An electric pump on the ground floor of a building takes 10 min to fill a tank of volume 2000 L with water. If the tank is 40 m above the ground and the efficiency of the pump is 40%, how much electric power is consumed by the pump in filling the tank?
Take `g=10 m//s^(2)`

Options

A. 2 kW
B. 3.33 kW
C. 4 kW
D. 6 kW

Explanation

To find the electric power consumed by the pump in filling the tank, we can follow these steps:

### Step 1: Calculate the Work Done by the Pump
The work done (W) by the pump to lift the water to a height (h) can be calculated using the formula:
\[ W = mgh \]
where:
- \(m\) is the mass of the water,
- \(g\) is the acceleration due to gravity,
- \(h\) is the height the water is lifted.

### Step 2: Convert Volume of Water to Mass
The volume of the tank is given as 2000 L. We need to convert this to cubic meters (m³):
\[ 2000 \, \text{L} = 2000 \times 10^{-3} \, \text{m}^3 = 2 \, \text{m}^3 \]
The density of water is \(1000 \, \text{kg/m}^3\), so the mass \(m\) of the water is:
\[ m = \text{density} \times \text{volume} = 1000 \, \text{kg/m}^3 \times 2 \, \text{m}^3 = 2000 \, \text{kg} \]

### Step 3: Calculate Work Done
Now we can calculate the work done:
\[ W = mgh = 2000 \, \text{kg} \times 10 \, \text{m/s}^2 \times 40 \, \text{m} = 800000 \, \text{J} = 8 \times 10^5 \, \text{J} \]

### Step 4: Calculate Time in Seconds
The time taken to fill the tank is given as 10 minutes. We need to convert this to seconds:
\[ 10 \, \text{min} = 10 \times 60 \, \text{s} = 600 \, \text{s} \]

### Step 5: Calculate Output Power
The output power (\(P_{\text{out}}\)) is given by the work done divided by the time taken:
\[ P_{\text{out}} = \frac{W}{t} = \frac{8 \times 10^5 \, \text{J}}{600 \, \text{s}} = \frac{8 \times 10^5}{600} \approx 1333.33 \, \text{W} \]

### Step 6: Calculate Input Power
The efficiency (\(η\)) of the pump is given as 40%, or 0.4. The input power (\(P_{\text{in}}\)) can be calculated using the relationship:
\[ η = \frac{P_{\text{out}}}{P_{\text{in}}} \]
Rearranging this gives:
\[ P_{\text{in}} = \frac{P_{\text{out}}}{η} = \frac{1333.33 \, \text{W}}{0.4} = 3333.33 \, \text{W} \]

### Step 7: Convert to Kilowatts
To express the power in kilowatts:
\[ P_{\text{in}} = 3.33 \, \text{kW} \]

### Final Answer
The electric power consumed by the pump in filling the tank is approximately:
\[ \boxed{3.33 \, \text{kW}} \]