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

$\frac{\frac{F}{A}}{\frac{ΔL}{L}} = Y$

$ΔL = \frac{FL}{AY }$

= $\frac{TavgL}{AY} = \frac{MgL}{2AY}$

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

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

= $2.5 \times 10^{-8}$

=$25 \times 10^{-9}$