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NEET 2026
Exam
NEET
Year
2026
Subject
Physics
Chapter
Chapter

NEET 2026 · Physics · Chapter

An aluminium rod is designed such that it can undergo a maximum...

An aluminium rod is designed such that it can undergo a maximum elongation of 5 mm5 \text{ mm} before permanent deformation sets in. If the elastic limit and Young's modulus of aluminium are 2.5×108 N m−22.5 \times 10^8 \text{ N m}^{-2} and 1.0×1011 N m−21.0 \times 10^{11} \text{ N m}^{-2} respectively, what is the original length of the rod?

Options

  1. A2 m2 \text{ m}
  2. B2000 m2000 \text{ m}
  3. C0.2 m0.2 \text{ m}
  4. D12.5 m12.5 \text{ m}

Correct answer

A. 2 m2 \text{ m}

Step-by-step solution

Given:

Maximum elongation, ΔL=5×10−3 m\Delta L = 5 \times 10^{-3} \text{ m}

Elastic limit (maximum stress), σ=2.5×108 N m−2\sigma = 2.5 \times 10^8 \text{ N m}^{-2}

Young's modulus, Y=1.0×1011 N m−2Y = 1.0 \times 10^{11} \text{ N m}^{-2}

Using the relation for Young's modulus:

Y=StressStrain=σΔLLY = \dfrac{\text{Stress}}{\text{Strain}} = \dfrac{\sigma}{\dfrac{\Delta L}{L}}

Rearranging for original length LL:

L=YΔLσL = \dfrac{Y \Delta L}{\sigma}

Substituting the given values:

L=1.0×1011×5×10−32.5×108L = \dfrac{1.0 \times 10^{11} \times 5 \times 10^{-3}}{2.5 \times 10^8}

L=5×1082.5×108L = \dfrac{5 \times 10^8}{2.5 \times 10^8}

L=2 mL = 2 \text{ m}

Answer: 2 m2 \text{ m}
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