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?

A pump on the ground floor of a building can pump up water to fill a tank of volume 30m³ in 15min. If the tank

Exercises · 5.15

30\ \text{m}^{3}

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Concept understanding — Power Against Gravity

Power Against Gravity: The Intuition

Imagine you're lifting a bucket of water from a well. The bucket is heavy — gravity is pulling it down with a force equal to its weight. To lift it, you must apply an upward force that exactly cancels gravity. If you lift it slowly, you feel the strain for a long time. If you yank it up quickly, you feel a burst of effort, but it's over fast.

That "burst of effort per unit time" is what we call power. When you're working against a constant force like gravity, power tells you how fast you're doing that work.

Note

Power is not the force itself, nor the work alone — it's the rate at which work is done. Lifting the same bucket to the same height requires the same total work, regardless of speed. But doing it faster requires more power.

The Precise Statement

When an object moves vertically against gravity (upward), the force you must supply is at least equal to the object's weight:

F = mg

where $m$ is mass and $g$ is acceleration due to gravity ( $9.8\ \text{m/s}^2$ on Earth).

The work done to lift it through a height $h$ is:

W = F \cdot h = mgh

Now, power $P$ is work per unit time. If you lift it in time $t$ :

P = \frac{W}{t} = \frac{mgh}{t}

But $\frac{h}{t}$ is just the upward speed $v$ (assuming constant speed). So we get the compact form:

P = mg \cdot v

This is power against gravity — the rate at which you must supply energy to lift a mass $m$ at constant speed $v$ against the pull of gravity.

What This Really Means

It's a minimum. If you accelerate the object upward, you need even more force (Newton's second law), and hence more power. The formula $P = mgv$ assumes you're lifting at steady speed — no acceleration.
Direction matters. If the object moves downward at constant speed, gravity does the work, and you (or a brake) must absorb power. The formula still gives the magnitude, but the sign flips.
It's independent of path. Only the vertical speed matters. Whether you lift straight up or along a ramp, the power against gravity depends only on the vertical component of velocity.

Tip

For a quick calculation: lifting a 10 kg mass at 0.5 m/s requires $P = 10 \times 9.8 \times 0.5 \approx 49$ watts. That's about the power of a dim incandescent bulb — and you'd feel it after a minute.

Common Mistake to Avoid …

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The pump must supply gravitational potential energy to the water at a certain rate. Accounting for 30% efficiency, the electric power consumed is $\boxed{43.6\ \text{kW}}$ .

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:

\dot{m} = \frac{m}{t} = \frac{30\,000}{900} = \frac{100}{3}\ \text{kg/s} \approx 33.33\ \text{kg/s}.

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:

P_{\text{useful}} = \dot{m} g h = \frac{100}{3} \times 9.8 \times 40.

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

Note

This is the mechanical power that actually lifts the water. If the pump were 100% efficient, this would be the electric power too.

3. Account for pump efficiency

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

\eta = \frac{\text{useful power output}}{\text{total power input}}.

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

So:

P_{\text{electric}} = \frac{P_{\text{useful}}}{\eta} = \frac{13\,066.67}{0.30} \approx 43\,555.56\ \text{W}.

4. Express in kilowatts

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

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Same / Similar Concept — real previous-year questions on the same or a closely similar concept, not this exact question.

Showing the 12 most recent of 13 on this concept.

AP EAPCET 2026Set eng-2026-05-15-FN1 markMCQ
$Q. A person lifts 60 kg load to a vertical height of 30 m over a duration of 20 seconds. If the power of the man is 1323 W, the mass of the man is (A) 30 kg (B) 40 kg (C) 50 kg (D) 60 kg$
›Reveal solutionSolution
The man's power output is spent lifting the total weight (himself + the load) through the given height in the given time. Solving $P = (M+60)gh/t$ for $M$ (using $g=9.8\,ms^{-2}$ ) gives $M=30\,kg$ .

Concept and Intuition

When a person carries a load up a height, the work done against gravity is on the combined mass — the person's own body plus whatever they're carrying — because both must be raised through the same vertical distance. Average power is just this total work divided by the time taken.

Step-by-Step Solution

Total work done $= (M + 60)\,g\,h$ , where $M$ is the man's mass, $60\,kg$ is the load, $h=30\,m$ .

Average power $P = \dfrac{\text{Work}}{\text{time}} = \dfrac{(M+60)gh}{t}$ , with $t = 20\,s$ .

Substitute the given power $P=1323\,W$ and $g=9.8\,ms^{-2}$ :

$1323 = \frac{(M+60)(9.8)(30)}{20} = (M+60)(14.7)$

Solve: $M + 60 = \dfrac{1323}{14.7} = 90 \Rightarrow M = 30\,kg$ .

Common Mistakes …
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AP EAPCET 2025Set eng-2025-05-21-FN1 markMCQ
$50\ \mathrm{m\,s^{-1}}$
›Reveal solutionSolution
This tests that at constant velocity, driving force balances friction, and power is force times speed; the answer is (A).

Concept and Intuition

When a vehicle moves at constant speed, its net force is zero, so the engine's tractive force must exactly balance the resistive (friction) force. Once you know that force, the power required to maintain the speed is simply $P=Fv$ .

Step-by-Step Solution

Friction force $=\mu_k mg$ , where $\mu_k=0.05$ , $m=3\times10^6$ kg, $g=10\ \mathrm{m/s^2}$ .

$F=0.05\times3\times10^6\times10=0.05\times3\times10^7=1.5\times10^6$ N.

Since speed is constant, the engine must supply exactly this force to overcome friction. …
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AP EAPCET 2025Set eng-2025-05-22-AN1 markMCQ
$10\ \mathrm{m\,s^{-2}}$
›Reveal solutionSolution
The useful output power (lifting work per unit time) is 2.4 kW; dividing by the 80% efficiency gives the crane's actual power rating, 3 kW.

Concept and Intuition

Efficiency $\eta = \dfrac{\text{useful output power}}{\text{power supplied/consumed}}$ . The work of lifting the coal against gravity is the useful output; because the crane is only 80% efficient, it must actually consume (draw) more power than this useful output to compensate for losses.

Step-by-Step Solution

Useful work done $=mgh = 8000\times10\times108 = 8{,}640{,}000\ \mathrm{J}$ .

Useful (output) power $=\dfrac{W}{t}=\dfrac{8{,}640{,}000}{3600}=2400\ \mathrm{W}=2.4\ \mathrm{kW}$ . …
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AP EAPCET 2025Set eng-2025-05-22-FN1 markMCQ
$10^6$
›Reveal solutionSolution
At constant speed the driving force equals friction; power is force times velocity, giving 150 kW.

Concept and Intuition

A train moving at constant speed has zero net force, so the engine's driving force must exactly balance friction. Once that force is known, power delivered is simply $P=Fv$ .

Step-by-Step Solution

Convert speed: $108\text{ km/h}=108\times\dfrac{5}{18}=30\ \mathrm{m/s}$ .

Total mass $=10^6$ kg. Friction is given as $0.5$ N per $100$ kg, so total friction: $f=\frac{10^6}{100}\times0.5=10^4\times0.5=5000\ \mathrm{N}.$

At constant velocity, the train's driving force equals this friction force: $F=5000$ N. …
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AP EAPCET 2025Set eng-2025-05-23-AN1 markMCQ
$= 10\ \mathrm{m\,s^{-2}}$
›Reveal solutionSolution
Tests work needed to pump water out of a well against gravity, using energy/power, where different layers of water are lifted through different heights. Answer: 2200 s.

Concept and Intuition

Unlike lifting a single mass through a fixed height, emptying a well means water at different depths must be lifted different distances — water near the bottom travels the full 14 m, water near the top travels almost nothing. So we must integrate over depth to get the total work, then use $P = W/t$ to find time, since the motor delivers energy at a constant rate.

Step-by-Step Solution

Well radius $= 1$ m (diameter 2 m), so cross-sectional area $A = \pi r^2 = \pi\ \mathrm{m^2}$ .

Consider a thin layer of water of thickness $dy$ at depth $y$ (measured from the top/ground level, $y$ from 0 to 14 m). Its mass is $dm = \rho A\, dy$ , and it must be raised through height $y$ to reach the ground.

Work to raise this layer: $dW = dm\cdot g \cdot y = \rho A g\, y\, dy$ .

Total work to empty the well:

$W = \rho A g \int_0^{14} y\, dy = \rho A g \cdot \frac{14^2}{2} = \rho A g \times 98$

Substituting $\rho=1000\ \mathrm{kg/m^3}$ , $A=\pi\ \mathrm{m^2}$ , $g=10\ \mathrm{m/s^2}$ : …
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AP EAPCET 2025Set eng-2025-05-26-AN1 markMCQ
$= 10\ \mathrm{m\,s^{-2}}$
›Reveal solutionSolution
Compute the useful gravitational power output, then divide by efficiency to get the pump's actual (input) power: 3 kW.

Concept and Intuition

The useful work done by the pump is lifting water against gravity; dividing this useful output rate by the given efficiency gives the actual power the pump consumes/delivers, since efficiency $=\dfrac{\text{output}}{\text{input}}$ .

Step-by-Step Solution

Mass of water pumped per second: $\dfrac{7560\ \text{kg}}{3600\ \text{s}} = 2.1\ \mathrm{kg\,s^{-1}}$ .

Useful power output $= \dfrac{mgh}{t} = 2.1\times10\times100 = 2100$ W. …
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AP EAPCET 2024Set ap-2024-05-17-AN1 markMCQ
$(g = 10\ \text{ms}^{-2})$
›Reveal solutionSolution
The engine must supply a force equal to the gravity component along the incline to keep the truck moving at constant speed; power is simply this force times speed.

Concept and Intuition

"Rising 1 in 20" describes the slope as a ratio: for every 20 m travelled along the incline, the truck rises 1 m, so $\sin\theta = 1/20$ (this is the conventional interpretation used for gentle inclines, where $\sin\theta \approx \tan\theta$ for small angles is not even needed — the ratio is $\sin\theta$ by the problem's own convention of rise per unit length along the slope). At constant velocity, net force is zero, so the engine's driving force must exactly balance the component of gravity pulling the truck back down the slope.

Step-by-Step Solution

Convert speed: $18\ \text{km/h} = \dfrac{18\times1000}{3600} = 5\ \text{m/s}$ .

Slope gives $\sin\theta = \dfrac{1}{20}$ . …
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AP EAPCET 2024Set ap-2024-05-17-FN1 markMCQ
$5 \times 10^3$
›Reveal solutionSolution
This tests power delivered by a motor against gravity and friction. Both the weight and the frictional resistance must be overcome at constant speed, giving 62.5 kW.

Concept and Intuition

For a body moving at constant velocity, the net force is zero, so the driving force from the motor must exactly balance all opposing forces — here, both the elevator's weight (gravity) and the frictional resistance to motion. Power delivered at constant velocity is $P = Fv$ , where $F$ is this total balancing force.

Step-by-Step Solution

Convert speed: $9\ \mathrm{kmph} = \dfrac{9\times1000}{3600} = 2.5\ \mathrm{ms^{-1}}$ .

Weight of elevator + passengers: $mg = 2000\times10 = 20000\ \mathrm N$ . …
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AP EAPCET 2024Set eng-2024-05-20-AN1 markMCQ
$m$
›Reveal solutionSolution
Average power of a constant force equals that force times the average velocity component along it; the vertical velocity drops linearly from $u\sin\theta$ to $0$ . Answer: magnitude $\dfrac{mgu\sin\theta}{2}$ .

Concept and Intuition

For a constant force like gravity, average power over an interval is simply $\vec F\cdot\vec v_{avg}$ , because power is linear in velocity and the force doesn't change. The vertical velocity of a projectile decreases uniformly (constant deceleration $g$ ) from $u\sin\theta$ at launch to $0$ at the top, so its time-average is just the arithmetic mean of the endpoints.

Step-by-Step Solution

Vertical velocity at launch: $u\sin\theta$ ; at the highest point: $0$ .

Time to reach the top: $t=\dfrac{u\sin\theta}{g}$ .

Average vertical velocity (linear decay): $\bar v_y=\dfrac{u\sin\theta+0}{2}=\dfrac{u\sin\theta}{2}$ .

Gravity's force is $-mg$ (taking upward positive). Average power $=F\cdot\bar v_y = -mg\cdot\dfrac{u\sin\theta}{2}=-\dfrac{mgu\sin\theta}{2}$ . …
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AP EAPCET 2024Set eng-2024-05-22-FN1 markMCQ
$= 10\ \mathrm{ms}^{-2}$
›Reveal solutionSolution
Average power is just the total work done (against gravity, climbing to height $h$ ) divided by the total time taken. Answer: 100 W.

Concept and Intuition

Average power over an interval is defined as total work (or energy transferred) divided by total time. Here the "work" is the gravitational potential energy gained in climbing height $h$ , and the time must be converted from minutes to seconds to keep units consistent (SI watts).

Step-by-Step Solution

Work done against gravity: $W = mgh = 90\times10\times600 = 540{,}000\,\mathrm J$ .

Convert time: $t = 90\,\text{min} = 90\times60 = 5400\,\mathrm s$ .

Average power: $P = \dfrac{W}{t} = \dfrac{540{,}000}{5400} = 100\,\mathrm W$ .

Common Mistakes …
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AP EAPCET 2023Set ap-2023-05-22-FN1 markMCQ
$= 10\ \mathrm{ms^{-2}}$
›Reveal solutionSolution
A straightforward work–power calculation for lifting water against gravity. Answer: (A) $1.5\ \mathrm{kW}$ .

Concept and Intuition

The minimum work required to raise a mass $m$ through height $h$ against gravity is $W = mgh$ . Power is the rate of doing this work, $P = W/t$ .

Step-by-Step Solution

Compute work: $W = mgh = 1000\ \mathrm{kg}\times10\ \mathrm{ms^{-2}}\times9\ \mathrm{m} = 90000\ \mathrm{J}$ .

Convert time: 1 minute $= 60\ \mathrm{s}$ .

Compute power: $P = \dfrac{90000\ \mathrm{J}}{60\ \mathrm{s}} = 1500\ \mathrm{W} = 1.5\ \mathrm{kW}$ . …
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AP EAPCET 2022Set eng-2022-07-07-FN1 markMCQ
$Q. A lift raises 50 passengers each having average weight 600 N to a height of 100 m at a constant speed in time T. If the average power of 15 kW is required by the lift, then the value of T in seconds is (A) 150 (B) 100 (C) 300 (D) 200$
›Reveal solutionSolution
$T=\text{Work}/\text{Power}=(50\times600\times100)/15000=200$ s.

Concept and Intuition

At constant speed, the lift does work against gravity equal to (total weight) $\times$ (height). Average power is that total work divided by the time taken, so time is just work divided by power.

Step-by-Step Solution

Total weight of passengers $=50\times600=30{,}000$ N.

Work done raising them by 100 m: $W=30{,}000\times100=3{,}000{,}000$ J $=3\times10^6$ J.

Average power $P=15$ kW $=15{,}000$ W.

$T=\dfrac{W}{P}=\dfrac{3\times10^6}{15{,}000}=200$ s. …
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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 understanding — Power Against Gravity

Power Against Gravity: The Intuition

The Precise Statement

What This Really Means

Common Mistake to Avoid …

Why This Approach Works

Step-by-Step Solution

Concept: Power Against Gravity with Efficiency

Concept and Intuition

Step-by-Step Solution

Common Mistakes …

Concept and Intuition

Step-by-Step Solution

Concept and Intuition

Step-by-Step Solution

Concept and Intuition

Step-by-Step Solution

Concept and Intuition

Step-by-Step Solution

Concept and Intuition

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Concept and Intuition

Step-by-Step Solution

Concept and Intuition

Step-by-Step Solution

Concept and Intuition

Step-by-Step Solution

Concept and Intuition

Step-by-Step Solution

Common Mistakes …

Concept and Intuition

Step-by-Step Solution

Concept and Intuition

Step-by-Step Solution

Q.A pump on the ground floor of a building can pump up water to fill a tank of volume 30 m330\ \text{m}^{3}30 m3 in 15 min15\ \text{min}15 min. If the tank is 40 m40\ \text{m}40 m above the ground, and the efficiency of the pump is 30%30\%30%, how much electric power is consumed by the pump?

Why This Approach Works

Step-by-Step Solution

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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?

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