Physics - Elasticity Question with Solution | TestHub

Here, Length of the rope, $L=10\mathrm{m}$ Diameter of the rope, $D=2\mathrm{cm}$ $\therefore$ Radius of the rope, $r=\frac{D}{2}=1\mathrm{cm}=10^{-2}\mathrm{m}$ Young's modulus of the rope $\begin{array}{rl}Y& =20\times 10^{11}\frac{\text{ dyne }}{{\mathrm{cm}}^2}=20\times 10^{11}\times \frac{10^{-5}}{{\left(10^{-2}\right)}^2}\frac{\mathrm{N}}{{\mathrm{m}}^2}\\ & =20\times 10^{11}\times \frac{10^{-5}\mathrm{N}}{10^{-4}{\mathrm{m}}^2}=20\times 10^{10}\frac{\mathrm{N}}{{\mathrm{m}}^2}\end{array}$ Elongation of rope, $\Delta L=1\mathrm{cm}=10^{-2}\mathrm{m}$ As $\begin{array}{rl}Y& =\frac{F/A}{\Delta L/L}\text{ or }F=Y\frac{A\Delta L}{L}=\frac{Y\pi r^2\Delta L}{L}\\ & =\frac{20\times 10^{10}\times 3.14\times {\left(10^{-2}\right)}^2\times 10^{-2}}{10}\\ & =6.28\times 10^4\mathrm{N}\end{array}$