Three solids of masses `m_(1),m_(2)` and `m_(3)` are connected with weightless string in succession and are placed on a frictionless table. If the mass `m_(3)` is dragged with a force T , the tension in the string between `m_(2)` and `m_(3)` is

Options

A. `(m_(2))/(m_(1)+m_(2)+m_(3))T`
B. `(m_(3))/(m_(1)+m_(2)+m_(3))T`
C. `(m_(1)+m_(2))/(m_(1)+m_(2)+m_(3))T`
D. `(m_(2)+m_(3))/(m_(1)+m_(2)+m_(3))T`

Answer

To find the tension in the string between masses \( m_2 \) and \( m_3 \) when mass \( m_3 \) is dragged with a force \( T \), we can follow these steps:

### Step 1: Understand the System
We have three masses \( m_1 \), \( m_2 \), and \( m_3 \) connected by weightless strings on a frictionless table. When a force \( T \) is applied to \( m_3 \), the entire system accelerates.

### Step 2: Write the Equation for Total Acceleration
The total mass being accelerated is \( m_1 + m_2 + m_3 \). According to Newton's second law, the net force applied to the system is equal to the total mass times the acceleration \( a \):
\[ T = (m_1 + m_2 + m_3) \cdot a \]
From this, we can express the acceleration \( a \) as:
\[ a = \frac{T}{m_1 + m_2 + m_3} \]

### Step 3: Analyze the Forces on Each Mass
1. **For mass \( m_1 \)**:
The only force acting on \( m_1 \) is the tension \( T_1 \) in the string connected to it. Thus, we can write:
\[ T_1 = m_1 \cdot a \]

2. **For mass \( m_2 \)**:
The forces acting on \( m_2 \) are the tension \( T_2 \) from the string connected to \( m_3 \) and the tension \( T_1 \) from the string connected to \( m_1 \). The equation becomes:
\[ T_2 - T_1 = m_2 \cdot a \]

3. **For mass \( m_3 \)**:
The forces acting on \( m_3 \) are the applied force \( T \) and the tension \( T_2 \) in the string connected to \( m_2 \):
\[ T - T_2 = m_3 \cdot a \]

### Step 4: Substitute \( a \) into the Equations
We can substitute \( a \) from Step 2 into the equations for \( T_1 \), \( T_2 \), and the equation for \( m_3 \):
1. From \( T_1 \):
\[ T_1 = m_1 \cdot \frac{T}{m_1 + m_2 + m_3} \]

2. Substitute \( T_1 \) into the equation for \( m_2 \):
\[ T_2 - m_1 \cdot \frac{T}{m_1 + m_2 + m_3} = m_2 \cdot \frac{T}{m_1 + m_2 + m_3} \]
Rearranging gives:
\[ T_2 = m_1 \cdot \frac{T}{m_1 + m_2 + m_3} + m_2 \cdot \frac{T}{m_1 + m_2 + m_3} \]
\[ T_2 = \frac{(m_1 + m_2)T}{m_1 + m_2 + m_3} \]

### Final Result
Thus, the tension in the string between \( m_2 \) and \( m_3 \) is:
\[ T_2 = \frac{(m_1 + m_2)T}{m_1 + m_2 + m_3} \]