A wire of resistance x ohm is draw out, so that its length in increased to twice its original length, and its new resistance becomes 20 `Omega` then x will be .

Options

A. `5 Omega`
B. `10 Omega`
C. `15 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 relationship between resistance, length, and area
The resistance \( R \) of a wire is given by the formula:
\[ R = \frac{\rho L}{A} \]
where \( \rho \) is the resistivity of the material, \( L \) is the length of the wire, and \( A \) is the cross-sectional area.

### Step 2: Define the initial conditions
Let the initial length of the wire be \( L \) and its initial resistance be \( x \) ohms. Thus, we can write:
\[ x = \frac{\rho L}{A} \]

### Step 3: Define the new conditions after the wire is drawn out
When the wire is drawn out to twice its original length, the new length \( L' \) becomes:
\[ L' = 2L \]
The new resistance \( R' \) is given as 20 ohms. Therefore, we have:
\[ R' = \frac{\rho L'}{A'} \]

### Step 4: Relate the new area to the original area
Since the volume of the wire remains constant when it is drawn out, we can express the relationship between the original and new areas. The volume \( V \) is given by:
\[ V = L \cdot A = L' \cdot A' \]
Substituting \( L' = 2L \):
\[ L \cdot A = 2L \cdot A' \]
From this, we can derive:
\[ A' = \frac{A}{2} \]

### Step 5: Substitute the new length and area into the resistance formula
Now substituting \( L' \) and \( A' \) into the resistance formula for the new resistance:
\[ R' = \frac{\rho (2L)}{\frac{A}{2}} = \frac{2\rho L}{\frac{A}{2}} = \frac{4\rho L}{A} \]
Thus, we can express the new resistance \( R' \) in terms of the original resistance \( x \):
\[ R' = 4 \cdot \frac{\rho L}{A} = 4x \]

### Step 6: Set the new resistance equal to the given value
We know that \( R' = 20 \) ohms, so we set up the equation:
\[ 4x = 20 \]

### Step 7: Solve for \( x \)
Now, we can solve for \( x \):
\[ x = \frac{20}{4} = 5 \text{ ohms} \]

### Conclusion
Thus, the original resistance \( x \) is 5 ohms.