Question

A uniform heavy rod of mass \(20\) kg, cross sectional area \(0.4\) \(m^2\) and length \(20\) \(m\) is hanging from a fixed support. Neglecting the lateral contraction, the elongation in the rod due to its own weight is \(x × 10^{–9}\) \(m\). The value of \(x\) is _____(Given Young’s modulus \(Y = 2 × 10^{11} Nm^{–2}\)and g = \(10\; ms^{–2})\)

Answer 1

Correct Answer: 25

\(\frac{\frac{F}{A}}{\frac{ΔL}{L}} = Y\)

\(ΔL = \frac{FL}{AY }\)

= \(\frac{TavgL}{AY} = \frac{MgL}{2AY}\)

= \(\frac{20 \times 10 \times 20}{2 \times .4 \times 2 \times 10^{11}}\)

=\(\frac{ 4 \times 10. \times 10^{-11}}{4 \times 0.4}\)

= \(2.5 \times 10^{-8}\)

=\( 25 \times 10^{-9}\)