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

Options

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

Explanation

To solve the problem step by step, we will use the principles of a meter bridge and the concept of balancing lengths.

### Step 1: Understand the Setup
We have two resistances, \( R_1 \) (the smaller resistance) and \( R_2 \) (the larger resistance), connected in a meter bridge. The balance point is initially at \( 20 \, \text{cm} \) from the zero end.

### Step 2: Apply the Balance Condition
Using the meter bridge principle, we can write the balance condition for the first scenario:

\[ \frac{R_1}{R_2} = \frac{L_1}{100 - L_1} \]

Where \( L_1 = 20 \, \text{cm} \).

Substituting the values:

\[ \frac{R_1}{R_2} = \frac{20}{100 - 20} = \frac{20}{80} = \frac{1}{4} \]

This implies:

\[ R_2 = 4R_1 \quad \text{(Equation 1)} \]

### Step 3: Introduce the Additional Resistance
Now, a \( 15 \, \Omega \) resistance is connected in series with \( R_1 \). The new resistance \( R_1' \) becomes:

\[ R_1' = R_1 + 15 \]

And \( R_2' \) remains:

\[ R_2' = R_2 = 4R_1 \]

### Step 4: Apply the New Balance Condition
In the second scenario, the balance point shifts to \( 40 \, \text{cm} \). We can write the new balance condition:

\[ \frac{R_1'}{R_2'} = \frac{L_2}{100 - L_2} \]

Where \( L_2 = 40 \, \text{cm} \).

Substituting the values:

\[ \frac{R_1 + 15}{4R_1} = \frac{40}{100 - 40} = \frac{40}{60} = \frac{2}{3} \]

### Step 5: Cross-Multiply and Simplify
Cross-multiplying gives:

\[ 3(R_1 + 15) = 2(4R_1) \]

Expanding both sides:

\[ 3R_1 + 45 = 8R_1 \]

### Step 6: Solve for \( R_1 \)
Rearranging the equation:

\[ 45 = 8R_1 - 3R_1 \]

This simplifies to:

\[ 45 = 5R_1 \]

Dividing both sides by 5:

\[ R_1 = 9 \, \Omega \]

### Conclusion
The value of the smaller resistance \( R_1 \) is \( 9 \, \Omega \).

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