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Chapter 14: Problem 35

A wire of resistance \(4 \Omega\) is stretched to twice its original length. What is the resistance of the wire now? (A) \(1 \Omega\) (B) \(14 \Omega\) (C) \(8 \Omega\) (D) \(16 \Omega\)

Short Answer

Expert verified
The resistance of the wire when stretched to twice its original length is \(16\,\Omega\), so the correct answer is (D) \(16\,\Omega\).

Step by step solution

01

Formula for resistance

The formula for resistance is given by: \(R = \rho \frac{L}{A}\) where \(R\) is the resistance, \(\rho\) is the resistivity of the material, \(L\) is the length of the wire, and \(A\) is the cross-sectional area of the wire.
02

Volume conservation

Since the volume of the wire doesn't change when it is stretched, \(V_1 = V_2\) or, \((L) (A) = (2L)(A')\) where \(V_1\) is the initial volume, \(V_2\) is the final volume, and \(A'\) is the new cross-sectional area. From this, we conclude that \(A' = \frac{1}{2}A\)
03

Find the new resistance

Now, we can substitute the new length and area into the resistance formula: \(R' = \rho \frac{2L}{\frac{1}{2}A}\)
04

Simplify the expression

Let's simplify the expression: \(R' = \rho \frac{4L}{A}\)
05

Compare the new and old resistances

Divide the new resistance by the old resistance: \(\frac{R'}{R} = \frac{\rho \frac{4L}{A}}{\rho \frac{L}{A}}\) Simplify the expression: \(\frac{R'}{R} = \frac{4L}{L}\) So, \(R' = 4R\)
06

Calculate the new resistance

Now that we know the relationship, plug in the given resistance value to find the new resistance: \(R' = 4(4\,\Omega)\) \(R' = 16\,\Omega\)
07

Choose the correct answer

The correct answer is: (D) \(16\,\Omega\)

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