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Properties of Matter question

2022 · 28 Jul · Shift 1 · Q52
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  5. /2022 · 28 Jul · Shift 1 · Q52

Properties of Matter question

2022 · 28 Jul · Shift 1 · Q52

JEE MainPhysicsProperties of MatterMCQ+4 / −1
The force required to stretch a wire of cross-section 1 cm21 \mathrm{~cm}^{2}1 cm2 to double its length will be : (Given Yong's modulus of the wire =2×1011 N/m2=2 \times 10^{11} \mathrm{~N} / \mathrm{m}^{2}=2×1011 N/m2)
  1. A
    1×107 N1 \times 10^{7} \mathrm{~N}1×107 N
  2. B
    1.5×107 N1.5 \times 10^{7} \mathrm{~N}1.5×107 N
  3. C
    2×107 N2 \times 10^{7} \mathrm{~N}2×107 N
  4. D
    2.5×107 N2.5 \times 10^{7} \mathrm{~N}2.5×107 N
View written solutionFree

Correct answer: C

  1. Use Young’s modulus definition

    Y=stressstrain=F/AΔL/LY = \frac{\text{stress}}{\text{strain}} = \frac{F/A}{\Delta L/L}Y=strainstress​=ΔL/LF/A​

  2. Condition: wire is stretched to double its length

    If final length is 2L2L2L, then

    ΔL=2L−L=L\Delta L = 2L - L = LΔL=2L−L=L

    So strain is

    ΔLL=LL=1\frac{\Delta L}{L} = \frac{L}{L} = 1LΔL​=LL​=1

  3. Rearrange formula for force

    Y=F/A1=FAY = \frac{F/A}{1} = \frac{F}{A}Y=1F/A​=AF​

    Hence,

    F=YAF = YAF=YA

  4. Substitute values

    Cross-sectional area:

    A=1 cm2=1×10−4 m2A = 1\,\text{cm}^2 = 1 \times 10^{-4}\,\text{m}^2A=1cm2=1×10−4m2

    Young’s modulus:

    Y=2×1011 N/m2Y = 2 \times 10^{11}\,\text{N/m}^2Y=2×1011N/m2

    Therefore,

    F=(2×1011)(1×10−4)F = (2 \times 10^{11})(1 \times 10^{-4})F=(2×1011)(1×10−4)

    F=2×107 NF = 2 \times 10^7\,\text{N}F=2×107N

  5. Match with options

    2×107 N2 \times 10^7\,\text{N}2×107N corresponds to Option C.

Comparison with stored answer: Stored correct answer is C, which matches the derived answer.

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