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

Options

A. `36.5 kW`
B. `44.4 kW`
C. `52.5 kW`
D. `60.5 kW`

Explanation

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

### Step 1: Calculate the mass of water to be pumped
The volume of water to be pumped is given as \( V = 30 \, m^3 \). The density of water is \( \rho = 1000 \, kg/m^3 \).

Using the formula for mass:
\[ m = V \times \rho \]
Substituting the values:
\[ m = 30 \, m^3 \times 1000 \, kg/m^3 = 30000 \, kg \]

### Step 2: Calculate the work done to lift the water
The work done \( W \) against gravity to lift the mass \( m \) to a height \( h \) is given by:
\[ W = m \times g \times h \]
Where \( g = 10 \, m/s^2 \) and \( h = 40 \, m \).

Substituting the values:
\[ W = 30000 \, kg \times 10 \, m/s^2 \times 40 \, m = 12000000 \, J \]

### Step 3: Calculate the output power of the pump
The time taken to fill the tank is given as \( t = 15 \, minutes = 15 \times 60 \, seconds = 900 \, seconds \).

Power \( P \) is defined as work done per unit time:
\[ P_{output} = \frac{W}{t} \]
Substituting the values:
\[ P_{output} = \frac{12000000 \, J}{900 \, s} = 13333.33 \, W \approx 13333 \, W \]

### Step 4: Calculate the input power using efficiency
The efficiency \( \eta \) of the pump is given as \( 30\% = 0.3 \).

Using the formula for efficiency:
\[ \eta = \frac{P_{output}}{P_{input}} \]
Rearranging to find \( P_{input} \):
\[ P_{input} = \frac{P_{output}}{\eta} \]
Substituting the values:
\[ P_{input} = \frac{13333 \, W}{0.3} = 44443.33 \, W \approx 44444 \, W \]

### Step 5: Convert to kilowatts
To convert watts to kilowatts:
\[ P_{input} = \frac{44444 \, W}{1000} = 44.44 \, kW \]

### Final Answer
The electric power consumed by the pump is approximately **44.44 kW**.

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