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Example · Example 1

Q.A steel rod of length 1.5 m1.5\ \text{m} and cross-sectional area 2.5×10−4 m22.5\times10^{-4}\ \text{m}^2 is stretched by a tensile force of 75,000 N75{,}000\ \text{N} applied along its length. Taking the Young's modulus of steel as Y=2.0×1011 PaY = 2.0\times10^{11}\ \text{Pa}, find

(a) the tensile stress in the rod,
(b) the longitudinal strain produced, and
(c) the elongation of the rod.
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✓ Free question

Given: L=1.5 mL = 1.5\ \text{m}, A=2.5×10−4 m2A = 2.5\times10^{-4}\ \text{m}^2, F=75,000 NF = 75{,}000\ \text{N}, Y=2.0×1011 PaY = 2.0\times10^{11}\ \text{Pa}.

  1. Tensile stress:

    Stress=FA=75,0002.5×10−4=3.0×108 Pa\text{Stress} = \frac{F}{A} = \frac{75{,}000}{2.5\times10^{-4}} = 3.0\times10^8\ \text{Pa}

  2. Longitudinal strain, using Y=stress/strainY = \text{stress}/\text{strain}:

    Strain=StressY=3.0×1082.0×1011=1.5×10−3\text{Strain} = \frac{\text{Stress}}{Y} = \frac{3.0\times10^8}{2.0\times10^{11}} = 1.5\times10^{-3}

  3. Elongation, using strain =ΔL/L= \Delta L/L:

    ΔL=Strain×L=1.5×10−3×1.5=2.25×10−3 m=2.25 mm\Delta L = \text{Strain} \times L = 1.5\times10^{-3} \times 1.5 = 2.25\times10^{-3}\ \text{m} = 2.25\ \text{mm}

    ✓Final answer

    Stress =3.0×108 Pa= 3.0\times10^8\ \text{Pa}; strain =1.5×10−3= 1.5\times10^{-3}; elongation =2.25 mm= 2.25\ \text{mm}.

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