The emf of a cell is 2V and its internal resistance is 2 ohm. A resistance of 8 ohm is joined to battery in parallel. This is contacted in secondary circuit of potentio meter. If IV standard cell balances for 100cm of potentiometer wire, the balance point of above cell is

Options

A. 120cm
B. 240cm
C. 160cm
D. 116cm

Answer

To solve the problem step by step, we will follow the information provided in the question and the video transcript.

### Step 1: Identify the given data
- EMF of the cell (E) = 2 V
- Internal resistance (r) = 2 ohms
- External resistance (R) = 8 ohms (connected in parallel)
- Length of potentiometer wire for 1 V standard cell (l1) = 100 cm

### Step 2: Calculate the equivalent resistance of the parallel circuit
When resistances are connected in parallel, the equivalent resistance (R_eq) can be calculated using the formula:
\[ \frac{1}{R_{eq}} = \frac{1}{r} + \frac{1}{R} \]
Substituting the values:
\[ \frac{1}{R_{eq}} = \frac{1}{2} + \frac{1}{8} \]
Finding a common denominator (8):
\[ \frac{1}{R_{eq}} = \frac{4}{8} + \frac{1}{8} = \frac{5}{8} \]
Thus,
\[ R_{eq} = \frac{8}{5} = 1.6 \, \text{ohms} \]

### Step 3: Calculate the total resistance in the circuit
The total resistance (R_total) in the circuit is the sum of the internal resistance and the equivalent resistance:
\[ R_{total} = r + R_{eq} = 2 + 1.6 = 3.6 \, \text{ohms} \]

### Step 4: Calculate the current flowing through the circuit
Using Ohm's law, the current (I) can be calculated as:
\[ I = \frac{E}{R_{total}} = \frac{2}{3.6} \approx 0.5556 \, \text{A} \]

### Step 5: Determine the potential difference across the external resistance
The potential difference (V) across the external resistance can be calculated using:
\[ V = I \times R = 0.5556 \times 8 \approx 4.4448 \, \text{V} \]

### Step 6: Calculate the balance point on the potentiometer
From the problem, we know that 1 V corresponds to 100 cm on the potentiometer. Therefore, we can find the length (l2) corresponding to the potential difference (V) calculated in the previous step:
\[ \text{Potential gradient (k)} = \frac{1 \, \text{V}}{100 \, \text{cm}} = 0.01 \, \text{V/cm} \]
Using the relationship:
\[ V = k \times l_2 \implies l_2 = \frac{V}{k} = \frac{4.4448}{0.01} \approx 444.48 \, \text{cm} \]

### Final Answer
The balance point of the above cell is approximately **444.48 cm**.

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