The resistances in the left and right gaps of a metre bridge are `10 Omega and 30 Omega` respectively. If the bridge is balanced, then the distance of the null point from the centre of the wire is

Options

A. 20 cm
B. 30 cm
C. 40 cm
D. 25 cm

Explanation

To solve the problem, we will use the principle of the meter bridge, which is based on the Wheatstone bridge concept. The meter bridge consists of a wire of length 1 meter, and we have two resistances connected in the left and right gaps.

### Step-by-Step Solution:

1. **Identify the Given Values:**
- Resistance in the left gap, \( R_1 = 10 \, \Omega \)
- Resistance in the right gap, \( R_2 = 30 \, \Omega \)

2. **Understand the Meter Bridge Setup:**
- The meter bridge is a wire of length 1 meter. The null point is the point where the bridge is balanced, meaning the ratio of the resistances is equal to the ratio of the lengths of the wire on either side of the null point.

3. **Define the Lengths:**
- Let \( L \) be the distance from the center of the wire to the null point.
- The length from the center to the left side of the null point is \( 50 - L \) cm (since the total length is 100 cm).
- The length from the center to the right side of the null point is \( 50 + L \) cm.

4. **Set Up the Balance Condition:**
- According to the meter bridge principle:
\[ \frac{R_1}{R_2} = \frac{50 - L}{50 + L} \]
Substituting the values of \( R_1 \) and \( R_2 \):
\[ \frac{10}{30} = \frac{50 - L}{50 + L} \]
Simplifying the left side:
\[ \frac{1}{3} = \frac{50 - L}{50 + L} \]

5. **Cross Multiply to Solve for L:**
\[ 1 \cdot (50 + L) = 3 \cdot (50 - L) \]
Expanding both sides:
\[ 50 + L = 150 - 3L \]
Rearranging gives:
\[ L + 3L = 150 - 50 \]
\[ 4L = 100 \]

6. **Calculate L:**
\[ L = \frac{100}{4} = 25 \, \text{cm} \]

7. **Determine the Distance from the Center:**
- Since we defined \( L \) as the distance from the center to the null point, the distance of the null point from the center of the wire is \( 25 \, \text{cm} \).

### Final Answer:
The distance of the null point from the center of the wire is **25 cm**.

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