Two masses 7 kg and 12 kg are connected at the two ends of a metal wire that goes ove a frictionless puulley. What should be in the minimum radius of the wire in order that the wire does not break,if the breaking stess of the metal is `1.3 xx 10^(8) N //m^(2)` ?

Explanation

To solve the problem, we need to find the minimum radius of the wire that will not break when two masses of 7 kg and 12 kg are connected over a frictionless pulley. The breaking stress of the metal wire is given as \(1.3 \times 10^8 \, \text{N/m}^2\).

### Step-by-Step Solution:

1. **Identify the Forces Acting on the System**:
- The two masses are \(m_1 = 7 \, \text{kg}\) and \(m_2 = 12 \, \text{kg}\).
- The weight of each mass can be calculated using \(W = mg\), where \(g = 9.8 \, \text{m/s}^2\).
- The tension in the wire will be influenced by the heavier mass, which will be \(m_2\).
...