A 40 Pa pressure is applied on the head of a nail placed perpendicular to the surface of a wall. If the area of cross-section of the tip of the nail is (1/0) times the area of cross-section of the head, then the pressure exerted on the wall is

Options

A. 10 Pa
B. 40 Pa
C. 400 Pa
D. None of these.

Explanation

To solve the problem, we need to understand how pressure is defined and how it relates to force and area. The pressure exerted on a surface is given by the formula:

\[ P = \frac{F}{A} \]

where:
- \( P \) is the pressure,
- \( F \) is the force applied,
- \( A \) is the area over which the force is applied.

### Step-by-Step Solution:

1. **Identify the Given Values**:
- Pressure applied on the head of the nail, \( P_{head} = 40 \, \text{Pa} \).
- The area of cross-section of the tip of the nail is \( \frac{1}{10} \) times the area of cross-section of the head of the nail.

2. **Let’s Define the Areas**:
- Let \( A_{head} \) be the area of the head of the nail.
- Then, the area of the tip of the nail, \( A_{tip} = \frac{1}{10} A_{head} \).

3. **Calculate the Force on the Head of the Nail**:
- The force exerted on the head of the nail can be calculated using the pressure formula rearranged:
\[ F = P \times A \]
- Therefore, the force on the head of the nail is:
\[ F_{head} = P_{head} \times A_{head} = 40 \, \text{Pa} \times A_{head} \]

4. **Determine the Pressure Exerted by the Tip of the Nail**:
- The same force \( F_{head} \) is exerted through the tip of the nail, but over a smaller area \( A_{tip} \).
- The pressure exerted on the wall by the tip of the nail can be calculated as:
\[ P_{tip} = \frac{F_{head}}{A_{tip}} \]
- Substituting \( F_{head} \):
\[ P_{tip} = \frac{40 \, \text{Pa} \times A_{head}}{A_{tip}} = \frac{40 \, \text{Pa} \times A_{head}}{\frac{1}{10} A_{head}} = 40 \, \text{Pa} \times 10 = 400 \, \text{Pa} \]

5. **Conclusion**:
- The pressure exerted on the wall by the tip of the nail is \( 400 \, \text{Pa} \).

### Final Answer:
The pressure exerted on the wall is **400 Pa**.