A metal wire of resistance `3 omega` is elongated to make a unform 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 centre, the equivalent resistance between these two points will be :

Options

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

Answer

To solve the problem step by step, we will follow the reasoning provided in the video transcript.

### Step 1: Determine the new resistance after elongation
The original resistance of the wire is given as \( R = 3 \, \Omega \). When the wire is elongated to double its length, the new length \( L' \) becomes \( 2L \). Since the volume of the wire remains constant, the cross-sectional area \( A' \) must change.

Using the formula for resistance:
\[ R = \rho \frac{L}{A} \]
If the length doubles, the area must halve to keep the volume constant:
\[ A' = \frac{A}{2} \]

Now, substituting the new values into the resistance formula:
\[ R' = \rho \frac{2L}{A/2} = \rho \frac{4L}{A} = 4R \]
Since \( R = 3 \, \Omega \):
\[ R' = 4 \times 3 = 12 \, \Omega \]

### Step 2: Understand the circular wire and angle
The wire is then bent to form a circle. The total resistance of the circular wire is \( 12 \, \Omega \). When two points on the circle make an angle of \( 60^\circ \) at the center, we need to find the equivalent resistance between these two points.

### Step 3: Calculate the lengths of the wire segments
The angle \( 60^\circ \) corresponds to \( \frac{1}{6} \) of the total circle (since \( 360^\circ \) is a full circle):
\[ \text{Length of arc PQ} = \frac{60}{360} \times \text{Circumference} = \frac{1}{6} \times 2\pi r = \frac{\pi r}{3} \]

The remaining arc (the rest of the circle) will be:
\[ \text{Remaining arc} = 2\pi r - \frac{\pi r}{3} = \frac{6\pi r}{3} - \frac{\pi r}{3} = \frac{5\pi r}{3} \]

### Step 4: Calculate the resistances of the segments
The resistance is proportional to the length of the wire:
- For the arc \( PQ \) (length \( \frac{\pi r}{3} \)):
\[ R_{PQ} = \frac{12 \, \Omega}{2\pi r} \times \frac{\pi r}{3} = \frac{12}{6} = 2 \, \Omega \]

- For the remaining arc (length \( \frac{5\pi r}{3} \)):
\[ R_{\text{remaining}} = \frac{12 \, \Omega}{2\pi r} \times \frac{5\pi r}{3} = \frac{12 \times 5}{6} = 10 \, \Omega \]

### Step 5: Calculate the equivalent resistance
The two resistances \( R_{PQ} \) and \( R_{\text{remaining}} \) are in parallel:
\[ \frac{1}{R_{\text{eq}}} = \frac{1}{R_{PQ}} + \frac{1}{R_{\text{remaining}}} = \frac{1}{2} + \frac{1}{10} \]

Finding a common denominator (10):
\[ \frac{1}{R_{\text{eq}}} = \frac{5}{10} + \frac{1}{10} = \frac{6}{10} = \frac{3}{5} \]
Thus,
\[ R_{\text{eq}} = \frac{5}{3} \, \Omega \]

### Final Answer
The equivalent resistance between the two points on the circle is:
\[ \boxed{\frac{5}{3} \, \Omega} \]