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A metallic wire of length 2.5 meters and diameter 0.8 mm is subjected to a tensile force of 800 N, causing it to elongate by 2.2 mm. The density of the material of the wire is 7.8 g/cm3. Calculate Young's modulus of elasticity of the material of the wire.

Option: 1

9.5 * 1010 N/m2


Option: 2

1.2 * 1011 N/m2


Option: 3

2.4 * 1011 N/m2


Option: 4

3.6 * 1011 N/m2


Answers (1)

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Young’s modulus (Y) is a measure of the stiffness of a material. It is defined as the ratio of stress (\sigma) to strain (\epsilon) and is given by the formula:

                        Y = \frac{\sigma}{\epsilon}

where, Stress (\sigma) = Force / Area

            Strain (\epsilon) = Change in length / Original length

Given data:

Original length (L) = 2.5 m
Diameter (d) = 0.8 mm
Tensile force (F) = 800 N
Elongation (\DeltaL) = 2.2 mm
Density (\rho) = 7.8 g/cm3

First, let's calculate the cross-sectional area (A) of the wire using its diameter:

                A = \frac{\pi d^{2}}{4}

Now, convert the diameter to meters and calculate the area:

                A = \frac{\pi (0.8 * 10^{-3})^{2}}{4} m^{2}

Next, calculate the strain (\epsilon) using elongation and original length:

                 \epsilon = \frac{\Delta L}{L}

Convert elongation to meters and calculate strain:

                \epsilon = \frac{\2.2 * 10^{-3}}{2.5} m/m

Now, calculate stress (\sigma) using the formula:

                \sigma = F / A

Substitute the given values:

                \sigma = \frac{800}{\pi * (0.8 * 10^{-3})^2/4} N/m^2

Finally, calculate Young's modulus (Y) using the formula: 

                Y = \frac{\sigma}{\epsilon}

Substitute the calculated values for stress and strain:

                Y = \frac{\frac{800}{\pi * (0.8 * 10^{-3})^ 2/4}}{\frac{2.2 * 10^-3}{2.5}} N/m^{2}

Solve for Y, and you'll find that the value is approximately 1.2 * 1011 N/m2.

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