Resistance in the two gaps of a meter bridge are `10 ohm` and `30 ohm` respectively. If the resistances are interchanged he balance point shifts by

Explanation

To solve the problem step by step, we will analyze the situation using the principles of a meter bridge.

### Step 1: Understand the setup
We have a meter bridge with two resistances in its gaps: \( R_1 = 10 \, \Omega \) and \( R_2 = 30 \, \Omega \). The balance point is determined by the ratio of the resistances.

### Step 2: Calculate the balance point for the first arrangement
Using the formula for the balance point in a meter bridge:
\[ \frac{R_1}{R_2} = \frac{L_1}{100 - L_1} \]
Substituting the values of \( R_1 \) and \( R_2 \):
\[ \frac{10}{30} = \frac{L_1}{100 - L_1} \]
This simplifies to:
\[ \frac{1}{3} = \frac{L_1}{100 - L_1} \]

### Step 3: Cross-multiply to find \( L_1 \)
Cross-multiplying gives:
\[ L_1 = \frac{1}{3}(100 - L_1) \]
Expanding this:
\[ 3L_1 = 100 - L_1 \]
Rearranging gives:
\[ 3L_1 + L_1 = 100 \]
\[ 4L_1 = 100 \]
\[ L_1 = 25 \, \text{cm} \]

### Step 4: Calculate the balance point for the second arrangement
Now, we interchange the resistances, so \( R_1 = 30 \, \Omega \) and \( R_2 = 10 \, \Omega \). Again, we use the balance point formula:
\[ \frac{R_1}{R_2} = \frac{L_2}{100 - L_2} \]
Substituting the new values:
\[ \frac{30}{10} = \frac{L_2}{100 - L_2} \]
This simplifies to:
\[ 3 = \frac{L_2}{100 - L_2} \]

### Step 5: Cross-multiply to find \( L_2 \)
Cross-multiplying gives:
\[ 3(100 - L_2) = L_2 \]
Expanding this:
\[ 300 - 3L_2 = L_2 \]
Rearranging gives:
\[ 300 = 4L_2 \]
\[ L_2 = 75 \, \text{cm} \]

### Step 6: Calculate the shift in the balance point
The shift in the balance point when the resistances are interchanged is given by:
\[ \text{Shift} = L_2 - L_1 \]
Substituting the values:
\[ \text{Shift} = 75 \, \text{cm} - 25 \, \text{cm} = 50 \, \text{cm} \]

### Final Answer
The balance point shifts by **50 cm**.

---