Showing posts with label NEET Physics Notes. Show all posts
Showing posts with label NEET Physics Notes. Show all posts

Sunday, July 26, 2026

Vectors and Scalars Notes — CBSE Class 11 & NEET Guide

 - Dr.Sanjaykumar Pawar  

Diagram of three vector arrows of different lengths and directions on a graph-paper background, illustrating magnitude and direction in physics.
Vectors are drawn as arrows — length shows magnitude, arrowhead shows direction.
 


Internal Links

Vectors & Scalars — NEET Field Notes
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.

3 m/s (tube) 4 m/s (ball) R = 5 m/s 53°
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 3 m/s 1 m/s 2.5 m/s 1 m/s Longer arrow = larger magnitude. Arrowhead = direction of travel.
FIG. A — vectors drawn to scale: 1 cm ≡ 1 m/s (arbitrary chosen scale)

Anatomy of a vector arrow

PartNameMeaning
Back endTailStarting point
Front endHeadPoints in the direction of the vector
LengthMagnitudeNumerical 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 A B draw AB (first vector)
Draw the first vector, A to B
step 2 A B C tail of BC starts at head B
From B, draw the second vector, B to C
step 3 — the resultant A B C AC = 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?

3 m (tube, t=1s) 4 m (ball) 5 m/s resultant θ = 53°
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

ConceptFormula
Resultant magnitudeR = √(A² + B² + 2AB cos θ)
Direction of resultanttan 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

PropertyScalarVector
NeedsMagnitude + unit onlyMagnitude + unit + direction
Addition ruleOrdinary algebraTriangle / parallelogram law
Examplesmass, speed, work, energy, current, chargedisplacement, velocity, acceleration, force, momentum
SignCan 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
Vectors & Scalars — CBSE Class 11 Question Bank
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.

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
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 AColumn 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 AColumn 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

Tuesday, July 7, 2026

Linear Magnification Formula, Examples & NEET Tricks | Physics Notes

 - Dr.Sanjaykumar Pawar  

Linear Magnification Class 11 Physics: Formula, Solved Questions & Mnemonics

Educational diagram explaining linear magnification in physics with slide, projector, screen image, area magnification formula, and NEET revision steps.
Linear Magnification Explained: Formula, Area Relation, and NEET Problem-Solving Method

 


Internal Links 
Ray Optics Notes
Magnification Concepts
Important Physics Formulas
Related Articles:
Magnification Formula in Ray Optics
 Complete Magnification Formula Guide
Mirror Formula and Lens Formula
 Ray Optics Numerical Problems
Sign Convention in Optics
Optics Sign Convention Explained
Important NEET Physics Formula Sheet
 NEET Physics Quick Revision Formula List
Class 11 Physics Numerical Tricks
Anchor Text: Physics Problem Solving Techniques

Linear Magnification - NEET Notes

Linear Magnification (NEET Study Notes)

Question

Q. The photograph of a house occupies an area of 1.75 cm² on a 35 mm slide. The slide is projected onto a screen, and the area of the house on the screen is 1.55 m².

Find the Linear Magnification of the projector-screen arrangement.

Step-by-Step Solution

Step 1 : Write the Given Data

Quantity Value
Area on Slide 1.75 cm²
Area on Screen 1.55 m²

Step 2 : Convert m² into cm²

Since 1 m = 100 cm

Therefore, 1 m² = 100 × 100 = 10,000 cm²

Hence, 1.55 m² = 1.55 × 10,000 = 15,500 cm²

Step 3 : Find Area Magnification

Formula

Area Magnification = Image Area / Object Area

= 15,500 / 1.75

= 8857.14

Step 4 : Find Linear Magnification

Area Magnification = (Linear Magnification)²

Linear Magnification

= √8857.14

= 94.1
Answer = 94 ×

Short Notes

What is Magnification?

Magnification tells us how many times bigger or smaller the image is compared to the object.

Types of Magnification

Type Formula
Linear Magnification Image Length / Object Length
Area Magnification Image Area / Object Area

Important Formula

Area Magnification = (Linear Magnification)²

Linear Magnification = √Area Magnification

Unit Conversion

Unit Conversion
1 m 100 cm
1 m² 10,000 cm²
1 cm² 100 mm²

Golden Rule

Always convert all units into the same unit before applying any formula.

Question Solving Flow

Question

Convert Units

Area Magnification

Take Square Root

Linear Magnification

Mnemonics & Memory Tricks

Mnemonic 1

AREA → ROOT → LENGTH

If Area is given, take the Square Root to get Linear Magnification.

Mnemonic 2

ADR Rule
  • A = Area
  • D = Divide
  • R = Root
Remember only ADR and solve the complete question.

Mnemonic 3

Square means Area. Whenever you see cm² or think Square Root for Linear Magnification.

Common Mistakes

  • Not converting m² into cm².
  • Using Length Formula instead of Area Formula.
  • Taking Square Root before dividing.
  • Ignoring unit conversion.

NEET Exam Trick

Whenever the question contains Area remember

Convert → Divide → Root → Answer

10 Second Revision Card

Area

Convert Unit

Divide

Square Root

Linear Magnification

Monday, June 15, 2026

Newton's Third Law Example 4.12 Explained for NEET Students

 

Dr.Sanjaykumar Pawar


 
Physics diagram showing a wooden block and iron cylinder on a soft floor with labeled forces, normal reaction, weight, acceleration, and action-reaction pairs explained for NEET students.
Example 4.12 illustrating Newton's Third Law, Free Body Diagrams, and Action-Reaction Pairs for NEET Physics preparation.

NEET Notes - Example 4.12

Example 4.12 – Newton's Laws & Action-Reaction Pairs

Example 4.12 See Fig. 4.15. A wooden block of mass 2 kg rests on a soft horizontal floor. When an iron cylinder of mass 25 kg is placed on top of the block, the floor yields steadily and the block and the cylinder together go down with an acceleration of 0.1 m s –2. What is the action of the block on the floor (a) before and (b) after the floor yields ? Take g = 10 m s–2. Identify the action-reaction pairs in the problem.

Given Data

  • Mass of wooden block = 2 kg
  • Mass of iron cylinder = 25 kg
  • Acceleration downward = 0.1 m s-2
  • Acceleration due to gravity, g = 10 m s-2

Part (a): Before the Floor Yields

Step 1: Forces Acting on the Block

  • Weight of block (downward)
  • Normal reaction of floor (upward)
Weight = mg
= 2 × 10
= 20 N

Step 2: Apply Newton's First Law

The block is at rest. Therefore, net force on it is zero.

R = 20 N

Step 3: Force Exerted by Block on Floor

According to Newton's Third Law:

Force by floor on block = Force by block on floor
Answer (a): 20 N downward

Part (b): After the Floor Starts Yielding

Step 1: Consider Block + Cylinder as One System

Total Mass = 2 + 25 = 27 kg

Step 2: Forces on the System

  • Weight of system = 270 N (downward)
  • Normal reaction R' (upward)
Weight = Mg
= 27 × 10
= 270 N

Step 3: Apply Newton's Second Law

W - R' = Ma
270 - R' = 27 × 0.1
270 - R' = 2.7
R' = 270 - 2.7
R' = 267.3 N

Step 4: Force Exerted on Floor

By Newton's Third Law:

Force on floor = 267.3 N downward
Answer (b): 267.3 N downward

Action-Reaction Pairs

Case (a): Block at Rest

Action Reaction
Earth pulls block downward (20 N) Block pulls Earth upward (20 N)
Block pushes floor downward Floor pushes block upward

Case (b): Block + Cylinder Moving Downward

Action Reaction
Earth pulls system downward (270 N) System pulls Earth upward (270 N)
System pushes floor downward (267.3 N) Floor pushes system upward (267.3 N)
Cylinder pushes block downward Block pushes cylinder upward

Important NEET Concept

Action-Reaction Forces:

  • Always equal in magnitude.
  • Always opposite in direction.
  • Act on different bodies.
  • Occur simultaneously.

Common NEET Mistake

Many students think:

  • Weight ↓
  • Normal Reaction ↑

form an Action-Reaction Pair.

Wrong!

Both forces act on the same object. Therefore, they cannot be an action-reaction pair.

NEET Shortcut

Ask this question:

Do these two forces act on different objects?

  • Yes → Action-Reaction Pair ✔
  • No → Not an Action-Reaction Pair ✘

Final Answers

Before floor yields = 20 N downward

After floor yields = 267.3 N downward

Exam Tip

  1. Always draw a Free Body Diagram (FBD).
  2. Mark all external forces.
  3. Apply Newton's First Law if the body is at rest.
  4. Apply Newton's Second Law if the body accelerates.
  5. Use Newton's Third Law only for identifying action-reaction pairs.

This is one of the most important NEET concepts from Newton's Laws of Motion.

Internal Links
Related Newton's Laws Articles
Newton's First Law of Motion Explained
Newton's Second Law Numerical Problems
Newton's Third Law Complete Notes
Free Body Diagram (FBD) Master Guide
Contact Forces and Normal Reaction Force
Weight vs Mass Explained
Common Mistakes in Newton's Laws
Laws of Motion NEET Revision Notes
Most Important NEET Physics Numericals
Mechanics Formula Sheet for NEET

Thursday, June 4, 2026

Friction Explained Simply | NEET Physics Notes for Beginners

Friction in Physics: Static & Kinetic Friction Easy Notes for NEET 


- Dr.Sanjaykumar pawar

Basic idea of forces on a body

  • A body of mass m is kept on a horizontal table.
  • Two vertical forces act on it:
    • Weight = mg (downward)
    • Normal reaction = N (upward)
  • These two forces cancel each other.
  • So, net vertical force = 0 → no vertical motion.

When horizontal force is applied

  • Now a horizontal force F is applied on the body.
  • We expect the body to move.
  • But sometimes it does not move immediately.

Why the body may not move

  • If only force F acted, acceleration would be:
    • a = F/m
  • But body stays at rest → contradiction.
  • So, another force must be acting opposite to F.

Introduction of friction

  • A force appears between surfaces in contact.
  • This force opposes motion.
  • It acts parallel to surface.
  • This force is called frictional force (f).

Types of friction

  • Static friction (fₛ): when body is not moving.
  • Kinetic friction (fₖ): when body is moving.

Static friction (fₛ)

Meaning

  • Acts when body is at rest.
  • Opposes impending motion (motion that is about to happen).

Important points

  • Static friction exists only when force is applied.
  • If no force is applied → fₛ = 0
  • It increases as applied force increases.
  • It always adjusts itself to balance applied force.

Condition

  • Until a limit:
    • fₛ = F (equal and opposite)
  • So net force = 0 → body remains at rest.

Limiting static friction

  • Maximum value of static friction is called limiting friction.
  • Formula:
    • (fₛ)max = μₛ N
  • Where:
    • μₛ = coefficient of static friction
    • depends on nature of surfaces
    • N = normal reaction

Law of static friction

  • fₛ ≤ μₛ N
  • Means static friction can vary from 0 to maximum value.

Kinetic friction (fₖ)

Meaning

  • Acts when body is already moving.
  • Opposes actual motion.

Important points

  • Always acts opposite to motion.
  • Independent of contact area.
  • Almost independent of speed.

Formula

  • fₖ = μₖ N
  • Where:
    • μₖ = coefficient of kinetic friction

Key comparison

  • μₖ < μₛ
  • So kinetic friction is less than maximum static friction.

Motion after overcoming friction

  • If applied force F > (fₛ)max → body starts moving.

  • During motion:

    • Net force = F − fₖ
    • Acceleration = (F − fₖ)/m

If force is removed

  • Only kinetic friction acts opposite motion.
  • Acceleration becomes negative:
    • a = −fₖ/m
  • Body slows down and eventually stops.

Nature of friction laws

  • These laws are not fundamental laws.
  • They are experimental (empirical).
  • They are approximate but very useful in physics problems.

Important concept

  • Friction acts on contact surfaces.
  • It is a component of contact force parallel to surface.
  • It opposes relative motion, not absolute motion.

Real-life example: Train and box

  • A train accelerates forward.
  • A box is kept inside it.

Without friction:

  • Box would stay at rest (due to inertia).
  • Train would move ahead.
  • Box would hit back wall.

With friction:

  • Static friction acts on box.
  • It pulls box forward with train.
  • So box accelerates with train.
  • Hence, box stays at rest relative to train.

Final NEET summary

  • Friction is a contact force opposing relative motion.
  • Two types:
    • Static friction (before motion)
    • Kinetic friction (during motion)
  • Key formulas:
    • fₛ ≤ μₛ N
    • fₖ = μₖ N
  • Always: μₖ < μₛ

Internal Links

  1. Newton's Laws of Motion Explained

  2. Contact and Non-Contact Forces

  3. Free Body Diagrams in Physics

  4. Force and Acceleration Relationship

  5. Circular Motion Fundamentals

  6. Work, Energy and Power

  7. Applications of Newton's Second Law

  8. Momentum and Impulse

  9. Laws of Motion Class 11 Notes

  10. Coefficient of Friction Numerical Problems

Friction Mind Map

Friction - Mind Map (NEET Level)

FRICTION
|
|-- Definition
|     |-- Force opposing relative motion between surfaces
|     |-- Acts parallel to surface of contact
|
|-- Types of Friction
|     |
|     |-- 1. Static Friction (fs)
|     |       |-- Acts when body is at rest
|     |       |-- Opposes impending motion
|     |       |-- Adjusts with applied force
|     |       |-- Range: 0 ≤ fs ≤ μs N
|     |
|     |-- 2. Kinetic Friction (fk)
|             |-- Acts when body is in motion
|             |-- Opposes actual motion
|             |-- fk = μk N
|             |-- μk < μs
|
|-- Laws of Friction
|     |-- Independent of area of contact
|     |-- Depends on nature of surfaces
|     |-- Proportional to normal reaction (N)
|
|-- Coefficients
|     |-- μs → coefficient of static friction
|     |-- μk → coefficient of kinetic friction
|
|-- Limiting Friction
|     |-- Maximum static friction
|     |-- (fs)max = μs N
|
|-- Motion Cases
|     |
|     |-- F ≤ (fs)max → body at rest
|     |
|     |-- F > (fs)max → motion starts
|     |       |-- Acceleration = (F - fk)/m
|     |
|     |-- Force removed
|             |-- Retarding force = fk
|             |-- Body stops eventually
|
|-- Important Concept
|     |-- Friction opposes relative motion, not absolute motion
|
|-- Example: Train and Box
      |-- Train accelerates
      |-- Box moves due to static friction
      |-- Without friction → box slips backward
Friction - Complete Question Bank (Class 11 / NEET)

Friction — Complete Question Bank
Class 11 Physics (CBSE / NEET Level)

✅ 1. Very Short Answer Questions (1 Mark)

Q1. What is friction?
Ans: Friction is the force that opposes the relative motion or the tendency of relative motion between two surfaces in contact.
Q2. Write the formula of limiting friction.
Ans: fs(max) = μsR (or μsN), where μs is the coefficient of static friction and R (or N) is the normal reaction.
Q3. Which is greater: μs or μk?
Ans: μs (coefficient of static friction) is greater than μk (coefficient of kinetic friction).
Q4. What is kinetic friction?
Ans: Kinetic friction is the opposing force that comes into play when there is actual relative motion between two surfaces in contact.
Q5. What is the direction of friction?
Ans: It acts tangential to the surfaces in contact, opposite to the direction of relative motion or impending relative motion.

✅ 2. Short Answer Questions (2–3 Marks)

Q1. Why does friction arise?
Ans: Friction arises due to two primary causes:
  • Interlocking of surface irregularities: No surface is perfectly smooth; microscopic hills and valleys interlock when surfaces press together.
  • Molecular Adhesion: Highly localized chemical bonding/attractive forces established at the actual contact points between the molecules of the two surfaces.
Q2. Differentiate between static and kinetic friction.
Ans:
  • Static friction operates when the body is at rest relative to the surface; Kinetic friction operates when the body is in relative motion.
  • Static friction is a self-adjusting variable force (0 ≤ fs} ≤ fs(max)), whereas kinetic friction is nearly constant for a given pair of surfaces.
  • The coefficient of static friction (μs) is always greater than the coefficient of kinetic friction (μk).
Q3. State the laws of limiting friction.
Ans:
  1. The magnitude of limiting friction depends entirely on the nature and roughness of the surfaces in contact.
  2. It acts tangentially and opposite to the direction of impending motion.
  3. The magnitude of limiting friction is directly proportional to the normal reaction (fs(max) ∝ R).
  4. It is independent of the apparent area of contact between the surfaces, as long as the normal reaction remains constant.
Q4. What is limiting friction?
Ans: Limiting friction is the maximum values of static frictional force that comes into action just before a body slides or begins to move over the surface of another body.

✅ 3. Long Answer Questions (5 Marks)

Q1. Explain static and kinetic friction with the help of a suitable graph.
Ans:

When an external force is applied to a body resting on a rough surface, the static friction increases linearly with the applied force to balance it (f = Fapplied). This continues up to a threshold limit called limiting friction.

Once the applied force crosses this threshold value, the molecular bonds break, the interlocking is partially overcome, and the body begins to slide. At this point, the friction drops slightly below the limiting value to a steady value called kinetic friction. Further increase in the applied force does not change the kinetic friction value.

(Graph Note: A plot of Friction Force vs. Applied Force shows a straight line at 45° representing the static region, peaks at the limiting friction value, takes a minor downward dip, and transitions into a flat horizontal line representing constant kinetic friction.)

Q2. Derive the mathematical expression for limiting friction.
Ans:

By experimental observation, the limiting friction force (fs(max)) is found directly proportional to the normal reaction force (R) pressing the surfaces together.

Mathematically:
fs(max) ∝ R

To eliminate the proportionality sign, we introduce a constant:

fs(max) = μs · R

Where μs is the dimensionless constant called the coefficient of static friction. It depends purely on the materials, temperature, and roughness conditions of the touching surfaces.

Q3. Explain why a box placed on the floor of an accelerating train moves forward along with the train. Identify the force responsible.
Ans:

When the train accelerates forward with an acceleration a, an observer inside the non-inertial frame views a pseudo force acting on the box in the backward direction. Relative to the floor of the train, the box has a tendency to slide backward due to inertia.

Because of this impending backward relative motion, a static frictional force acts on the box in the forward direction (tangential to the floor). If this static friction is large enough (fs = ma) and does not exceed the maximum limiting value (μsmg), it prevents relative slipping. Therefore, static friction acts as the accelerating force that moves the box forward alongside the train.

✅ 4. Multiple Choice Questions (1 Mark Each)

Q1. Friction always acts:
  • A) In the direction of motion
  • B) Opposite to the direction of relative motion
  • C) Perpendicular to the surface
  • D) In a random direction
Ans: B
Q2. Limiting friction is:
  • A) Minimum friction
  • B) Maximum value of static friction
  • C) Kinetic friction
  • D) Zero friction
Ans: B
Q3. The coefficient of kinetic friction (μk) is generally:
  • A) Greater than μs
  • B) Equal to μs
  • C) Less than μs
  • D) Independent of the nature of surfaces
Ans: C
Q4. Frictional force between two solid surfaces depends directly on:
  • A) Apparent area of contact
  • B) Normal reaction
  • C) Speed of sliding only
  • D) Mass of the earth
Ans: B
Q5. Kinetic friction acts when:
  • A) The body is at rest
  • B) There is relative motion between surfaces
  • C) No external force is applied
  • D) The body moves through deep space
Ans: B

✅ 5. Assertion and Reason Questions

Directions: Choose Option (A) if both Assertion and Reason are true and Reason is correct explanation; Option (B) if both are true but Reason is not correct explanation; Option (C) if Assertion is true but Reason is false; Option (D) if Assertion is false but Reason is false.

Q1.
Assertion (A): Static friction is a self-adjusting force.
Reason (R): It changes its magnitude and direction according to the applied external force up to its maximum threshold limit.
Ans: Both A and R are true, and R is the correct explanation of A.
Q2.
Assertion (A): Kinetic friction is greater than static friction.
Reason (R): Mechanical interlocking between surface irregularities increases once the relative motion starts.
Ans: Both A and R are false. (Kinetic friction is less than static friction, and interlocking decreases during motion).
Q3.
Assertion (A): Friction always opposes the relative motion between surfaces.
Reason (R): Friction always acts opposite to the absolute velocity vector of the body.
Ans: A is true but R is false. (Friction opposes *relative* motion, not necessarily the actual direction of velocity—e.g., in walking or an accelerating train box).

✅ 6. Fill in the Blanks

Q1. Frictional force acts _________ to the surfaces in contact.
Ans: parallel (or tangentially)
Q2. Limiting friction value is given by = _________ × Normal Reaction.
Ans: μs (coefficient of static friction)
Q3. For any two given surfaces, μk is always _________ μs.
Ans: less than
Q4. Friction opposes _________ motion between contact points.
Ans: relative
Q5. Kinetic friction is also broadly referred to as _________ friction when a body slides.
Ans: sliding

✅ 7. Match the Column

Column A Column B
(1) Static friction (A) Ratio of limiting friction to normal reaction
(2) Kinetic friction (B) Force perpendicular to contact plane
(3) μs (C) Operates under relative motion conditions
(4) μk (D) Operates under relative rest conditions
(5) N (or R) (E) Ratio of sliding friction to normal reaction
Correct Match Answers:
(1) → D
(2) → C
(3) → A
(4) → E
(5) → B

✅ 8. Case Study Based Question

Case Background: A heavy block of mass 5 kg is kept stationary on a rough horizontal track surface. A pulling horizontal force is applied to it gradually. It is observed that the block refuses to shift initially, but just starts moving the instant the applied force crosses exactly 20 N.
Q1. What specific term is given to this threshold value of 20 N?
Ans: Limiting friction (fs(max)).
Q2. What is the value of the frictional force acting when the applied force is only 12 N?
Ans: 12 N. (Before motion starts, static friction is self-adjusting and perfectly balances the applied force).
Q3. What kind of frictional force acts once the block starts moving across the track?
Ans: Kinetic (sliding) friction.
Q4. If the block is in steady sliding motion, will the required force to maintain velocity be less than, equal to, or greater than 20 N?
Ans: Less than 20 N (since kinetic friction is slightly less than limiting static friction).

✅ 9. Statement Based Questions

Q1.
Statement I: Frictional force depends heavily on the visible apparent area of contact.
Statement II: This area dependency law holds true for all macroscopically rigid engineering surfaces.
Ans: Both Statement I and Statement II are false. (Friction is independent of apparent area).
Q2.
Statement I: Friction is a necessary evil that allows humans to walk safely on ground platforms.
Statement II: Without any friction force components acting, walking on a surface is completely impossible.
Ans: Both Statement I and Statement II are true.

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