A potentiometer wire has resistance `40Omega` and its length is 10m. It is connected by a resistance of `760Omega` in series if emf of battery is 2V then potential gradient is:-

Explanation

To solve the problem, we will follow these steps:

### Step 1: Calculate the resistance per unit length of the potentiometer wire.
The resistance of the potentiometer wire is given as \( R_w = 40 \, \Omega \) and its length is \( L = 10 \, m \).

\[ \text{Resistance per unit length} = \frac{R_w}{L} = \frac{40 \, \Omega}{10 \, m} = 4 \, \Omega/m \]

### Step 2: Calculate the total resistance in the circuit.
The potentiometer wire is connected in series with a resistance of \( R_s = 760 \, \Omega \).

\[ R_{\text{total}} = R_w + R_s = 40 \, \Omega + 760 \, \Omega = 800 \, \Omega \]

### Step 3: Calculate the current flowing through the circuit.
The electromotive force (emf) of the battery is given as \( E = 2 \, V \).

Using Ohm's law, the current \( I \) can be calculated as:

\[ I = \frac{E}{R_{\text{total}}} = \frac{2 \, V}{800 \, \Omega} = \frac{1}{400} \, A \]

### Step 4: Calculate the potential gradient.
The potential gradient \( PG \) can be calculated using the formula:

\[ PG = I \times \text{Resistance per unit length} \]

Substituting the values we have:

\[ PG = \left(\frac{1}{400} \, A\right) \times (4 \, \Omega/m) = \frac{4}{400} \, V/m = \frac{1}{100} \, V/m \]

### Step 5: Convert the potential gradient to standard form.
\[ PG = 0.01 \, V/m = 10^{-2} \, V/m \]

Thus, the potential gradient is \( 10^{-2} \, V/m \).

### Final Answer:
The potential gradient is \( 10^{-2} \, V/m \).

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