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AIIMS 2015
Exam
AIIMS
Year
2015
Subject
Physics
Chapter
Mechanical Properties of Solids

AIIMS 2015 · Physics · Mechanical Properties of Solids

Young's modulus for a steel wire is 2 × 10^ 11 ~Pa and its elastic...

Young's modulus for a steel wire is 2×1011 Pa2 \times 10^{11} \mathrm{~Pa} and its elastic limit is 2.5×108 Pa2.5 \times 10^8 \mathrm{~Pa}. By how much can a steel wire 3 m3 \mathrm{~m} long and 2 mm2 \mathrm{~mm} in diameter be stretched before the elastic limit is exceeded?

Options

  1. A3.75 mm3.75 \mathrm{~mm}
  2. B7.50 mm7.50 \mathrm{~mm}
  3. C4.75 mm4.75 \mathrm{~mm}
  4. D4.00 mm4.00 \mathrm{~mm}

Correct answer

A. 3.75 mm3.75 \mathrm{~mm}

Step-by-step solution

The cross-sectional area of a wire of radius r=1 mm=10−3 mr=1 \mathrm{~mm}=10^{-3} \mathrm{~m} is A=πr2=3.14×10−6 m2A=\pi r^2=3.14 \times 10^{-6} \mathrm{~m}^2 The maximum force that can be applied without exceeding the elastic limit is therefore F=(FA)max⁡×(A)=(2.5×108 N m−2)(3.14×10−6 m2)=785 N\begin{aligned}& F=\left(\frac{F}{A}\right)_{\max } \times(A) \\ & =\left(2.5 \times10^8 \mathrm{~N} \mathrm{~m}^{-2}\right)\left(3.14 \times10^{-6} \mathrm{~m}^2\right)=785 \mathrm{~N} \end{aligned}When this force is applied, the wire will stretch by \DeltaL=LYFA=(3 m)(785 N)(2×1011 N/m2)(3.14×10−6 m2)=3.75×10−3 m=3.75 mm\begin{aligned}& \DeltaL=\frac{L}{Y} \frac{F}{A}=\frac{(3 \mathrm{~m})(785 \mathrm{~N})}{\left(2 \times 10^{11} \mathrm{~N} / \mathrm{m}^2\right)\left(3.14 \times 10^{-6} \mathrm{~m}^2\right)}\\ & =3.75 \times10^{-3} \mathrm{~m}=3.75 \mathrm{~mm} \end{aligned}
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