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.

TG EAPCET 2025Set ap-2025-04-29-FN1 markMCQ
$20\,\mathrm{m}^2$
›Reveal solutionSolution

Wind power $= \tfrac{1}{2}\rho A v^3 = 12\ \text{kW}$ ; at $25\%$ efficiency the electrical output is $3\ \text{kW}$ — option (D).

Convert the wind speed: $v = 36\ \text{kmph} = 36 \times \tfrac{5}{18} = 10\ \text{ms}^{-1}$ .

The kinetic power carried by the wind passing through the swept area $A$ is
$P_{\text{wind}} = \frac{1}{2}\,\rho\, A\, v^3$
Substituting $\rho = 1.2\ \text{kg m}^{-3}$ , $A = 20\ \text{m}^2$ , $v = 10\ \text{ms}^{-1}$ : …
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TG EAPCET 2024Set ap-2024-05-07-FN1 markMCQ
$25\%$
›Reveal solutionSolution
The output power is the electrical power delivered, which is the thermal power from fission multiplied by the efficiency. Thermal power = (fissions per second) × (energy per fission). With 25% efficiency, output power = 400 W → option (D).

The key idea is to separate thermal power (the raw energy released per second by fission) from output power (the useful electrical power after efficiency losses). The efficiency tells us what fraction of the thermal power is converted to electrical output.

Find the thermal power produced by fission. Each fission releases $200\ \text{MeV}$ of energy. The number of fissions per second is $5 \times 10^{13}$ . So the energy released per second (thermal power) is:

$P_{\text{thermal}} = (5 \times 10^{13}) \times (200\ \text{MeV})$
Convert MeV to joules: $1\ \text{MeV} = 1.6 \times 10^{-13}\ \text{J}$ .
$P_{\text{thermal}} = 5 \times 10^{13} \times 200 \times 1.6 \times 10^{-13}\ \text{J/s}$
Simplify:
$P_{\text{thermal}} = 5 \times 200 \times 1.6 \times 10^{0} = 5 \times 320 = 1600\ \text{W}$
So the reactor’s thermal power is $1600\ \text{W}$ .

Apply the efficiency to get output power. Efficiency $\eta = 25\% = 0.25$ . Output power is the useful electrical power:
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TG EAPCET 2023Set eng-2023-05-12-AN1 markMCQ
$Q. While a person climbs stairs, the gravitational potential energy of the person increases. The source of this energy is (A) Work done by normal force from the steps (B) Work done by frictional force from the steps (C) Work done by air resistance (D) Work done by internal forces within the person's body$
›Reveal solutionSolution
The increase in gravitational potential energy when climbing stairs comes from chemical energy converted by the person’s muscles — internal forces do the work, not external forces like the normal or friction from the steps. The correct answer is (D).

Why this approach works

The key is to ask: What force actually does the work that increases the person’s gravitational potential energy?

Gravitational potential energy changes when the center of mass moves upward against gravity. The work done against gravity is $mgh$ . But the force that directly lifts the person is not gravity — it’s the force the person’s legs exert on their own body. The stairs only provide a support (normal force), but that normal force does no net work on the person because the point of application (the foot) doesn’t move relative to the step while the force is applied (the step is stationary). So the real source is internal: the person’s muscles convert chemical energy into mechanical work.

Step-by-step reasoning

Identify the energy change

As the person climbs, their gravitational potential energy increases by $\Delta U = mgh$ . This energy must come from work done on the person by some force.

Check the normal force from the steps

The normal force acts perpendicular to the step’s surface. While the foot is in contact, the step does not move — the normal force is applied at a stationary point. Work is $W = \vec{F} \cdot \Delta \vec{r}$ , and here $\Delta \vec{r} = 0$ for the point of application. So the normal force does zero work.

Watch out

A common mistake is to think the normal force “pushes you up.” But the step doesn’t move with you — your foot pushes down on the step, and the step pushes back, but that push does no work because the contact point is stationary. The work is done by your muscles lifting your body.

Check the frictional force

Friction prevents your foot from slipping, but it acts horizontally (parallel to the step surface). It does no vertical work. Even if friction helps you move forward, it doesn’t directly increase gravitational potential energy.

Check air resistance …
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TG EAPCET 2023Set ap-2023-05-11-FN1 markMCQ
$^{-1}$
›Reveal solutionSolution
Friction provides exactly the centripetal force needed at any speed below the maximum; at 5 m/s the car requires 2500 N toward the centre, so static friction supplies precisely that amount.

Why friction adjusts to what the car needs

Static friction is a responsive force. It doesn't always act at its maximum value $\mu_s N$ ; instead, it supplies exactly the force required to prevent slipping, up to that maximum limit. When a car rounds a curve, the tyres need an inward (centripetal) force to keep the car on the circular path. Static friction between the tyres and the road provides this force—as much as needed, but no more—until the demand exceeds $\mu_s N$ , at which point the tyres slip.

The maximum speed tells us the radius of the track. At that speed, friction is pushed to its limit: $f_{\text{max}} = \mu_s m g = \frac{m v_{\text{max}}^2}{r}$ . Once we know $r$ , we can find the actual friction required at any lower speed.

Finding the friction at 5 m/s

Extract the radius from the maximum-speed condition.

At $v_{\text{max}} = 7 \, \text{m/s}$ , static friction is at its limit:

$\mu_s m g = \frac{m v_{\text{max}}^2}{r}.$
Cancel $m$ and solve for $r$ :
$r = \frac{v_{\text{max}}^2}{\mu_s g} = \frac{(7)^2}{0.5 \times 10} = \frac{49}{5} = 9.8 \, \text{m}.$

Calculate the centripetal force needed at 5 m/s.

At $v = 5 \, \text{m/s}$ on the same track (same $r = 9.8 \, \text{m}$ ), the required centripetal force is

$F_c = \frac{m v^2}{r} = \frac{1000 \times (5)^2}{9.8} = \frac{1000 \times 25}{9.8} = \frac{25\,000}{9.8} \approx 2551 \, \text{N}.$

Recognize that static friction supplies exactly this force. …
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TG EAPCET 2023Set eng-2023-05-12-FN1 markMCQ
$s = \frac{t^3}{3}$
›Reveal solutionSolution
The work done is the change in kinetic energy. Using the given displacement $s = t^3/3$ , we find velocity $v = t^2$ , then kinetic energy at $t=2$ s is $24$ J, so the answer is 24 J.

Concept & Intuition

Work done by a force on a body equals the change in its kinetic energy (work–energy theorem). Since we are given displacement as a function of time, we can differentiate to get velocity, then compute kinetic energy at the start and end of the interval. No need to find the force explicitly — that’s the elegant shortcut.

Step-by-step solution

Find velocity from displacement Displacement: $s = \frac{t^3}{3}$ . Velocity is the time derivative:

$v = \frac{ds}{dt} = \frac{d}{dt}\left(\frac{t^3}{3}\right) = t^2 \ \text{m/s}.$

Kinetic energy at any time Mass $m = 3$ kg. Kinetic energy:

$K = \frac{1}{2} m v^2 = \frac{1}{2} \cdot 3 \cdot (t^2)^2 = \frac{3}{2} t^4 \ \text{J}.$

Work done in first two seconds At $t = 0$ : $K(0) = 0$ (body starts from rest). At $t = 2$ s: $K(2) = \frac{3}{2} \cdot (2)^4 = \frac{3}{2} \cdot 16 = 24$ J. …
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TG EAPCET 2022Set eng-2022-07-20-AN1 markMCQ
$36\, \mathrm{m}^3$
›Reveal solutionSolution
The efficiency of a pump is the ratio of the useful power output (rate of increasing water's potential energy) to the total electrical power input. By calculating the potential energy gained by the water and the time taken, we find the useful power output, which leads to an efficiency of $\boxed{25\%}$ .

The efficiency of any machine, including a pump, tells us how effectively it converts the energy supplied to it into useful work. In this problem, the pump consumes electrical energy (input power) to lift water against gravity, thereby increasing its potential energy (useful output power). The efficiency is simply the ratio of this useful output power to the total input power.

Here's how we calculate the pump's efficiency:

Identify the given quantities and the goal.

We are given:

Volume of water ( $V$ ) = $36\, \mathrm{m}^3$

Time taken ( $t$ ) = $30\, \mathrm{min}$

Height the water is lifted ( $h$ ) = $50\, \mathrm{m}$

Electric power consumed by the pump ( $P_{input}$ ) = $40\, \mathrm{kW}$

Acceleration due to gravity ( $g$ ) = $10\, \mathrm{m/s}^2$

Density of water ( $\rho$ ) = $1000\, \mathrm{Kg/m}^3$

Our goal is to find the efficiency ( $\eta$ ) of the pump.

Convert all units to the SI system for consistency.

Time: $t = 30\, \mathrm{min} = 30 \times 60\, \mathrm{s} = 1800\, \mathrm{s}$

Input Power: $P_{input} = 40\, \mathrm{kW} = 40 \times 10^3\, \mathrm{W} = 40000\, \mathrm{W}$

Calculate the mass of water pumped.

The mass ( $m$ ) of the water can be found using its volume and density:

$m = \rho \times V$ $m = 1000\, \mathrm{Kg/m}^3 \times 36\, \mathrm{m}^3 = 36000\, \mathrm{Kg}$

Calculate the potential energy gained by the water. When water is lifted to a height $h$ , its gravitational potential energy ( $PE$ ) increases. This is the useful work done by the pump.

$PE = mgh$
$$PE = 36000\, \mathrm{Kg} \times 10\, \mathrm{m/s}^2 \times 50\, \mathrm{m}$$ …
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TG EAPCET 2021Set ap-2021-08-09-FN1 markMCQ
$2\,\mathrm{g}$
›Reveal solutionSolution
The power of the gun equals the rate at which it gives kinetic energy to the bullets. Using $P = \frac{1}{2} n m v^2$ with $n = 6$ bullets per second, $v = 500$ m/s, and $P = 4500$ W, we find $m = 0.006$ kg = 6 g, so the correct option is (C).

The key idea is that the machine gun’s power is the rate at which it does work — here, that work goes entirely into the kinetic energy of the bullets. Since each bullet starts from rest inside the gun and leaves at 500 m/s, the energy per bullet is $\frac{1}{2} m v^2$ . The power tells us how much energy is transferred per second, so if we know how many bullets are fired per second, we can solve for the mass of one bullet.

Find the number of bullets per second. The gun fires 360 bullets per minute.

$n = \frac{360}{60} = 6 \text{ bullets per second}.$

Express the power in terms of bullet mass and speed. The kinetic energy of one bullet is $\frac{1}{2} m v^2$ . In one second, $n$ bullets are fired, so the energy given to them per second (which is the power) is:

$P = n \cdot \frac{1}{2} m v^2.$

Substitute the known values. $P = 4.5 \text{ kW} = 4500 \text{ W}$ , $v = 500 \text{ m/s}$ , $n = 6$ .

$4500 = 6 \cdot \frac{1}{2} \cdot m \cdot (500)^2.$

Simplify and solve for $m$ . First, $(500)^2 = 250000$ . Then $\frac{1}{2} \times 250000 = 125000$ . So:

$4500 = 6 \times 125000 \times m = 750000 \, m.$
$m = \frac{4500}{750000} = \frac{45}{7500} = \frac{3}{500} = 0.006 \text{ kg}.$ …
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TG EAPCET 2021Set eng-2021-08-06-AN1 markMCQ
$15\,\mathrm{kg}$
›Reveal solutionSolution
For a body starting from rest under a constant force, the instantaneous power is directly proportional to time. Using this relationship, the power at the end of the fourth second is $6.67\,\mathrm{W}$ .

The problem asks for the instantaneous power at a specific time, given the power at another time, for a body starting from rest under the action of a force. To solve this, we need to understand how instantaneous power, force, and velocity are related, and how these quantities change with time when a body starts from rest.

Concept and Intuition

Instantaneous power ( $P$ ) is defined as the rate at which work is done by a force, and it can be expressed as the dot product of the force vector ( $\vec{F}$ ) and the instantaneous velocity vector ( $\vec{v}$ ):
$P = \vec{F} \cdot \vec{v}$
If the force and velocity are in the same direction (as is the case when a force acts on a body initially at rest and causes it to move in the direction of the force), this simplifies to $P = Fv$ .

When a constant force $F$ acts on a body of mass $m$ , it produces a constant acceleration $a$ given by Newton's second law:
$F = ma \implies a = \frac{F}{m}$
If the body starts from rest (initial velocity $u=0$ ), its velocity $v$ at any time $t$ is given by the kinematic equation:
$v = u + at \implies v = at$
Now, we can substitute these relationships into the power equation. Since $a = F/m$ , we have $v = (F/m)t$ .

Substituting this into $P = Fv$ :
$P = F \left(\frac{F}{m}\right)t$ $P = \left(\frac{F^2}{m}\right)t$
This crucial relationship shows that for a constant force acting on a body initially at rest, the instantaneous power is directly proportional to time ( $P \propto t$ ). This is the key insight to solve the problem.

Step-by-step Derivation

Identify the given information:

Mass of the body, $m = 15\,\mathrm{kg}$ .

Initial state: at rest ( $u = 0$ ).

Instantaneous power at the end of the third second, $P_3 = 5\,\mathrm{W}$ (at $t_3 = 3\,\mathrm{s}$ ).

We need to find the instantaneous power at the end of the fourth second, $P_4$ (at $t_4 = 4\,\mathrm{s}$ ).

Assume a constant force:

In the absence of information about how the force varies, it is standard to assume that the force acting on the body is constant. This simplifies the problem significantly and allows us to use the kinematic equations for constant acceleration.

Relate force, acceleration, and velocity:

Since the force $F$ is constant and the mass $m$ is constant, the acceleration $a = F/m$ is also constant.

As the body starts from rest ( $u=0$ ), its velocity $v$ at time $t$ is given by:

$v = at$

Derive the expression for instantaneous power: The instantaneous power $P$ due to the force is given by:

$P = Fv$
Substitute the expression for $v$ from the previous step:
$P = F(at)$
Now, substitute $a = F/m$: $$P = F \left(\frac{F}{m}\right)t$$ …
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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

Why this approach works

Step-by-step reasoning

Why friction adjusts to what the car needs

Finding the friction at 5 m/s

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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