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Numerical · Q42

Q.A wire of mild steel has initial length 1.5 m and diameter 0.60 mm and is extended by 6.3 mm when a certain force is applied to it. If Young's modulus of mild steel is 2.1×10¹¹ N/m², calculate the force applied.

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Step 1. Given: L=1.5L = 1.5 m, diameter =0.60= 0.60 mm so radius r=0.30 mm=3×10−4r = 0.30\ \text{mm} = 3\times10^{-4} m, ΔL=6.3 mm=6.3×10−3\Delta L = 6.3\ \text{mm} = 6.3\times10^{-3} m, Y=2.1×1011 N/m2Y = 2.1\times10^{11}\ \text{N/m}^2.

Step 2. Cross-sectional area: A=πr2=π(3×10−4)2=2.827×10−7 m2A = \pi r^2 = \pi (3\times10^{-4})^2 = 2.827\times10^{-7}\ \text{m}^2.

Step 3. From Y=FLA ΔLY = \dfrac{FL}{A\,\Delta L}, rearrange: F=YA ΔLL=2.1×1011×2.827×10−7×6.3×10−31.5F = \dfrac{YA\,\Delta L}{L} = \dfrac{2.1\times10^{11}\times 2.827\times10^{-7}\times 6.3\times10^{-3}}{1.5}. …

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