A thick uniform rubber rope of density 1.5 g cm^-3 and Young's modulus $5\times {10}^{6 }N{m}^{-2}$  has a length of 8 m. When hung from the ceiling of a room, the increase in length of the rope due to its own weight  will be

The weight of the rope can be assumed to act at its mid-point. Now, the extension I is proportional to original length L. If the weight of the rope acts at its mid-point, the extension will be that produced by half the rope. So, replacing£ by L/2 in the expression for Young's modulus, we have

                    $$\begin{gathered} Y=\frac{F L/2}{Al}=\frac{F L}{2Al} \\ l=\frac{F L}{2AY}=\frac{g{L}^2\rho }{2Y} \\ (·: F = mg and m = \rho V= \rho AL) \\ Now \rho = 1.5 g c{m}^{-3} = 1500 kg m^{-3} , therefore \\ l=\frac{10\times {\left(8\right)}^2\times 1500}{2\times 5\times {10}^6}=9.6\times {10}^{-2}m \end{gathered}$$