A steel wire of mass `mu` per unit length with a circular cross-section has a radius of `0.1cm`. The wire is of length `10m` when measured lying horizontal, and hangs from a hook on the wall. A mass fo `25kg` is hung from the free end of the wire. Assume the wire to be uniform and laterla strain `lt lt` logitudinal strain. If density of steel is `7860 kg m^(-3)` and Young's modulus is `2xx10^(11) N//m^(2)` then the extension in the length fo the wire is

Explanation

Consider the diagram when a small element of length dx is considered at x from the load (x = 0).
(a) Let T (x) and T (x + dx) are tensions on the two cross-sections a distance dx apart, then -- - T (x+dx) + T(x) = dmg = `mu dxg` (where `mu` is the mass/length). `" "(becausedm=mudx)`
`" "dT = mugdx " "[because dT = T (x+dx)-T(x)]`
`implies " "T(x) = mugx+C " (on integrating)"`
At `x = 0, T (0) = Mg " "implies " "C = Mg`
`therefore" "T(x) = mugx+ Mg`
Let the length dx at x increase by dr, then
Young's modulus Y = `("Stress")/("Strain")`
`" "(T(x)//A)/(dr//dx)=Y`
`implies " "(dr)/(dx) = (1)/(YA)T(x)`
`implies " "r = (1)/(YA) int_(0)^(L)(mugx + Mg)dx`
`" "=(1)/(YA)[(mugx^(2))/(2)+Mgx]_(0)^(L)`
`" "=(1)/(YA)[(mgL^(2))/(2) + MgL]`
`" "A = pi xx(10^-3)^(2)m^(2)`
`" "Y= 200 xx10^(9) Nm^(-2)`
`" "m = pixx(10^(-3))^(2)xx10xx7860 kg`
`therefore" "r=(1)/(2xx10^(11)xxpixx10^(-6))" "[(pixx786xx10^(-3)xx10xx10)/(2)+25 xx 10 xx 10]`
`=[196.5 xx 10^(-6)+ 3.98 xx 10^(-3)]~~4xx10^(-3)m`

(b) Clearly tension will be maximum at x = L
`therefore" "T = mu gL + Mg = (m + M)g" "[becausem=muL]`
The yield force = (Yield strength Y) area `= 250 xx 10^(6) xx pi xx (10^(-3))^(2) = 250 xx piN`
AT yield point T = yield force
`implies " "(m+M)g=250 xxpi`
`" "m = pixx(10^(-3))^(2)xx10xx7860 ltlt M`
`therefore" "Mg ~~ 250 xxpi`
Hence, `" "M = (250 xxpi)/(10)=25 xxpi ~~ 75kg`.