Solution:
Formula: \(Delta L = frac{FL}{AY}\). Area \(A = pi r^2 = 3.14 times 10^{-4}\text{ m}^2\. Substituting the given values yields \(Delta L = frac{10^5 times 1}{3.14 times 10^{-4} times 2 times 10^{11}} approx 1.59\text{ mm}\.
A steel rod has a radius of 10 mm and a length of 1 m. A 100 kN force stretches it along its length. The elongation in rod is (Assume young's modulus \(Y = 2 times 10^{11}\text{ N/m}^2\))
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