The length of potentiometer wire is 200 cm and the emf of standard cell is primary circuit is E volts. It is employed to a battery of emf 0.4 v the balance point is obtained at l = 40 cm from positive end, the E of the battery is (cell in primary is ideal and series resistance is zone

Options

A. 4v
B. 2v
C. 5v
D. 7v

Answer

To solve the problem step by step, we will use the concept of a potentiometer and the relationship between the electromotive force (emf) and the length of the wire.

### Step 1: Understand the given data
- Length of the potentiometer wire (L) = 200 cm
- Emf of the standard cell (E) = unknown
- Emf of the battery (E_battery) = 0.4 V
- Balance point length (l) = 40 cm

### Step 2: Calculate the potential gradient (K) for both cases
The potential gradient (K) is defined as the voltage per unit length of the wire.

1. For the standard cell:
\[ K_1 = \frac{E}{L} = \frac{E}{200 \text{ cm}} \quad \text{(in volts per cm)} \]

2. For the battery:
\[ K_2 = \frac{E_battery}{l} = \frac{0.4 \text{ V}}{40 \text{ cm}} = 0.01 \text{ V/cm} \]

### Step 3: Set the potential gradients equal
Since the potentiometer is in equilibrium at the balance point, the potential gradients for both cases are equal:
\[ K_1 = K_2 \]
This gives us the equation:
\[ \frac{E}{200} = 0.01 \]

### Step 4: Solve for E
To find the value of E, we can rearrange the equation:
\[ E = 0.01 \times 200 \]
Calculating this gives:
\[ E = 2 \text{ V} \]

### Final Answer
The emf of the standard cell (E) is **2 V**.

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