Q.A pump on the ground floor of a building can pump up water to fill a tank of volume $30\ \text{m}^{3}$ in $15\ \text{min}$. If the tank is $40\ \text{m}$ above the ground, and the efficiency of the pump is $30\%$, how much electric power is consumed by the pump?

Concept: Power Against Gravity — the pump must supply power to raise water against gravity, and the input electric power is larger due to efficiency losses.

Step 1 — Work done against gravity

Mass of water: $m = \rho V = 1000 \times 30 = 3 \times 10^{4}\ \text{kg}$

Work done: $W = mgh = 3 \times 10^{4} \times 9.8 \times 40 = 1.176 \times 10^{7}\ \text{J}$

(Using $g = 9.8\ \text{m/s}^2$, the standard NCERT value.)

Step 2 — Output power of pump

Time: $t = 15\ \text{min} = 900\ \text{s}$

$P_{\text{out}} = \frac{W}{t} = \frac{1.176 \times 10^{7}}{900} \approx 1.307 \times 10^{4}\ \text{W} = 13.07\ \text{kW}$ …

Why This Approach Works

The pump's job is to lift water against gravity. Every kilogram of water raised to height $h$ gains gravitational potential energy $mgh$. The pump doesn't create this energy — it converts electrical energy into mechanical work, but only 30% of the electrical input actually goes into lifting water. The rest is lost as heat, noise, etc.

So the chain is: electric power → mechanical power (30% efficient) → rate of gaining potential energy. We know the volume flow rate and the height, so we can find the required mechanical power, then back-calculate the electrical power.

Step-by-Step Solution

1. Find the mass flow rate of water

Water density is $\rho = 1000\ \text{kg/m}^3$. Volume $V = 30\ \text{m}^3$ is pumped in time $t = 15\ \text{min} = 15 \times 60 = 900\ \text{s}$.

Mass of water: $m = \rho V = 1000 \times 30 = 30\,000\ \text{kg}$.

Mass flow rate:

2. Calculate the rate of potential energy gain (useful power)

Height $h = 40\ \text{m}$, $g = 9.8\ \text{m/s}^2$.

Each second, the water gains potential energy at the rate:

Compute stepwise:

$\frac{100}{3} \times 9.8 = \frac{980}{3}$,

then $\frac{980}{3} \times 40 = \frac{39\,200}{3} \approx 13\,066.67\ \text{W}$.

So $P_{\text{useful}} \approx 13.07\ \text{kW}$.

3. Account for pump efficiency

Efficiency $\eta = 30\% = 0.30$. Efficiency is defined as:

Here, useful output is $P_{\text{useful}}$, and input is the electric power $P_{\text{electric}}$ we need.

So:

4. Express in kilowatts

$P_{\text{electric}} \approx 43.6\ \text{kW}.$ …

Concept: Power Against Gravity with Efficiency

Step 1: Mass flow rate of water

Step 2: Useful (mechanical) power delivered to the water

Step 3: Apply pump efficiency (input power must be larger than useful output) …