In a meter bridge experiment the ratio of left gap resistance to right gap resistance is `2:3` the balance point from is

Options

A. 60cm
B. 50cm
C. 40cm
D. 20cm

Explanation

To solve the problem regarding the meter bridge experiment, we will follow these steps:

### Step 1: Understand the Setup
In a meter bridge, we have a wire of length 100 cm, and we connect two resistances on either side of a bridge. The left gap has an unknown resistance \(X\) and the right gap has a known resistance \(R\).

### Step 2: Set Up the Ratio
According to the problem, the ratio of the left gap resistance to the right gap resistance is given as \(2:3\). This can be expressed mathematically as:
\[ \frac{X}{R} = \frac{2}{3} \]
From this, we can express \(X\) in terms of \(R\):
\[ X = \frac{2}{3} R \]

### Step 3: Apply the Principle of the Meter Bridge
The principle of the meter bridge states that the ratio of the resistances is equal to the ratio of the lengths of the bridge. If \(L_1\) is the length from point A to the balance point B, then the remaining length from B to the end of the bridge (100 cm) is \(100 - L_1\).

According to the principle:
\[ \frac{X}{R} = \frac{L_1}{100 - L_1} \]

### Step 4: Substitute the Value of \(X\)
Substituting \(X = \frac{2}{3} R\) into the equation gives:
\[ \frac{\frac{2}{3} R}{R} = \frac{L_1}{100 - L_1} \]
This simplifies to:
\[ \frac{2}{3} = \frac{L_1}{100 - L_1} \]

### Step 5: Cross-Multiply to Solve for \(L_1\)
Cross-multiplying gives:
\[ 2(100 - L_1) = 3L_1 \]
Expanding this results in:
\[ 200 - 2L_1 = 3L_1 \]

### Step 6: Combine Like Terms
Rearranging the equation yields:
\[ 200 = 3L_1 + 2L_1 \]
\[ 200 = 5L_1 \]

### Step 7: Solve for \(L_1\)
Dividing both sides by 5 gives:
\[ L_1 = \frac{200}{5} = 40 \text{ cm} \]

### Conclusion
Thus, the balance point from point A is at 40 cm.