`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`

Answer

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

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

### Step 2: Define the length of wire R
According to the problem, the length of wire R is twice the length of wire P:
\[ L_R = 2L \]

### Step 3: Calculate the area of wire R
Since wire R is made from part P, we need to consider the volume conservation. The volume of wire P is equal to the volume of wire R:
\[ \text{Volume of P} = \text{Volume of R} \]
\[ A_P \cdot L = A_R \cdot L_R \]
Substituting \( L_R = 2L \):
\[ A_P \cdot L = A_R \cdot 2L \]
Cancelling \( L \) from both sides (assuming \( L \neq 0 \)):
\[ A_P = 2A_R \]
Thus, the area of wire R is:
\[ A_R = \frac{A_P}{2} \]

### Step 4: Calculate the resistance of wires P and R
The resistance \( R \) of a wire is given by the formula:
\[ R = \frac{\rho L}{A} \]
Where \( \rho \) is the resistivity, \( L \) is the length, and \( A \) is the cross-sectional area.

#### Resistance of wire P:
\[ R_P = \frac{\rho L}{A_P} \]

#### Resistance of wire R:
Substituting \( L_R = 2L \) and \( A_R = \frac{A_P}{2} \):
\[ R_R = \frac{\rho (2L)}{A_R} = \frac{\rho (2L)}{\frac{A_P}{2}} = \frac{4\rho L}{A_P} \]

### Step 5: Find the ratio of the resistances \( R_P \) and \( R_R \)
Now we can find the ratio of the resistances:
\[ \frac{R_P}{R_R} = \frac{\frac{\rho L}{A_P}}{\frac{4\rho L}{A_P}} = \frac{1}{4} \]

### Step 6: Find the ratio of lengths \( P \) and \( Q \)
From our earlier definition, we have:
\[ L_Q = 1 - L \]
Now we can express the ratio of lengths \( P \) and \( Q \):
\[ \frac{L_P}{L_Q} = \frac{L}{1 - L} \]
Since we know the resistance of wire R is equal to the resistance of wire Q, we can set up the equation:
\[ R_R = R_Q \]
Substituting the expressions for resistance:
\[ \frac{4\rho L}{A_P} = \frac{\rho (1 - L)}{A_Q} \]
Assuming the areas are equal (for simplicity), we can cancel \( \rho \) and rearrange:
\[ 4L = 1 - L \]
Solving for \( L \):
\[ 5L = 1 \implies L = \frac{1}{5} \]
Thus, the length of Q is:
\[ L_Q = 1 - L = 1 - \frac{1}{5} = \frac{4}{5} \]
Now we can find the ratio:
\[ \frac{L_P}{L_Q} = \frac{\frac{1}{5}}{\frac{4}{5}} = \frac{1}{4} \]

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