$ ext{The equivalent resistance of the resistances joined in parallel is 6/5 } \Omega. ext{ When one of the resistance wire is broken, the effective resistance becomes 2 } \Omega. ext{ The resistance of wire that got broken was :}$

Q: $\text{The equivalent resistance of the resistances joined in parallel is 6/5 } \Omega. \text{ When one of the resistance wire is broken, the effective resistance becomes 2 } \Omega. \text{ The resistance of wire that got broken was :}$

$\text{The equivalent resistance of the resistances joined in parallel is 6/5 } \Omega. \text{ When one of the resistance wire is broken, the effective resistance becomes 2 } \Omega. \text{ The resistance of wire that got broken was :}$

Question

$	ext{The equivalent resistance of the resistances joined in parallel is 6/5 } \Omega. 	ext{ When one of the resistance wire is broken, the effective resistance becomes 2 } \Omega. 	ext{ The resistance of wire that got broken was :}$

Accepted Answer

To determine the resistance of the wire that got broken, we need to analyze the situation using the concept of resistances in parallel.

### Explanation:

1. **Initial Condition:**
   - The equivalent resistance of the resistances joined in parallel is given as $ \frac{6}{5} \, \Omega $.

2. **Condition After One Wire Breaks:**
   - When one of the resistance wires is broken, the effective resistance becomes $ 2 \, \Omega $.

### Analysis:

Let's denote the resistances in parallel as $ R_1, R_2, \ldots, R_n $. The formula for the equivalent resistance $ R_{\text{eq}} $ of resistances in parallel is:

$$
\frac{1}{R_{\text{eq}}} = \frac{1}{R_1} + \frac{1}{R_2} + \ldots + \frac{1}{R_n}
$$

Initially, the equivalent resistance is $ \frac{6}{5} \, \Omega $:

$$
\frac{1}{R_{\text{eq}}} = \frac{5}{6}
$$

After one resistance, say $ R_x $, is broken, the new equivalent resistance becomes $ 2 \, \Omega $:

$$
\frac{1}{R_{\text{new}}} = \frac{1}{R_1} + \frac{1}{R_2} + \ldots + \frac{1}{R_{x-1}} + \frac{1}{R_{x+1}} + \ldots + \frac{1}{R_n} = \frac{1}{2}
$$

### Solving for the Broken Resistance:

The change in the equivalent resistance indicates that the broken resistance $ R_x $ was contributing to the initial equivalent resistance. We can set up the equation:

$$
\frac{5}{6} - \frac{1}{R_x} = \frac{1}{2}
$$

Solving for $ \frac{1}{R_x} $:

$$
\frac{5}{6} - \frac{1}{2} = \frac{1}{R_x}
$$

$$
\frac{5}{6} - \frac{3}{6} = \frac{1}{R_x}
$$

$$
\frac{2}{6} = \frac{1}{R_x}
$$

$$
\frac{1}{3} = \frac{1}{R_x}
$$

Thus, $ R_x = 3 \, \Omega $.

### Conclusion:

- **Option 2: $ 3 \, \Omega $** is the correct answer because the calculations show that the resistance of the wire that got broken was indeed $ 3 \, \Omega $.

### Explanation for Incorrect Options:

- **Option 1: $ 2 \, \Omega $** is incorrect because substituting $ 2 \, \Omega $ into the equation does not satisfy the condition for the change in equivalent resistance.
- **Option 3: $ 5 \, \Omega $** is incorrect because it does not satisfy the equation derived from the change in equivalent resistance.
- **Option 4: $ 7 \, \Omega $** is incorrect because it also does not satisfy the equation derived from the change in equivalent resistance.

By following the logical steps and calculations, we confirm that Option 2 is indeed the correct answer.