A load of 10 kg is suspended by a metal wire that is 3 m long and possesses a cross-sectional area of 4 mm^2 . Determine the strain. The Young modulus of the metal is 2.0 × 10^ 11 N/m^2 .

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Quantrex Academy · Accepted Answer

Correct answer: B. 1.225 × 10^ -4

The force applied on the wire is the weight of the load:

F = mg = 10 \times 9.8 = 98 \text{ N}

The cross-sectional area is:

A = 4 \text{ mm}^2 = 4 \times 10^{-6} \text{ m}^2

The formula for Young's modulus is:

Y = \dfrac{\text{Stress}}{\text{Strain}} = \dfrac{F / A}{\text{Strain}}

Rearranging for strain:

\text{Strain} = \dfrac{F}{A \cdot Y}

Substituting the given values:

\text{Strain} = \dfrac{98}{4 \times 10^{-6} \times 2.0 \times 10^{11}}

\text{Strain} = \dfrac{98}{8 \times 10^5}

\text{Strain} = 12.25 \times 10^{-5} = 1.225 \times 10^{-4}

A load of 10 kg is suspended by a metal wire that is 3 m long and...

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