The diameters of neck and bottom of a bottle are 3 cm and 18 cm, respectively. What will be the force exerted on the bottom of a bottle if a force of 0.9 kgf is applied at the neck of the bottle?

Answer

To solve the problem of finding the force exerted on the bottom of the bottle when a force of 0.9 kgf is applied at the neck, we can follow these steps:

### Step-by-Step Solution:

1. **Identify Given Data:**
- Diameter of the neck (d_A) = 3 cm
- Diameter of the bottom (d_B) = 18 cm
- Force applied at the neck (F_A) = 0.9 kgf

2. **Convert the Force from kgf to Newtons:**
- 1 kgf = 9.81 N
- Therefore, F_A in Newtons = 0.9 kgf × 9.81 N/kgf = 8.829 N

3. **Calculate the Radius of the Neck and Bottom:**
- Radius of the neck (r_A) = d_A / 2 = 3 cm / 2 = 1.5 cm
- Radius of the bottom (r_B) = d_B / 2 = 18 cm / 2 = 9 cm

4. **Calculate the Area of the Neck and Bottom:**
- Area of the neck (A_A) = π × (r_A)^2 = π × (1.5 cm)^2 = π × 2.25 cm²
- Area of the bottom (A_B) = π × (r_B)^2 = π × (9 cm)^2 = π × 81 cm²

5. **Use the Principle of Constant Pressure:**
- Since the pressure at both the neck and the bottom is the same, we can write:
\[ P_A = P_B \]
- This gives us:
\[ \frac{F_A}{A_A} = \frac{F_B}{A_B} \]

6. **Rearranging the Equation to Find F_B:**
- Rearranging gives:
\[ F_B = F_A \times \frac{A_B}{A_A} \]

7. **Substituting the Areas:**
- Substitute the areas calculated:
\[ F_B = F_A \times \frac{\pi \times 81}{\pi \times 2.25} \]
- The π cancels out:
\[ F_B = F_A \times \frac{81}{2.25} \]

8. **Calculate the Value of F_B:**
- Calculate the fraction:
\[ \frac{81}{2.25} = 36 \]
- Therefore:
\[ F_B = 8.829 N \times 36 = 317.844 N \]

9. **Final Answer:**
- The force exerted on the bottom of the bottle is approximately **317.84 N**.