A cubical vessel of height `1` m is full of water. What is the workdone in pumping water out of the vessel?

Explanation

To find the work done in pumping water out of a cubical vessel of height 1 m, we can follow these steps:

### Step 1: Determine the Volume of the Cubical Vessel
Since the vessel is cubical and has a height of 1 m, the volume \( V \) of the vessel can be calculated using the formula for the volume of a cube:
\[ V = L^3 \]
where \( L \) is the length of a side of the cube. Here, \( L = 1 \, \text{m} \).
\[ V = 1^3 = 1 \, \text{m}^3 \]

### Step 2: Calculate the Mass of Water in the Vessel
The mass \( m \) of the water can be found using the density \( \rho \) of water, which is approximately \( 1000 \, \text{kg/m}^3 \):
\[ m = \rho \times V \]
Substituting the values:
\[ m = 1000 \, \text{kg/m}^3 \times 1 \, \text{m}^3 = 1000 \, \text{kg} \]

### Step 3: Determine the Height of the Center of Mass
For a uniform cube, the center of mass is located at half the height of the cube. Thus, the height \( h \) to which we need to pump the water is:
\[ h = \frac{L}{2} = \frac{1}{2} \, \text{m} = 0.5 \, \text{m} \]

### Step 4: Calculate the Work Done to Pump the Water
The work done \( W \) in lifting the water to the top of the vessel can be calculated using the formula:
\[ W = m \cdot g \cdot h \]
where \( g \) is the acceleration due to gravity, approximately \( 9.8 \, \text{m/s}^2 \). Substituting the values:
\[ W = 1000 \, \text{kg} \times 9.8 \, \text{m/s}^2 \times 0.5 \, \text{m} \]
\[ W = 1000 \times 9.8 \times 0.5 = 4900 \, \text{J} \]

### Final Answer
The work done in pumping the water out of the vessel is \( 4900 \, \text{J} \).

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