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Chapter 7: Problem 904

Two wires of same diameter of the same material having the length \(\ell\) and \(2 \ell\). If the force \(\mathrm{F}\) is applied on each, what will be the ratio of the work done in the two wires? (A) \(1: 2\) (B) \(1: 4\) (C) 2: 1 (D) \(1: 1\)

Short Answer

Expert verified
The short answer is: The ratio of the work done in the two wires is 1:1 (option D).

Step by step solution

01

Identify the Variables

Let's identify the given variables: Length of Wire 1: \( \ell \) Length of Wire 2: \( 2 \ell \) Force Applied on each wire: \(F\) Material and diameter of the wires are the same. Now, we need to find the ratio of the work done in the two wires, denoted as \(W_1\) and \(W_2\).
02

Find the elongation of the wires

According to Hooke's Law, the force applied on the wire is proportional to its elongation. \[F = k \Delta L\] Since, both wires have the same material and diameter, we can assume the same constant of proportionality. Rearranging to find the elongation of each wire: \[\Delta L_1 = \frac{F}{k} \text{ and } \Delta L_2 = \frac{F}{k}\] Next, we'll find the work done on both wires.
03

Determine the work done

To calculate the work done on a wire, the formula is: \[W = \frac{1}{2} F \cdot \Delta L\] We'll plug-in the elongation value into the work formula for Wire 1 and Wire 2: \[W_1 = \frac{1}{2} F \cdot \frac{F}{k}\] \[W_2 = \frac{1}{2} F \cdot \frac{F}{k}\] After calculating the work done, we notice that the values for both wires are the same. Thus, the ratio between both works is 1:1.
04

Compare the work done

The ratio of the work done in the two wires is: \[\frac{W_1 }{W_2}\] Plugging in the values: \[\frac{ \frac{1}{2} F \cdot \frac{F}{k} }{ \frac{1}{2} F \cdot \frac{F}{k} }\] On simplifying the equation, makes us conclude that: \[\frac{W_1}{W_2} = \frac{1}{1}\]
05

Conclusion

The ratio of the work done in the two wires is 1:1 (option D).

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