A cylindrical steel wire of 3 m length is to stretch no more than 0.2 cm when a tensile force of 400 N is applied to each end of the wire. What minimum diameter is required for the wire?
$Y_{s t e e l} = 2.1 \times 10^{11} N / m^{2}$

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Solution

Given , $l = 3 m, Δ l = 0.2 c m = 0.2 \times 10^{- 2} m, F = 400 N, Y = 2.1 \times 10^{11} N / m^{2}$
let radius of wire is $r$ .
thus, $Y = \frac{F l}{A Δ l} = \frac{F l}{π r^{2} Δ l}$
$r^{2} = \frac{F l}{π Δ l Y} = \frac{400 \times 3}{3.14 \times 0.2 \times 10^{- 2} \times 2.1 \times 10^{11}} = 90.99 \times 10^{- 8}$
$r = 9.53 \times 10^{- 4} m$
Diameter of the wire is $d = 2 r = 1.91 \times 10^{- 3} m = 1.91 m m \approx 2 m m$

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