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Properties of Matter question

2021 · 24 Feb · Shift 2 · Q61
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Properties of Matter question

2021 · 24 Feb · Shift 2 · Q61

JEE MainPhysicsProperties of MatterNumerical+4 / −1
A uniform metallic wire is elongated by 0.04 m when subjected to a linear force F. The elongation, if its length and diameter is doubled and subjected to the same force will be ‾\underline{\hspace{2cm}}​ cm.
Numerical answer
View written solutionFree

Correct answer: 2

  1. Use the formula for elongation of a wire

For a wire under tensile force,

ΔL=FLAY\Delta L = \frac{F L}{A Y}ΔL=AYFL​

where:

  • FFF = applied force
  • LLL = original length
  • AAA = cross-sectional area
  • YYY = Young's modulus
  1. Given initial elongation

The wire elongates by

ΔL1=0.04 m\Delta L_1 = 0.04\text{ m}ΔL1​=0.04 m

under force FFF.

So,

ΔL1=FLAY=0.04\Delta L_1 = \frac{F L}{A Y} = 0.04ΔL1​=AYFL​=0.04
  1. Now length and diameter are both doubled
  • New length:
L2=2LL_2 = 2LL2​=2L
  • Diameter doubled means area becomes 4 times:
A2=4AA_2 = 4AA2​=4A
  1. Find new elongation

Using the same formula,

ΔL2=FL2A2Y=F(2L)(4A)Y=12⋅FLAY\Delta L_2 = \frac{F L_2}{A_2 Y} = \frac{F(2L)}{(4A)Y} = \frac{1}{2}\cdot \frac{FL}{AY}ΔL2​=A2​YFL2​​=(4A)YF(2L)​=21​⋅AYFL​

Thus,

ΔL2=12ΔL1\Delta L_2 = \frac{1}{2} \Delta L_1ΔL2​=21​ΔL1​ ΔL2=12×0.04=0.02 m\Delta L_2 = \frac{1}{2}\times 0.04 = 0.02\text{ m}ΔL2​=21​×0.04=0.02 m
  1. Convert to cm
0.02 m=2 cm0.02\text{ m} = 2\text{ cm}0.02 m=2 cm

Final Answer

2\boxed{2}2​

The derived answer matches the stored correct answer.

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