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JEE Main
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Physics
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Mechanical Properties of Solids

JEE Main · Physics · Mechanical Properties of Solids

An iron wire AB has diameter of 0.6 mm and length 3 m at 0 ^ ∘ C...

An iron wire AB has diameter of 0.6 mm\text{0}\text{.6 mm} and length 3 m\text{3 m} at 0∘C{0}^{∘} \text{C}.The wire is now stretched between the opposite walls of a brass casing at 0∘C{0}^{∘} \text{C}. What is the extra tension that will be set up in the wire when the temperature of the system is raised to 40∘C{40}^{∘} \text{C}?

Given αbrass=18×10–6/K{α}_{brass} = 18 \times {10}^{– 6} / \text{K}

αiron=12×10–6/K{α}_{iron} = 12 \times {10}^{– 6} / \text{K}

Yiron=21×1010N/m2{Y}_{iron} = 21 \times {10}^{10} {\text{N/m}}^{2}

An iron wire AB has diameter of 0.6 mm and length 3 m at 0 ^ ∘ C .The wire is now stretched between the opposite walls

Options

  1. A14.2N14 . 2 N
  2. B13.8N13 . 8N
  3. C16.3N16 . 3N
  4. D21.7N21 . 7N

Correct answer

A. 14.2N14 . 2 N

Step-by-step solution

Increase in length of the iron wire =LαΔt=L×12×10−6×40=48L×10−5m= \text{L} \alpha \Delta \text{t} = \text{L} \times \text{12} \times {10}^{- 6} \times \text{40} = \text{48L} \times {10}^{- 5} m

Increase in length of the brass tube =LαΔt=L×18×10−6×40=72L×10−5m= \text{L} \alpha \Delta \text{t} = \text{L} \times \text{18} \times {10}^{- 6} \times \text{40} = \text{72L} \times {10}^{- 5} m

Since the increase in length of the brass tube is greater than that of the iron wire, the latter will be tightened. Let T be the tension in the wire. Then

Stress=Tπr2=Tπ(3×10−4)2=T9π×10−8\text{Stress} = \dfrac{T }{\pi {r }^{2}} = \dfrac{T }{\pi {\left(3 \times 1 {0}^{- 4}\right)}^{2}} = \dfrac{T }{9 \pi \times 1 {0}^{- 8}}

Strain=(72L−48L)×10−5L=24×10−5\text{Strain} = \dfrac{\left(7 2 L - 4 8 L \right) \times 1 {0}^{- 5}}{L } = 2 4 \times 1 {0}^{- 5}

Since stress=Y×strain,T9π×10−8=21×1010×24×10−5\text{Since stress} = Y \times \text{strain}\text{,}\dfrac{T }{9\pi \times 1{0}^{- 8}} = 21 \times 1{0}^{10} \times 24 \times 1{0}^{- 5}

T=21×1010×24×10−5×9π×10−8=14.2 N\text{T} = 2 1 \times 1 {0}^{10} \times 2 4 \times 1 {0}^{- 5} \times 9 \pi \times 1 {0}^{- 8} = \text{14.2 N}
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