Q.Match the column - select the correct definition (from Column B) for the term 'Power' (Column A):
🔒You're viewing a preview — the full solution, concept, methods & PYQ mapping are locked.
🔒 Start your 14-day free trial to unlock the full solution →Concept understanding — Power Time Relation
Power and Time: The Intuition
Think about lifting a heavy box. If you lift it slowly, you feel tired but you manage. If you lift it fast — the same box to the same height — you feel a much greater strain. The work done (force × distance) is identical in both cases. So why does the fast lift feel harder?
The answer is power. Power tells you not just how much work is done, but how quickly it is done. The fast lift requires more power because the same work is compressed into a shorter time.
Everyday language often confuses power with energy. A "powerful" car isn't one that uses more fuel (energy) — it's one that can accelerate faster or climb a hill at higher speed, meaning it delivers energy more quickly.
The Precise Statement
Power is defined as the rate at which work is done (or energy is transferred). Mathematically:
P=tW
where:
- P = power (watts, W)
- W = work done (joules, J)
- t = time taken (seconds, s)
This is the power-time relation in its simplest form: power is work divided by time.
P=tW
If you rearrange it, you get two other useful forms:
W=P×tandt=PW
These tell you:
- To do a fixed amount of work, using more power means less time.
- To run a device for a fixed time, more power means more work (and more energy consumed).
A Concrete Example
Suppose you need to lift a 10 kg mass to a height of 2 metres. The work done against gravity is:
W=mgh=10×9.8×2=196 J
Now consider two cases:
| Case | Time taken | Power required |
|---|---|---|
| Slow lift | 4 seconds | P=4196=49 W |
| Fast lift | 1 second | P=1196=196 W |
The fast lift requires four times the power — that's why it feels much harder, even though the work is the same.
A common mistake is to think power and work are the same thing. They are not. Work is the total energy transferred; power is the rate of that transfer. A 100 W bulb uses 100 J of energy every second, but if you leave it on for an hour, the total work (energy used) is 100×3600=360,000 J.
Why This Matters for Exams
The power-time relation appears in two main forms in problems:
- Direct calculation: Given work and time, find power (or any missing variable). …
Power is defined as the rate at which work is done (or energy is transferred), P = dW/dt. …
Power is the rate of doing work.
Power is defined as the time rate at which work is done by a force, or equivalently the rate of transfer of energy:
P = dW/dt (instantaneous power), or P_avg = W/t for a constant rate …
Showing the 12 most recent of 22 on this concept.
- CBSE 2026Set ANNUAL1 markMCQQ.Match the column - select the correct definition (from Column B) for the term 'Power' (Column A):(a) change in linear momentum(b) motion opposing force(c) loss of energy(d) rate of change of momentum(e) ability of doing work(f) rate of doing work
›Reveal solutionSolution
Power is the rate of doing work.
Power is defined as the time rate at which work is done by a force, or equivalently the rate of transfer of energy:
P = dW/dt (instantaneous power), or P_avg = W/t for a constant rate …
- CBSE 2026Set ANNUAL1 markMCQQ.A body which is initially at rest, undergoes one-dimensional motion with constant acceleration. The power delivered to it at time t is proportional to(a) t^2(b) t(c) t^3/2(d) t^1/2
›Reveal solutionSolution
P = Fv = (ma)(at) is proportional to t. Answer (B).
With constant acceleration a from rest, the velocity at time t is v = at.
The force is constant: F = ma.
…
- CBSE 2026Set ANNUAL1 markMCQQ.Kilowatt-hour (kWh) is the unit of(a) power(b) momentum(c) energy(d) electric current
›Reveal solutionSolution
kWh = power x time = energy. Answer (C).
Energy = power x time. A kilowatt-hour is the energy delivered by a 1 kilowatt device running for 1 hour:
1 kWh = 1000 W x 3600 s = 3.6 x 10^6 J.
…
- CBSE 2025Set ANNUAL1 markMCQQ.If a force F is applied on a body and it moves with velocity v, then its power will be(a) Fv(b) Fv^2(c) F / v(d) F / v^2
›Reveal solutionSolution
Power delivered by a force is the rate of doing work, P = dW/dt = F.v, which for force and velocity along the same direction is simply Fv.
Work done, dW = F.ds (force dotted with small displacement).
Power P = dW/dt = F.(ds/dt) = F.v
…
- CBSE 2025Set ANNUAL1 markMCQQ.2hp (horse power) is equal to -(a) 746 watt(b) 373 watt(c) 2238 watt(d) 1492 watt
›Reveal solutionSolution
2 horsepower equals 1492 watt.
Horsepower is a non-SI unit of power, with the standard conversion:
1 hp=746 W
Therefore:
…
- CBSE 2025Set ANNUAL1 markMCQQ.10 Horse power is equal to -(a) 746 watt(b) 74.6 watt(c) .746 watt(d) 7460 watt
›Reveal solutionSolution
10 horsepower converts to 7460 watts using 1 HP = 746 W.
The horsepower (HP) is a traditional unit of power, related to the SI unit watt by the standard conversion:
1 HP = 746 W. …
- CBSE 2025Set ANNUAL1 markMCQQ.Match the following - Column A item: Power. Pick the matching relation from Column B.(a) F.V (Force . Velocity)(b) T is proportional to sqrt(l)(c) eta is proportional to 1/(dv/dx)(d) mu_s = tan(theta)(e) Y is proportional to 1/l(f) v^2/r
›Reveal solutionSolution
Power is the scalar (dot) product of force and velocity, P = F.V.
Power is defined as the rate at which work is done: P = dW/dt. Since infinitesimal work dW = F.dr (force dotted with infinitesimal displacement), dividing by dt gives P = F.(dr/dt) = F.V, where V is the instantaneous velocity of the point of application of the force. This dot-product form …
- CBSE 2024Set ANNUAL1 markMCQQ.If an athlete does 500 joules of work in 4 seconds, what is his power output?(a) 200 watts(b) 125 watts(c) 1250 watts(d) 4 watts
›Reveal solutionSolution
Power = Work done / Time taken = 500/4 = 125 W.
Given: Work done, W = 500 J; Time taken, t = 4 s
…
- CBSE 2024Set ANNUAL1 markMCQQ.A body is initially at rest. It undergoes one-dimensional motion with constant acceleration. The power delivered to it at time t is proportional to:(a) t^(1/2)(b) t(c) t^(3/2)(d) t^2
›Reveal solutionSolution
For a body starting from rest under constant acceleration, instantaneous power P is directly proportional to t (P is proportional to t^1).
Since the body starts from rest (u = 0) with constant acceleration a: v(t) = u + at = at.
The constant force acting is F = ma.
Instantaneous power P = F.v = (ma)(at) = m a^2 t. …
- CBSE 2023Set ANNUAL1 markMCQQ.If F is the force applied on the object and V is the velocity, then it's power will be:(a) F × V(b) F/V^2(c) F/V(d) F × V^2
›Reveal solutionSolution
Power P = F·V.
Power is work done per unit time: P = W/t. For a force F acting over displacement d in time t, P = F·d/t = F·(d/t) = F·V, where V = d/t is the velocity.
…
- CBSE 2022Set TERM11 markMCQQ.Which of the following units is a unit of power?(1) kilowatt-hour(2) watt(3) erg(4) calorie
›Reveal solutionSolution
Power = work/time, with SI unit watt (1 W = 1 J/s). The other three options -- kilowatt-hour, erg, calorie -- are all units of energy (work), not of the rate at which work is done.
Power is defined as the rate of doing work: P = W/t. Its SI unit is the WATT, where 1 watt = 1 joule/second.
Checking the distractors:
- Kilowatt-hour (kWh): a unit of ENERGY (power x time), commonly used for electricity billing -- not a unit of power itself. …
- CBSE 2022Set ANNUAL1 markQ.The rate of doing work by a machine is called ............. of the machine.
›Reveal solutionSolution
Rate of doing work is called power.
Power is defined as:
P=tW(average power),P=dtdW(instantaneous power) …
🎓Unlock everything free for 14 days
- ✓Full step-by-step solutions
- ✓Concept-first explanations
- ✓Methods, shortcuts & mistakes
- ✓PYQ mapping + timed mock tests
Full access for 14 days. No credit card required.