A box 'A' is lying on the horizontal floor the compartment of a train running along horizontal ralls from left to right. At time 't', it decelerates. Then the resultant contact force R by the floor on the box is given best by :

Options

A.
B.
C.
D.

Explanation

To solve the problem of determining the resultant contact force \( R \) exerted by the floor on the box \( A \) when the train compartment decelerates, we can follow these steps:

### Step-by-Step Solution:

1. **Identify the Forces Acting on the Box:**
- The weight of the box \( A \) acting downwards, which is given by \( mg \) (where \( m \) is the mass of the box and \( g \) is the acceleration due to gravity).
- The normal force \( N \) acting upwards from the floor of the train on the box.
- A pseudo force acting on the box due to the deceleration of the train. Since the train is decelerating to the left, the pseudo force \( F_{\text{pseudo}} = ma \) acts to the right (where \( a \) is the deceleration of the train).

2. **Set Up the Force Diagram:**
- In the vertical direction, the forces are balanced:
\[ N = mg \]
- In the horizontal direction, the pseudo force acts to the right:
\[ F_{\text{pseudo}} = ma \]

3. **Determine the Resultant Force:**
- The resultant contact force \( R \) is the vector sum of the normal force \( N \) and the pseudo force \( F_{\text{pseudo}} \). This can be represented as:
\[ R = \sqrt{N^2 + F_{\text{pseudo}}^2} \]
- Substituting the values of \( N \) and \( F_{\text{pseudo}} \):
\[ R = \sqrt{(mg)^2 + (ma)^2} \]

4. **Final Expression for Resultant Contact Force:**
- Thus, the resultant contact force \( R \) is given by:
\[ R = \sqrt{(mg)^2 + (ma)^2} \]

5. **Conclusion:**
- The resultant contact force \( R \) by the floor on the box \( A \) when the train decelerates is expressed as above.