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Which combination of properties would produce the smallest extension of a wire when the same tensile force is applied to the wire? | Cross-sectional area | Length | Young modulus of material | |---------------------|--------|--------------------------| | A | X | 3L | E | | B | 2X | L | E | | C | X | 3L | 4E | | D | 2X | L | 4E | - AQA - A-Level Physics - Question 20 - 2019 - Paper 1

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Which-combination-of-properties-would-produce-the-smallest-extension-of-a-wire-when-the-same-tensile-force-is-applied-to-the-wire?--|-Cross-sectional-area-|-Length-|-Young-modulus-of-material-|-|---------------------|--------|--------------------------|-|-A-------------------|---X---|---3L----|---E--------------------|-|-B-------------------|-2X----|----L----|---E--------------------|-|-C-------------------|---X---|---3L----|---4E-------------------|-|-D-------------------|-2X----|----L----|---4E-------------------|-AQA-A-Level Physics-Question 20-2019-Paper 1.png

Which combination of properties would produce the smallest extension of a wire when the same tensile force is applied to the wire? | Cross-sectional area | Length |... show full transcript

Worked Solution & Example Answer:Which combination of properties would produce the smallest extension of a wire when the same tensile force is applied to the wire? | Cross-sectional area | Length | Young modulus of material | |---------------------|--------|--------------------------| | A | X | 3L | E | | B | 2X | L | E | | C | X | 3L | 4E | | D | 2X | L | 4E | - AQA - A-Level Physics - Question 20 - 2019 - Paper 1

Step 1

Determine the Variables Affecting Extension

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Answer

The extension of a wire can be determined by the formula derived from Hooke’s Law:

extExtension(x)=FLAY ext{Extension} (x) = \frac{F \cdot L}{A \cdot Y}

where:

  • FF = tensile force applied (constant for this question)
  • LL = original length of the wire
  • AA = cross-sectional area
  • YY = Young's modulus of the material

Step 2

Evaluate Each Option

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Answer

Substituting values from each option:

  • Option A: xA=F3LXEx_A = \frac{F \cdot 3L}{X \cdot E}
  • Option B: xB=FL2XEx_B = \frac{F \cdot L}{2X \cdot E}
  • Option C: xC=F3LX4Ex_C = \frac{F \cdot 3L}{X \cdot 4E}
  • Option D: xD=FL2X4Ex_D = \frac{F \cdot L}{2X \cdot 4E}

Step 3

Calculate Extension for Each Option and Compare

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Answer

  1. For Option A: xA=3FLXEx_A = \frac{3FL}{XE}

  2. For Option B: xB=FL2XEx_B = \frac{FL}{2XE}

  3. For Option C: xC=3FL4XEx_C = \frac{3FL}{4XE}

  4. For Option D: xD=FL8XEx_D = \frac{FL}{8XE}

The smallest extension occurs in Option D (xDx_D), where the multiple of the Young's modulus is higher, and the cross-sectional area is larger.

Step 4

Conclusion

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Answer

After evaluating all options, the combination of properties that would produce the smallest extension of the wire when the same tensile force is applied is Option D: 2X, L, 4E.

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