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A steel wire of mass mu per unit length ...

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

Text Solution

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(a) Let T (x) be the tension in the wire at a distance x from the lower end of the wire. Then `T(x) = mu gx + Mg` Let dl be the increase in length dx at disntance x from lower end. Then `Y = (T (x)//A)/(dl //dx) = (T (x)dx)/(A dl) or dl = (T(x) dx)/(YA)`
Total extension l of lenght L is given by ` = int_0^l dl = int_0^L (T(x)dx)/(YA)`
`or l = (1)/(YA) int_0^L (mu g x +Mg) dx = (1)/(YA) [(mug L^2)/(2)+ mgL] = (1)/(YA) [(mgL)/(2) + Mg L] ( :. mu = (m)/(L))`
there m is the mass of wire of length L. Here, `Y = 2xx10^(11) N//m^2, A = pi r^2 = 3.14 xx 10^(-6) m^2 , L = 10m , g = 10 m//s^2`
m = area `xx` length `xx` density =` (3.14 xx10^(-6))xx10 xx (7860) = 0.256 kg , M = 25 kg`
`I = (1)/((2xx10^(11))xx(3.14 xx10^(-6)) [(0.25 xx 10xx10)/(2) + 25 xx10 xx10]) = 4.0 xx 10^(-3)m`
(b) There will be maximum tension in the wire at location for which x = L. Then maximum tension is `T_(max) = mu gL+Mg = mg + Mg = (m + M)g` Yield force = Yield strength` xx `area of cross - section = `(2.5xx10^8)xxpi r^2`
=` 2.5 xx10^8 xx 3.14 xx10^(-6) = 785 N`
At yield `(m + M)g = 785 ,` Here, `m = 0.25 kg`
`:. (0.25 + M)xx10 = 785` or `M = (785)/(10)- 0.25 = 78.25 kg`
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