Uniformly Accelerated Motion Class 11 Physics Notes | NEET & JEE MCQs

 - Dr.Sanjaykumar Pawar  

Uniformly Accelerated Motion (1-D) Physics Notes, Formulas & NEET Questions 

Uniformly Accelerated Motion (1-D): Equations of Motion, Graphs and NEET Physics Concepts

Internal Links

1. Motion in One Dimension Foundation

 Motion in One Dimension Class 11 Physics Notes

Before introducing uniformly accelerated motion, link readers to the basic concepts of displacement, velocity and speed.

2. Instantaneous Velocity & Acceleration

Instantaneous Velocity and Acceleration Explained

NCERT Physics Class 11 Chapter 2: Instantaneous Velocity & Acceleration

Under "Important Terms" section after explaining acceleration.

3. Vectors in Physics

Vectors Class 11 Physics Notes

CBSE Class 11 Physics Vectors Notes, MCQs, Questions & Answers | NEET Preparation

  explaining displacement as a vector quantity.

4. Newton's Laws of Motion

Newton's Laws of Motion Class 11 Physics

Block and Trolley System NEET Solution | Acceleration & Tension Explained

acceleration concepts because force causes acceleration.

5. Free Fall Motion

Free Fall and Acceleration Due to Gravity

NEET Tips section where free fall problems are mentioned.

6. Graphical Motion Analysis

Velocity-Time and Acceleration-Time Graphs

 diagram-based questions.

Uniform acceleration formula

Equations of motion derivation

Motion in one dimension notes

Class 11 Physics chapter 2 notes

NEET physics motion questions

JEE physics kinematics notes

Velocity time graph questions

Acceleration numericals with solutions

CBSE Physics important questions

Physics revision notes for NEET

FAQ Schema Questions

Q1. What is uniformly accelerated motion?

Uniformly accelerated motion is motion in which acceleration remains constant with time.

Q2. What are the three equations of motion?

The three equations are v = u + at, s = ut + ½at² and v² = u² + 2as.

Q3. What does the slope of a velocity-time graph represent?

The slope of a velocity-time graph represents acceleration.

Q4. What does the area under a velocity-time graph represent?

The area represents displacement.

Q5. Is uniformly accelerated motion important for NEET and JEE?

Yes, it is a fundamental topic used in many mechanics problems.

Uniformly Accelerated Motion (1-D)

1. Meaning of Uniformly Accelerated Motion

Uniformly accelerated motion means an object is moving in a straight line and its acceleration remains constant with time.

• The velocity changes by the same amount in equal intervals of time.
• The motion takes place in one dimension.

Example: A bike increases its speed by 5 m/s every second.

Important Terms

1. Initial Velocity (u)

Initial velocity is the velocity of an object at the starting time. It is represented by u.

At starting time:
t = 0, velocity = u

2. Final Velocity (v)

Final velocity is the velocity of an object after a certain time. It is represented by v.

3. Acceleration (a)

Acceleration is the rate of change of velocity.

a = (v - u) / t

Unit of acceleration = m/s²

4. Displacement (s)

Displacement is the distance travelled by an object in a particular direction.

Unit = metre (m)

First Equation of Motion

v = u + at

Derivation:

Acceleration:

a = dv/dt

Rearranging:

a dt = dv

Integrating:

∫a dt = ∫dv

a(t - 0) = v - u

at = v - u

v = u + at

Final velocity = Initial velocity + Change in velocity

Second Equation of Motion

s = ut + 1/2 at²

Derivation:

Velocity:

v = ds/dt

Therefore:

ds = v dt

Using:
v = u + at

ds = (u + at)dt

After integration:

s = ut + 1/2 at²

Displacement = Distance due to initial velocity + Distance due to acceleration

Third Equation of Motion

v² = u² + 2as

Derivation:

From first equation:

v = u + at

Rearranging:

t = (v - u)/a

Using second equation:

s = ut + 1/2 at²

After simplification:

v² = u² + 2as

This equation is useful when time is not given.

Three Equations of Motion Summary

1. Velocity Equation

v = u + at

Used to find final velocity when time is given.

2. Displacement Equation

s = ut + 1/2 at²

Used to find displacement when time is given.

3. Time Independent Equation

v² = u² + 2as

Used when time is not given.

Easy Memory Trick

• V-U-AT: v = u + at
• S-U-T-A-T: s = ut + 1/2 at²
• V-U-AS: v² = u² + 2as

Symbols at a Glance

Symbol	Meaning	Unit
u	Initial Velocity	m/s
v	Final Velocity	m/s
a	Acceleration	m/s²
t	Time	second
s	Displacement	metre

Conclusion

The three equations of motion are used to solve problems involving constant acceleration in one-dimensional motion.

NEET Physics Practice Questions

Chapter: Uniformly Accelerated Motion (1-D)

NEET Question Types

• Direct MCQs (Single Correct Option)
• Statement Based Questions
• Assertion and Reason
• Match the Columns
• Diagram Based / Graphical Questions

PART 1: Direct MCQs (Single Correct Option)

Q1. A car starts from rest and accelerates uniformly at 4 m/s². Its velocity after 5 seconds will be:

A) 10 m/s
B) 20 m/s
C) 25 m/s
D) 40 m/s

Solution:
u = 0
a = 4 m/s²
t = 5 s

v = u + at
v = 0 + 4 × 5
v = 20 m/s

Answer: B) 20 m/s

Q2. The SI unit of acceleration is:

A) m/s
B) m²/s
C) m/s²
D) km/h

Answer: C) m/s²

Q3. A body moving with velocity 20 m/s is brought to rest in 5 seconds. The acceleration is:

A) +4 m/s²
B) -4 m/s²
C) +5 m/s²
D) -5 m/s²

a = (v-u)/t
a = (0-20)/5
a = -4 m/s²

Answer: B) -4 m/s²

Q4. The velocity-time graph for uniformly accelerated motion is:

A) Straight line
B) Circle
C) Parabola
D) Hyperbola

Answer: A) Straight line

Q5. A particle starts from rest and travels 100 m in 5 seconds with uniform acceleration. Find acceleration.

A) 4 m/s²
B) 6 m/s²
C) 8 m/s²
D) 10 m/s²

s = ut + 1/2 at²
100 = 0 + 1/2 × a × 25
a = 8 m/s²

Answer: C) 8 m/s²

PART 2: Statement Based Questions

Options:
A) Both statements true and II explains I
B) Both true but II does not explain I
C) I true, II false
D) I false, II true

Q6.

Statement I: Velocity-time graph for uniformly accelerated motion is a straight line.

Statement II: Acceleration is constant in uniformly accelerated motion.

Answer: A

Q7.

Statement I: Displacement can be zero even if distance travelled is not zero.

Statement II: Displacement depends only on initial and final position.

Answer: A

Q8.

Statement I: Acceleration due to gravity is constant near Earth's surface.

Statement II: Value of g is approximately 9.8 m/s².

Answer: A

PART 3: Assertion and Reason

Options:
A) Both true and R explains A
B) Both true but R does not explain A
C) A true, R false
D) A false, R true

Q9.

Assertion: A body moving with constant velocity has zero acceleration.

Reason: Acceleration is the rate of change of velocity.

Answer: A

Q10.

Assertion: Area under velocity-time graph gives displacement.

Reason: Velocity is displacement divided by time.

Answer: A

Q11.

Assertion: A body can have zero velocity and non-zero acceleration.

Reason: At highest point of upward motion, velocity is zero but acceleration acts downward.

Answer: A

PART 4: Match the Columns

Column I	Column II
Velocity	m/s
Acceleration	m/s²
Displacement	m
Time	second

Answer: Velocity-m/s, Acceleration-m/s², Displacement-m, Time-second

PART 5: Diagram Based / Graphical Questions

Q14. The slope of velocity-time graph represents:

A) Distance
B) Velocity
C) Acceleration
D) Displacement

Answer: C) Acceleration

Q15. Area under velocity-time graph represents:

A) Acceleration
B) Displacement
C) Force
D) Momentum

Answer: B) Displacement

Q16. Velocity-time graph:

v
|
|        /
|       /
|      /
|_____/________ t

The particle has:

A) Constant velocity
B) Constant acceleration
C) Zero acceleration
D) Variable acceleration

Answer: B) Constant acceleration

Hard Numerical Practice

Q17. A train moving at 72 km/h stops in 10 seconds. Find retardation.

72 km/h = 20 m/s a = (0-20)/10 a = -2 m/s² Retardation = 2 m/s²

Q18. A particle has initial velocity 5 m/s and acceleration 2 m/s². Find distance in 10 seconds.

s = ut + 1/2at² s = 5×10 + 1/2×2×100 s = 150 m Answer: 150 m

NEET Formula Revision

v = u + at

s = ut + 1/2 at²

v² = u² + 2as

s = ((u+v)/2)t

NEET Tips

• Practice velocity-time graphs.
• Remember sign convention.
• Understand distance and displacement difference.
• Use correct equation according to given data.
• Practice free fall problems.

Power Class 11 Physics Notes for NEET | Complete Revision Guide

 - Dr.Sanjaykumar Pawar  

Power Physics Notes PDF | NEET Work, Energy and Power Chapter

Power in Physics – Complete NEET Notes with Formulas, Units, Examples and Quick Revision

Internal Links

Work and Energy Notes

Kinetic Energy Complete Notes

Potential Energy Explained

Work-Energy Theorem

Conservation of Energy

Mechanical Energy Notes

Collisions in Physics

Circular Motion Notes

Laws of Motion Complete Notes

Newton's Laws Explained

Units and Dimensions Notes

Vectors for NEET

Scalars and Vectors

Friction Complete Notes

Gravitation Notes

Oscillations and SHM 

Mechanical Properties of Solids

Mechanical Properties of Fluids

Complete Class 11 Physics Notes

NEET Physics Formula Handbook

Work, Energy and Power · Unit IV

Power — Complete Question Bank

A full CBSE Class 11 Physics practice set on Power: MCQs, assertion–reason, case studies, match-the-columns, numericals, and a quick answer key — click any answer to reveal it.

Class 11 · CBSE Chapter: Work, Energy & Power 100+ Questions Answer Key Included

Jump to a section

1. A · Multiple Choice Questions
2. B · Very Short Answer (1 mark)
3. C · Short Answer (2 marks)
4. D · Long Answer (3–5 marks)
5. E · Assertion & Reason
6. F · Fill in the Blanks
7. G · Case Study Based
8. H · Statement Based
9. I · Match the Columns
10. J · True or False
11. K · Numerical Problems
12. Answer Key (Quick Reference)
13. Exam Tips

Section A Multiple Choice Questions

Tap "Show Answer" under any question to reveal the solution.

Level · Easy

Q1.Power is defined as:

Easy

• (a) Total work done
• (b) Rate of doing work
• (c) Energy stored in a body
• (d) Force applied per unit area

Show Answer

Answer: (b) Rate of doing work

Q2.The SI unit of power is:

Easy

• (a) Joule
• (b) Newton
• (c) Watt
• (d) Pascal

Show Answer

Answer: (c) Watt

Q3.1 horsepower is equal to:

Easy

• (a) 500 W
• (b) 746 W
• (c) 1000 W
• (d) 980 W

Show Answer

Answer: (b) 746 W

Q4.A machine does 1000 J of work in 10 s. Its power is:

Easy

• (a) 10 W
• (b) 100 W
• (c) 1000 W
• (d) 10000 W

Show Answer

Answer: (b) 100 W

Q5.The dimensional formula of power is:

Easy

• (a) [ML²T⁻²]
• (b) [ML²T⁻³]
• (c) [MLT⁻²]
• (d) [M²L²T⁻²]

Show Answer

Answer: (b) [ML²T⁻³]

Q6.Power is a:

Easy

• (a) Vector quantity
• (b) Scalar quantity
• (c) Neither scalar nor vector
• (d) Tensor quantity

Show Answer

Answer: (b) Scalar quantity

Q7.A pump lifts 200 kg of water to a height of 5 m in 10 s (g = 10 m/s²). The power of the pump is:

Easy

• (a) 100 W
• (b) 500 W
• (c) 1000 W
• (d) 2000 W

Show Answer

Answer: (c) 1000 W

Q8.If force and velocity are perpendicular to each other, the power is:

Easy

• (a) Maximum
• (b) Minimum
• (c) Zero
• (d) Infinite

Show Answer

Answer: (c) Zero

Level · Medium

Q9.A car engine exerts a force of 500 N while moving at a constant speed of 20 m/s. The power of the engine in hp is approximately:

Medium

• (a) 10.7 hp
• (b) 13.4 hp
• (c) 15.2 hp
• (d) 20.0 hp

Show Answer

Answer: (b) 13.4 hpP = Fv = 500 × 20 = 10000 W = 10000/746 ≈ 13.4 hp

Q10.The position of a particle is given by x = 3t² + 2t (x in m, t in s). A force of 6 N acts on it. The power at t = 2 s is:

Medium

• (a) 72 W
• (b) 84 W
• (c) 96 W
• (d) 108 W

Show Answer

Answer: (b) 84 Wv = dx/dt = 6t + 2. At t = 2s, v = 14 m/s. P = Fv = 6 × 14 = 84 W

Q11.A body of mass 2 kg is moved by a force of 10 N at constant velocity of 5 m/s at 60° to the direction of force. The power is:

Medium

• (a) 50 W
• (b) 25 W
• (c) 43.3 W
• (d) 0 W

Show Answer

Answer: (b) 25 WP = Fv cosθ = 10 × 5 × cos60° = 50 × 0.5 = 25 W

Q12.An engine of power 2 kW can do how much work in 1 minute?

Medium

• (a) 120 J
• (b) 2000 J
• (c) 120000 J
• (d) 20000 J

Show Answer

Answer: (c) 120000 JW = P × t = 2000 × 60 = 120000 J

Q13.A force F acts on a body moving with velocity v. If the angle between F and v is 120°, the power is:

Medium

• (a) Fv
• (b) Fv/2
• (c) Zero
• (d) −Fv/2

Show Answer

Answer: (d) −Fv/2P = Fv cos120° = Fv × (−1/2) = −Fv/2

Q14.The power of a pump that can lift 5000 kg of water per minute to a height of 20 m is: (g = 10 m/s²)

Medium

• (a) 16.67 kW
• (b) 10 kW
• (c) 100 kW
• (d) 1.67 kW

Show Answer

Answer: (a) 16.67 kWP = mgh/t = (5000 × 10 × 20)/60 = 1,000,000/60 ≈ 16,667 W ≈ 16.67 kW

Q15.A man of mass 60 kg climbs up a staircase carrying a load of 20 kg. If the total height gained is 10 m in 20 s, the average power is: (g = 10 m/s²)

Medium

• (a) 200 W
• (b) 300 W
• (c) 400 W
• (d) 800 W

Show Answer

Answer: (c) 400 WTotal mass = 80 kg. P = mgh/t = (80 × 10 × 10)/20 = 400 W

Level · Hard

Q16.A particle of mass m moves along a circular path of radius r with uniform speed v. The power delivered by the centripetal force is:

Hard

• (a) mv²/r
• (b) mv³/r
• (c) Zero
• (d) mv²r

Show Answer

Answer: (c) ZeroCentripetal force is always perpendicular to velocity (θ = 90°), so P = Fv cos90° = 0

Q17.The power delivered to a body moving in a straight line is given by P = 3t² + 2t (in watts). The work done in the first 2 seconds is:

Hard

• (a) 10 J
• (b) 12 J
• (c) 14 J
• (d) 16 J

Show Answer

Answer: (d) 16 JW = ∫P dt = ∫(3t² + 2t)dt = t³ + t². At t = 2: W = 8 + 8 = 16 J

Q18.A body of mass 1 kg is thrown vertically upward with initial velocity 20 m/s. The instantaneous power due to gravity at t = 1 s is: (g = 10 m/s²)

Hard

• (a) 100 W
• (b) −100 W
• (c) 200 W
• (d) −200 W

Show Answer

Answer: (b) −100 Wv = u − gt = 20 − 10 = 10 m/s (upward). F = mg = 10 N (downward). P = Fv cos180° = 10 × 10 × (−1) = −100 W

Q19.A pump motor is rated at 5 hp. How many kilograms of water can it raise in 1 minute through a height of 10 m? (g = 10 m/s², 1 hp = 746 W)

Hard

• (a) 1492 kg
• (b) 2238 kg
• (c) 2984 kg
• (d) 3730 kg

Show Answer

Answer: (b) 2238 kgP = 5 × 746 = 3730 W. W = P × t = 3730 × 60 = 223,800 J. m = W/(gh) = 223,800/(10×10) = 2238 kg

Q20.A vehicle of mass m accelerates uniformly from rest to velocity v in time t. The instantaneous power delivered by the engine at time t/2 is:

Hard

• (a) mv²/2t
• (b) mv²/4t
• (c) mv²/t
• (d) 3mv²/4t

Show Answer

Answer: (a) mv²/2ta = v/t (constant). At t/2, instantaneous velocity v′ = a(t/2) = v/2. F = ma = mv/t. P = F·v′ = (mv/t)(v/2) = mv²/2t.

Section B Very Short Answer Questions (1 Mark Each)

Q1.Define power.

Show Answer

Power is defined as the rate of doing work or the rate of energy transfer. Mathematically, P = W/t

Q2.Write the SI unit of power.

Show Answer

Watt (W), where 1 W = 1 J/s.

Q3.What is 1 horsepower in watts?

Show Answer

1 hp = 746 W.

Q4.Is power a scalar or vector quantity? Why?

Show Answer

Power is a scalar quantity because it has only magnitude and no direction. It is the ratio of two scalars (work and time).

Q5.Write the dimensional formula of power.

Show Answer

[M¹L²T⁻³]

Q6.A force acts perpendicular to the velocity of a body. What is the power?

Show Answer

Zero, because P = Fv cos90° = 0.

Q7.What is the relation between power, force, and velocity?

Show Answer

P = Fv (when force and velocity are in the same direction).

Q8.A machine does no work. Can it have power?

Show Answer

No. Since P = W/t, if W = 0, then P = 0.

Q9.What is the commercial unit of power?

Show Answer

Horsepower (hp).

Q10.Write the expression for instantaneous power.

Show Answer

P = dW/dt or P = F·v (dot product of force and instantaneous velocity).

Section C Short Answer Questions (2 Marks Each)

Q1.Distinguish between average power and instantaneous power.

Show Answer

Average Power	Instantaneous Power
Defined as total work divided by total time	Defined as power at a particular instant
Formula: Pavg = W/t	Formula: Pinst = F·v
Used when total work and time are given	Used when force or velocity varies with time

Q2.Derive the relation P = Fv.

Show Answer

We know power P = W/t. Work done W = F × s (for constant force in direction of displacement). Therefore, P = (F × s)/t = F × (s/t) = F × v. Hence, P = Fv

Q3.A pump delivers 1000 liters of water per minute to a tank at a height of 20 m. Find the power of the pump. (Density of water = 1000 kg/m³, g = 10 m/s²)

Show Answer

Volume per second = 1000/60 = 50/3 L/s = (50/3) × 10⁻³ m³/s. Mass per second = ρ × volume/s = 1000 × (50/3) × 10⁻³ = 50/3 kg/s. Power = (m/t) × gh = (50/3) × 10 × 20 = 10000/3 ≈ 3333.33 W ≈ 3.33 kW

Q4.Show that the power delivered by the centripetal force in uniform circular motion is zero.

Show Answer

In uniform circular motion, the centripetal force is always directed towards the center, while the velocity is always tangential. So θ = 90° always. P = Fv cosθ = Fv cos90° = 0.

Q5.The power of an engine is 5 kW. How much work can it do in 10 minutes?

Show Answer

Given: P = 5 kW = 5000 W, t = 10 min = 600 s. W = P × t = 5000 × 600 = 3,000,000 J = 3 × 10⁶ J

Q6.A car of mass 1000 kg moves up an incline of 1 in 20 at a constant speed of 10 m/s. Find the power of the engine. (g = 10 m/s², neglect friction)

Show Answer

Slope = 1/20, so sinθ = 1/20. Force required F = mg sinθ = 1000 × 10 × (1/20) = 500 N. Power P = Fv = 500 × 10 = 5000 W = 5 kW

Q7.Why is the power of a body moving with constant velocity on a frictionless horizontal surface zero?

Show Answer

On a frictionless horizontal surface, no external force is required to maintain constant velocity (Newton's first law). Since F = 0, power P = Fv = 0 × v = 0.

Q8.A 2 kW motor pump is used to pump water from a well 10 m deep. How much water can be pumped per minute? (g = 10 m/s²)

Show Answer

P = 2 kW = 2000 W, t = 60 s. W = P × t = 2000 × 60 = 120,000 J. m = W/(gh) = 120,000/(10 × 10) = 1200 kg

Section D Long Answer Questions (3–5 Marks Each)

Q1.(a) Define power and derive its SI unit. (2 marks)
(b) A pump can throw 8000 kg of water per minute to a height of 15 m. Calculate the power of the pump. (g = 9.8 m/s²) (3 marks)

Show Answer

(a) Power is defined as the rate of doing work. If W is the work done in time t, then P = W/t. The SI unit of work is Joule (J) and time is second (s). Therefore, SI unit of power = J/s = Watt (W). 1 Watt = 1 Joule per second.

(b) Given: m = 8000 kg (per minute), h = 15 m, t = 60 s, g = 9.8 m/s². Work done per minute = mgh = 8000 × 9.8 × 15 = 1,176,000 J. Power P = W/t = 1,176,000/60 = 19,600 W = 19.6 kW

Q2.(a) Derive the expression for instantaneous power. (2 marks)
(b) The position of a body of mass 2 kg is given by x = 2t³ + 3t² + 5, where x is in meters and t in seconds. A constant force of 12 N acts on the body in the direction of motion. Calculate the power delivered at t = 2 s. (3 marks)

Show Answer

(a) Consider a small amount of work dW done in a small time interval dt. Instantaneous power P = dW/dt. Since dW = F·ds, we get P = (F·ds)/dt = F·(ds/dt) = F·v. Therefore, P = F·v (dot product of force and instantaneous velocity).

(b) x = 2t³ + 3t² + 5 → v = dx/dt = 6t² + 6t. At t = 2 s: v = 6(4) + 6(2) = 24 + 12 = 36 m/s. Power P = F × v = 12 × 36 = 432 W

Q3.(a) Prove that power can also be expressed as the scalar product of force and velocity. (2 marks)
(b) An engine of power 10 hp is used to pump water from a well 8 m deep. How many kilograms of water can be pumped in 1 hour? (1 hp = 746 W, g = 9.8 m/s²) (3 marks)

Show Answer

(a) Work done by a force F in displacing a body by ds is dW = F·ds. Power is rate of doing work: P = dW/dt = (F·ds)/dt = F·(ds/dt) = F·v. Hence, P = F·v.

(b) P = 10 hp = 10 × 746 = 7460 W. t = 1 hour = 3600 s. Total work W = P × t = 7460 × 3600 = 26,856,000 J. m = W/(gh) = 26,856,000/(9.8 × 8) = 26,856,000/78.4 ≈ 342,551 kg

Q4.(a) What is the difference between kilowatt and kilowatt-hour? (2 marks)
(b) A family uses a 2 kW electric heater for 4 hours daily. Calculate the energy consumed in 30 days in kWh and joules. (3 marks)

Show Answer

(a)

Kilowatt (kW)	Kilowatt-hour (kWh)
Unit of power	Unit of energy
1 kW = 1000 W	1 kWh = 1000 W × 3600 s = 3.6 × 10⁶ J
Measures rate of energy consumption	Measures total energy consumed

(b) Power P = 2 kW, daily usage t = 4 h. Daily energy = 2 × 4 = 8 kWh. 30-day energy = 8 × 30 = 240 kWh. In joules: 240 kWh = 240 × 3.6 × 10⁶ = 8.64 × 10⁸ J

Section E Assertion and Reason Questions

Choose: (a) Both true, Reason correctly explains Assertion  |  (b) Both true, Reason does NOT explain Assertion  |  (c) Assertion true, Reason false  |  (d) Assertion false, Reason true

Q1.Assertion: Power is a scalar quantity.
Reason: Power is the ratio of two scalar quantities, work and time.

Show Answer

Answer: (a) Both true and Reason correctly explains Assertion.

Q2.Assertion: When a body moves in a circular path with uniform speed, the power delivered by the centripetal force is zero.
Reason: The centripetal force is always perpendicular to the velocity of the body.

Show Answer

Answer: (a) Both true and Reason correctly explains Assertion.

Q3.Assertion: A body moving with constant velocity on a frictionless horizontal surface has zero power.
Reason: No force is required to maintain constant velocity on a frictionless surface.

Show Answer

Answer: (a) Both true and Reason correctly explains Assertion.

Q4.Assertion: 1 horsepower is equal to 1000 watts.
Reason: Horsepower is the commercial unit of power.

Show Answer

Answer: (c) Assertion is false but Reason is true. (1 hp = 746 W, not 1000 W)

Q5.Assertion: The instantaneous power of a body can be negative.
Reason: Power is always positive because it is the rate of doing work.

Show Answer

Answer: (c) Assertion is true but Reason is false. (Power can be negative when force opposes motion, e.g., friction doing negative work)

Q6.Assertion: Average power is always equal to instantaneous power.
Reason: Average power is calculated over a time interval while instantaneous power is at a specific instant.

Show Answer

Answer: (d) Assertion is false but Reason is true.

Q7.Assertion: The power delivered by gravity to a body thrown vertically upward is negative.
Reason: The gravitational force acts opposite to the direction of motion during upward journey.

Show Answer

Answer: (a) Both true and Reason correctly explains Assertion.

Q8.Assertion: A pump requires more power to lift the same amount of water to a greater height.
Reason: Power is directly proportional to the height through which water is lifted.

Show Answer

Answer: (a) Both true and Reason correctly explains Assertion.

Section F Fill in the Blanks

Q1.Power is defined as the ______ of doing work.

Show Answer

rate

Q2.The SI unit of power is ______, named after the scientist ______.

Show Answer

Watt, James Watt

Q3.1 horsepower = ______ watts.

Show Answer

746

Q4.The dimensional formula of power is ______.

Show Answer

[M¹L²T⁻³]

Q5.When force and velocity are perpendicular to each other, the power is ______.

Show Answer

zero

Q6.The commercial unit of power is ______.

Show Answer

horsepower (hp)

Q7.For a body moving with constant velocity on a frictionless surface, the power is ______.

Show Answer

zero

Q8.Instantaneous power is given by the scalar product of ______ and ______.

Show Answer

force, velocity

Q9.1 kilowatt = ______ watts.

Show Answer

1000

Q10.The power of a pump lifting mass m through height h in time t is given by P = ______.

Show Answer

mgh/t

Section G Case Study Based Questions

Case Study 1 · Hydroelectric Power Plant

A hydroelectric power plant uses water falling from a height to generate electricity. Water from a reservoir at a height of 100 m flows down through penstocks to turn turbines. The plant has a capacity to process 5000 kg of water per second. (Take g = 10 m/s²)

Q1.What is the power generated by the falling water?

Show Answer

P = mgh/t = (5000 × 10 × 100)/1 = 5,000,000 W = 5 MW

Q2.If the efficiency of the turbine-generator system is 80%, what is the actual electrical power output?

Show Answer

Actual power = 80% of 5 MW = 0.8 × 5 = 4 MW

Q3.How much energy is produced in 1 hour at this actual power output?

Show Answer

E = P × t = 4 MW × 1 h = 4 MWh = 4 × 10⁶ Wh = 1.44 × 10¹⁰ J

Case Study 2 · Electric Vehicle

An electric car of mass 1500 kg accelerates from rest to a speed of 30 m/s in 10 seconds. The motor delivers constant power during this time.

Q1.What is the acceleration of the car?

Show Answer

a = (v − u)/t = (30 − 0)/10 = 3 m/s²

Q2.What is the average power delivered by the motor during acceleration?

Show Answer

Work done = ΔKE = ½mv² = ½ × 1500 × 30² = 675,000 J. P_avg = W/t = 675,000/10 = 67,500 W = 67.5 kW

Q3.If the car maintains a constant speed of 30 m/s on a level road with a frictional force of 500 N, what power is required to overcome friction?

Show Answer

P = Fv = 500 × 30 = 15,000 W = 15 kW

Case Study 3 · Human Power Output

A person of mass 70 kg climbs a staircase of 50 steps, each 20 cm high, in 20 seconds. (g = 9.8 m/s²)

Q1.What is the total height climbed?

Show Answer

h = 50 × 0.20 = 10 m

Q2.What is the work done against gravity?

Show Answer

W = mgh = 70 × 9.8 × 10 = 6860 J

Q3.What is the average power output of the person?

Show Answer

P = W/t = 6860/20 = 343 W

Section H Statement Based Questions

Passage 1

"Power is the rate at which work is done. It is a scalar quantity. The SI unit of power is watt. When a force acts on a body in the direction of its motion, the power is given by P = Fv. If the force makes an angle θ with the velocity, then P = Fv cosθ."

Q1.Why is power called a scalar quantity?

Show Answer

Power is a scalar quantity because it has only magnitude and no direction. It is derived as the ratio of work (scalar) and time (scalar), and the dot product of two vectors (F·v) also yields a scalar.

Q2.A force of 20 N acts on a body moving with a velocity of 4 m/s. The force makes an angle of 60° with the direction of motion. Calculate the power.

Show Answer

P = Fv cosθ = 20 × 4 × cos60° = 80 × 0.5 = 40 W

Passage 2

"A pump is used to lift water from a well. The power of the pump depends on the mass of water lifted, the height to which it is lifted, and the time taken. The efficiency of the pump is defined as the ratio of useful power output to the total power input."

Q1.Write the expression for the power of a pump lifting water.

Show Answer

P = mgh/t

Q2.A pump of power 2 kW and efficiency 75% is used to lift water through 10 m. How much water can it lift in 1 minute? (g = 10 m/s²)

Show Answer

Useful power = 75% of 2 kW = 0.75 × 2000 = 1500 W. Work done in 1 min = 1500 × 60 = 90,000 J. m = W/(gh) = 90,000/(10 × 10) = 900 kg

Section I Match the Columns

Match the Following 1

Column A	Column B
(a) Unit of power	(p) [M¹L²T⁻³]
(b) Dimensional formula of power	(q) Joule
(c) Unit of work	(r) Watt
(d) 1 hp	(s) 746 W

Show Answer

(a) → (r)  |  (b) → (p)  |  (c) → (q)  |  (d) → (s)

Match the Following 2

Column A (Physical Quantity)	Column B (Expression)
(a) Average power	(p) F·v
(b) Instantaneous power	(q) W/t
(c) Power in lifting	(r) mgh/t
(d) Power when F ⊥ v	(s) Zero

Show Answer

(a) → (q)  |  (b) → (p)  |  (c) → (r)  |  (d) → (s)

Match the Following 3

Column A (Situation)	Column B (Power)
(a) Body moving with constant velocity on frictionless surface	(p) Positive
(b) Body thrown upward against gravity	(q) Negative
(c) Body falling freely under gravity	(r) Zero
(d) Body in uniform circular motion	(s) Variable

Show Answer

(a) → (r) Zero, as F = 0  |  (b) → (q) Negative, gravity opposes motion  |  (c) → (p) Positive, gravity aids motion  |  (d) → (r) Zero, centripetal force ⊥ velocity

Match the Following 4

Column A	Column B
(a) James Watt	(p) Unit of energy
(b) Joule	(q) Commercial unit of power
(c) Horsepower	(r) SI unit of power
(d) Kilowatt-hour	(s) Improved steam engine

Show Answer

(a) → (s)  |  (b) → (p)  |  (c) → (q)  |  (d) → (p) — kWh is a unit of energy, like Joule

Match the Following 5

Column A (Value)	Column B (Equivalent)
(a) 1 kW	(p) 3.6 × 10⁶ J
(b) 1 kWh	(q) 1000 W
(c) 1 hp	(r) 746 W
(d) 1 W	(s) 1 J/s

Show Answer

(a) → (q)  |  (b) → (p)  |  (c) → (r)  |  (d) → (s)

Section J True or False

Q1.Power and work have the same dimensions.

Show Answer

False. [Work] = [ML²T⁻²], [Power] = [ML²T⁻³]

Q2.1 kilowatt-hour is a unit of power.

Show Answer

False. It is a unit of energy.

Q3.The power of a body can be negative.

Show Answer

True. When force opposes motion (e.g., friction), power is negative.

Q4.A body in uniform circular motion has zero power due to centripetal force.

Show Answer

True. Centripetal force is perpendicular to velocity.

Q5.The SI unit of power is named after James Prescott Joule.

Show Answer

False. It is named after James Watt.

Section K Numerical Problems (With Detailed Solutions)

Q1.A pump lifts 1200 kg of water per minute to a height of 25 m. Calculate the power of the pump in kW. (g = 10 m/s²)

Show Solution

m = 1200 kg, h = 25 m, t = 60 s, g = 10 m/s². P = mgh/t = (1200 × 10 × 25)/60 = 300,000/60 = 5000 W = 5 kW

Q2.A car of mass 1000 kg moves on a level road at a constant speed of 20 m/s against a resistance of 400 N. Find the power of the engine in horsepower.

Show Solution

At constant speed, engine force = resistance = 400 N. P = Fv = 400 × 20 = 8000 W. In hp: 8000/746 ≈ 10.7 hp

Q3.The power of an engine is 5 kW. How much time will it take to lift a load of 500 kg to a height of 40 m? (g = 10 m/s²)

Show Solution

W = mgh = 500 × 10 × 40 = 200,000 J. P = 5000 W. t = W/P = 200,000/5000 = 40 s

Q4.A force F = (3i + 4j) N acts on a body moving with velocity v = (4i − 3j) m/s. Calculate the power.

Show Solution

P = F·v = (3)(4) + (4)(−3) = 12 − 12 = 0 W

Q5.A motor pump is rated at 3 hp. Calculate the maximum mass of water it can lift in 2 minutes through a height of 15 m. (1 hp = 746 W, g = 9.8 m/s²)

Show Solution

P = 3 × 746 = 2238 W. t = 120 s. W = P × t = 2238 × 120 = 268,560 J. m = W/(gh) = 268,560/(9.8 × 15) = 268,560/147 ≈ 1827 kg

Answer Key (Quick Reference — MCQs & Assertion-Reason)

Section	Q. No.	Answer
MCQ (Easy)	1	(b)
	2	(c)
	3	(b)
	4	(b)
	5	(b)
	6	(b)
	7	(c)
	8	(c)
MCQ (Medium)	9	(b)
	10	(b)
	11	(b)
	12	(c)
	13	(d)
	14	(a)
	15	(c)
MCQ (Hard)	16	(c)
	17	(d)
	18	(b)
	19	(b)
	20	(a)
Assertion-Reason	1	(a)
	2	(a)
	3	(a)
	4	(c)
	5	(c)
	6	(d)
	7	(a)
	8	(a)

Exam Tips for CBSE Class 11

1. Formula Sheet: Memorize P = W/t, P = Fv, P = Fv cosθ, and P = mgh/t
2. Unit Conversions: Always convert to SI units first (especially hp → W, min → s)
3. Sign Convention: Power is positive when force aids motion, negative when it opposes
4. Scalar Nature: Remember power is scalar — never add vectorially
5. Time Management: Most Power questions take 1–2 minutes max

Best of luck for your CBSE Class 11 exams — master these questions and you're set for full marks on Power. 🎓

NEET Physics Notes
Chapter: POWER

Definition:
Power is the rate at which work is done or energy is transferred.

Simple Meaning

Work tells us how much work is done, while Power tells us how fast the work is done.

Example

• Student A climbs 4 floors in 20 seconds.
• Student B climbs 4 floors in 40 seconds.

Both students do the same work, but Student A has more power because he completes the work in less time.

Average Power

P = W / t

Where

• P = Average Power (Watt)
• W = Work Done (Joule)
• t = Time Taken (Second)

More Work + Less Time = More Power

Instantaneous Power

The power at a particular instant of time is called Instantaneous Power.

P = dW / dt

Power in Terms of Force

We know

dW = F · dr

Therefore

P = dW / dt

Since

dr / dt = v

Hence

P = F · v

General Formula

P = Fv cosθ

Special Cases

Case 1: Force and Velocity in Same Direction

θ = 0°

P = Fv

Power is Maximum.

---

Case 2: Force Opposite to Velocity

θ = 180°

P = -Fv

Negative power means energy is removed from the body.

Example: Braking a moving car.

---

Case 3: Force Perpendicular to Velocity

θ = 90°

P = 0

Example: Uniform Circular Motion

The centripetal force changes only the direction of velocity, not its speed.

Nature of Power

Power is a Scalar Quantity.

Reason: It is obtained from the dot product of force and velocity.

Dimensions of Power

[ML²T⁻³]

SI Unit

Watt (W)

Named after James Watt, who improved the steam engine.

Definition of One Watt

1 Watt = 1 Joule / Second

A power of one watt means one joule of work is done every second.

Other Units

1 kW = 1000 W

1 hp = 746 W

Horsepower is commonly used for engines, cars and motorcycles.

Electrical Energy

Electrical appliances are rated in watts.

• 60 W Bulb
• 100 W Bulb
• 1000 W Heater

Higher wattage means the appliance consumes energy faster.

Kilowatt-hour (kWh)

Electricity bills are measured in kilowatt-hour (kWh).

Energy = Power × Time

Conversion

1 kWh = 1000 W × 3600 s

= 3.6 × 10⁶ J

Example

A 100 W bulb runs for 10 hours.

100 × 10 = 1000 Wh

= 1 kWh

A 100 W bulb used for 10 hours consumes 1 Unit of electricity.

Important Fact

1 Unit of Electricity = 1 kWh = 3.6 × 10⁶ Joules

Remember:
kWh is a unit of Energy, NOT Power.

Difference Between Power and Energy

Power	Energy
Rate of doing work	Capacity to do work
P = W / t	W = Pt
Unit = Watt (W)	Unit = Joule (J)
Scalar Quantity	Scalar Quantity
Measures speed of work	Measures amount of work

Graph Concept

Slope of Work-Time Graph = Power

NEET Formula Sheet

Formula	Expression
Average Power	P = W / t
Instantaneous Power	P = dW / dt
Power by Force	P = F · v
General Formula	P = Fv cosθ
Parallel Force	P = Fv
Perpendicular Force	P = 0
Opposite Force	P = -Fv
1 Watt	1 J/s
Horsepower	1 hp = 746 W
1 kWh	3.6 × 10⁶ J

NEET Quick Revision

• Power = Rate of doing work.
• Power = Speed of energy transfer.
• Power is a scalar quantity.
• P = Fv cosθ.
• Maximum power when θ = 0°.
• Zero power when θ = 90°.
• Negative power when θ = 180°.
• 1 Watt = 1 Joule/second.
• 1 Horsepower = 746 W.
• 1 Unit of electricity = 1 kWh = 3.6 × 10⁶ J.
• kWh is a unit of Energy, not Power.

Frequently Asked NEET Questions

Q1. What is Power?

Power is the rate at which work is done or energy is transferred.

Q2. Why is Power a Scalar Quantity?

Because Power is obtained from the dot product of Force and Velocity.

Q3. What is SI Unit of Power?

Watt (W).

Q4. Is kWh a unit of Power?

No. It is a unit of Energy.

Q5. What is the power when Force is perpendicular to Velocity?

P = 0

End of Chapter - POWER
Complete Beginner-Friendly NEET Physics Notes

Displacement in the nth Second Formula | Complete Notes with Derivation

Displacement during the nth second under constant acceleration explained with formula and illustration.

-  Dr.Sanjaykumar Pawar

 nth Second Displacement Formula | Class 11 Physics Notes

---

 Internal Links

• Motion in a Straight Line Notes

• Distance and Displacement Explained

• Speed, Velocity and Acceleration

• Equations of Motion (SUVAT)

• Graphs of Motion (Distance-Time & Velocity-Time)

• Average Velocity Formula

• Relative Motion Notes

• Uniform and Non-Uniform Motion

• Free Fall Motion

• Projectile Motion Basics

• Kinematics Formula Sheet

• Motion Formula Cheat Sheet

• Class 11 Physics Chapter 3 Notes

• JEE Kinematics Questions

• NEET Physics Motion Practice Problems

---

FAQ 

Q1. What is displacement in the nth second?
A. It is the displacement covered by an object during only the nth second of its motion.

Q2. What is the formula for displacement in the nth second?
A. The formula is:
Sₙₜₕ = u + (a/2)(2n−1)

Q3. When can this formula be used?
A. It is applicable only when the acceleration is constant.

Q4. Is nth second displacement the same as total displacement?
A. No. Total displacement is measured from the start of motion, whereas nth second displacement is measured only during one specific second.

Q5. Which exams include questions on the nth second formula?
A. This topic is important for Class 11 Physics, JEE Main, JEE Advanced, NEET, and other competitive examinations. 

Displacement in the n^th Second

1. What is Displacement in the n^th Second?

Displacement in the n^th second means the displacement covered during only one particular second.

Examples

• 1st second → 0 s to 1 s
• 2nd second → 1 s to 2 s
• 3rd second → 2 s to 3 s
• 5th second → 4 s to 5 s

It does not mean the total displacement from the starting point.

2. Total Displacement After n Seconds

The displacement after n seconds is given by:

S_n = un + ½ an²

Where:

• u = Initial velocity
• a = Constant acceleration
• n = Time in seconds
• S_n = Total displacement after n seconds

3. Total Displacement After (n−1) Seconds

The displacement up to the end of the previous second is:

S_n−1 = u(n−1) + ½ a(n−1)²

This represents the total displacement before the n^th second begins.

4. Formula for Displacement in the n^th Second

Displacement during only the n^th second is:

S_nth = S_n − S_n−1

This means:

Displacement in one interval = Total displacement at the end − Total displacement at the beginning.

5. Substitute the Values

S_nth = (un + ½an²) − [u(n−1) + ½a(n−1)²]

Now simplify the expression.

6. Final Formula

S_nth = u + (a/2)(2n − 1)

This is the standard formula used to calculate displacement during the n^th second.

Meaning of Symbols

Symbol	Meaning
Snth	Displacement during the nth second
u	Initial velocity
a	Constant acceleration
n	Number of the second

Example

Given:

• u = 5 m/s
• a = 2 m/s²
• Find displacement in the 4th second.

S_4th = 5 + (2/2)(2×4 −1)

= 5 + 7

= 12 m

Answer: The object moves 12 m during the 4th second.

Important Points

• Acceleration must remain constant.
• The formula gives displacement, not total displacement.
• The time interval is always 1 second.
• The n^th second starts at (n−1) s and ends at n s.
• Examples:
  • 3rd second → 2 s to 3 s
  • 6th second → 5 s to 6 s
  • 10th second → 9 s to 10 s

Total Displacement vs Displacement in the n^th Second

Total Displacement	Displacement in nth Second
Measured from 0 s to n s	Measured only during one second
S = ut + ½at²	Snth = u + (a/2)(2n−1)
Gives total position change	Gives position change in one specific second

Quick Revision

1. Find total displacement after n seconds.
2. Find total displacement after (n−1) seconds.
3. Subtract both values.
4. Use the final formula:

S_nth = u + (a/2)(2n−1)

Memory Trick:

n^th second = Total displacement till n − Total displacement till (n−1).

NEET Practice Questions – Displacement in the n^th Second

Formula: S_nth = u + ^a⁄₂(2n − 1)

Type 1: Direct MCQs

Q1 (Easy)
Displacement in the n^th second is:

A. u + a/2(2n+1)      B. u + a/2(2n−1)      C. ut + ½at²      D. un + ½an²

Answer: B

Q2
u = 8 m/s, a = 4 m/s². Find displacement in 3rd second.

Answer: 18 m

Q3
Starts from rest, a = 6 m/s². Displacement in 5th second?

Answer: 27 m

Q4
6th second displacement = 29 m, a = 2 m/s². Find u.

Answer: 18 m/s

Q5 (Hard)
8th second displacement = 40 m, a = 4 m/s². Find u.

Answer: 10 m/s

Type 2: Statement Based

Q6
Statement I: S_nth = S_n − S_(n−1).
Statement II: nth-second displacement equals difference of total displacements.

Answer: Both True.

Q7
Statement I: Formula works for variable acceleration.
Statement II: Constant acceleration is required.

Answer: I False, II True.

Q8
Statement I: Formula gives displacement.
Statement II: Distance and displacement are always equal.

Answer: I True, II False.

Type 3: Assertion & Reason

Q9
Assertion (A): 4th-second displacement = S₄ − S₃.
Reason (R): 4th second is from 3 s to 4 s.

Answer: Both true and R explains A.

Q10
Assertion (A): Formula works for changing acceleration.
Reason (R): Derivation assumes constant acceleration.

Answer: A False, R True.

Q11
Assertion (A): If u = 0, S_nth = (a/2)(2n−1).
Reason (R): Body starts from rest.

Answer: Both true and R explains A.

Type 4: Match the Columns

Column I	Column II
A. 1st second	P. 4–5 s
B. 2nd second	Q. 1–2 s
C. 5th second	R. 0–1 s
D. 8th second	S. 7–8 s

Correct Match: A-R, B-Q, C-P, D-S

Type 5: Diagram Based

Q12

0s ---- 1 ---- 2 ---- 3 ---- 4
4th second = 3 s to 4 s

Answer: 3 s to 4 s

Q13

S
|
|      *
|   *
| *
+---------------> t

Correct equation: S = ut + ½at²

Answer Key

1-B, 2-18 m, 3-27 m, 4-18 m/s, 5-10 m/s, 6-Both True, 7-I False, II True, 8-I True, II False, 9-Both true and R explains A, 10-A False, R True, 11-Both true and R explains A, 12-3 s to 4 s, 13-S = ut + ½at²

Column Match: A-R, B-Q, C-P, D-S

Units and Measurements Class 11 Physics NCERT Solutions | Step-by-Step Answers, Short Notes & Mnemonics for NEET

- Dr.Sanjaykumar pawar 

Units and Measurements Class 11 Physics – Easy NCERT Solutions, Short Notes, Formulas, and NEET Revision Tricks

 Internal Links

Physics Class 11 Chapter 1 Physical World Notes

Measurement Errors Complete Notes

Significant Figures Explained

Vernier Calipers Complete Guide

Screw Gauge Complete Guide

Least Count Formula Explained

Dimensional Analysis Notes

NCERT Class 11 Physics Solutions

NEET Physics Short Notes

NEET Physics Formula Handbook

Class 11 Physics Important Questions

Physics MCQs with Solutions

NCERT Exemplar Physics Solutions

JEE Physics Revision Notes

Measurement Instruments Comparison

UNITS AND MEASUREMENTS

Question (a)

You are given a thread and a metre scale. How will you estimate the diameter of the thread?

Step-by-Step Answer

Step 1: Take a pencil or a cylindrical rod.

Step 2: Wind the thread closely around the pencil without leaving any gap.

Step 3: Make 20–50 turns of the thread.

Step 4: Measure the total length (L) of all turns using the metre scale.

Step 5: Count the total number of turns (N).

Diameter of Thread = Total Length (L) / Number of Turns (N)

Answer: Divide the total length of all turns by the number of turns to obtain the diameter of the thread.

Question (b)

A screw gauge has a pitch of 1.0 mm and 200 divisions on the circular scale. Do you think it is possible to increase the accuracy of the screw gauge arbitrarily by increasing the number of divisions on the circular scale?

Step-by-Step Answer

Step 1: Use the formula

Least Count = Pitch / Circular Scale Divisions

Step 2:

Least Count = 1.0 / 200 = 0.005 mm

Step 3: Increasing the number of divisions decreases the least count.

Step 4: However, accuracy cannot increase forever because of:

• Human error
• Instrument error
• Manufacturing limitations
• Backlash error

Final Answer:

No. Accuracy cannot be increased arbitrarily by increasing the number of divisions.

Question (c)

The mean diameter of a thin brass rod is to be measured using Vernier Calipers. Why is a set of 100 measurements more reliable than only 5 measurements?

Step-by-Step Answer

Step 1: Every measurement contains small random errors.

Step 2: With only 5 readings, random errors affect the average more.

Step 3: With 100 readings, positive and negative errors cancel each other.

Step 4: Therefore, the mean value becomes closer to the true value.

More Measurements → Less Random Error → Better Mean Value

Short Notes (Quick Revision)

Topic	Key Point
Thread Diameter	Wind the thread around a pencil and use Length ÷ Turns.
Screw Gauge	Measures very small thickness like wire diameter.
Least Count	Least Count = Pitch ÷ Circular Divisions.
Vernier Calipers	Measures external diameter, internal diameter and depth.
Repeated Measurements	More readings reduce random errors.

Remember:

• More Readings = More Accuracy
• Smaller Least Count = Better Precision
• Screw Gauge is more accurate than Vernier Calipers.

Mnemonics (Easy Memory Tricks)

1. Instrument Order

Mnemonic:

Meter → Vernier → Screw

Sentence:

"My Very Smart Friend"

Word	Meaning
My	Meter Scale
Very	Vernier Calipers
Smart	Screw Gauge

2. Accuracy Order

Mnemonic:

Meter < Vernier < Screw

Sentence:

"Accuracy Climbs Up"

Instrument	Accuracy
Meter Scale	Low
Vernier Calipers	Medium
Screw Gauge	Highest

3. Formula Memory Trick

L ÷ N = Diameter

Mnemonic:

"Length Needs Number"

Formula	Meaning
L	Total Length
N	Number of Turns
L ÷ N	Diameter

4. NEET MCQ Trick

If Question Says	Remember
More Readings	Better Mean Value
Increase Circular Divisions	Accuracy improves only up to practical limits.
Least Count	Smaller LC → Better Precision