A uniform wire of resistance `=3Omega` and length `l` is stretched to double its length. Now it is bent to form a circular loop and two point `P & Q` lies on the loop such that they subtend `60^(@)` angle at centre. The equivalent resistance between two point P & Q is:

Options

A. `(5)/(3)Omega`
B. `12Omega`
C. `(3)/(5)Omega`
D. `(1)/(12)Omega`

Answer

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

### Step 1: Determine the new length and area of the wire after stretching
The original length of the wire is \( L \) and the original resistance is \( R = 3 \, \Omega \). When the wire is stretched to double its length, the new length \( L' \) becomes:
\[ L' = 2L \]
Since the volume of the wire remains constant, we have:
\[ A \cdot L = A' \cdot L' \]
where \( A \) is the original cross-sectional area and \( A' \) is the new cross-sectional area. Therefore:
\[ A \cdot L = A' \cdot (2L) \implies A' = \frac{A}{2} \]

### Step 2: Calculate the new resistance of the stretched wire
The resistance \( R' \) of the wire after stretching can be calculated using the formula:
\[ R' = \rho \frac{L'}{A'} \]
Substituting \( L' = 2L \) and \( A' = \frac{A}{2} \):
\[ R' = \rho \frac{2L}{\frac{A}{2}} = \rho \frac{4L}{A} \]
Since the original resistance \( R \) is given by:
\[ R = \rho \frac{L}{A} = 3 \, \Omega \]
we can substitute this into the equation for \( R' \):
\[ R' = 4 \cdot R = 4 \cdot 3 = 12 \, \Omega \]

### Step 3: Form a circular loop and determine the arc length
When the wire is bent to form a circular loop, the total length of the wire is equal to the circumference of the circle:
\[ C = 2\pi r = 2L \]
From this, we can find the radius \( r \):
\[ r = \frac{L}{\pi} \]

### Step 4: Calculate the length of the arc between points P and Q
The angle subtended by points P and Q at the center is \( 60^\circ \) or \( \frac{\pi}{3} \) radians. The length of the arc \( PQ \) can be calculated as:
\[ \text{Length of arc PQ} = r \cdot \theta = r \cdot \frac{\pi}{3} = \frac{L}{\pi} \cdot \frac{\pi}{3} = \frac{L}{3} \]

### Step 5: Calculate the resistance of the arc PQ
The resistance \( R_1 \) of the arc \( PQ \) is proportional to its length:
\[ R_1 = R' \cdot \frac{\text{Length of arc PQ}}{C} = 12 \, \Omega \cdot \frac{\frac{L}{3}}{2L} = 12 \, \Omega \cdot \frac{1}{6} = 2 \, \Omega \]

### Step 6: Calculate the resistance of the remaining part of the loop
The remaining length of the wire (the rest of the loop) is:
\[ \text{Remaining length} = 2L - \frac{L}{3} = \frac{6L}{3} - \frac{L}{3} = \frac{5L}{3} \]
The resistance \( R_2 \) of this remaining part is:
\[ R_2 = R' \cdot \frac{\text{Remaining length}}{C} = 12 \, \Omega \cdot \frac{\frac{5L}{3}}{2L} = 12 \, \Omega \cdot \frac{5}{6} = 10 \, \Omega \]

### Step 7: Calculate the equivalent resistance between points P and Q
The resistances \( R_1 \) and \( R_2 \) are in parallel:
\[ \frac{1}{R_{\text{eq}}} = \frac{1}{R_1} + \frac{1}{R_2} = \frac{1}{2} + \frac{1}{10} \]
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 points P and Q is:
\[ \boxed{\frac{5}{3} \, \Omega} \]