A metal wire of resistance `3Omega` is elongated to make a uniform wire of double its previous length. This new wire is now bent and the ends joined to make a circle. If two points on this circle make an angle `60^(@)` at the center, the equivalent resistance between these two points will be :

Options

A. `(5)/(3)Omega`
B. `(12)/(5)Omega`
C. `(7)/(2)Omega`
D. `(5)/(2)Omega`

Answer

To solve the problem step by step, let's break it down:

### Step 1: Understand the Initial Resistance
The initial resistance of the metal wire is given as \( R = 3 \, \Omega \).

### Step 2: Determine the New Length and Area
When the wire is elongated to double its length, the new length \( L' \) becomes:
\[ L' = 2L \]
Since the volume of the wire is conserved, we can express the relationship between the initial and final lengths and areas:
\[ A \cdot L = A' \cdot L' \]
Substituting \( L' \):
\[ A \cdot L = A' \cdot (2L) \]
From this, we can derive the new area \( A' \):
\[ A' = \frac{A}{2} \]

### Step 3: Calculate the New Resistance
Using the formula for resistance:
\[ R = \rho \frac{L}{A} \]
For the new wire, the resistance \( R' \) is:
\[ R' = \rho \frac{L'}{A'} = \rho \frac{2L}{A/2} = \rho \frac{4L}{A} \]
Since the initial resistance \( R = \rho \frac{L}{A} = 3 \, \Omega \), we find:
\[ R' = 4R = 4 \times 3 \, \Omega = 12 \, \Omega \]

### Step 4: Forming a Circle
The new wire of resistance \( 12 \, \Omega \) is bent into a circle. The total length of the wire is now the circumference of the circle:
\[ C = 2\pi r = 2L' = 2 \times 2L = 4L \]

### Step 5: Determine the Length of the Arc
If two points on the circle make an angle of \( 60^\circ \) at the center, the fraction of the circle corresponding to this angle is:
\[ \text{Fraction} = \frac{60^\circ}{360^\circ} = \frac{1}{6} \]
Thus, the length of the arc \( L_{\text{arc}} \) between these two points is:
\[ L_{\text{arc}} = \frac{1}{6} \times C = \frac{1}{6} \times 4L = \frac{2L}{3} \]

### Step 6: Calculate the Equivalent Resistance
Since the wire is uniform, the resistance of the arc can be calculated using:
\[ R_{\text{arc}} = R' \times \frac{L_{\text{arc}}}{C} \]
Substituting the values:
\[ R_{\text{arc}} = 12 \, \Omega \times \frac{\frac{2L}{3}}{4L} = 12 \, \Omega \times \frac{1}{6} = 2 \, \Omega \]

### Final Answer
The equivalent resistance between the two points on the circle is:
\[ \boxed{2 \, \Omega} \]