A mete bridge is used to determine the resistance of an unknown wire by measuring the balance point length l. If the wire is replaced by another wire of same material but with double the length and half the thickness the balancing point is expected to be

Options

A. `1/(8l)`
B. `1/(4l)`
C. `8l`
D. 16l

Answer

To solve the problem, we need to determine the new balancing point length \( L' \) when the wire is replaced by another wire of the same material but with double the length and half the thickness.

### Step-by-Step Solution:

1. **Understanding Resistance of the Original Wire**:
- The resistance \( R \) of a wire is given by the formula:
\[ R = \frac{\rho L}{A} \]
where \( \rho \) is the resistivity, \( L \) is the length, and \( A \) is the cross-sectional area.
- For a wire of diameter \( D \), the radius \( r \) is \( \frac{D}{2} \), and the cross-sectional area \( A \) is:
\[ A = \pi r^2 = \pi \left(\frac{D}{2}\right)^2 = \frac{\pi D^2}{4} \]
- Therefore, the resistance of the original wire can be expressed as:
\[ R = \frac{\rho L}{\frac{\pi D^2}{4}} = \frac{4\rho L}{\pi D^2} \]

2. **Resistance of the New Wire**:
- The new wire has double the length \( L' = 2L \) and half the thickness, which means the new diameter \( D' = \frac{D}{2} \).
- The new radius \( r' \) is \( \frac{D'}{2} = \frac{D}{4} \).
- The cross-sectional area \( A' \) of the new wire is:
\[ A' = \pi \left(\frac{D}{4}\right)^2 = \frac{\pi D^2}{16} \]
- The resistance \( R' \) of the new wire is:
\[ R' = \frac{\rho (2L)}{\frac{\pi D^2}{16}} = \frac{16\rho L}{\pi D^2} \]

3. **Finding the Ratio of Resistances**:
- The ratio of the resistances \( R \) and \( R' \) is:
\[ \frac{R}{R'} = \frac{\frac{4\rho L}{\pi D^2}}{\frac{16\rho L}{\pi D^2}} = \frac{4}{16} = \frac{1}{4} \]

4. **Using the Meter Bridge Principle**:
- According to the meter bridge principle, the ratio of resistances is equal to the ratio of the lengths at the balance point:
\[ \frac{R}{R'} = \frac{L}{L'} \]
- From the previous step, we have \( \frac{R}{R'} = \frac{1}{4} \), thus:
\[ \frac{1}{4} = \frac{L}{L'} \]
- Rearranging gives:
\[ L' = 4L \]

5. **Conclusion**:
- The new balancing point length \( L' \) is \( 4L \).

### Final Answer:
The balancing point is expected to be \( 4L \).