Two wires of copper having the length in the ratio \(4: 1\) and their radii are as \(1: 4\) are stretched by the same force. What will be the ratio of longitudinal strain in the two wires? (A) \(1: 16\) (B) \(16: 1\) (C) 1:64 (D) \(64: 1\)

Short Answer

Expert verified
The ratio of longitudinal strain in the two wires is \(16:1\).

Step by step solution

01

List the given information

- Length ratio: \(L_1:L_2 = 4:1\) - Radii ratio: \(r_1:r_2 = 1:4\) - Same force F is applied on both wires
02

Define longitudinal strain and Young's modulus (Y)

Longitudinal strain, \(\epsilon\), is the fractional change in length (\(\frac{\Delta L}{L}\)) when the force is applied. Young's modulus (Y) is a property of the material that relates the longitudinal stress and strain: \(Y = \frac{σ}{\epsilon}\) Where σ is the longitudinal stress, given by \(\frac{F}{A}\), and A is the cross-sectional area of the wire.
03

Find the expression for the longitudinal strain

For the two copper wires, the Young's Modulus is the same, as their material is identical. Thus, \(Y_1 = Y_2\) Using \(Y = \frac{σ}{\epsilon}\), \(\frac{σ_1}{\epsilon_1} = \frac{σ_2}{\epsilon_2}\) Then, stress can be written in terms of force and area. Since both wires experience the same force, F, \(\frac{F/A_1}{\epsilon_1} = \frac{F/A_2}{\epsilon_2}\) We simplify this expression to obtain the ratio of \(\epsilon\): \(\frac{\epsilon_1}{\epsilon_2} = \frac{A_2}{A_1}\)
04

Calculate the cross-sectional area ratio

The cross-sectional area of a wire is given by \(A = πr^2\). Using the given radius ratio, we can calculate the area ratio: \(\frac{A_1}{A_2} = \frac{πr_1^2}{πr_2^2} = \frac{(1)^2}{(4)^2} = \frac{1}{16}\)
05

Conclude the ratio of strain

Now, we can use the obtained area ratio to find the ratio of strains: \(\frac{\epsilon_1}{\epsilon_2} = \frac{A_2}{A_1} = \frac{16}{1}\) This corresponds to option (B) \(16:1\). So, the ratio of longitudinal strain in the two wires is \(16:1\).

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