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  • A copper wire of length 2.0 m and diameter 1.2 mm is suspended vertically from a ceiling. A mass of 4.5 kg is attached to the lower end of the wire. Given that the density of copper is 8.92 g/cm<su

A copper wire of length 2.0 m and diameter 1.2 mm is suspended vertically from a ceiling. A mass of 4.5 kg is attached to the lower end of the wire. Given that the density of copper is 8.92 g/cm3 , calculate the elongation of the wire due to the weight of the attached mass.

 

Option: 1

0.000631m


Option: 2

170 m


Option: 3

200 m


Option: 4

20 m


Answers (1)

best_answer

The elongation of the wire can be calculated using Hooke’s law, which states that the elongation of a material is directly proportional to the applied force (stress) and inversely proportional to the material’s Young’s modulus.

The formula for elongation (?L) is given by:

\Delta L=\frac{F \cdot L}{A \cdot Y}

Where:

F is the force applied (weight of the mass)

L is the original length of the wire

A is the cross-sectional area of the wire

Y is the Young’s modulus of the material

Given data:

Original length (L) = 2.0 m

Diameter (d) = 1.2 mm

Mass (m) = 4.5 kg

Density of copper (ρ) = 8.92 g/cm3

Young’s modulus of copper (Y ) = 1.25 × 1011 N/m2

First, calculate the cross-sectional area (A) using the diameter:

A=\frac{\pi d^2}{4}=\frac{\pi\left(1.2 \times 10^{-3}\right)^2}{4} \approx 1.1309 \times 10^{-6} \mathrm{~m}^2

Next, calculate the force (F) due to the weight:

F = m · g

Where g is the acceleration due to gravity (9.81 m/s2 ).

F = 4.5 · 9.81 ≈ 44.145 N

Now, plug in the values into the elongation formula:

\Delta L=\frac{44.145 \cdot 2.0}{1.1309 \times 10^{-6} \cdot 1.25 \times 10^{11}} \approx 0.000631 \mathrm{~m}The elongation of the wire due to the weight of the attached mass is approximately 0.000631 m. Therefore, the correct answer is A.

Posted by

Devendra Khairwa

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