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 concept of balancing the circuit.

### Step 1: Understand the Setup
We have two resistances, let's denote them as:
- \( R_1 = X \) (the smaller resistance)
- \( R_2 = Y \) (the larger resistance)

The balance point is initially at 20 cm from the zero end of the meter bridge.

### Step 2: Apply the Balance Condition
According to the meter bridge principle, when the bridge is balanced, the ratio of the resistances is equal to the ratio of the lengths from the zero end:
\[ \frac{X}{20} = \frac{Y}{80} \]
This can be rearranged to give:
\[ X = \frac{Y}{4} \quad \text{(Equation 1)} \]

### Step 3: Introduce the Additional Resistance
Now, when a resistance of \( 15 \, \Omega \) is connected in series with the smaller resistance \( X \), the new balance point shifts to 40 cm.

### Step 4: Apply the Balance Condition Again
Using the new balance point, we can write:
\[ \frac{X + 15}{40} = \frac{Y}{60} \]
Rearranging gives:
\[ X + 15 = \frac{2Y}{3} \quad \text{(Equation 2)} \]

### Step 5: Substitute Equation 1 into Equation 2
Now we substitute \( X = \frac{Y}{4} \) from Equation 1 into Equation 2:
\[ \frac{Y}{4} + 15 = \frac{2Y}{3} \]

### Step 6: Solve for Y
To eliminate the fractions, multiply the entire equation by 12 (the least common multiple of 4 and 3):
\[ 12 \left(\frac{Y}{4}\right) + 12 \times 15 = 12 \left(\frac{2Y}{3}\right) \]
This simplifies to:
\[ 3Y + 180 = 8Y \]
Rearranging gives:
\[ 180 = 8Y - 3Y \]
\[ 180 = 5Y \]
Thus, we find:
\[ Y = \frac{180}{5} = 36 \, \Omega \]

### Step 7: Find X
Now, we can find \( X \) using Equation 1:
\[ X = \frac{Y}{4} = \frac{36}{4} = 9 \, \Omega \]

### Final Answer
The smaller of the two resistances \( X \) is \( 9 \, \Omega \).

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