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We have two vertically stacked water-filled cylinders. One has a diameter of 0.25 m and height of 1.2 m, while the other has a diameter of 0.9 m and height of 0.8 m. The goal is to calculate:

1. The **total pressure** at the bottom of the larger cylinder.
2. The **weight** of the total volume of water.

Let's break these calculations down.

### 1. **Total Pressure at the Bottom of the Larger Cylinder**

The pressure at a given depth in a fluid is given by:

\[
P = \rho \cdot g \cdot h
\]

where:
- \( \rho \) = density of water (\( 1000 \, \text{kg/m}^3 \)),
- \( g \) = gravitational acceleration (\( 9.81 \, \text{m/s}^2 \)),
- \( h \) = total depth of water above the point in question (in meters).

The height of water above the bottom of the larger cylinder consists of the height of both cylinders:

- The smaller cylinder height = 1.2 m
- The larger cylinder height = 0.8 m

Thus, the total depth \( h = 1.2 + 0.8 = 2.0 \, \text{m} \).

Now, calculating the pressure:

\[
P = 1000 \times 9.81 \times 2.0
\]

\[
P = 19620 \, \text{Pa}
\]

So, the total pressure at the bottom of the larger cylinder is \( 19620 \, \text{Pa} \) (Pascals).

### 2. **Weight of Total Volume of Water**

The volume of water in both cylinders needs to be calculated first:

- **Smaller cylinder** volume \( V_1 \):
  \[
  V_1 = \pi \times \left(\frac{d_1}{2}\right)^2 \times h_1
  \]
  where \( d_1 = 0.25 \, \text{m} \) and \( h_1 = 1.2 \, \text{m} \).
  
- **Larger cylinder** volume \( V_2 \):
  \[
  V_2 = \pi \times \left(\frac{d_2}{2}\right)^2 \times h_2
  \]
  where \( d_2 = 0.9 \, \text{m} \) and \( h_2 = 0.8 \, \text{m} \).

The total volume \( V_{\text{total}} = V_1 + V_2 \).

Finally, the weight of the water can be calculated using:

\[
W = \rho \times V_{\text{total}} \times g
\]

Let me calculate these step-by-step.Here are the calculations for the volume of water in each cylinder and the total weight:

- **Volume of water in the smaller cylinder**: \( 0.0589 \, \text{m}^3 \)
- **Volume of water in the larger cylinder**: \( 0.5089 \, \text{m}^3 \)
- **Total volume of water**: \( 0.5678 \, \text{m}^3 \)

Now, the **weight of the total volume of water** is:

\[
W = 5570.54 \, \text{N} \, (\text{Newtons})
\]

### Summary:
1. **Total pressure** at the bottom of the larger cylinder: \( 19620 \, \text{Pa} \)
2. **Weight** of the total volume of water: \( 5570.54 \, \text{N} \)

Would you like further details or clarifications?

### Follow-up questions:
1. How would the pressure change if the height of the smaller cylinder increased?
2. What happens to the total pressure if the fluid is replaced by a denser liquid?
3. How is the water pressure distributed in the smaller cylinder?
4. What if the larger cylinder's diameter doubled? How would the volume change?
5. How would atmospheric pressure affect the total pressure calculation?

**Tip:** When calculating pressure at a depth, it's important to consider both the height of the liquid and the properties of the liquid itself (e.g., density).

A cylinder of 0.25 m diameter and 1.2 m height is fixed centrally on top of a large cylinder of 0.9 m diameter and 0.8 m height. Both the cylinders are filled with water. Calculate:
1. Total pressure at the bottom of the bigger cylinder
2. Weight of total volume of water

Learn how to calculate the total pressure at the bottom of a large water-filled cylinder and the weight of water in two stacked cylinders of different dimensions. This fluid mechanics problem is ideal for students in Grades 10-12.

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