The area of base of a cylindrical vessel is `300 cm^(2)`. Water `("density"=1000 kg m^(-3))` is poured into it up to a depth of 6 cm. Calculate : (a) the pressure and (b) the thrust of water on the base. `(g=10ms^(-2))`.

Answer

To solve the problem, we will follow these steps:

### Step 1: Convert the given measurements to appropriate units
- The area of the base of the cylindrical vessel is given as \(300 \, \text{cm}^2\). We need to convert this to square meters.

\[ \text{Area} = 300 \, \text{cm}^2 = 300 \times 10^{-4} \, \text{m}^2 = 0.03 \, \text{m}^2 \]

- The depth of water is given as \(6 \, \text{cm}\). We need to convert this to meters as well.

\[ \text{Height} = 6 \, \text{cm} = 6 \times 10^{-2} \, \text{m} = 0.06 \, \text{m} \]

### Step 2: Calculate the pressure at the depth of water
- The formula for pressure \(P\) at a certain depth in a fluid is given by:

\[ P = \text{density} \times g \times h \]

where:
- Density of water (\(\rho\)) = \(1000 \, \text{kg/m}^3\)
- Gravitational acceleration (\(g\)) = \(10 \, \text{m/s}^2\)
- Height of water (\(h\)) = \(0.06 \, \text{m}\)

Substituting the values into the formula:

\[ P = 1000 \, \text{kg/m}^3 \times 10 \, \text{m/s}^2 \times 0.06 \, \text{m} = 600 \, \text{Pa} \]

### Step 3: Calculate the thrust of water on the base
- The thrust \(F\) on the base can be calculated using the formula:

\[ F = P \times A \]

where:
- Pressure \(P\) = \(600 \, \text{Pa}\)
- Area \(A\) = \(0.03 \, \text{m}^2\)

Substituting the values:

\[ F = 600 \, \text{Pa} \times 0.03 \, \text{m}^2 = 18 \, \text{N} \]

### Final Answers:
(a) The pressure at the depth of 6 cm is \(600 \, \text{Pa}\).

(b) The thrust of water on the base is \(18 \, \text{N}\).

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