Two resistances `R_1` and another `R_2` of the same material but twice the length and half the thickness are connected in series with a standard battery E of internal resistance r. The balancing point is-

Options

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

Explanation

To solve the problem, we need to find the balancing point for two resistances \( R_1 \) and \( R_2 \) connected in series with a battery of EMF \( E \) and internal resistance \( r \). The resistances are made of the same material, but \( R_2 \) is twice the length and half the thickness of \( R_1 \).

### Step-by-Step Solution:

1. **Determine the Resistance of Each 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.

2. **Calculate the Cross-Sectional Area**:
The cross-sectional area \( A \) of a wire with radius \( r \) is:
\[ A = \pi r^2 \]
If the thickness is halved, the new radius \( r_2 \) becomes \( \frac{r_1}{2} \), hence:
\[ A_2 = \pi \left(\frac{r_1}{2}\right)^2 = \frac{\pi r_1^2}{4} = \frac{A_1}{4} \]

3. **Express the Resistances**:
For \( R_1 \):
\[ R_1 = \frac{\rho L_1}{A_1} \]
For \( R_2 \) (twice the length and half the thickness):
\[ R_2 = \frac{\rho (2L_1)}{\frac{A_1}{4}} = \frac{8\rho L_1}{A_1} = 8R_1 \]

4. **Total Resistance in Series**:
The total resistance \( R_t \) in the circuit is:
\[ R_t = R_1 + R_2 + r = R_1 + 8R_1 + r = 9R_1 + r \]

5. **Potential Gradient**:
The potential gradient \( \phi \) is given by:
\[ \phi = \frac{V}{L} \]
where \( V \) is the potential difference across the resistances.

6. **Balancing Point**:
We can set up the equations for the potential drops across \( R_1 \) and \( R_2 \) when the galvanometer shows zero deflection (balancing point):
\[ \phi L_1 = \frac{E R_1}{9R_1 + r} \]
\[ \phi L_2 = \frac{E (8R_1)}{9R_1 + r} \]

7. **Setting Up the Ratio**:
From the balancing point condition:
\[ \frac{\phi L_1}{\phi L_2} = \frac{L_1}{L_2} = \frac{R_1}{8R_1} = \frac{1}{8} \]
Thus, \( L_2 = 8L_1 \).

8. **Conclusion**:
The balancing point occurs at a distance \( L_1 \) from one end, which corresponds to \( \frac{1}{8} \) of the total length \( L \) of the wire.

### Final Answer:
The balancing point is at \( \frac{1}{8}L \).