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.

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

 - Dr.Sanjaykumar Pawar  

Vectors Class 11 Physics Notes, MCQs, Assertion Reason, Case Study & CBSE Questions

Complete Class 11 Physics Vectors Notes with formulas, diagrams, MCQs, assertion-reason, case studies and CBSE exam questions.

Internal Links

• Class 11 Physics Units and Measurements
• Motion in a Straight Line Notes
• Motion in a Plane
• Projectile Motion
• Laws of Motion
• Work, Energy and Power
• System of Particles
• Circular Motion
• Kinematics Formula Sheet
• Physics Formula Handbook
• NEET Physics Notes
• CBSE Class 11 Physics MCQs
• Class 11 Physics Previous Year Questions
• Class 11 Physics Sample Papers
• NCERT Solutions for Class 11 Physics
• Important Physics Derivations
• Physics Practical Experiments
• Physics Revision Notes
• Physics Chapter-wise Question Bank
• CBSE Class 11 Study Material

NEET Physics Chapter : Vectors

Vectors are one of the most important topics in NEET Physics. Almost every chapter uses vectors. Therefore understanding vectors properly makes Mechanics very easy.

---

1. Equality of Vectors

Two vectors are called equal if

• Magnitude is same.
• Direction is same.

Position does NOT matter. Even if vectors are shifted parallel, they are still equal.

Equal vectors ⇒ Same Magnitude + Same Direction

---

2. Addition of Vectors

Vector addition means combining two vectors to get one resultant vector.

Triangle Law

Place the tail of second vector at the head of first vector. Join the starting point to the final point. That gives resultant.

Remember: Head to Tail Rule

---

3. Parallelogram Law

If two vectors start from the same point, complete a parallelogram. Diagonal gives resultant vector.

Resultant = Diagonal of Parallelogram

---

4. Magnitude of Resultant

Suppose

First Vector = a

Second Vector = b

Angle between them = θ

R = √(a² + b² + 2ab cosθ)

This is one of the MOST IMPORTANT formulas for NEET. Learn it perfectly.

---

5. Direction of Resultant

tanα = (b sinθ)/(a + b cosθ)

α = angle made by resultant with first vector.

---

6. Special Cases

Angle	Magnitude
0°	a+b
180°	|a-b|
90°	√(a²+b²)

---

7. Example

Question: Two vectors have equal magnitude A. Angle between them is θ. Find resultant.

Solution

R = √(A²+A²+2A²cosθ)

= √(2A²(1+cosθ))

Using 1+cosθ=2cos²(θ/2)

R = 2A cos(θ/2)

Resultant = 2A cos(θ/2)

Direction:

α = θ/2

The resultant bisects the angle between two equal vectors.

---

8. Memory Tricks

✔ Triangle Rule → Head to Tail

✔ Parallelogram Rule → Diagonal

✔ Equal Vectors → Same Magnitude + Same Direction

✔ 90° → Pythagoras

✔ 180° → Subtraction

✔ 0° → Addition

---

9. NEET Important Points

• Magnitude is always positive.
• Direction decides vector.
• Vectors obey triangle law.
• Resultant depends on angle.
• Equal vectors can have different positions.
• Parallelogram law is frequently asked in NEET.

---

10. Practice Questions

1. Define equal vectors.
2. State triangle law.
3. State parallelogram law.
4. Write magnitude formula.
5. Write direction formula.
6. Find resultant when angle is 90°.
7. Find resultant when angle is 180°.
8. Two vectors 10 N each make 60°. Find resultant.
9. Two vectors 5 N each make 120°. Find resultant.
10. Why does the resultant bisect equal vectors?

---

Summary

• Equal vectors → Same magnitude + same direction
• Triangle Law → Head to Tail
• Parallelogram Law → Diagonal
• Magnitude → √(a²+b²+2abcosθ)
• Direction → tanα=(bsinθ)/(a+bcosθ)
• Equal vectors → Resultant = 2Acos(θ/2)
• Direction = θ/2

CBSE Class 11 Physics

Chapter : Vectors Question Bank

1. Multiple Choice Questions (MCQs)

1. A vector quantity has

1. Only magnitude
2. Only direction
3. Magnitude and direction
4. None

Answer: C

2. Equal vectors have

1. Equal magnitude only
2. Equal direction only
3. Equal magnitude and direction
4. Different directions

Answer: C

3. The diagonal of a parallelogram represents

1. Difference of vectors
2. Resultant vector
3. Unit vector
4. Zero vector

Answer: B

2. Very Short Answer Questions (1 Mark)

Q1. Define a vector.

A quantity having both magnitude and direction is called a vector.

Q2. Give one example of a vector.

Velocity.

Q3. What is a zero vector?

A vector whose magnitude is zero.

3. Short Answer Questions (2-3 Marks)

Q1. Define equal vectors.

Two vectors having equal magnitude and same direction are called equal vectors.

Q2. State the triangle law of vector addition.

If two vectors are represented by two sides of a triangle taken in order, the third side taken in opposite order represents the resultant.

4. Long Answer Questions (5 Marks)

Q1. Explain the parallelogram law of vector addition with diagram.

If two vectors acting simultaneously are represented by two adjacent sides of a parallelogram, then the diagonal passing through the common point represents the resultant vector. Magnitude: R = √(A² + B² + 2AB cosθ) Direction: tanα = (B sinθ)/(A + B cosθ)

5. Assertion and Reason

Assertion: Equal vectors may have different positions.

Reason: A vector depends only on magnitude and direction.

Answer: Both Assertion and Reason are true, and Reason correctly explains Assertion.

Assertion: Scalar quantities have direction.

Reason: Scalars possess only magnitude.

Answer: Assertion is false. Reason is true.

6. Fill in the Blanks

Question	Answer
A vector has ______ and ______.	Magnitude, Direction
The diagonal of a parallelogram gives the ______.	Resultant
A quantity having only magnitude is called ______.	Scalar
The SI unit of displacement is ______.	metre

7. Match the Columns

Column A	Column B
Velocity	Vector
Mass	Scalar
Acceleration	Vector
Time	Scalar

Answers Velocity → Vector Mass → Scalar Acceleration → Vector Time → Scalar

8. Statement Based Questions

Statement I: The resultant of two equal vectors bisects the angle between them.

Statement II: The magnitudes of both vectors are equal.

Both statements are true. Statement II explains Statement I.

9. Case Study Questions

Rahul pushes a box with force 20 N towards east. Aman pushes the same box with force 20 N making an angle of 60°. Answer the following.

1. Which law is used?
2. Write the magnitude formula.
3. If angle becomes 180°, what happens?

1. Parallelogram law.
2. R = √(A²+B²+2ABcosθ)
3. Resultant = |A−B|

10. Numerical Questions

Two vectors of magnitude 5 N each make an angle of 60°. Find the resultant.

R = √(25+25+50×0.5) = √75 = 8.66 N

11. HOTS Questions

Why can two vectors be equal even if they are drawn at different places?

Because a vector depends only on magnitude and direction, not on position.

12. Competency Based Questions

A boat is moving across a river. Which physical quantities should be treated as vectors?

Velocity, displacement and acceleration.

13. One Word Questions

Question	Answer
Quantity having direction	Vector
Quantity having only magnitude	Scalar
Magnitude zero vector	Zero Vector
Vector of magnitude one	Unit Vector

14. Important CBSE Questions

1. Define vector.
2. State triangle law.
3. State parallelogram law.
4. Define equal vectors.
5. What is a unit vector?
6. What is a null vector?
7. Derive the magnitude formula.
8. Derive the direction formula.
9. Differentiate scalar and vector.
10. Give five examples each of scalars and vectors.

Vectors and Scalars Notes — CBSE Class 11 & NEET Guide

 - Dr.Sanjaykumar Pawar  

Vectors are drawn as arrows — length shows magnitude, arrowhead shows direction.

 

Internal Links

	Links To
parallelogram law of vector addition	Parallelogram Law — Class 11 Physics Notes
resolution of vectors into components	Resolution of Vectors & Rectangular Components
dot product and cross product	Scalar and Vector Product of Two Vectors
motion in a plane	Motion in a Plane — Full Chapter Notes (NCERT Ch. 4)
projectile motion numericals	Projectile Motion — Solved Problems for NEET
CBSE Class 11 Physics MCQs	Class 11 Physics — Chapter-wise MCQ Bank
NEET previous year vector questions	NEET PYQs — Vectors & Motion in a Plane

NEET Physics · Chapter 2

Vectors & Scalars

The quantities that need a direction, the ones that don't, and the one rule that tells them apart. Complete beginner notes with worked diagrams.

THE TUBE-AND-BALL PROBLEM — a right triangle hiding in a physics question

01 — Foundations

Why physics needs vectors

Mathematics is the language of physics. Some quantities are fully described by just a number. Others refuse to make sense without a direction attached. Splitting these two apart is the entire point of this chapter — and it shows up in almost every numerical on the NEET paper afterward, from projectile motion to electric fields.

02 — The simple ones

Scalars

Definition: quantities completely described by a numerical value (with a unit) alone. No direction is involved, and they combine using ordinary algebra.

Worked example

A system made of two bodies — one of mass 5 kg, the other 2 kg — has a combined mass of:

5 kg + 2 kg = 7 kg

No angles, no diagrams. Just addition. That is the signature of a scalar.

Common scalars: mass, time, temperature, speed, energy, work, power, distance, charge, density.

03 — The directional ones

Vectors

Definition: quantities that need both magnitude and direction for a complete description, and which add according to the geometric triangle law — not plain algebra.

representation

FIG. A — vectors drawn to scale: 1 cm ≡ 1 m/s (arbitrary chosen scale)

Anatomy of a vector arrow

Part	Name	Meaning
Back end	Tail	Starting point
Front end	Head	Points in the direction of the vector
Length	Magnitude	Numerical size, drawn to scale

Notation: written with an arrow on top — $\vec{AB}$, $\vec{v}$ — or in bold print: AB, v, F.

TRAP

Having a direction is not enough. Electric current flows through a wire in a direction — but current does not add up by the triangle rule. Two currents meeting at a junction just add algebraically (Kirchhoff's rule), not geometrically. So current is a scalar, despite having a direction. This exact question appears repeatedly in NEET-level papers.

04 — The addition rule

Triangle law of vector addition

Statement: if two vectors are represented, in magnitude and direction, by two sides of a triangle taken in order, the resultant is given by the third side, taken in the reverse order.

step 1

Draw the first vector, A to B

step 2

From B, draw the second vector, B to C

step 3 — the resultant

Join A to C. This is $\vec{AB} + \vec{BC} = \vec{AC}$

RULE

The resultant always runs from the tail of the first vector to the head of the last vector — regardless of how many vectors you chain together, tail-to-head.

05 — Solved numerical

The tube-and-ball problem

A small ball moves inside a long tube at 3 m/s while the tube itself moves at 4 m/s perpendicular to its own length. What is the ball's resultant velocity, as seen from the room?

Pythagoras hiding inside a physics question: 3-4-5 triangle

Solution

In 1 second: tube carries the ball 3 m along its length; the ball also moves 4 m perpendicular to it. The two displacements form a right triangle, so:

R = √(3² + 4²) = √25 = 5 m

Since this happened in 1 s, the resultant velocity is 5 m/s, directed at

θ = tan⁻¹(4/3) = 53°

…from the direction of the tube.

MEMORIZE

The 3–4–5 right triangle (and its cousin 5–12–13) shows up constantly in NEET vector numericals. Spotting it instantly saves calculator time in the exam.

06 — Reference

Quick formula box

Concept	Formula
Resultant magnitude	R = √(A² + B² + 2AB cos θ)
Direction of resultant	tan α = B sin θ / (A + B cos θ)
Maximum resultant (θ = 0°)	R = A + B
Minimum resultant (θ = 180°)	R = |A − B|
Perpendicular vectors (θ = 90°)	R = √(A² + B²)

This is the parallelogram-law version of the same triangle rule — NEET numericals usually hand you this formula directly, so both pictures are worth knowing.

07 — Side by side

Scalar vs vector

Property	Scalar	Vector
Needs	Magnitude + unit only	Magnitude + unit + direction
Addition rule	Ordinary algebra	Triangle / parallelogram law
Examples	mass, speed, work, energy, current, charge	displacement, velocity, acceleration, force, momentum
Sign	Can be negative (e.g. temperature)	Magnitude always positive; direction shows sense

Last-minute recall

• Scalar = magnitude + unit. Vector = magnitude + unit + direction + triangle law.
• Triangle law: tail-to-head arrangement; resultant runs from the very first tail to the very last head.
• Current has a direction but is still a scalar — it fails the triangle law test.
• 3-4-5 right triangle → resultant 5, angle 53°. A recurring numerical pattern — recognize it on sight.
• Vector addition is commutative: $\vec{A}+\vec{B} = \vec{B}+\vec{A}$.

NEET PHYSICS — VECTORS & SCALARS · FIELD NOTES

CBSE · Class 11 Physics · Ch. Vectors & Scalars

Complete Question Bank

Every CBSE exam format in one place — MCQs, assertion-reason, fill-in-the-blanks, match-the-column, case study, and short/long answers. Tap any question to reveal the answer.

Very Short Short Answer Long Answer MCQs Assertion-Reason Fill in Blanks Match Column Statement-Based Case Study

Section A

Very Short Answer Questions

1 mark each — one line / one word answers

1 markQ1. Define a scalar quantity.

Show answer

A scalar is a physical quantity that is completely described by its magnitude (with a proper unit) alone; it has no associated direction. Example: mass, time.

1 markQ2. Define a vector quantity.

Show answer

A vector is a physical quantity that requires both magnitude and direction for its complete description, and which obeys the triangle law of addition.

1 markQ3. Give one example each of a scalar and a vector quantity other than mass and velocity.

Show answer

Scalar: electric charge (or work, energy). Vector: force (or momentum, acceleration).

1 markQ4. Is electric current a vector quantity? Justify in one line.

Show answer

No. Although current has a direction of flow, it does not add according to the triangle law of vector addition, so it is treated as a scalar.

1 markQ5. What is a unit vector?

Show answer

A vector having a magnitude of exactly one, used only to indicate direction. Example: î, ĵ, k̂ along the x, y, z axes.

1 markQ6. What is meant by a null (zero) vector?

Show answer

A vector whose magnitude is zero and whose direction is indeterminate. Example: the resultant of two equal and opposite vectors.

1 markQ7. Can the magnitude of a vector be negative?

Show answer

No. The magnitude of a vector is always a non-negative real number; a negative sign only reverses its direction.

1 markQ8. State whether displacement is a scalar or a vector.

Show answer

Displacement is a vector quantity — it has both magnitude (shortest distance) and direction (from initial to final position).

1 markQ9. Two vectors are said to be equal when — complete the statement.

Show answer

…when they have the same magnitude and the same direction, regardless of their initial points (position).

1 markQ10. What is the angle between two vectors for their resultant to be maximum?

Show answer

0° (vectors acting in the same direction); the resultant magnitude is then A + B.

Section B

Short Answer Questions

2–3 marks each

2 marksQ1. Distinguish between scalar and vector quantities with one example of each.

Show answer

Answer A scalar has magnitude and unit only, and adds by ordinary algebra (e.g., mass: 2 kg + 3 kg = 5 kg). A vector has magnitude, unit, and direction, and adds by the triangle/parallelogram law (e.g., velocity: two velocities at an angle combine geometrically, not by simple addition).

2 marksQ2. Why is electric current not considered a vector quantity even though it has direction?

Show answer

Answer Current has magnitude and a sense of direction along a wire, but two currents meeting at a junction combine algebraically (Kirchhoff's current law), not by the triangle law. Since a valid vector must obey vector addition rules, current fails this test and is classified as a scalar.

3 marksQ3. State the triangle law of vector addition and mention one limitation of representing vectors only graphically.

Show answer

Answer Triangle law: if two vectors are represented in magnitude and direction by two sides of a triangle taken in order, their resultant is represented by the third side taken in the reverse order (tail of first to head of second).

Limitation: a purely graphical (scale-drawing) method is time-consuming and gives limited accuracy compared to the analytical formula R = √(A² + B² + 2AB cosθ), especially for angles that are not simple values.

2 marksQ4. What are equal and negative vectors? Give an example of each.

Show answer

Answer Equal vectors: same magnitude and same direction (e.g., two cars moving at 40 km/h due north). Negative vectors: same magnitude but opposite direction (e.g., $\vec{A}$ and $-\vec{A}$ — a vector and its reverse).

3 marksQ5. Explain resolution of a vector into rectangular components with a labelled reasoning (no diagram needed, describe it).

Show answer

Answer Any vector $\vec{A}$ in the xy-plane can be broken into two mutually perpendicular components: $A_x = A\cos\theta$ along the x-axis and $A_y = A\sin\theta$ along the y-axis, where θ is the angle the vector makes with the x-axis. These components, added vectorially, reproduce the original vector: $\vec{A} = A_x\hat{i} + A_y\hat{j}$. This makes vector algebra (addition/subtraction) far simpler because components along the same axis just add algebraically.

2 marksQ6. Two forces of 3 N and 4 N act on a body at right angles to each other. Find the magnitude of the resultant.

Show answer

Answer Since θ = 90°, R = √(3² + 4²) = √25 = 5 N, directed at tan⁻¹(4/3) = 53° from the 3 N force.

Section C

Long Answer Questions

5 marks each — full derivations expected in exam

5 marksQ1. State and derive the expression for the magnitude and direction of the resultant of two vectors using the parallelogram law of vector addition.

Show answer

Answer (outline) Let $\vec{A}$ and $\vec{B}$ act at angle θ, represented as two adjacent sides OP and OQ of a parallelogram OPRQ from a common point O. The diagonal OR represents the resultant $\vec{R}$.

Drop a perpendicular from R to the extended OP, meeting it at N. In right triangle ONR: ON = A + B cosθ, and NR = B sinθ.

By Pythagoras: R² = (A + Bcosθ)² + (Bsinθ)² ⇒ R = √(A² + B² + 2AB cosθ).

Direction: tanα = NR / ON = B sinθ / (A + B cosθ), where α is the angle the resultant makes with $\vec{A}$.

Special cases: θ=0° gives R=A+B (maximum); θ=180° gives R=|A−B| (minimum); θ=90° gives R=√(A²+B²).

5 marksQ2. Explain the resolution of a vector in a plane into two mutually perpendicular components, and use it to derive the formula for the resultant of two vectors by the component method.

Show answer

Answer (outline) A vector $\vec{A}$ making angle θ with the x-axis has components $A_x = A\cos\theta$, $A_y = A\sin\theta$, so $\vec{A} = A_x\hat{i} + A_y\hat{j}$, and $A = \sqrt{A_x^2+A_y^2}$.

For two vectors $\vec{A} = A_x\hat{i}+A_y\hat{j}$ and $\vec{B}=B_x\hat{i}+B_y\hat{j}$, the resultant is found by adding components along each axis separately:
$R_x = A_x + B_x$, $R_y = A_y + B_y$
$\vec{R} = R_x\hat{i} + R_y\hat{j}$, with magnitude $R = \sqrt{R_x^2 + R_y^2}$ and direction $\theta = \tan^{-1}(R_y/R_x)$.

This component method avoids drawing diagrams for every problem and is the standard technique used in numericals involving 3 or more vectors.

5 marksQ3. Distinguish between scalar (dot) product and vector (cross) product of two vectors, giving their definitions, formulae, and one physical example of each.

Show answer

Answer Scalar (dot) product: $\vec{A}\cdot\vec{B} = AB\cos\theta$, a scalar result. It represents the component of one vector along another. Physical example: Work done, $W = \vec{F}\cdot\vec{d} = Fd\cos\theta$.

Vector (cross) product: $\vec{A}\times\vec{B} = AB\sin\theta\,\hat{n}$, a vector result, where $\hat{n}$ is perpendicular to the plane of $\vec{A}$ and $\vec{B}$ (direction by the right-hand rule). Physical example: Torque, $\vec{\tau} = \vec{r}\times\vec{F}$.

Key differences: dot product is commutative ($\vec{A}\cdot\vec{B}=\vec{B}\cdot\vec{A}$), cross product is anti-commutative ($\vec{A}\times\vec{B}=-\vec{B}\times\vec{A}$); dot product is maximum when vectors are parallel (θ=0°), cross product is maximum when perpendicular (θ=90°).

Section D

Multiple Choice Questions

1 mark each — correct option highlighted in the reveal

Q1. Which of the following is a scalar quantity?

• A. Momentum
• B. Electric current
• C. Force
• D. Displacement

Show answer

Correct option: B. Electric current — has direction but does not obey the triangle law, so it is a scalar.

Q2. The resultant of two vectors of magnitude 3 and 4 units acting at 90° to each other is:

• A. 1 unit
• B. 7 units
• C. 5 units
• D. 25 units

Show answer

Correct option: C. 5 units — √(3²+4²) = √25 = 5.

Q3. Two vectors are equal if they have:

• A. the same magnitude only
• B. the same direction only
• C. the same magnitude and direction
• D. the same initial point

Show answer

Correct option: C. the same magnitude and direction.

Q4. The maximum number of components a vector can be resolved into is:

• A. Exactly two
• B. Exactly three
• C. Any number, but two mutually perpendicular components are most commonly used
• D. Only one

Show answer

Correct option: C — a vector can be resolved into any number of components, but resolving into two (or three, in 3D) mutually perpendicular components is standard practice.

Q5. If $\vec{A} + \vec{B} = \vec{A} - \vec{B}$, then:

• A. $\vec{A} = 0$
• B. $\vec{B} = 0$
• C. Both are zero
• D. $\vec{A} = \vec{B}$

Show answer

Correct option: B. $\vec{B}=0$ — the equation simplifies to 2$\vec{B}$ = 0.

Q6. A unit vector has:

• A. Magnitude 1 and no unit
• B. Magnitude equal to the vector it represents direction for
• C. Zero magnitude
• D. Magnitude 10

Show answer

Correct option: A. Magnitude 1 and no unit — it is dimensionless and purely indicates direction.

Q7. The dot product of two mutually perpendicular vectors is:

• A. Maximum
• B. Equal to AB
• C. Zero
• D. Negative

Show answer

Correct option: C. Zero — since cos 90° = 0.

Section E

Assertion & Reason

Each question has an Assertion (A) and a Reason (R). Choose the correct option:

• (a) Both A and R are true, and R is the correct explanation of A
• (b) Both A and R are true, but R is NOT the correct explanation of A
• (c) A is true, R is false
• (d) A is false, R is true

Q1. Assertion (A): Electric current is not a vector quantity.
Reason (R): Electric current does not obey the triangle law of vector addition.

Show answer

Correct option: (a) — both true, and R correctly explains A; current has direction but fails the addition test required of vectors.

Q2. Assertion (A): The magnitude of the resultant of two vectors can never be less than the difference of their magnitudes.
Reason (R): The resultant is minimum when the two vectors act in the same direction.

Show answer

Correct option: (c) — A is true (minimum resultant = |A−B|), but R is false: the resultant is minimum when vectors act in opposite directions (θ = 180°), not the same direction.

Q3. Assertion (A): Two vectors of unequal magnitude can never give a zero resultant.
Reason (R): A zero resultant requires the two vectors to be exactly equal in magnitude and opposite in direction.

Show answer

Correct option: (a) — both true and R correctly explains A. Only two vectors of equal magnitude acting in exactly opposite directions can cancel to give a null vector.

Q4. Assertion (A): A physical quantity having both magnitude and direction is always a vector.
Reason (R): Electric current has both magnitude and direction, yet it is a scalar.

Show answer

Correct option: (d) — A is false (having magnitude and direction alone doesn't guarantee vector status; it must also obey the triangle law); R is a true, independent statement that in fact contradicts A.

Q5. Assertion (A): The scalar (dot) product of two vectors can be negative.
Reason (R): cos θ is negative for angles between 90° and 180°.

Show answer

Correct option: (a) — both true, and R correctly explains why A holds.

Section F

Fill in the Blanks

Q1. A quantity having magnitude only and no direction is called a .

Show answer

scalar

Q2. The rear end of a vector arrow is called the , and the front end is called the .

Show answer

tail; head

Q3. Two vectors acting in exactly opposite directions are called vectors.

Show answer

negative

Q4. The resultant of two vectors is maximum when the angle between them is .

Show answer

0°

Q5. The resultant of two vectors is minimum when the angle between them is .

Show answer

180°

Q6. A vector whose magnitude is zero is called a vector.

Show answer

null (zero)

Q7. The dot product of two vectors is also known as the product.

Show answer

scalar

Q8. The cross product of two vectors is also known as the product, and its result is always a .

Show answer

vector; vector (perpendicular to the plane of the two vectors)

Q9. $\hat{i}, \hat{j}, \hat{k}$ are examples of vectors along the x, y, z axes.

Show answer

unit

Q10. The law used to find the resultant of two vectors represented as adjacent sides of a figure from a common point is called the law.

Show answer

parallelogram

Section G

Match the Column

Match Column A (quantity) with Column B (type):

Column A	Column B
1. Mass	(p) Vector
2. Displacement	(q) Scalar
3. Electric current	(r) Vector
4. Force	(s) Scalar

Show answer

1 → (q) Scalar  |  2 → (p) Vector  |  3 → (s) Scalar  |  4 → (r) Vector

Match the angle between two vectors (Column A) with the type of resultant (Column B):

Column A	Column B
1. θ = 0°	(p) R = |A − B| (minimum)
2. θ = 90°	(q) R = A + B (maximum)
3. θ = 180°	(r) R = √(A² + B²)

Show answer

1 → (q)  |  2 → (r)  |  3 → (p)

Section H

Statement-Based Questions

Read the statements and choose: (a) Both true, (b) Statement I true, II false, (c) Statement I false, II true, (d) Both false

Q1.

Statement I: Every vector has both magnitude and direction.
Statement II: Every quantity with magnitude and direction is a vector.

Show answer

Correct option: (b) — Statement I is true by definition. Statement II is false, since current disproves it (it fails the triangle law).

Q2.

Statement I: The scalar product of two perpendicular vectors is zero.
Statement II: The vector product of two parallel vectors is zero.

Show answer

Correct option: (a) — Both true. Dot product ∝ cosθ = 0 at 90°; cross product ∝ sinθ = 0 at 0°/180° (parallel).

Q3.

Statement I: Vector addition is commutative.
Statement II: Vector subtraction is commutative.

Show answer

Correct option: (b) — $\vec{A}+\vec{B}=\vec{B}+\vec{A}$ is true, but $\vec{A}-\vec{B} \neq \vec{B}-\vec{A}$ in general, so Statement II is false.

Section I

Case Study Based Question

A student is studying a small ball moving inside a long straight tube. While the ball moves along the length of the tube at a steady speed, the tube itself is being carried across the room, moving in a direction perpendicular to its own length, at a different steady speed. The student wants to determine the actual velocity of the ball as observed by someone standing still in the room (not moving with the tube).

(i) Which law of vector addition should the student use to find the ball's resultant velocity?

Show answer

The triangle law of vector addition (equivalently, the parallelogram law), since the two velocities act at an angle to each other, not along the same line.

(ii) If the tube moves at 3 m/s and the ball moves at 4 m/s relative to the tube, find the magnitude of the resultant velocity.

Show answer

Since the two velocities are perpendicular: R = √(3² + 4²) = √25 = 5 m/s.

(iii) Find the angle the resultant velocity makes with the direction of the tube's motion.

Show answer

θ = tan⁻¹(4/3) = 53° from the direction of the tube's velocity.

(iv) If instead the ball's velocity along the tube were reversed in sense (but same magnitude), would the magnitude of the resultant velocity change?

Show answer

No. Reversing one component's sense changes the resultant's direction, but since the two velocities remain perpendicular with the same magnitudes (3 and 4), the resultant magnitude stays 5 m/s.

CBSE CLASS 11 PHYSICS · VECTORS & SCALARS · COMPLETE QUESTION BANK

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

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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