To solve the problem step-by-step, we will use the principles of the meter bridge and the concept of parallel resistances.
### Step 1: Understand the initial setup
We have an unknown resistance \( R_x \) and a known resistance of \( 4 \, \Omega \) connected in a meter bridge. The balance point is at \( 50 \, cm \).
### Step 2: Apply the meter bridge formula
Using the meter bridge formula, we can write:
\[
\frac{R_x}{L} = \frac{4}{100 - L}
\]
Here, \( L = 50 \, cm \) and \( 100 - L = 50 \, cm \).
### Step 3: Substitute the values into the equation
Substituting \( L \) into the equation:
\[
\frac{R_x}{50} = \frac{4}{50}
\]
### Step 4: Solve for the unknown resistance \( R_x \)
Cross-multiplying gives:
\[
R_x \cdot 50 = 4 \cdot 50
\]
\[
R_x = 4 \, \Omega
\]
### Step 5: Analyze the new configuration
Now, we connect another \( 4 \, \Omega \) resistance in parallel with the existing \( 4 \, \Omega \) resistance in the right gap.
### Step 6: Calculate the equivalent resistance of the parallel combination
The equivalent resistance \( R_{eq} \) of two resistances \( R_1 \) and \( R_2 \) in parallel is given by:
\[
R_{eq} = \frac{R_1 \cdot R_2}{R_1 + R_2}
\]
Substituting \( R_1 = 4 \, \Omega \) and \( R_2 = 4 \, \Omega \):
\[
R_{eq} = \frac{4 \cdot 4}{4 + 4} = \frac{16}{8} = 2 \, \Omega
\]
### Step 7: Set up the new balance point equation
Now, we will find the new balance point \( L' \) using the new equivalent resistance:
\[
\frac{R_x}{L'} = \frac{R_{eq}}{100 - L'}
\]
Substituting \( R_x = 4 \, \Omega \) and \( R_{eq} = 2 \, \Omega \):
\[
\frac{4}{L'} = \frac{2}{100 - L'}
\]
### Step 8: Cross-multiply to solve for \( L' \)
Cross-multiplying gives:
\[
4(100 - L') = 2L'
\]
\[
400 - 4L' = 2L'
\]
\[
400 = 6L'
\]
\[
L' = \frac{400}{6} = \frac{200}{3} \approx 66.67 \, cm
\]
### Step 9: Calculate the shift in the balance point
The initial balance point was at \( 50 \, cm \). The new balance point is \( \frac{200}{3} \, cm \):
\[
\text{Shift} = L' - L = \frac{200}{3} - 50 = \frac{200}{3} - \frac{150}{3} = \frac{50}{3} \approx 16.67 \, cm
\]
### Final Answer
The shift in the balance point is approximately \( 16.67 \, cm \).
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