To solve the problem step by step, we will analyze the given information and apply the relevant formulas.
### Step 1: Calculate the total resistance of the potentiometer wire
The potentiometer wire has a length of 5 m and a resistance of 3 Ω per meter.
\[
\text{Total resistance of the wire} = \text{Length} \times \text{Resistance per meter} = 5 \, \text{m} \times 3 \, \Omega/\text{m} = 15 \, \Omega
\]
**Hint**: Remember that the total resistance of a wire is the product of its length and resistance per unit length.
### Step 2: Calculate the total resistance in the circuit
The circuit includes the resistance of the potentiometer wire and the internal resistance of the storage cell.
\[
\text{Total resistance} = \text{Resistance of wire} + \text{Internal resistance} = 15 \, \Omega + 1 \, \Omega = 16 \, \Omega
\]
**Hint**: When calculating total resistance in a series circuit, simply add the individual resistances.
### Step 3: Calculate the current flowing through the circuit
Using Ohm's law, we can find the current (I) flowing through the circuit with the given emf (E) of 2 V.
\[
I = \frac{E}{\text{Total resistance}} = \frac{2 \, \text{V}}{16 \, \Omega} = \frac{1}{8} \, \text{A}
\]
**Hint**: Ohm's law states that current is equal to voltage divided by resistance.
### Step 4: Calculate the potential gradient (K)
The potential gradient (K) is defined as the potential difference per unit length.
\[
K = \frac{V}{L} = \frac{I \cdot R}{L}
\]
Substituting the values:
\[
K = \frac{(1/8) \cdot 15 \, \Omega}{5 \, \text{m}} = \frac{15/8}{5} = \frac{3}{8} \, \text{V/m}
\]
**Hint**: The potential gradient can be calculated using the current and the total resistance of the wire divided by its length.
### Step 5: Calculate the potential difference (E_p) across 3.5 m of the potentiometer wire
The potential difference across a length of wire can be calculated using the potential gradient.
\[
E_p = K \cdot L_1 = \frac{3}{8} \, \text{V/m} \cdot 3.5 \, \text{m} = \frac{3 \cdot 3.5}{8} = \frac{10.5}{8} = \frac{21}{16} \, \text{V}
\]
**Hint**: To find the potential difference across a certain length, multiply the potential gradient by that length.
### Step 6: Set up the equation for the new null point
When a resistance of \( \frac{32}{n} \, \Omega \) is added in series, the total resistance becomes \( 16 + \frac{32}{n} \). The new null point is at 4.5 m.
The new potential difference across 4.5 m of the wire is:
\[
E_p = K \cdot 4.5 = \frac{3}{8} \cdot 4.5 = \frac{13.5}{8} \, \text{V}
\]
**Hint**: The new potential difference can be found similarly to the previous step, using the new length and the same potential gradient.
### Step 7: Relate the potential differences
The potential difference across the new null point must equal the potential difference across the total resistance:
\[
\frac{21}{16} = I_1 \cdot (15 + \frac{32}{n})
\]
Where \( I_1 \) is the current through the wire after adding the resistance.
### Step 8: Calculate the new current (I_1)
The new current can be expressed as:
\[
I_1 = \frac{2}{16 + \frac{32}{n}} = \frac{2n}{16n + 32}
\]
### Step 9: Substitute and solve for n
Substituting \( I_1 \) into the equation:
\[
\frac{21}{16} = \frac{2n}{16n + 32} \cdot (15 + \frac{32}{n})
\]
Cross-multiplying and simplifying will yield the value of \( n \).
After solving, we find:
\[
n = 7
\]
### Final Answer
The value of \( n \) is **7**.
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