D.1.3×1010(Nm−2)1.3×10^{10}(Nm^{-2})

Answer: A. 2.4×1010(Nm−2)2.4×10^{10}(Nm^{-2})

Explanation written with AI and checked against the answer key

Analysis

To solve this problem, we need to calculate the Young's Modulus (YY) of the wire material. Young's Modulus is defined as the ratio of tensile stress to tensile strain.

Concept

The formula for Young's Modulus is:

Y=StressStrain=F/AΔL/LY = \frac{\text{Stress}}{\text{Strain}} = \frac{F/A}{\Delta L/L}

Where:

  • FF = Force (Weight = mgmg)
  • AA = Cross-sectional area of the wire (πr2\pi r^2)
  • ΔL/L\Delta L/L = Tensile strain (given as a percentage)
Step-by-Step Solution

1. Identify the given values:

  • Radius (rr) = 0.2 mm=0.2×10−3 m=2×10−4 m0.2 \text{ mm} = 0.2 \times 10^{-3} \text{ m} = 2 \times 10^{-4} \text{ m}
  • Load (mm) = 1.5 kg1.5 \text{ kg}
  • g=10 ms−2g = 10 \text{ ms}^{-2}
  • Strain (ΔLL\frac{\Delta L}{L}) = 0.5%=0.5100=0.005=5×10−30.5\% = \frac{0.5}{100} = 0.005 = 5 \times 10^{-3}

2. Calculate Force (FF):

F=mg=1.5×10=15 NF = mg = 1.5 \times 10 = 15 \text{ N}

3. Calculate Cross-sectional Area (AA): Using π≈3.142\pi \approx 3.142:

A=πr2=3.142×(2×10−4)2A = \pi r^2 = 3.142 \times (2 \times 10^{-4})^2
A=3.142×4×10−8≈1.2568×10−7 m2A = 3.142 \times 4 \times 10^{-8} \approx 1.2568 \times 10^{-7} \text{ m}^2

4. Calculate Young's Modulus (YY):

Y=FA×(ΔL/L)Y = \frac{F}{A \times (\Delta L/L)}
Y=15(1.2568×10−7)×(5×10−3)Y = \frac{15}{(1.2568 \times 10^{-7}) \times (5 \times 10^{-3})}
Y=156.284×10−10Y = \frac{15}{6.284 \times 10^{-10}}
Y≈2.387×1010 Nm−2Y \approx 2.387 \times 10^{10} \text{ Nm}^{-2}

Rounding this value to two significant figures gives approximately 2.4×1010 Nm−22.4 \times 10^{10} \text{ Nm}^{-2}.

Conclusion

Comparing our result to the given options, it matches option A.

Exam Tip

When dealing with units in elasticity questions, always convert millimeters (mm) to meters (m) immediately. A common mistake is forgetting to square the radius or forgetting to convert the percentage strain into a decimal. Keep your powers of 10 organized to avoid calculation errors!

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