`1` meter long metallic wire is broken into two unequal parts `P and Q P` part of the wire in uniformly extended into another wire `R`. Length of `R` is twice the length of `P` and the resistance of `R` is equal to that of `Q`. Find the ratio of the resistance `P` and `R` and also the ratio of the length `P and Q`

Explanation

To solve the problem step by step, we will use the concepts of resistance and the relationship between length, area, and resistance.

### Step 1: Define the lengths of the wire parts
Let the length of wire part \( P \) be \( L \) meters. Since the total length of the wire is 1 meter, the length of wire part \( Q \) will be:
\[ L_Q = 1 - L \]

### Step 2: Define the relationship of wire \( R \)
The wire \( R \) is formed by extending part \( P \). The length of wire \( R \) is given as twice the length of \( P \):
\[ L_R = 2L \]

### Step 3: Determine the area relationships
Since the volume of the wire remains constant during extension, we have:
\[ L_P \cdot A_P = L_R \cdot A_R \]
Substituting the values:
\[ L \cdot A_P = 2L \cdot A_R \]
This simplifies to:
\[ A_R = \frac{A_P}{2} \]

### Step 4: Calculate the resistances
The resistance \( R_P \) of part \( P \) is given by:
\[ R_P = \frac{\rho L}{A_P} \]
The resistance \( R_Q \) of part \( Q \) is:
\[ R_Q = \frac{\rho (1 - L)}{A} \]
And the resistance \( R_R \) of wire \( R \) is:
\[ R_R = \frac{\rho (2L)}{A_R} = \frac{\rho (2L)}{\frac{A_P}{2}} = \frac{4\rho L}{A_P} \]

### Step 5: Set the resistances equal
According to the problem, the resistance of wire \( R \) is equal to the resistance of wire \( Q \):
\[ R_R = R_Q \]
Substituting the expressions we derived:
\[ \frac{4\rho L}{A_P} = \frac{\rho (1 - L)}{A} \]
Cancelling \( \rho \) and rearranging gives:
\[ 4L \cdot A = (1 - L) \cdot A_P \]

### Step 6: Express \( A_P \) in terms of \( A \)
From the area relationship we found earlier, \( A_P = 2A_R \). Therefore:
\[ A_P = 2 \cdot \frac{A_P}{2} = A \]
Substituting this back:
\[ 4L \cdot A = (1 - L) \cdot A \]
Dividing both sides by \( A \) (assuming \( A \neq 0 \)):
\[ 4L = 1 - L \]
Solving for \( L \):
\[ 5L = 1 \implies L = \frac{1}{5} = 0.2 \text{ meters} \]

### Step 7: Find the length of \( Q \)
Now we can find the length of \( Q \):
\[ L_Q = 1 - L = 1 - 0.2 = 0.8 \text{ meters} \]

### Step 8: Calculate the ratio of the lengths
The ratio of the lengths \( P \) and \( Q \) is:
\[ \frac{L_P}{L_Q} = \frac{0.2}{0.8} = \frac{1}{4} \]

### Step 9: Calculate the ratio of the resistances
The resistance of \( R \) is:
\[ R_R = \frac{4\rho L}{A_P} \]
And the resistance of \( P \) is:
\[ R_P = \frac{\rho L}{A_P} \]
The ratio of the resistances \( R_P \) and \( R_R \) is:
\[ \frac{R_P}{R_R} = \frac{\frac{\rho L}{A_P}}{\frac{4\rho L}{A_P}} = \frac{1}{4} \]

### Final Results
1. The ratio of the lengths \( P \) and \( Q \) is \( \frac{1}{4} \).
2. The ratio of the resistances \( P \) and \( R \) is \( \frac{1}{4} \).