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Chapter 7: Problem 915

When a force is applied on a wire of uniform cross-sectional area $3 \times 10^{-6} \mathrm{~m}^{2}\( and length \)4 \mathrm{~m}$, the increase in length is \(1 \mathrm{~mm}\). what will be energy stored in it ? [ $\mathrm{Y}=2 \times 10^{11}\left(\mathrm{~N} / \mathrm{m}^{2}\right)$ (A) \(62.50 \mathrm{~J}\) (B) \(0.177 \mathrm{~J}\) (C) \(0.075 \mathrm{~J}\) (D) \(0.150 \mathrm{~J}\)

Short Answer

Expert verified
The energy stored in the wire is \(0.075 J\).

Step by step solution

01

List the given information and required formula.

We are given the following: 1. Cross-sectional area (A) = \(3 \times 10^{-6} m^2\) 2. Length of the wire (L) = 4 m 3. Increase in length, ∆L = 1 mm = \(1 \times 10^{-3} m\) 4. Young's modulus (Y) = \(2 \times 10^{11} N/m^2\) We need to find the energy stored (U) in the wire. The formula to calculate energy is: \(U = \frac{1}{2} \times \frac{F^2}{Y} \times \frac{\Delta L}{A}\)
02

Calculate the applied force (F) on the wire.

From Young's modulus, we can find the applied force using the formula: \(Y = \frac{F \times L}{A \times \Delta L}\) Now, we solve for F: \(F = \frac{Y \times A \times \Delta L}{L}\) Plugging in the given values: \(F = \frac{(2 \times 10^{11} N/m^2)(3 \times 10^{-6} m^2)(1 \times 10^{-3} m)}{4 m}\) Calculating the value of F: \(F = 15 \times 10^5 N\)
03

Calculate the energy stored in the wire.

Now we can use the formula for the potential energy stored in the wire: \(U = \frac{1}{2} \times \frac{F^2}{Y} \times \frac{\Delta L}{A}\) Plugging in the values we've found and the given values: \(U = \frac{1}{2} \times \frac{(15 \times 10^5 N)^2}{(2 \times 10^{11} N/m^2)} \times \frac{(1 \times 10^{-3} m)}{(3 \times 10^{-6} m^2)}\) Calculating the value of U: \(U = 0.075 J\) Therefore, the energy stored in the wire is \(0.075 J\), which corresponds to answer choice (C).

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