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NEET PHYSICSEasy

The maximum elongation of a steel wire of 1 m1 \text{ m} length if the elastic limit of steel and its Young's modulus, respectively, are 8×108 N m−28 \times 10^8 \text{ N m}^{-2} and 2×1011 N m−22 \times 10^{11} \text{ N m}^{-2}, is:

A

0.4 mm0.4 \text{ mm}

B

40 mm40 \text{ mm}

C

8 mm8 \text{ mm}

D

4 mm4 \text{ mm}

Step-by-Step Solution

  1. Identify Given Values: Length (LL) = 1 m1 \text{ m} Elastic Limit (Maximum Stress, σ\sigma) = 8×108 N m−28 \times 10^8 \text{ N m}^{-2}
  • Young's Modulus (YY) = 2×1011 N m−22 \times 10^{11} \text{ N m}^{-2}

  1. Formula Application: Young's Modulus is defined as the ratio of longitudinal stress to longitudinal strain within the elastic limit. Y=Stress(σ)Strain(ε)=σΔL/LY = \frac{\text{Stress} (\sigma)}{\text{Strain} (\varepsilon)} = \frac{\sigma}{\Delta L / L} Rearranging the formula to solve for elongation (ΔL\Delta L): ΔL=σ⋅LY\Delta L = \frac{\sigma \cdot L}{Y}

  2. Calculation: Substitute the values into the equation: ΔL=(8×108)×12×1011\Delta L = \frac{(8 \times 10^8) \times 1}{2 \times 10^{11}} ΔL=4×108−11 m\Delta L = 4 \times 10^{8-11} \text{ m} ΔL=4×10−3 m\Delta L = 4 \times 10^{-3} \text{ m}

  3. Unit Conversion: Since 10−3 m=1 mm10^{-3} \text{ m} = 1 \text{ mm}: ΔL=4 mm\Delta L = 4 \text{ mm}

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