Question

A uniform wire of length l = 800 m and cross-sectional area $s=1m{m}^2$ is made of a material of density $P=8\times {10}^{3}kg/m^3$ and young's modulus $Y=2\times {10}^{11}N/m^2$. Wire is placed on a frictionless horizontal surface and an end is pulled by a constant horizontal force( F = 1 N) then extention in front half of wire in mm is (Write upto two digits after the decimal point.)

Answer 1

Say $\lambda$ is the linear density of the wire $=\frac{m}{L}$

$a=\frac{F}{M}$ $T=ma=\lambda (L-x)a$

By $Y=\frac{T}{s}\frac{dx}{dl}\Rightarrow dl=\frac{T}{sY}dx=\frac{\lambda (L-x)a}{sY}dx$

$\Delta l={\int }_0^{\frac{L}{2}}\frac{\lambda (L-x)a}{sY}dx=\frac{F}{LsY}(\frac{L^2}{2}-\frac{L^2}{8})=\frac{3}{8}\frac{FL}{5Y}=1.5mm$