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Chapter 7: Problem 882

A wire of length \(2 \mathrm{~m}\) is made from \(10 \mathrm{~cm}^{3}\) of copper. A force \(\mathrm{F}\) is applied so that its length increases by $2 \mathrm{~mm}\(. Another wire of length \)8 \mathrm{~m}$ is made from the same volume of copper. If the force \(\mathrm{F}\) is applied to it, its length will increase by......... (A) \(0.8 \mathrm{~cm}\) (B) \(1.6 \mathrm{~cm}\) (C) \(2.4 \mathrm{~cm}\) (D) \(3.2 \mathrm{~cm}\)

Short Answer

Expert verified
The increase in length of the second wire is \(0.8 \mathrm{~cm}\).

Step by step solution

01

Find the cross-sectional area of the first wire.

The volume of the first wire is given as 10 cm³, and its length is 2 m or 200 cm. To find the area A, we can use the formula: \[ Volume = Area \cdot Length \] \[ A = \frac{Volume}{Length} \] \[ A = \frac{10}{200} = 0.05 cm^{2} \]
02

Find the cross-sectional area of the second wire.

Since the volume of copper used for both wires is the same and the second wire's length is 8 m or 800 cm, we can find its cross-sectional area A' using the same formula: \[ A' = \frac{Volume}{Length} \] \[ A' = \frac{10}{800} = 0.0125 cm^{2}\]
03

Determine the ratio of the cross-sectional areas.

Now, let's find the ratio of the cross-sectional areas of the two wires, which we will call R: \[R = \frac{A'}{A} \] \[R = \frac{0.0125}{0.05} \] \[R = 0.25\]
04

Calculate the increase in length of the second wire.

We know that the length of the first wire increases by 2 mm or 0.2 cm. We can now use the ratio R to find the increase in length of the second wire (ΔL): \[ \Delta L = \frac{\Delta L_{1}}{R}\] \[ \Delta L = \frac{0.2}{0.25} = 0.8 cm\] The increase in length of the second wire is 0.8 cm. Therefore, the correct answer is (A) \(0.8 \mathrm{~cm}\).

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