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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

Cross-sectional area: 2X, Length: L, Young modulus of material: 4E

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Answer

To determine which combination produces the smallest extension, we can use the formula for extension:

extExtension=FLAY ext{Extension} = \frac{F L}{A Y}

where:

  • FF is the tensile force,
  • LL is the length,
  • AA is the cross-sectional area,
  • YY is the Young's modulus of the material.

From the options:

  • In option D: The cross-sectional area is doubled (2X2X), the length is LL, and the Young modulus is quadrupled (4E4E). This maximizes the resistance to extension. The resulting extension willbe as follows:

extExtensionD=FL2X4E=F8XE ext{Extension}_D = \frac{F \cdot L}{2X \cdot 4E} = \frac{F}{8X E}

  • In contrast, the other options will yield greater extensions due to lower areas or lower Young's moduli. Thus, option D gives the smallest extension under the same tensile force.

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