A water filled cylinder of height 50 cm and base area `20cm^(2)` is placed on a table with the base on the table. The thrust offered by water on the table is

Options

A. 98 N
B. 49 N
C. 9.8 N
D. 4.9 N

Explanation

To solve the problem of finding the thrust offered by the water on the table, we will follow these steps:

### Step 1: Understand the Concept of Thrust
Thrust is the force exerted by the water on the table due to the weight of the water column above it. The thrust can be calculated using the formula:
\[ \text{Thrust} = \text{Pressure} \times \text{Area} \]

### Step 2: Calculate the Height of the Water Column
The height of the water column (h) is given as 50 cm. We need to convert this into meters for consistency in SI units:
\[ h = 50 \, \text{cm} = 0.50 \, \text{m} \]

### Step 3: Determine the Density of Water
The density of water (\( \rho \)) is approximately:
\[ \rho = 1000 \, \text{kg/m}^3 \]

### Step 4: Calculate the Pressure Exerted by the Water Column
The hydrostatic pressure (P) at the base of the cylinder can be calculated using the formula:
\[ P = h \cdot \rho \cdot g \]
Where \( g \) (acceleration due to gravity) is approximately \( 9.81 \, \text{m/s}^2 \).

Substituting the values:
\[ P = 0.50 \, \text{m} \cdot 1000 \, \text{kg/m}^3 \cdot 9.81 \, \text{m/s}^2 \]
\[ P = 0.50 \cdot 1000 \cdot 9.81 \]
\[ P = 4905 \, \text{Pa} \] (Pascals)

### Step 5: Calculate the Area of the Base of the Cylinder
The base area (A) is given as \( 20 \, \text{cm}^2 \). We need to convert this into square meters:
\[ A = 20 \, \text{cm}^2 = 20 \times 10^{-4} \, \text{m}^2 = 0.0020 \, \text{m}^2 \]

### Step 6: Calculate the Thrust
Now, we can calculate the thrust using the pressure and area:
\[ \text{Thrust} = P \cdot A \]
\[ \text{Thrust} = 4905 \, \text{Pa} \cdot 0.0020 \, \text{m}^2 \]
\[ \text{Thrust} = 9.81 \, \text{N} \]

### Final Answer
The thrust offered by the water on the table is approximately:
\[ \text{Thrust} \approx 9.81 \, \text{N} \]

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