The resistance of a 50 cm long wire is `10 Omega`. The wire is stretched of uniform wire of length 100 cm. The resistance now will be

Options

A. `15 Omega`
B. `30 Omega`
C. `20 Omega`
D. `40 Omega`

Explanation

To solve the problem, we need to determine the new resistance of a wire after it has been stretched. Here’s a step-by-step solution:

### Step 1: Understand the initial conditions
We are given:
- Initial length of the wire, \( L_1 = 50 \, \text{cm} = 0.5 \, \text{m} \)
- Initial resistance of the wire, \( R_1 = 10 \, \Omega \)

### Step 2: Identify the new length of the wire
The wire is stretched to a new length:
- New length of the wire, \( L_2 = 100 \, \text{cm} = 1.0 \, \text{m} \)

### Step 3: Use the concept of volume conservation
When the wire is stretched, its volume remains constant. The volume of the wire can be expressed as:
\[ V = A_1 \cdot L_1 = A_2 \cdot L_2 \]
where \( A_1 \) and \( A_2 \) are the cross-sectional areas of the wire before and after stretching, respectively.

### Step 4: Relate the areas and lengths
From the volume conservation:
\[ A_1 \cdot L_1 = A_2 \cdot L_2 \]
Rearranging gives:
\[ A_2 = \frac{A_1 \cdot L_1}{L_2} \]

### Step 5: Calculate the new area
Substituting the known values:
\[ A_2 = \frac{A_1 \cdot 0.5}{1.0} = 0.5 A_1 \]

### Step 6: Relate the resistances
The resistance of a wire is given by the formula:
\[ R = \rho \frac{L}{A} \]
where \( \rho \) is the resistivity of the material. Therefore, the initial and new resistances can be expressed as:
\[ R_1 = \rho \frac{L_1}{A_1} \]
\[ R_2 = \rho \frac{L_2}{A_2} \]

### Step 7: Substitute for \( R_2 \)
Substituting \( A_2 \) into the equation for \( R_2 \):
\[ R_2 = \rho \frac{L_2}{0.5 A_1} \]
\[ R_2 = 2 \cdot \rho \frac{L_2}{A_1} \]
Since \( L_2 = 1.0 \, \text{m} \), we can express \( R_2 \) in terms of \( R_1 \):
\[ R_2 = 2 \cdot \frac{L_2}{L_1} \cdot R_1 \]
Substituting \( L_2 = 1.0 \, \text{m} \) and \( L_1 = 0.5 \, \text{m} \):
\[ R_2 = 2 \cdot \frac{1.0}{0.5} \cdot 10 \]
\[ R_2 = 2 \cdot 2 \cdot 10 = 40 \, \Omega \]

### Final Answer
The new resistance of the wire after being stretched to 100 cm is \( R_2 = 40 \, \Omega \).

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