Shown in the figure is a system of three particles of mass `1 kg, 2 kg` and `4 kg` connected by two springs. The acceleration of `A. B` and `C` at any instant are `1 ms^(-2), 2 ms^(-2)` and `1//2 ms^(-2)` respectively directed as shown in the figure external force acting on the system is

Options

A. `1N`
B. `7N`
C. `3N`
D. none of these

Explanation

To solve the problem, we need to find the external force acting on a system of three particles connected by springs. The given masses and their respective accelerations are as follows:

- Mass \( m_A = 1 \, \text{kg} \) with acceleration \( a_A = 1 \, \text{m/s}^2 \)
- Mass \( m_B = 2 \, \text{kg} \) with acceleration \( a_B = 2 \, \text{m/s}^2 \)
- Mass \( m_C = 4 \, \text{kg} \) with acceleration \( a_C = 0.5 \, \text{m/s}^2 \)

### Step-by-Step Solution:

1. **Calculate the Total Mass of the System:**
\[ m_{\text{total}} = m_A + m_B + m_C = 1 \, \text{kg} + 2 \, \text{kg} + 4 \, \text{kg} = 7 \, \text{kg} \]

2. **Calculate the Center of Mass Acceleration:**
The acceleration of the center of mass \( a_{\text{cm}} \) can be calculated using the formula:
\[ a_{\text{cm}} = \frac{m_A \cdot a_A + m_B \cdot a_B + m_C \cdot a_C}{m_{\text{total}}} \]
Substituting the values:
\[ a_{\text{cm}} = \frac{(1 \, \text{kg} \cdot 1 \, \text{m/s}^2) + (2 \, \text{kg} \cdot 2 \, \text{m/s}^2) + (4 \, \text{kg} \cdot 0.5 \, \text{m/s}^2)}{7 \, \text{kg}} \]
\[ = \frac{1 + 4 + 2}{7} = \frac{7}{7} = 1 \, \text{m/s}^2 \]

3. **Calculate the External Force Acting on the System:**
According to Newton's second law, the external force \( F_{\text{external}} \) can be calculated as:
\[ F_{\text{external}} = m_{\text{total}} \cdot a_{\text{cm}} \]
Substituting the values:
\[ F_{\text{external}} = 7 \, \text{kg} \cdot 1 \, \text{m/s}^2 = 7 \, \text{N} \]

### Final Answer:
The external force acting on the system is \( 7 \, \text{N} \).

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