# a) A steel wire of mass μ per unit length with a circular cross-section has a radius of 0.1 cm. The wire is of length 10 m when measured lying horizontal, and hangs from a hook on the wall. A mass of 25 kg is hung from the free end of the wire. Assuming the wire to be uniform and lateral strains << longitudinal strains, find the extension in the length of the wire. The density of steel is 7860 kg/m3.b) If the yield strength of steel is 2.5 × 108 N/m2, what is the maximum weight that can be hung at the lower end of the wire?

An element dx is of mass dm taken from a wire of length L from a distance x. mass per unit length here is u.

M = 25 kg, L =10m

a) The force that acts in the downward direction on dx = wt

dr is the increase in the length of wire

integrating both sides we get,

r (change in length of wire) = 1/AY [ug x2/g + Mgx] L0

extension in wire of length L =

$\frac{gL}{2 \pi r^{2}Y}[ul + 2M]$

extension=

$10\times 10 [0.25 + 2 \times 0.25] / (2\times 3.14 \times 10^{-6} \times 2 \times 10^{11})$

extension = $400 \times 10^{-5}$ m mass m = volume x density =

b) at x = L, the wire experiences maximum tension

T(x) = ugx + Mg

T(L) = ugL + Mg

T = (m + M) g {as uL = m}

Yield force = Yield strength x A

$= 250 \times 3.14$

Max tension = yield force

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