Question
How to calculate the distance the water level falls when a 10cm diameter ball is removed from a 14cm diameter cylindrical container after being covered with water?
Answers
GPT-4o mini
To calculate the distance the water level falls when a 10 cm diameter ball is removed from a 14 cm diameter cylindrical container, you can use the principle of displacement.
### Steps to calculate:
**1. Calculate the volume of the ball:**
The formula for the volume \( V \) of a sphere is given by:
\[
V = \frac{4}{3} \pi r^3
\]
The radius \( r \) of the ball is half its diameter:
\[
r_{\text{ball}} = \frac{10 \text{ cm}}{2} = 5 \text{ cm}
\]
Now, substituting this into the volume formula:
\[
V_{\text{ball}} = \frac{4}{3} \pi (5 \text{ cm})^3
\]
\[
V_{\text{ball}} = \frac{4}{3} \pi (125 \text{ cm}^3)
\]
\[
V_{\text{ball}} = \frac{500}{3} \pi \text{ cm}^3 \approx 523.6 \text{ cm}^3
\]
**2. Calculate the cross-sectional area of the cylindrical container:**
The radius of the cylindrical container is:
\[
r_{\text{cylinder}} = \frac{14 \text{ cm}}{2} = 7 \text{ cm}
\]
The area \( A \) of the cylinder's base is:
\[
A = \pi r^2
\]
\[
A = \pi (7 \text{ cm})^2
\]
\[
A = \pi (49 \text{ cm}^2) \approx 153.9 \text{ cm}^2
\]
**3. Calculate the height (distance) the water level falls:**
When the ball is removed, it displaces an equivalent volume of water. The fall in the water level \( h \) can be determined using the formula:
\[
\text{Volume of water displaced} = \text{Base Area} \times \text{Height}
\]
\[
V_{\text{ball}} = A \times h
\]
Rearranging gives:
\[
h = \frac{V_{\text{ball}}}{A}
\]
Substituting the known values:
\[
h = \frac{\frac{500}{3} \pi \text{ cm}^3}{\pi (49 \text{ cm}^2)}
\]
The \(\pi\) cancels out:
\[
h = \frac{\frac{500}{3}}{49}
\]
\[
h = \frac{500}{147} \approx 3.40 \text{ cm}
\]
### Conclusion:
The distance the water level falls when the 10 cm diameter ball is removed from the 14 cm diameter cylindrical container is approximately **3.40 cm**.
### Steps to calculate:
**1. Calculate the volume of the ball:**
The formula for the volume \( V \) of a sphere is given by:
\[
V = \frac{4}{3} \pi r^3
\]
The radius \( r \) of the ball is half its diameter:
\[
r_{\text{ball}} = \frac{10 \text{ cm}}{2} = 5 \text{ cm}
\]
Now, substituting this into the volume formula:
\[
V_{\text{ball}} = \frac{4}{3} \pi (5 \text{ cm})^3
\]
\[
V_{\text{ball}} = \frac{4}{3} \pi (125 \text{ cm}^3)
\]
\[
V_{\text{ball}} = \frac{500}{3} \pi \text{ cm}^3 \approx 523.6 \text{ cm}^3
\]
**2. Calculate the cross-sectional area of the cylindrical container:**
The radius of the cylindrical container is:
\[
r_{\text{cylinder}} = \frac{14 \text{ cm}}{2} = 7 \text{ cm}
\]
The area \( A \) of the cylinder's base is:
\[
A = \pi r^2
\]
\[
A = \pi (7 \text{ cm})^2
\]
\[
A = \pi (49 \text{ cm}^2) \approx 153.9 \text{ cm}^2
\]
**3. Calculate the height (distance) the water level falls:**
When the ball is removed, it displaces an equivalent volume of water. The fall in the water level \( h \) can be determined using the formula:
\[
\text{Volume of water displaced} = \text{Base Area} \times \text{Height}
\]
\[
V_{\text{ball}} = A \times h
\]
Rearranging gives:
\[
h = \frac{V_{\text{ball}}}{A}
\]
Substituting the known values:
\[
h = \frac{\frac{500}{3} \pi \text{ cm}^3}{\pi (49 \text{ cm}^2)}
\]
The \(\pi\) cancels out:
\[
h = \frac{\frac{500}{3}}{49}
\]
\[
h = \frac{500}{147} \approx 3.40 \text{ cm}
\]
### Conclusion:
The distance the water level falls when the 10 cm diameter ball is removed from the 14 cm diameter cylindrical container is approximately **3.40 cm**.