With two resistance `R_1 and R_2 (>R_1)` in the two gaps of a metre bridge the balance was found to be 1/3 m from the zero end. When a `6 Omega `resistance is connected in series with the smaller of the two resistance, the point is shifted to 2/3 m from the same end, then `R_1 and R_2` are

Options

A. 4,2
B. 2,4
C. Both (a) and (b)
D. Neither (a) nor (b)

Answer

To solve the problem, we will use the principles of a meter bridge and the concept of a balanced Wheatstone bridge.

### Step-by-Step Solution:

1. **Understanding the Setup**:
We have a meter bridge with two resistances \( R_1 \) and \( R_2 \) placed in its gaps. The balance point is initially at \( \frac{1}{3} \) m from the zero end. This means that the lengths of the bridge wire on either side of the galvanometer are \( \frac{1}{3} \) m and \( \frac{2}{3} \) m.

2. **Applying the Wheatstone Bridge Principle**:
According to the principle of the Wheatstone bridge, we have:
\[ \frac{R_1}{R_2} = \frac{x}{L - x} \]
where \( x \) is the length from the zero end (which is \( \frac{1}{3} \) m) and \( L \) is the total length of the bridge (1 m). Thus, we can write:
\[ \frac{R_1}{R_2} = \frac{\frac{1}{3}}{\frac{2}{3}} = \frac{1}{2} \]
From this, we can express \( R_2 \) in terms of \( R_1 \):
\[ R_2 = 2R_1 \quad \text{(Equation 1)} \]

3. **Introducing the 6 Ohm Resistor**:
When a 6 Ohm resistor is connected in series with the smaller resistance \( R_1 \), the effective resistance becomes \( R_1' = R_1 + 6 \). The new balance point is now at \( \frac{2}{3} \) m from the zero end.

4. **Applying the Wheatstone Bridge Principle Again**:
Now, we apply the Wheatstone bridge principle again:
\[ \frac{R_1'}{R_2} = \frac{\frac{2}{3}}{\frac{1}{3}} = 2 \]
Substituting \( R_1' \):
\[ \frac{R_1 + 6}{R_2} = 2 \]
Rearranging gives us:
\[ R_1 + 6 = 2R_2 \quad \text{(Equation 2)} \]

5. **Substituting Equation 1 into Equation 2**:
Now, we substitute \( R_2 = 2R_1 \) from Equation 1 into Equation 2:
\[ R_1 + 6 = 2(2R_1) \]
Simplifying this gives:
\[ R_1 + 6 = 4R_1 \]
Rearranging gives:
\[ 6 = 4R_1 - R_1 \]
\[ 6 = 3R_1 \]
Thus, we find:
\[ R_1 = 2 \, \Omega \]

6. **Finding \( R_2 \)**:
Now, substituting \( R_1 \) back into Equation 1:
\[ R_2 = 2R_1 = 2 \times 2 = 4 \, \Omega \]

### Final Answer:
The values of the resistances are:
- \( R_1 = 2 \, \Omega \)
- \( R_2 = 4 \, \Omega \)