The resistance of a wire of 100 cmlength is `10Omega`. Now, it is cut into 10 equal parts and all of them are twisted to form a single bundle. Its resistances is

Options

A. `1Omega`
B. `0.5Omega`
C. `5Omega`
D. `0.1Omega`

Explanation

To solve the problem, we need to find the resistance of a wire after it has been cut into 10 equal parts and then twisted together to form a single bundle.

### Step-by-Step Solution:

1. **Identify the Initial Resistance and Length**:
- The initial resistance \( R \) of the wire is given as \( 10 \, \Omega \).
- The length \( L \) of the wire is \( 100 \, \text{cm} \).

2. **Cut the Wire into Equal Parts**:
- The wire is cut into 10 equal parts.
- Therefore, the length of each part \( L' \) will be:
\[ L' = \frac{L}{10} = \frac{100 \, \text{cm}}{10} = 10 \, \text{cm} \]

3. **Calculate the Resistance of Each Part**:
- The resistance of a wire is given by the formula:
\[ R = \rho \frac{L}{A} \]
where \( \rho \) is the resistivity, \( L \) is the length, and \( A \) is the cross-sectional area.
- When the wire is cut into 10 parts, the resistance of each part \( R' \) can be calculated as:
\[ R' = \frac{R}{n} = \frac{10 \, \Omega}{10} = 1 \, \Omega \]
- Here, \( n \) is the number of parts (which is 10).

4. **Combine the Resistors in Parallel**:
- When all 10 parts are twisted together, they are effectively connected in parallel.
- The equivalent resistance \( R_{eq} \) of resistors in parallel is given by:
\[ \frac{1}{R_{eq}} = \frac{1}{R_1} + \frac{1}{R_2} + \ldots + \frac{1}{R_n} \]
- Since all resistances are equal:
\[ \frac{1}{R_{eq}} = \frac{10}{R'} = \frac{10}{1 \, \Omega} = 10 \]
- Therefore, the equivalent resistance \( R_{eq} \) is:
\[ R_{eq} = \frac{1}{10} = 0.1 \, \Omega \]

5. **Final Answer**:
- The resistance of the twisted bundle of wires is \( 0.1 \, \Omega \).