To solve the problem step by step, we will follow these steps:
### Step 1: Understand the initial conditions
We have a conductor with an initial resistance \( R = 3 \, \Omega \). When the conductor is stretched uniformly, its length is doubled.
### Step 2: Determine the new resistance after stretching
When the length of the conductor is doubled, the cross-sectional area \( A \) will decrease. Since the volume of the conductor remains constant, we can use the relationship:
\[
\text{Volume} = \text{Length} \times \text{Area}
\]
Let the original length be \( L \) and the original area be \( A \). The new length is \( 2L \) and the new area \( A' \) can be calculated as:
\[
L \cdot A = 2L \cdot A' \implies A' = \frac{A}{2}
\]
### Step 3: Calculate the new resistance
The resistance \( R' \) of the conductor can be calculated using the formula:
\[
R' = \rho \frac{L'}{A'}
\]
Substituting the new length and area:
\[
R' = \rho \frac{2L}{A/2} = \rho \frac{4L}{A} = 4R
\]
Since the original resistance \( R = 3 \, \Omega \):
\[
R' = 4 \times 3 \, \Omega = 12 \, \Omega
\]
### Step 4: Form the equilateral triangle
Now, the wire with resistance \( 12 \, \Omega \) is bent into the form of an equilateral triangle. Each side of the triangle will have equal resistance. Since the total length of the wire is now \( 2L \), and it is divided into 3 equal sides, the resistance of each side \( R_{\text{side}} \) is:
\[
R_{\text{side}} = \frac{R'}{3} = \frac{12 \, \Omega}{3} = 4 \, \Omega
\]
### Step 5: Calculate the effective resistance between two points
To find the effective resistance between the ends of any side of the triangle, we note that when measuring between two vertices of the triangle, we have two resistors of \( 4 \, \Omega \) in parallel (the two other sides of the triangle). The formula for the equivalent resistance \( R_{\text{eq}} \) of two resistors \( R_1 \) and \( R_2 \) in parallel is given by:
\[
\frac{1}{R_{\text{eq}}} = \frac{1}{R_1} + \frac{1}{R_2}
\]
Substituting \( R_1 = R_2 = 4 \, \Omega \):
\[
\frac{1}{R_{\text{eq}}} = \frac{1}{4} + \frac{1}{4} = \frac{2}{4} = \frac{1}{2}
\]
Thus,
\[
R_{\text{eq}} = 2 \, \Omega
\]
### Final Answer
The effective resistance between the ends of any side of the triangle is \( 2 \, \Omega \).
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