Question

A wire of length 2 m, radius of cross-section 20 mm and Young's Modulus 2×1011 N/m is subjected to a force of 62.8 kN.The change in length of the wire is p×10–5 (in m). Find P.

Answer 1

The correct answer is 50.
$\Delta L=\frac{FL}{AY}=\frac{62.8\times 1000\times 2}{3.14\times 20\times 20\times 10^{-6}\times 2\times 10^{11}}=50\times 10^{-5} m$
Hence, the value of P= 50