A pump on the ground floor of a building can pump of water to fill a tank of volume `30 ms^(3)` in `15 min`. If the tank is `40m` above the ground and the efficiency of the pump is `30 %` , how much electric power is consumed by the pump? `("Take" g = 10 ms^(2))`

Options

A. 36.5 kW
B. 50 kW
C. 52.5 kW
D. 60.5 kW

Explanation

To solve the problem step by step, we will follow the given information and apply the relevant formulas.

### Step 1: Calculate the mass of water
We know the volume of water (V) is 30 m³ and the density of water (ρ) is \(1000 \, \text{kg/m}^3\).

\[ \text{Mass of water (m)} = \text{Volume} \times \text{Density} = V \times \rho = 30 \, \text{m}^3 \times 1000 \, \text{kg/m}^3 = 30000 \, \text{kg} = 3 \times 10^4 \, \text{kg} \]

### Step 2: Calculate the work done to lift the water
The work done (W) against gravity to lift the water to a height (h) of 40 m is given by:

\[ W = m \cdot g \cdot h \]

Where \(g = 10 \, \text{m/s}^2\).

\[ W = 3 \times 10^4 \, \text{kg} \times 10 \, \text{m/s}^2 \times 40 \, \text{m} = 1.2 \times 10^6 \, \text{J} \]

### Step 3: Calculate the output power of the pump
The output power (P_output) is the work done divided by the time (t) taken to do that work. The time given is 15 minutes, which we convert to seconds:

\[ t = 15 \, \text{min} = 15 \times 60 \, \text{s} = 900 \, \text{s} \]

Now we can calculate the output power:

\[ P_{\text{output}} = \frac{W}{t} = \frac{1.2 \times 10^6 \, \text{J}}{900 \, \text{s}} = \frac{1.2 \times 10^6}{900} \approx 1333.33 \, \text{W} \approx 1.33 \, \text{kW} \]

### Step 4: Calculate the input power of the pump
The efficiency (η) of the pump is given as 30%, or 0.3 in decimal form. The relationship between output power and input power is given by:

\[ \eta = \frac{P_{\text{output}}}{P_{\text{input}}} \]

Rearranging this gives:

\[ P_{\text{input}} = \frac{P_{\text{output}}}{\eta} \]

Substituting the values we have:

\[ P_{\text{input}} = \frac{1333.33 \, \text{W}}{0.3} \approx 4444.44 \, \text{W} \approx 4.44 \, \text{kW} \]

### Final Answer
The electric power consumed by the pump is approximately **4.44 kW**.

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