Definition
Power
Power P is the time rate of doing work: P_avg = W / t (average) and P_inst = dW/dt = F · v (instantaneous, where v is the velocity). SI unit: watt (W) = J/s. 1 horsepower (hp) ≈ 746 W.
-- NCERT Class 11 Physics, Ch. 5, p. 83Power
8 MCQs1 revision card9-step worked example
Source: NCERT Work, Energy and PowerPYQ coverage: NEET 2021, 2022, 2024Official key: NTA-verifiedLast updated: 24 Sep 2026
Power, explained for NEET
Three power questions in three recent papers, three different ways to lose the mark — and none of them is the formula.
Trap 1: x is not v. A question hands you a displacement function like x = 2t − 1 and a constant force, then asks for instantaneous power. P = F·v, and v = dx/dt. Differentiate first. Plugging x straight into F·x is the single most-rewarded wrong answer in these stems, because it needs no work and lands on an option that is printed.
Trap 2: the efficiency factor. Turbine and motor questions state a loss percentage or an efficiency η somewhere in the sentence — often after the numbers. Compute the ideal power, then multiply by η (or by 1 − loss). The distractor is always the un-multiplied ideal value.
Trap 3: the friction term. A lift rising at constant speed has zero net force, so cable tension is Mg + f, not Mg. Power = (Mg + f)v. Dropping f gives a printed option too.
The definition. NCERT Class 11 Physics Chapter 5, page 77, defines power as the time rate at which work is done. Average power is W/t over an interval. Instantaneous power is dW/dt, which equals the dot product F·v at that instant. SI unit: the watt, 1 W = 1 J/s. Mechanical horsepower ≈ 746 W.
Two consequences worth holding. First, because it is a dot product, a force perpendicular to velocity delivers zero power however large it is. Second, at constant speed the power you need is set entirely by the opposing forces — gravity, friction, drag — because the kinetic energy is not changing.
Watch-out. Power questions here are labelled easy and carry medium negative-marking risk. That combination is exact: the physics is one line, so the mark is lost in the reading. Before you compute, finish the sentence and ask which forces oppose the motion, and whether an efficiency was stated.
Can you answer these Power MCQs?
Select an option to see the explanation. Wrong answers show why your choice was tempting — and name the exact trap it exploits.
MCQ 1Easy RecallPractice
In SI units, power is measured in:
Show answer and why every option is right or wrong
Answer: C. C is correct. NCERT Class 11 Physics Chapter 5, page 77, gives the SI unit of power as the watt, where 1 W = 1 J/s.
Why A is wrong: A is wrong: the joule is the unit of work and energy, not of the rate at which work is done.
Why B is wrong: B is wrong: the newton-metre is dimensionally a joule — again work, not power.
Why D is wrong: D is wrong: the joule-second is energy multiplied by time, the inverse of what power requires; power is energy divided by time.
MCQ 2Easy RecallPractice
According to the definition given in NCERT Class 11 Physics Chapter 5, instantaneous power is:
Show answer and why every option is right or wrong
Answer: D. D is correct. Page 77 of NCERT Class 11 Physics Chapter 5 defines instantaneous power as P = dW/dt, and shows it equals F·v.
Why A is wrong: A is wrong: W/t is the average power over an interval. Instantaneous power is the limiting rate at one instant, and the two are equal only when the power is constant.
Why B is wrong: B is wrong: dF/dt is a rate of change of force, with units N/s. It has no standard role here.
Why C is wrong: C is wrong: force times displacement is work, measured in joules. Power is work per unit time (trap: substituting x where v belongs).
MCQ 3Easy RecallPractice
One mechanical horsepower is approximately equal to:
Show answer and why every option is right or wrong
Answer: D. D is correct. 1 hp ≈ 746 W, which is 7.46 × 10² W — the conversion noted alongside the power definition in NCERT Class 11 Physics Chapter 5, page 77.
Why A is wrong: A is wrong: 1.00 × 10³ W is one kilowatt, a different unit entirely — a horsepower is roughly three-quarters of it.
Why B is wrong: B is wrong: 1.00 × 10² W is not a standard conversion for horsepower; it appears to come from confusing the number with a round hundred.
Why C is wrong: C is wrong: 4.2 × 10³ relates to the calorie-to-joule factor (1 cal ≈ 4.2 J), not to horsepower.
MCQ 4Direct ApplicationPractice
A constant force of 5.0 N acts on a particle whose position along the line of the force varies as x = 2t − 1, with x in metres and t in seconds. The instantaneous power delivered by the force at t = 3.0 s is:
Show answer and why every option is right or wrong
Answer: A. A is correct. v = dx/dt = 2 m/s (constant), so P = Fv = 5.0 × 2 = 10 W. NCERT Class 11 Physics Chapter 5, page 77.
Why B is wrong: B is wrong: 75 W is F·x·t or a similar compounding of the same substitution error; power is not proportional to elapsed time here.
Why C is wrong: C is wrong: 25 W comes from using x at t = 3.0 s, that is x = 5 m, and computing F·x = 25 — displacement substituted for velocity (trap: instantaneous power x vs v).
Why D is wrong: D is wrong: 15 W is F·t = 5.0 × 3.0, multiplying force by time instead of by velocity.
MCQ 5Direct ApplicationPractice
A lift of total mass 2.0 × 10³ kg moves upward at a constant speed of 1.5 m/s. A constant frictional force of 4.0 × 10³ N opposes the motion. Taking g = 10 m/s² (exact), the power delivered by the cable is:
Show answer and why every option is right or wrong
Answer: C. C is correct. At constant speed the net force is zero, so tension T = Mg + f = 2.0 × 10⁴ + 4.0 × 10³ = 2.4 × 10⁴ N, and P = Tv = 2.4 × 10⁴ × 1.5 = 3.6 × 10⁴ W. NCERT Class 11 Physics Chapter 5, page 77.
Why A is wrong: A is wrong: 3.0 × 10⁴ W is Mgv with the friction term dropped — the tension is Mg + f, not Mg (trap: lift-power friction term).
Why B is wrong: B is wrong: 6.0 × 10³ W is fv alone, the power spent against friction only, ignoring the weight the cable is also holding up.
Why D is wrong: D is wrong: 2.4 × 10⁴ N is the tension, not the power; it has not been multiplied by the speed.
MCQ 6Direct ApplicationPractice
Water falls from a height of 60 m at a rate of 15 kg/s onto a turbine. Ten per cent of the energy is lost in transmission. Taking g = 10 m/s² (exact), the usable power output is:
Show answer and why every option is right or wrong
Answer: B. B is correct. Ideal power = (dm/dt)gh = 15 × 10 × 60 = 9.0 kW; usable = 9.0 × 0.90 = 8.1 kW. NCERT Class 11 Physics Chapter 5, page 77.
Why A is wrong: A is wrong: 9.0 kW is the ideal power with the 10 % loss factor never applied (trap: power efficiency factor dropped).
Why C is wrong: C is wrong: 0.90 kW is the power lost, that is 10 % of the ideal value, rather than the 90 % that remains usable.
Why D is wrong: D is wrong: 9.9 kW adds 10 % instead of subtracting it; a loss cannot raise the output above the ideal value.
MCQ 7Concept TrapPractice
A satellite moves in a circular orbit at constant speed under the gravitational pull of the Earth. Regarding the power delivered by the gravitational force, which statement is correct?
Show answer and why every option is right or wrong
Answer: B. B is correct. P = F·v = Fv cos θ; in a circular orbit θ = 90° at every instant, so cos θ = 0 and the power is zero however large F is. The dot-product form is given in NCERT Class 11 Physics Chapter 5, page 77.
Why A is wrong: A is wrong: the gravitational force is emphatically not zero — it is the force providing the centripetal acceleration. The power is zero because of the angle, not the magnitude.
Why C is wrong: C is wrong: a non-zero force does not guarantee non-zero power. Power vanishes whenever the force is perpendicular to the velocity, and here it always is.
Why D is wrong: D is wrong: negative power needs an obtuse angle between force and velocity. Here the angle is exactly 90°, so the power is zero, not negative — and the orbital speed is indeed constant.
MCQ 8CalculationPractice
A pump raises water from a well and delivers it at the surface with a speed of 2.0 m/s. The water is lifted through a vertical height of 10.0 m at a rate of 5.0 kg/s. The pump is 80 % efficient. Taking g = 10 m/s² (exact), the input power required is:
Show answer and why every option is right or wrong
Answer: A. A is correct. Useful output per second = (dm/dt)gh + ½(dm/dt)v² = 5.0 × 10 × 10.0 + ½ × 5.0 × (2.0)² = 500 + 10 = 5.10 × 10² W; input = 510/0.80 = 6.4 × 10² W. NCERT Class 11 Physics Chapter 5, page 77.
Why B is wrong: B is wrong: 6.3 × 10² W divides only the lifting term by the efficiency (500/0.80), omitting the 10 W needed to give the water its exit speed.
Why C is wrong: C is wrong: 5.1 × 10² W is the useful output power. The question asks for the input, which must be larger because the pump is only 80 % efficient (trap: efficiency factor dropped).
Why D is wrong: D is wrong: 5.0 × 10² W keeps only the lifting term, dropping both the kinetic-energy term and the efficiency division.
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Power: quick recall before you leave
How do you solve a Power question? A worked example
Pattern: hydroelectric / turbine power with a stated loss factor.
- 1
Given
• Height of fall, h = 60 m• Mass flow rate, dm/dt = 15 kg/s• Transmission loss = 10 %, so efficiency η = 0.90• g = 10 m/s² (exact, as specified by the problem)
- 2
Required
The usable power output of the turbine, in kW.
- 3
Concept
Falling water converts gravitational potential energy to kinetic energy, which the turbine converts to electrical work. Power is the rate of doing work, so the ideal power available equals the rate at which potential energy is delivered — the potential energy lost per second. A stated loss then removes a fixed fraction of that.
- 4
Formula
Ideal power: P_ideal = (dm/dt) × g × h
Usable power: P_usable = η × P_ideal = P_ideal × (1 − loss fraction) - 5
Substitution
P_ideal = 15 × 10 × 60
P_usable = (15 × 10 × 60) × 0.90 - 6
Calculation
P_ideal = 15 × 10 = 1.5 × 10²; 1.5 × 10² × 60 = 9.0 × 10³ W = 9.0 kW
P_usable = 9.0 × 10³ × 0.90 = 8.1 × 10³ W = 8.1 kW
Note on exact values: g = 10 m/s² is specified as exact by the problem, and the efficiency 0.90 is an exact defined fraction. Neither constrains the significant-figure count. The measured inputs — 60 m and 15 kg/s, both two significant figures — set the precision, so the answer is quoted to two significant figures. - 7
Final answer
P_usable = 8.1 kW (8.1 × 10³ W).
- 8
Common trap
The printed distractor is 9.0 kW: the ideal power, with the loss factor never applied. The loss is stated in a short clause that is easy to read past once the numbers have been spotted. Make a habit of reading to the full stop and asking "was an efficiency or a loss given?" before you write the answer down. A second, subtler slip is applying the factor the wrong way — 10 % loss means keeping 90 %, so you multiply by 0.90, not by 0.10 and not by 1.10.
- 9
Similar NEET-style question
A wind turbine intercepts air delivering kinetic energy at a rate of 40 kW. If 25 % of this is lost to drag and generator heating, what is the electrical power output? *(Answer: 40 × 0.75 = 30 kW.)*
What to remember before solving Power questions
Which Power formulas do you need for NEET?
1 formula — click to collapse
Instantaneous power
Instantaneous power is the time rate of doing work; equivalently, the dot product of the force and velocity. Average power over an interval is W/t.
| Symbol | Quantity | SI Unit |
|---|---|---|
| P | Power (instantaneous) | W |
| F | Force | N |
| v | Velocity | m/s |
Valid when
- F.v form when force and velocity are both known at the instant
Where do students lose marks on Power?
These are the exact patterns that cause wrong answers in NEET. Each trap includes when it triggers and how to avoid it.
3 items — click to collapse
Category: Overthinking
Student computes P = Mgv (just lifting against gravity) and ignores the friction-opposing-motion term.
When it triggers
Question describes a lift moving at constant speed with explicit friction force on cable or guides.
How to avoid
At constant speed, net force = 0, so cable tension T = Mg + f_friction. Power = T × v = (Mg + f) × v. Always add friction when stated.
Category: Overthinking
Student computes ideal power and forgets the (1 - loss_fraction) or efficiency multiplier.
When it triggers
Question gives turbine, motor, or transformer with stated efficiency or loss percentage.
How to avoid
Always read the question for efficiency η or loss%. Useful power P_useful = η × P_input or P_input × (1 - loss). Don't drop the factor even if the rest of the calc is in the unrelated parts of the problem.
Category: Similar Terms
Student plugs displacement x into P = F·v formula instead of velocity v.
When it triggers
Question gives displacement x(t) explicitly and a constant force; asks for instantaneous power.
How to avoid
P = F · v where v = dx/dt. Compute v first (differentiate x(t) once), THEN plug into F·v. P is NOT F·x.
More in Work, Energy and Power: 9 exam traps and mistakes · 9 formulas · 4 question patterns from its other lessons.
Power questions from past NEET papers
3 questions from NEET 2021, 2022, 2024. Answers verified against NTA official keys. — click to collapse
NEET 2024Revised key
110
25
37
46
NTA Answer: Option 1(revised_final)
NEET 2022
123500
223000
320000
434500
NTA Answer: Option 4(final)
NEET 2021
17.0 kW
210.2 kW
38.1 kW
412.3 kW
NTA Answer: Option 3(final)
How does NEET ask about Power?
3 recurring patterns from past papers — click to collapse
Water falls from height h at flow rate dm/dt (kg/s); turbine efficiency or loss factor given; find usable power. Power = dm/dt × g × h × (1 - loss_fraction). Common shape: 60 m, 15 kg/s, 10% loss → P = 15 × 10 × 60 × 0.9 = 8.1 kW.
Multi StepEasy
Common distractors
forgets loss factor
Default to ideal-case formula
Given displacement x(t) and net force F (constant); find instantaneous power P = F · v(t) where v(t) = dx/dt. Common shape: x = 2t-1, F = 5 N → v = 2 m/s, P = 10 W.
Direct ApplicationEasy
Common distractors
uses x instead of v
Default plugging x into the wrong formula
Lift moving up at constant speed v with total mass M; friction force f opposes motion. Power required from cable = (Mg + f) × v. Common shape: M = 2000 kg, v = 1.5 m/s, f given; find motor power.
Multi StepEasy
Common distractors
forgets friction term
Ignoring opposing force
Sources
NCERT refs: Class 11 Physics Chapter 5, p.77
Page numbers are the ones printed in the current NCERT textbook (2023 rationalised edition), unless marked pre-2023. The books are free at ncert.nic.in.
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