A rectangular wooden box of dimensions `10 cm xx 20 cm xx 40cm` is kept on a horizontal surface with its face of smallest area on the surface. A minimum force of 12 N applied parallel to the surface sets the box in sliding motion along the surface. If the box is now kept with its face of larger area in contact with the surface, the minimum force applied parallel to the surface, to set the box in motion, is

Options

A. Less than 12 N
B. May be greater or less than 12 N
C. Greater than 12 N
D. Equal to 12 N.

Explanation

To solve the problem, we need to analyze the situation regarding the frictional force acting on the box when it is placed with different faces in contact with the horizontal surface.

### Step-by-Step Solution:

1. **Identify the dimensions of the box**:
The dimensions of the box are given as 10 cm x 20 cm x 40 cm. The smallest face area is 10 cm x 20 cm.

2. **Calculate the area of the smallest face**:
The area of the smallest face (10 cm x 20 cm) is:
\[ A_{\text{small}} = 10 \, \text{cm} \times 20 \, \text{cm} = 200 \, \text{cm}^2 \]

3. **Identify the force required to move the box with the smallest face down**:
It is given that a minimum force of 12 N is required to set the box in motion when the smallest face is in contact with the surface.

4. **Understand the relationship between force and friction**:
The force required to overcome friction is given by:
\[ F = \mu \cdot N \]
where \( F \) is the frictional force, \( \mu \) is the coefficient of friction, and \( N \) is the normal force (which is equal to the weight of the box in this case).

5. **Determine the normal force**:
Since the box is resting on a horizontal surface, the normal force \( N \) is equal to the weight of the box. The weight can be calculated from the volume and density of the wood, but since we are looking for the force required to move the box, we can use the given information directly.

6. **Calculate the coefficient of friction**:
From the first scenario, we can express the coefficient of friction as:
\[ \mu = \frac{F}{N} \]
Since we don't know \( N \), we will keep it as is for now.

7. **Identify the area of the larger face**:
When the box is placed with its larger face down, the dimensions of the larger face are 20 cm x 40 cm. The area of the larger face is:
\[ A_{\text{large}} = 20 \, \text{cm} \times 40 \, \text{cm} = 800 \, \text{cm}^2 \]

8. **Understand that the weight of the box remains the same**:
The weight of the box does not change regardless of the orientation. Thus, the normal force \( N \) remains the same.

9. **Determine the force required to move the box with the larger face down**:
The frictional force will still be given by:
\[ F = \mu \cdot N \]
Since the coefficient of friction \( \mu \) and the normal force \( N \) remain unchanged, the force required to overcome friction will also remain the same.

10. **Conclusion**:
Therefore, the minimum force required to set the box in motion when it is placed with its larger face in contact with the surface is still **12 N**.

### Final Answer:
The minimum force required to set the box in motion with its larger face in contact with the surface is **12 N**.