Understanding the Problem: Stretched Metallic Wire

The question asks how the resistance, resistivity, and power rating of a metallic wire change when it is uniformly stretched to reduce its radius to half the original value, while keeping the voltage constant. We are given the initial resistance (\(R\)), resistivity (\(\rho\)), and power rating (\(P\)) at voltage (\(V\)). We need to find the new values \(R'\), \(\rho'\), and \(P'\) at the same voltage \(V\) and determine the correct relationships.

Analyzing the Material Properties: Resistivity

Resistivity (\(\rho\)) is an intrinsic property of the material itself. It depends on the type of material and its temperature. When a wire is stretched, the material composition does not change, and the temperature is assumed to be constant unless otherwise stated. Therefore, the resistivity of the wire remains unchanged.

So, the new resistivity \(\rho'\) is equal to the original resistivity \(\rho\):

Calculating the New Resistance: Effect of Stretching

The resistance (\(R\)) of a wire is given by the formula:

where \(\rho\) is the resistivity, \(L\) is the length, and \(A\) is the area of cross-section.

The original wire has resistance \(R\), length \(L\), and area of cross-section \(A\). The area is given by \(A = \pi r^2\), where \(r\) is the original radius. So, \(R = \rho \frac{L}{\pi r^2}\).

When the wire is uniformly stretched, its volume remains constant. Let the new length be \(L'\) and the new radius be \(r'\). The new area of cross-section is \(A' = \pi (r')^2\). The problem states that the new radius is half the original radius, so \(r' = \frac{r}{2}\).

The volume of the original wire is \(V = A \times L = \pi r^2 L\).

The volume of the stretched wire is \(V' = A' \times L' = \pi (r')^2 L' = \pi \left(\frac{r}{2}\right)^2 L' = \pi \frac{r^2}{4} L'\).

Since the volume is constant, \(V = V'\):

We can cancel \(\pi r^2\) from both sides:

This implies the new length \(L'\) is four times the original length \(L\):

Now let's find the new area \(A'\):

Since the original area \(A = \pi r^2\), the new area is:

The new resistance \(R'\) is given by:

Substitute the values we found: \(\rho' = \rho\), \(L' = 4L\), and \(A' = \frac{A}{4}\).

Since the original resistance \(R = \rho \frac{L}{A}\), we can substitute \(R\) into the equation for \(R'\):

So, the new resistance is 16 times the original resistance.

Determining the New Power Rating

The power rating (\(P\)) of a resistor connected to a voltage source (\(V\)) is given by the formula:

The original power rating is \(P = \frac{V^2}{R}\).

The new power rating \(P'\) at the same voltage \(V\) with the new resistance \(R'\) is:

We found that \(R' = 16R\). Substitute this into the formula for \(P'\):

We can rewrite this as:

Since \(P = \frac{V^2}{R}\), we have:

So, the new power rating is \(1/16\) times the original power rating.

Summarizing the Results

Based on our calculations, the relationships between the original and new values are:

• New resistivity \(\rho'\) = Original resistivity \(\rho\)
• New resistance \(R'\) = 16 times Original resistance \(R\)
• New power rating \(P'\) = \(1/16\) times Original power rating \(P\)

Comparing with Options

Let's compare our derived relationships (\(\rho' = \rho\), \(R' = 16R\), \(P' = (1/16)P\)) with the given options:

1. \(\rho' = 2\rho\), \(R' = 2R\), \(P' = 2P\) - Incorrect.
2. \(\rho' = (1/2)\rho\), \(R' = (1/2)R\), \(P' = (1/2)P\) - Incorrect.
3. \(\rho' = \rho\), \(R' = 16R\), \(P' = (1/16)P\) - Matches our results.
4. \(\rho' = \rho\), \(R' = (1/16)R\), \(P' = 16P\) - Incorrect.

The correct relationships are \(\rho' = \rho\), \(R' = 16R\), and \(P' = (1/16)P\).

Revision Table: Stretched Wire Properties

Additional Information: Mechanics of Stretching and Electrical Properties

When a wire is stretched uniformly, the material flows along the length. The key assumption is that the total volume of the material remains constant. This volume conservation is crucial for determining the change in length and area when one dimension (like radius) is changed.

If a wire is stretched such that its length becomes \(n\) times the original length (\(L' = nL\)), assuming constant volume, its area of cross-section must decrease. The new area \(A'\) will be \(A/n\). Consequently, the new resistance \(R'\) becomes \(\rho \frac{nL}{A/n} = n^2 \rho \frac{L}{A} = n^2 R\). In our case, the length became 4 times, so \(n=4\), and \(R' = 4^2 R = 16R\).

Alternatively, if the radius changes such that the new radius is \(1/n\) times the original radius (\(r' = r/n\)), then the new area \(A'\) is \(\pi (r/n)^2 = \pi r^2/n^2 = A/n^2\). For volume conservation \(V=V'\), \(AL = A'L'\), so \(AL = (A/n^2)L'\), which implies \(L' = n^2 L\). In our problem, the radius was reduced to half, so \(n=2\) (since \(r' = r/2 = r/n\)), making \(n=2\). The length becomes \(L' = 2^2 L = 4L\), and the area becomes \(A' = A/2^2 = A/4\). The resistance then becomes \(R' = \rho \frac{n^2 L}{A/n^2} = \rho \frac{n^4 L}{A}\). Oh wait, the formula is \(R' = \rho' \frac{L'}{A'}\). Substituting \(L' = n^2 L\) and \(A' = A/n^2\), and \(\rho'=\rho\), we get \(R' = \rho \frac{n^2 L}{A/n^2} = \rho \frac{n^4 L}{A}\) - No, that's not right. Let's re-check the steps.

From \(L' = n^2 L\) and \(A' = A/n^2\) (when \(r' = r/n\)), the new resistance is \(R' = \rho \frac{L'}{A'} = \rho \frac{n^2 L}{A/n^2}\). This requires careful fraction handling: \(R' = \rho \frac{n^2 L}{1} \times \frac{n^2}{A} = \rho \frac{n^4 L}{A}\). Oh, this formula \(R' = \rho \frac{n^4 L}{A}\) seems incorrect. Let's re-derive the relationship between \(R\) and \(R'\) directly using the radius change.

Original resistance: \(R = \rho \frac{L}{\pi r^2}\). Thus, \(L = \frac{R \pi r^2}{\rho}\).

New resistance: \(R' = \rho' \frac{L'}{\pi (r')^2}\). We know \(\rho' = \rho\) and \(r' = r/2\).

From volume conservation, \(\pi r^2 L = \pi (r')^2 L'\). Substitute \(r' = r/2\):

Substitute \(L' = 4L\) into the equation for \(R'\):

Now substitute \(L = \frac{R \pi r^2}{\rho}\) into the equation for \(R'\):

Cancel \(\rho\), \(\pi\), and \(r^2\):

Okay, the derivation \(R' = 16R\) is correct and consistent.

The relation \(R' = n^2 R\) is valid when the length is stretched to \(n\) times its original length. The relation \(R' = n^4 R\) is valid when the radius is reduced to \(1/n\) times its original value (so \(r' = r/n\)). In this problem, \(r' = r/2\), so \(n=2\). Using \(R' = n^4 R\) with \(n=2\) gives \(R' = 2^4 R = 16R\). This formula is a shortcut if you remember it, but deriving it from volume conservation is more fundamental.

The power relationship \(P' = (1/16)P\) directly follows from \(P = V^2/R\) and \(R' = 16R\), keeping \(V\) constant. If the current were constant instead of voltage, the power relationship would be \(P = I^2 R\), so \(P' = I^2 R' = I^2 (16R) = 16 (I^2 R) = 16P\). The question specifies power rating at V volts, meaning voltage is constant.