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Properties of Matter question

2019 · 10 Apr · Shift 2 · Q55
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Properties of Matter question

2019 · 10 Apr · Shift 2 · Q55

JEE MainPhysicsProperties of MatterMCQ+4 / −1
In an experiment, brass and steel wires of length 1 m each with areas of cross section 1mm2 are used. The wires are connected in series and one end of the combined wire is connected to a rigid support and other end is subjected to elongation. The stress required to produce a net elongation of 0.2 mm is, [Given, the Young's Modulus for steel and brass are, respectively, 120 × 109 N/m2 and 60 × 109 N/m2]
  1. A
    8.0 × 106 N/m2
  2. B
    1.2 × 106 N/m2
  3. C
    0.2 × 106 N/m2
  4. D
    1.8 × 106 N/m2
View written solutionFree

Correct answer: A

  1. Given data
  • Length of steel wire: Ls=1 mL_s = 1\,\text{m}Ls​=1m
  • Length of brass wire: Lb=1 mL_b = 1\,\text{m}Lb​=1m
  • Area of cross section of each wire: A=1 mm2=1×10−6 m2A = 1\,\text{mm}^2 = 1\times 10^{-6}\,\text{m}^2A=1mm2=1×10−6m2
  • Young's modulus of steel: Ys=120×109 N/m2Y_s = 120\times 10^9\,\text{N/m}^2Ys​=120×109N/m2
  • Young's modulus of brass: Yb=60×109 N/m2Y_b = 60\times 10^9\,\text{N/m}^2Yb​=60×109N/m2
  • Total elongation: ΔL=0.2 mm=2×10−4 m\Delta L = 0.2\,\text{mm} = 2\times 10^{-4}\,\text{m}ΔL=0.2mm=2×10−4m

The two wires are connected in series, so the same force acts in both. Since their cross-sectional areas are equal, the stress is also the same in both wires.

  1. Elongation of each wire

For a wire,

ΔL=stressY L\Delta L = \frac{\text{stress}}{Y}\,LΔL=Ystress​L

Let the common stress be σ\sigmaσ.

Then for steel,

ΔLs=σYsLs\Delta L_s = \frac{\sigma}{Y_s}L_sΔLs​=Ys​σ​Ls​

For brass,

ΔLb=σYbLb\Delta L_b = \frac{\sigma}{Y_b}L_bΔLb​=Yb​σ​Lb​

So total elongation is

ΔL=ΔLs+ΔLb\Delta L = \Delta L_s + \Delta L_bΔL=ΔLs​+ΔLb​ 2×10−4=σ(LsYs+LbYb)2\times 10^{-4} = \sigma\left(\frac{L_s}{Y_s} + \frac{L_b}{Y_b}\right)2×10−4=σ(Ys​Ls​​+Yb​Lb​​)

Substitute values:

2×10−4=σ(1120×109+160×109)2\times 10^{-4} = \sigma\left(\frac{1}{120\times 10^9} + \frac{1}{60\times 10^9}\right)2×10−4=σ(120×1091​+60×1091​)
  1. Simplify the bracket
1120×109+160×109=1120×109+2120×109=3120×109=140×109\frac{1}{120\times 10^9} + \frac{1}{60\times 10^9} = \frac{1}{120\times 10^9} + \frac{2}{120\times 10^9} = \frac{3}{120\times 10^9} = \frac{1}{40\times 10^9}120×1091​+60×1091​=120×1091​+120×1092​=120×1093​=40×1091​

Thus,

2×10−4=σ⋅140×1092\times 10^{-4} = \sigma \cdot \frac{1}{40\times 10^9}2×10−4=σ⋅40×1091​

So,

σ=2×10−4×40×109\sigma = 2\times 10^{-4} \times 40\times 10^9σ=2×10−4×40×109 σ=8×106 N/m2\sigma = 8\times 10^6\,\text{N/m}^2σ=8×106N/m2
  1. Match with options
σ=8.0×106 N/m2\boxed{\sigma = 8.0\times 10^6\,\text{N/m}^2}σ=8.0×106N/m2​

This corresponds to Option A.

  1. Comparison with stored answer

Stored correct answer: A

Derived answer: A

So the derived answer agrees with the stored correct answer.

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