A vessel contains water up to a height of `1.5` m. Taking the density of water `10^(3)kg m^(-3)`, acceleration due to gravity `9.8 ms^(-2)` and area of base of vessel `100 cm^(2)` calculate : (a) the pressure and (b) the thrust, at the base of vessel.

Answer

To solve the problem step by step, we will calculate the pressure and thrust at the base of the vessel containing water.

### Step 1: Calculate the Pressure at the Base of the Vessel

The formula for pressure (P) at a certain depth in a fluid is given by:

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

Where:
- \( \rho \) = density of the fluid (water in this case) = \( 1000 \, \text{kg/m}^3 \)
- \( g \) = acceleration due to gravity = \( 9.8 \, \text{m/s}^2 \)
- \( h \) = height of the water column = \( 1.5 \, \text{m} \)

Substituting the values into the formula:

\[ P = 1000 \, \text{kg/m}^3 \cdot 9.8 \, \text{m/s}^2 \cdot 1.5 \, \text{m} \]

Calculating the multiplication step-by-step:

1. Calculate \( 9.8 \cdot 1.5 = 14.7 \)
2. Now, multiply by the density: \( 14.7 \cdot 1000 = 14700 \)

Thus, the pressure at the base of the vessel is:

\[ P = 14700 \, \text{Pa} \, (\text{Pascals}) \]

### Step 2: Calculate the Thrust at the Base of the Vessel

The thrust (F) at the base of the vessel can be calculated using the formula:

\[ F = P \cdot A \]

Where:
- \( P \) = pressure calculated previously = \( 14700 \, \text{Pa} \)
- \( A \) = area of the base of the vessel

The area is given as \( 100 \, \text{cm}^2 \). We need to convert this to \( \text{m}^2 \):

\[ A = 100 \, \text{cm}^2 = \frac{100}{10000} \, \text{m}^2 = 0.01 \, \text{m}^2 \]

Now substituting the values into the thrust formula:

\[ F = 14700 \, \text{Pa} \cdot 0.01 \, \text{m}^2 \]

Calculating this gives:

\[ F = 147 \, \text{N} \]

### Final Answers

(a) The pressure at the base of the vessel is \( 14700 \, \text{Pa} \).

(b) The thrust at the base of the vessel is \( 147 \, \text{N} \).

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