A metallic bar 50 cm long has a uniform cross-sectional area of 4.0 cm\(^2\). If a tensile force of 35 kN produces an extension of 0.25 mm, calculate the value of Young’s modulus
Explanation
E \(\frac{Fl}{Ae}\)
= \(\frac{35 \times 10^3 \times 50 \times 10^{-2}}{4 \times 10^{-4} \times 0.25 \times 10 ^{-3}}\)
= 1.75 x 10\(^{11}\)Nm\(^{-2}\)