Two resistances are connected in the two gaps of a meter bridge. The balance point is `20 cm` from the zero end. When a resistance `15 Omega` is connected in series with the smaller of two resistance, the null point+ shifts to `40 cm`. The smaller of the two resistance has the value.

Options

A. 3
B. 6
C. 9
D. 12

Answer

To solve the problem step by step, we will use the principles of a meter bridge and the balance condition.

### Step-by-Step Solution:

1. **Understanding the Meter Bridge Principle**:
In a meter bridge, when two resistances \( R \) and \( S \) are connected, the balance point \( L \) is given by the formula:
\[ R = \frac{S \cdot L}{100 - L} \]
where \( L \) is the distance from the zero end to the balance point.

2. **First Condition**:
Given that the balance point is at \( L = 20 \, \text{cm} \):
\[ R = \frac{S \cdot 20}{100 - 20} = \frac{S \cdot 20}{80} = \frac{S}{4} \]
This gives us our first equation:
\[ R = \frac{S}{4} \quad \text{(Equation 1)} \]

3. **Second Condition**:
When a resistance of \( 15 \, \Omega \) is connected in series with the smaller resistance \( R \), the new balance point shifts to \( L' = 40 \, \text{cm} \):
\[ R' = R + 15 \]
The new balance condition can be expressed as:
\[ R' = \frac{S \cdot L'}{100 - L'} = \frac{S \cdot 40}{100 - 40} = \frac{S \cdot 40}{60} = \frac{2S}{3} \]
This gives us our second equation:
\[ R + 15 = \frac{2S}{3} \quad \text{(Equation 2)} \]

4. **Substituting Equation 1 into Equation 2**:
From Equation 1, we have \( S = 4R \). Substituting this into Equation 2:
\[ R + 15 = \frac{2(4R)}{3} \]
Simplifying this:
\[ R + 15 = \frac{8R}{3} \]

5. **Clearing the Fraction**:
Multiply through by 3 to eliminate the fraction:
\[ 3R + 45 = 8R \]

6. **Rearranging the Equation**:
Rearranging gives:
\[ 45 = 8R - 3R \]
\[ 45 = 5R \]

7. **Solving for R**:
Dividing both sides by 5:
\[ R = 9 \, \Omega \]

8. **Conclusion**:
The smaller of the two resistances \( R \) has a value of \( 9 \, \Omega \).

### Final Answer:
The smaller of the two resistances has the value \( 9 \, \Omega \).