A resistance wire has a resistance R. Half of this wire is stretched to double its length and half is twisted so double its thickness, then its new resistance becomes

Options

A. `(17R)/(8)`
B. `(17R)/(16)`
C. `(65R)/(32)`
D. `(65R)/(16)`

Explanation

To solve the problem, we will analyze the changes in resistance for both parts of the wire separately and then combine the results.

### Step-by-Step Solution:

1. **Understanding the Initial Resistance**:
- Let the initial resistance of the wire be \( R \).
- The resistance \( R \) can be expressed using 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.

2. **Dividing the Wire**:
- The wire is divided into two halves:
- Half of the wire (length \( L/2 \)) will be stretched to double its length.
- The other half (length \( L/2 \)) will be twisted to double its thickness.

3. **Analyzing the First Half (Stretched)**:
- When the first half is stretched to double its length, its new length \( L_1 \) becomes:
\[ L_1 = 2 \times \frac{L}{2} = L \]
- The volume of the wire remains constant, so:
\[ A \cdot \frac{L}{2} = A_1 \cdot L_1 \]
\[ A \cdot \frac{L}{2} = A_1 \cdot L \implies A_1 = \frac{A}{2} \]
- The resistance of this part \( R_1 \) is given by:
\[ R_1 = \frac{\rho L_1}{A_1} = \frac{\rho L}{\frac{A}{2}} = \frac{2\rho L}{A} = 2R \]

4. **Analyzing the Second Half (Twisted)**:
- The second half is twisted, which doubles its thickness. Since thickness is doubled, the area becomes:
\[ A_2 = 2A \]
- The length of this half remains \( L/2 \), so the new length \( L_2 \) is:
\[ L_2 = \frac{L}{2} \]
- The resistance of this part \( R_2 \) is given by:
\[ R_2 = \frac{\rho L_2}{A_2} = \frac{\rho \cdot \frac{L}{2}}{2A} = \frac{\rho L}{4A} = \frac{R}{4} \]

5. **Combining the Resistances**:
- Since both parts are in series, the total resistance \( R_{total} \) is:
\[ R_{total} = R_1 + R_2 = 2R + \frac{R}{4} \]
- To combine these, convert \( 2R \) to a fraction:
\[ 2R = \frac{8R}{4} \]
- Therefore:
\[ R_{total} = \frac{8R}{4} + \frac{R}{4} = \frac{9R}{4} \]

### Final Answer:
The new resistance of the wire after stretching and twisting is:
\[ R_{total} = \frac{9R}{4} \]