Question
A cylindrical tank of radius 1.5 m and height 3.5 m is full of water. Due to a crack, water starts leaking from the tank at a rate of 5.5 L per min. How long will it take for the tank to get empty? (1 L = 1000 cubic cm , pi = 22/7)
Answers
GPT-4o mini
To find out how long it will take for the cylindrical tank to empty, we first need to calculate the total volume of water in the tank.
The volume \( V \) of a cylinder is given by the formula:
\[
V = \pi r^2 h
\]
where
- \( r \) is the radius of the cylinder, and
- \( h \) is the height of the cylinder.
Given:
- Radius \( r = 1.5 \) m
- Height \( h = 3.5 \) m
- \( \pi \approx \frac{22}{7} \)
Now, let's calculate the volume of the tank.
1. Plug in the values into the volume formula:
\[
V = \frac{22}{7} \times (1.5)^2 \times 3.5
\]
2. Calculate \( (1.5)^2 \):
\[
(1.5)^2 = 2.25
\]
3. Now substitute back into the equation:
\[
V = \frac{22}{7} \times 2.25 \times 3.5
\]
4. Calculate \( 2.25 \times 3.5 \):
\[
2.25 \times 3.5 = 7.875
\]
5. Finally, substitute that back into the volume equation:
\[
V = \frac{22}{7} \times 7.875
\]
6. Perform the multiplication:
\[
V = 22 \times 1.125 = 24.75 \, \text{m}^3
\]
Now, since \( 1 \, \text{m}^3 = 1000 \, \text{L} \):
\[
V = 24.75 \times 1000 = 24750 \, \text{L}
\]
Now we need to determine how long it takes to leak this volume of water at a rate of 5.5 L/min.
To find the time \( t \) in minutes, we use the formula:
\[
t = \frac{\text{Total Volume}}{\text{Leak Rate}} = \frac{24750 \, \text{L}}{5.5 \, \text{L/min}}
\]
Calculating that gives:
\[
t = \frac{24750}{5.5} \approx 4500 \, \text{minutes}
\]
To convert minutes into hours:
\[
4500 \div 60 = 75 \, \text{hours}
\]
Thus, it will take approximately **4500 minutes**, or **75 hours**, for the tank to empty completely.
The volume \( V \) of a cylinder is given by the formula:
\[
V = \pi r^2 h
\]
where
- \( r \) is the radius of the cylinder, and
- \( h \) is the height of the cylinder.
Given:
- Radius \( r = 1.5 \) m
- Height \( h = 3.5 \) m
- \( \pi \approx \frac{22}{7} \)
Now, let's calculate the volume of the tank.
1. Plug in the values into the volume formula:
\[
V = \frac{22}{7} \times (1.5)^2 \times 3.5
\]
2. Calculate \( (1.5)^2 \):
\[
(1.5)^2 = 2.25
\]
3. Now substitute back into the equation:
\[
V = \frac{22}{7} \times 2.25 \times 3.5
\]
4. Calculate \( 2.25 \times 3.5 \):
\[
2.25 \times 3.5 = 7.875
\]
5. Finally, substitute that back into the volume equation:
\[
V = \frac{22}{7} \times 7.875
\]
6. Perform the multiplication:
\[
V = 22 \times 1.125 = 24.75 \, \text{m}^3
\]
Now, since \( 1 \, \text{m}^3 = 1000 \, \text{L} \):
\[
V = 24.75 \times 1000 = 24750 \, \text{L}
\]
Now we need to determine how long it takes to leak this volume of water at a rate of 5.5 L/min.
To find the time \( t \) in minutes, we use the formula:
\[
t = \frac{\text{Total Volume}}{\text{Leak Rate}} = \frac{24750 \, \text{L}}{5.5 \, \text{L/min}}
\]
Calculating that gives:
\[
t = \frac{24750}{5.5} \approx 4500 \, \text{minutes}
\]
To convert minutes into hours:
\[
4500 \div 60 = 75 \, \text{hours}
\]
Thus, it will take approximately **4500 minutes**, or **75 hours**, for the tank to empty completely.