A battery of e.m.f. 10 V and internal resistance `2 Omega` is connected in primary circuit with a uniform potentiometer wire and a rheostat whose resistance is fixed at `998 Omega`.A battery of unknown e.m.f. is being balanced on this potentiometer wire and balancing length is found to be 50 cm.When area of cross section of potentiometer wire is doubled, then balancing length is found to be 75 cm.
(i)Calculate e.m.f. of the battery.
(ii)Calculate resistivity of potentiometer wire if length of wire is 100 cm and area of cross-section (initially) is `100 cm^2`.

Answer

To solve the problem step by step, we will follow the outlined procedure:

### Step 1: Understand the given data
- E.M.F. of the primary battery (E) = 10 V
- Internal resistance of the battery (r) = 2 Ω
- Fixed resistance of the rheostat (R) = 998 Ω
- Balancing length for the unknown battery (L1) = 50 cm = 0.5 m
- Balancing length when area of cross-section is doubled (L2) = 75 cm = 0.75 m
- Length of potentiometer wire (L) = 100 cm = 1 m
- Initial area of cross-section (A) = 100 cm² = 1 × 10⁻² m²

### Step 2: Calculate the total resistance in the circuit
The total resistance in the primary circuit is given by:
\[ R_{total} = R + r = 998 \, \Omega + 2 \, \Omega = 1000 \, \Omega \]

### Step 3: Calculate the potential difference across the potentiometer wire
Using Ohm's law, the potential difference (V) across the potentiometer wire can be calculated as:
\[ V = E \times \frac{R}{R + r} \]
Substituting the values:
\[ V = 10 \, V \times \frac{998 \, \Omega}{1000 \, \Omega} = 9.98 \, V \]

### Step 4: Relate the balancing lengths to the E.M.F. of the unknown battery
The potential difference across the potentiometer wire is proportional to the balancing length:
\[ \frac{E_{unknown}}{V} = \frac{L_{unknown}}{L} \]
Where \( L \) is the total length of the potentiometer wire.

For the first balancing length:
\[ E_{unknown} = V \times \frac{L_1}{L} = 9.98 \, V \times \frac{0.5 \, m}{1 \, m} = 4.99 \, V \]

### Step 5: Calculate the new potential difference when the area is doubled
When the area is doubled, the resistance of the wire is halved. Thus, the new total resistance becomes:
\[ R_{new} = \frac{R}{2} + r = \frac{998 \, \Omega}{2} + 2 \, \Omega = 499 \, \Omega + 2 \, \Omega = 501 \, \Omega \]

The new potential difference across the potentiometer wire is:
\[ V_{new} = E \times \frac{R_{new}}{R_{new} + r} \]
Substituting the values:
\[ V_{new} = 10 \, V \times \frac{501 \, \Omega}{503 \, \Omega} \approx 9.95 \, V \]

Using the new balancing length:
\[ E_{unknown} = V_{new} \times \frac{L_2}{L} = 9.95 \, V \times \frac{0.75 \, m}{1 \, m} = 7.4625 \, V \]

### Step 6: Calculate the resistivity of the potentiometer wire
Using the formula for resistivity:
\[ \rho = R \times \frac{A}{L} \]
Where:
- \( R = 1000 \, \Omega \)
- \( A = 1 \times 10^{-2} \, m^2 \)
- \( L = 1 \, m \)

Substituting the values:
\[ \rho = 1000 \, \Omega \times \frac{1 \times 10^{-2} \, m^2}{1 \, m} = 10 \, \Omega \cdot m \]

### Final Answers:
(i) The E.M.F. of the unknown battery is approximately **4.99 V**.

(ii) The resistivity of the potentiometer wire is **10 Ω·m**.