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 can use the relationship between resistance, length, and cross-sectional area of a wire.

### Step 1: Understand the relationship between resistance and length
The resistance \( R \) of a wire is given by the formula:
\[ R = \rho \frac{L}{A} \]
where:
- \( R \) is the resistance,
- \( \rho \) is the resistivity of the material,
- \( L \) is the length of the wire,
- \( A \) is the cross-sectional area.

### Step 2: Set up the initial conditions
Let the initial length of the wire be \( L \) and the initial resistance be \( x \) ohms. So, we have:
\[ x = \rho \frac{L}{A} \]

### Step 3: Analyze the new conditions
When the wire is drawn out to double its length, the new length becomes \( 2L \). The new resistance is given as \( 20 \) ohms. The new resistance can be expressed as:
\[ 20 = \rho \frac{2L}{A'} \]
where \( A' \) is the new cross-sectional area.

### Step 4: Relate the areas
When the length of the wire is doubled, the volume of the wire remains constant. Thus, we have:
\[ L \cdot A = 2L \cdot A' \]
From this, we can derive:
\[ A' = \frac{A}{2} \]

### Step 5: Substitute \( A' \) into the new resistance equation
Substituting \( A' \) into the new resistance equation gives:
\[ 20 = \rho \frac{2L}{A/2} = \rho \frac{4L}{A} \]

### Step 6: Relate the two resistance equations
Now we have two expressions for resistance:
1. \( x = \rho \frac{L}{A} \)
2. \( 20 = \rho \frac{4L}{A} \)

From the second equation, we can express \( \rho \frac{L}{A} \):
\[ \rho \frac{L}{A} = \frac{20}{4} = 5 \]

### Step 7: Set the two equations equal
Now, we can set the two expressions for resistance equal to each other:
\[ x = 5 \]

### Conclusion
Thus, the initial resistance \( x \) is \( 10 \) ohms.

### Final Answer
The value of \( x \) is \( 10 \, \Omega \).

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