`4m^(3)` of water is to be pumped to a height of 20 m and forced into a reservoir at a pressure of `2xx10^(5)Nm^(-2)`. The work done by the motor is (external pressure `= 10^(5)Nm^(-2)`)

Options

A. (a)`8xx10^(5)J`
B. (b)`16xx10^(5)J`
C. (c)`12xx10^(5)J`
D. (d)`32xx10^(5)J`

Answer

To solve the problem of calculating the work done by the motor in pumping water, we will break it down into several steps.

### Step-by-Step Solution:

1. **Identify Given Values**:
- Volume of water, \( V = 4 \, \text{m}^3 \)
- Height to which water is pumped, \( h = 20 \, \text{m} \)
- Pressure at the reservoir, \( P = 2 \times 10^5 \, \text{N/m}^2 \)
- External pressure, \( P_{\text{external}} = 1 \times 10^5 \, \text{N/m}^2 \)
- Density of water, \( \rho = 10^3 \, \text{kg/m}^3 \)
- Acceleration due to gravity, \( g = 10 \, \text{m/s}^2 \)

2. **Calculate the Mass of Water**:
\[ m = \rho \times V = 10^3 \, \text{kg/m}^3 \times 4 \, \text{m}^3 = 4000 \, \text{kg} \]

3. **Calculate Work Done Against Gravity (Potential Energy)**:
\[ W_{\text{gravity}} = mgh = 4000 \, \text{kg} \times 10 \, \text{m/s}^2 \times 20 \, \text{m} = 800000 \, \text{J} = 8 \times 10^5 \, \text{J} \]

4. **Calculate Work Done Against External Pressure**:
\[ W_{\text{pressure}} = P_{\text{external}} \times V = 1 \times 10^5 \, \text{N/m}^2 \times 4 \, \text{m}^3 = 400000 \, \text{J} = 4 \times 10^5 \, \text{J} \]

5. **Total Work Done by the Motor**:
\[ W_{\text{total}} = W_{\text{gravity}} + W_{\text{pressure}} = 8 \times 10^5 \, \text{J} + 4 \times 10^5 \, \text{J} = 12 \times 10^5 \, \text{J} \]

### Final Answer:
The work done by the motor is \( 12 \times 10^5 \, \text{J} \).

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