Two blocks of masses 2 kg and 1 kg are in contact with each other on a frictionless table. When a horizontal force of `3.0` N is applied to the block of mass 2 kg the value of the force of contact between the two blocks is:

Options

A. 4 N
B. 3 N
C. 2 N
D. 1 N

Explanation

To solve the problem, we need to determine the force of contact between the two blocks when a horizontal force of 3 N is applied to the block of mass 2 kg.

### Step-by-Step Solution:

1. **Identify the system**: We have two blocks in contact on a frictionless table: one block has a mass of 2 kg and the other has a mass of 1 kg.

2. **Calculate the total mass**: The total mass of the system (both blocks) is:
\[ m_{\text{total}} = m_1 + m_2 = 2 \, \text{kg} + 1 \, \text{kg} = 3 \, \text{kg} \]

3. **Determine the acceleration of the system**: The net external force acting on the system is the applied force of 3 N. Using Newton's second law (F = ma), we can calculate the acceleration (a) of the system:
\[ a = \frac{F}{m_{\text{total}}} = \frac{3 \, \text{N}}{3 \, \text{kg}} = 1 \, \text{m/s}^2 \]

4. **Analyze the 1 kg block**: Now, we will focus on the 1 kg block. The only horizontal force acting on it is the contact force (C) exerted by the 2 kg block. According to Newton's second law:
\[ F_{\text{net}} = m \cdot a \]
For the 1 kg block, we can write:
\[ C = m_2 \cdot a = 1 \, \text{kg} \cdot 1 \, \text{m/s}^2 = 1 \, \text{N} \]

5. **Conclusion**: The force of contact between the two blocks is 1 N.

### Final Answer:
The value of the force of contact between the two blocks is **1 N**.