## In This post you will get answer of Q.

The Young’s modulus of a rope of $ 10text{ }m $ length and having diameter of $ 2text{ }cm $ is $ 200times {{10}^{11}},dyne/c{{m}^{2}} $ . If the elongation produced in the rope is $ 1text{ }cm, $ the force applied on the rope is

Solution:

## Young’s modulus of a rope

$ Y=frac{FL}{ADelta l} $

Given, $ L=10m,,,A=pi {{r}^{2}}=pi {{(1)}^{2}}=pi ; $

$ Y=20times {{10}^{11}},dyne/c{{m}^{2}},,Delta l=1cm, $

$ F=frac{Y.A.Delta l}{L} $

$ F=frac{20times {{10}^{11}}times 1times 1}{10times {{10}^{2}}} $ $ F=6.28times {{10}^{9}}dyne $

$ F=6.28times {{10}^{4}}N $