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A concrete piling of 50 kg is suspended from a steel wire of diameter 1.0 mm and length 11.2 m. How much will the wire stretch?

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Explanation:

It is given that,

Mass of concrete pilling, m = 50 kg

Diameter of wire, d = 1 mm

Radius of wire, r = 0.0005 m

Length of wire, L = 11.2

Young modulus of steel, $Y=20\times 10^{10}\ N/m^2$

The young modulus of a wire is given by :

$Y=\dfrac{\dfrac{F}{A}}{\dfrac{\Delta L}{L}}$

$Y=\dfrac{F.L}{A\Delta L}$

$\Delta L=\dfrac{F.L}{A.Y}$

$\Delta L=\dfrac{50\ kg\times 9.8\ m/s^2\times 11.2\ m}{\pi (0.0005\ m)^2\times 20\times 10^{10}\ N/m^2}$

$\Delta L=0.034\ m$

So, the wire will stretch 0.034 meters. Hence, this is the required solution.

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