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. `8 Omega`
B. `9 Omega`
C. `10 Omega`
D. `12 Omega`

Explanation

To solve the problem step by step, we will use the principles of a meter bridge and the concept of resistances in series and parallel.

### Step 1: Understanding the Meter Bridge
A meter bridge is based on the principle of the Wheatstone bridge. When two resistances are connected in the two gaps of the meter bridge, the balance point (null point) is determined by the ratio of the resistances.

### Step 2: Initial Setup
Let:
- \( R_1 \) = smaller resistance
- \( R_2 \) = larger resistance

From the problem, we know that the balance point is at 20 cm from the zero end. This means:
\[ \frac{R_1}{R_2} = \frac{20}{100 - 20} = \frac{20}{80} = \frac{1}{4} \]
Thus, we can express this as:
\[ R_1 = \frac{1}{4} R_2 \quad \text{(Equation 1)} \]

### Step 3: Adding Series Resistance
Next, when a resistance of 15 Ω is connected in series with \( R_1 \), the new balance point shifts to 40 cm. Now, we can write the equation for the new balance point:
\[ \frac{R_1 + 15}{R_2} = \frac{40}{100 - 40} = \frac{40}{60} = \frac{2}{3} \]
This can be rearranged to:
\[ R_1 + 15 = \frac{2}{3} R_2 \quad \text{(Equation 2)} \]

### Step 4: Solving the Equations
Now we have two equations:
1. \( R_1 = \frac{1}{4} R_2 \)
2. \( R_1 + 15 = \frac{2}{3} R_2 \)

Substituting Equation 1 into Equation 2:
\[ \frac{1}{4} R_2 + 15 = \frac{2}{3} R_2 \]

### Step 5: Clear the Fractions
To eliminate the fractions, we can multiply the entire equation by 12 (the least common multiple of 4 and 3):
\[ 12 \left(\frac{1}{4} R_2\right) + 12 \cdot 15 = 12 \left(\frac{2}{3} R_2\right) \]
This simplifies to:
\[ 3 R_2 + 180 = 8 R_2 \]

### Step 6: Rearranging the Equation
Now, rearranging gives:
\[ 180 = 8 R_2 - 3 R_2 \]
\[ 180 = 5 R_2 \]
\[ R_2 = \frac{180}{5} = 36 \, \Omega \]

### Step 7: Finding \( R_1 \)
Now, substituting \( R_2 \) back into Equation 1 to find \( R_1 \):
\[ R_1 = \frac{1}{4} R_2 = \frac{1}{4} \times 36 = 9 \, \Omega \]

### Conclusion
The smaller of the two resistances \( R_1 \) is \( 9 \, \Omega \).

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