The length of a potentiometer wire is 10 m and carries steady current on connecting the sliding jockey to standard cell of 1.018 volt, the null point is obtained at the distance of 850 cm.
If the length of potentiowire is doubled, then the e.m.f. will be:

Options

A. halved
B. doubled
C. not change
D. none of these

Explanation

To solve the problem, we need to understand the relationship between the electromotive force (e.m.f.) of the standard cell, the length of the potentiometer wire, and the potential gradient.

### Step-by-Step Solution:

1. **Identify the Given Values:**
- Length of the potentiometer wire (L₁) = 10 m = 1000 cm
- E.m.f. of the standard cell (E) = 1.018 V
- Null point distance (d) = 850 cm

2. **Calculate the Potential Gradient (k):**
The potential gradient (k) is defined as the e.m.f. per unit length of the wire. It can be calculated using the formula:
\[ k = \frac{E}{L} \]
Here, E is the e.m.f. and L is the length of the wire.
\[ k = \frac{1.018 \text{ V}}{1000 \text{ cm}} = 0.001018 \text{ V/cm} \]

3. **Doubling the Length of the Potentiometer Wire:**
If the length of the potentiometer wire is doubled, the new length (L₂) will be:
\[ L₂ = 2 \times L₁ = 2 \times 1000 \text{ cm} = 2000 \text{ cm} \]

4. **Calculate the New E.m.f. (E₂):**
Since the potential gradient remains constant (as the current is steady), the new e.m.f. can be calculated using the same formula:
\[ E₂ = k \times L₂ \]
Substitute the values:
\[ E₂ = 0.001018 \text{ V/cm} \times 2000 \text{ cm} = 2.036 \text{ V} \]

5. **Conclusion:**
The e.m.f. will be doubled when the length of the potentiometer wire is doubled. Therefore, the answer is:
\[ E₂ = 2 \times 1.018 \text{ V} = 2.036 \text{ V} \]

### Final Answer:
The e.m.f. will be doubled.

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