The emf of a standard cell is 1.5 volt and its balancing length is 7.5 m. The balancing length in meters for a 3.5 ohm resistance, through which a current of 0.2 A, flows will be

Options

A. 3.5
B. 5
C. 5.7
D. 6.5

Explanation

To solve the problem step by step, we will follow these calculations:

### Step 1: Calculate the potential difference (V) across the 3.5 ohm resistor using Ohm's Law.
Ohm's Law states that \( V = I \times R \).

Given:
- Current \( I = 0.2 \, \text{A} \)
- Resistance \( R = 3.5 \, \Omega \)

Calculating the potential difference:
\[ V = 0.2 \, \text{A} \times 3.5 \, \Omega = 0.7 \, \text{V} \]

### Step 2: Calculate the potential gradient (k) for the standard cell.
The potential gradient is defined as the EMF of the cell divided by the balancing length.

Given:
- EMF of the standard cell \( E = 1.5 \, \text{V} \)
- Balancing length \( L_0 = 7.5 \, \text{m} \)

Calculating the potential gradient:
\[ k = \frac{E}{L_0} = \frac{1.5 \, \text{V}}{7.5 \, \text{m}} = 0.2 \, \text{V/m} \]

### Step 3: Relate the potential difference to the new balancing length (L).
The potential difference across the resistor can be expressed in terms of the new balancing length and the potential gradient:
\[ V = k \times L \]

Where \( L \) is the new balancing length we want to find. We can rearrange this to find \( L \):
\[ L = \frac{V}{k} \]

### Step 4: Substitute the values into the equation.
Substituting the values we found:
- \( V = 0.7 \, \text{V} \)
- \( k = 0.2 \, \text{V/m} \)

Calculating the new balancing length:
\[ L = \frac{0.7 \, \text{V}}{0.2 \, \text{V/m}} = 3.5 \, \text{m} \]

### Conclusion
The balancing length for the 3.5 ohm resistance through which a current of 0.2 A flows is **3.5 meters**.

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