A resistor of 25 cm length and 4 ohm is stretched to a uniform wire of 50 cm length. The resistance now is

Options

A. `1 Omega`
B. `12 Omega`
C. `16 Omega`
D. `20 Omega`

Answer

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

### Step 1: Understand the initial conditions
We have a resistor of length \( L_1 = 25 \, \text{cm} \) and resistance \( R_1 = 4 \, \Omega \).

### Step 2: Identify the final conditions after stretching
After stretching, the new length of the wire is \( L_2 = 50 \, \text{cm} \). We need to find the new resistance \( R_2 \).

### Step 3: Use the formula for resistance
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.

### Step 4: Relate the initial and final conditions
Since the wire is stretched uniformly, the volume of the wire remains constant. Therefore, we can write:
\[ A_1 L_1 = A_2 L_2 \]
where \( A_1 \) and \( A_2 \) are the cross-sectional areas before and after stretching, respectively.

### Step 5: Substitute known values
Substituting the known values:
\[ A_1 \cdot 25 = A_2 \cdot 50 \]
This gives:
\[ A_2 = \frac{A_1 \cdot 25}{50} = \frac{A_1}{2} \]

### Step 6: Relate the resistances
Using the relationship of resistances:
\[ \frac{R_1}{R_2} = \frac{L_1}{L_2} \cdot \frac{A_2}{A_1} \]
Substituting the values we have:
\[ \frac{4}{R_2} = \frac{25}{50} \cdot \frac{A_2}{A_1} \]
Since \( A_2 = \frac{A_1}{2} \), we can substitute this into the equation:
\[ \frac{R_1}{R_2} = \frac{25}{50} \cdot \frac{1/2}{1} = \frac{25}{100} = \frac{1}{4} \]

### Step 7: Solve for \( R_2 \)
From the equation:
\[ \frac{4}{R_2} = \frac{1}{4} \]
Cross-multiplying gives:
\[ 4 = \frac{R_2}{4} \]
Thus:
\[ R_2 = 4 \cdot 4 = 16 \, \Omega \]

### Final Answer
The new resistance \( R_2 \) after stretching the wire is \( 16 \, \Omega \).

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