Showing posts with label CBSE Class 11 Physics. Show all posts
Showing posts with label CBSE Class 11 Physics. 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

Monday, June 29, 2026

NEET Physics Trick: Speed of Light = 1 Unit | Distance = Time Shortcut Explained

 - Dr.Sanjaykumar Pawar 

Physics concept showing speed of light = 1 unit system where distance equals time in seconds for NEET exam shortcut learning.
When speed of light becomes 1 unit/s, distance turns equal to time in seconds — the fastest NEET shortcut!


🔗 INTERNAL LINKS 

Units and Measurements Full Chapter Notes

NEET Physics Formula Sheet PDF

Speed, Velocity & Time MCQ Practice Set

Assertion Reason Questions NEET Physics

Dimensional Analysis Tricks

NEET Physics Important Shortcuts (All Chapters)

Previous Year Questions (PYQ) Physics Class 11

CBSE Physics Solution - Units and Measurements

CBSE Class 11 Physics

Chapter: Units and Measurements

Question

A new unit of length is chosen such that the speed of light in vacuum is unity. What is the distance between the Sun and the Earth in terms of the new unit if light takes 8 min 20 s to cover this distance?

Step-by-Step Solution (CBSE Toppers Style)

Given:

  • Speed of light in the new system = 1 unit/s
  • Time taken by light = 8 min 20 s

Step 1: Convert time into seconds

8 minutes = 8 × 60 = 480 seconds

Total time = 480 + 20 = 500 seconds

Step 2: Apply the formula

Distance = Speed × Time

Since,

Speed = 1 unit/s

Therefore,

Distance = 1 × 500 = 500 new units

Final Answer

Distance between the Sun and the Earth = 500 new units.

Short Notes (Easy for Beginners)

  • A unit is a standard way of measuring a physical quantity.
  • Scientists sometimes define new units to simplify calculations.
  • In this question, the speed of light is taken as 1 unit per second.
  • This means light travels 1 new unit in 1 second.
  • The basic formula is:
Distance = Speed × Time
  • If speed = 1 unit/s, then
Distance = Time (in seconds)

How to Solve Similar Questions

  1. Convert the given time into seconds.
  2. Use the formula: Distance = Speed × Time.
  3. If speed = 1 unit/s, then distance equals the time in seconds.

Example

Time = 8 min 20 s

= (8 × 60) + 20

= 480 + 20

= 500 seconds

Therefore,

Distance = 500 new units

Exam Tip

  • Always convert minutes into seconds first.
  • Remember the formula: Distance = Speed × Time.
  • If speed is 1 unit/s, then distance is simply the time in seconds.
  • This shortcut helps solve such questions quickly in CBSE examinations.
``` NEET Smart Learning Structure

NEET Smart Learning Structure

1. Concept Snapshot

Topic: New Unit of Length (Speed of Light = 1 unit/s)

Key Idea: When speed = 1, distance = time (in seconds)

Formula: Distance = Speed × Time

Special Case: If Speed = 1 → Distance = Time

2. Smart Table

Parameter Value Memory Trick
Speed of Light 1 unit/s (new system) "Light = 1 Rule"
Time Given 8 min 20 s Convert to seconds
Total Time 500 s 8×60 + 20
Distance 500 units Time = Distance

3. Mnemonic

"LIGHT = ONE, TIME = DONE"

If speed is 1 unit/sec → Distance is always time in seconds.

4. Flow Chart

Question ↓ Convert minutes → seconds ↓ Apply formula D = S × T ↓ S = 1 unit/s ↓ Distance = Time (sec) ↓ Final Answer

5. NEET Shortcut

If Speed = 1 → Directly write Distance = Time (seconds)

No calculation of velocity needed.

6. Exam Trap

❌ Mistake: Using minutes directly as distance

✔ Correct: Always convert into seconds first

7. Practice

Q: Light takes 10 min. Find distance.

Solution: 10 × 60 = 600 sec → 600 units

NEET Practice Questions - Units & Light

NEET Practice Questions

Topic: Speed of Light & New Unit System

1. Direct MCQs (Single Correct Option)

Q1 (Easy): If speed of light is 1 unit/s, then distance equals:
A) Speed × Time
B) Time (seconds)
C) Speed only
D) Zero
Answer: B
Q2 (Moderate): Light takes 3 min to reach a point. Distance is:
A) 180 units
B) 3 units
C) 60 units
D) 1800 units
Answer: A (3×60=180 s)
Q3 (Hard): If time is doubled, distance becomes:
A) Half
B) Double
C) Same
D) Zero
Answer: B
PYQ Type: In a system where c = 1, unit of length is:
A) Meter
B) Time unit × c
C) Second
D) Joule
Answer: B

2. Statement-Based Questions

Q: Statement I: In new unit system, speed of light is 1.
Statement II: Distance becomes numerically equal to time in seconds.
Answer: Both I and II are true, and II is correct explanation of I.

3. Assertion and Reason (A & R)

Assertion (A): Distance between Sun and Earth becomes 500 units.
Reason (R): Light takes 500 seconds to reach Earth.
Answer: Both A and R are true, and R explains A.
Assertion (A): If speed = 1, conversion is not required.
Reason (R): Distance equals time directly.
Answer: Both true, R explains A.

4. Match the Columns

Match Column A with B:
Column A Column B
Speed of light = 1 A. 500 units
8 min 20 s B. New unit system
Distance C. 500 s
Answer:
Speed = 1 → B
8 min 20 s → C
Distance → A

5. Diagram / Flow Based Questions

Study the flow chart:
Time (min) → Convert to seconds → Multiply by speed → Distance
Q: If speed = 1 unit/s, what is final step?
Answer: Distance = Time (in seconds)
Graph Type Question: If distance vs time graph is straight line passing through origin, speed is:
A) Increasing
B) Constant
C) Decreasing
D) Zero
Answer: B (Constant speed)

Monday, June 22, 2026

CBSE Class 11 Physics Chapter 2 Motion in a Straight Line Exercise 2.1 to 2.7 Solutions

-  Dr.Sanjaykumar Pawar  

Student studying CBSE Class 11 Physics Motion in a Straight Line chapter with graphs, velocity equations, acceleration concepts and solved NCERT exercises.

Student studying CBSE Class 11 Physics Motion in a Straight Line chapter with graphs, velocity equations, acceleration concepts and solved NCERT exercises.
CBSE Class 11 Physics Chapter 2 Motion in a Straight Line NCERT Exercise 2.1 to 2.7 Solved Answers.


 Internal Links

CBSE Class 11 Physics Chapter 1 Physical World NCERT Solutions

CBSE Class 11 Physics Chapter 2 Motion in a Straight Line Notes

CBSE Class 11 Physics Chapter 3 Motion in a Plane Solutions

Class 11 Physics Important Formula Sheet PDF

CBSE Class 11 Physics MCQs Chapter 2

Motion in a Straight Line Revision Notes

NCERT Class 11 Physics All Chapters Solutions

CBSE Class 11 Physics Sample Questions with Answers

Kinematics Complete Study Guide for Class 11

Previous Year CBSE Class 11 Physics Important Questions

NCERT Class 11 Physics Chapter 2 Exercise Solutions

NCERT Class 11 Physics
Chapter 2: Motion in a Straight Line
Exercise Solutions

Question 2.1

In which of the following examples of motion can the body be considered approximately a point object?

  1. A railway carriage moving without jerks between two stations.
  2. A monkey sitting on top of a man cycling smoothly on a circular track.
  3. A spinning cricket ball that turns sharply on hitting the ground.
  4. A tumbling beaker that has slipped off the edge of a table.
A body can be treated as a point object if its size is negligible compared to the distance travelled.
  • (a) Yes. The distance between stations is much greater than the size of the carriage.
  • (b) Yes. The radius of the circular track is much larger than the dimensions of the monkey-man system.
  • (c) No. Rotation and spin are important.
  • (d) No. The tumbling motion depends on the size and shape of the beaker.

Final Answer: (a) and (b)

Question 2.2

The position-time (x-t) graphs for two children A and B returning from school to their homes are shown in Fig. 2.9. Choose the correct entries in the brackets.

(a) (A/B) lives closer to the school than (B/A).

Since OP is less than OQ, A lives closer to the school.

Answer: A lives closer to the school than B.

(b) (A/B) starts from the school earlier than (B/A).

A starts at t = 0 while B starts later.

Answer: A starts earlier than B.

(c) (A/B) walks faster than (B/A).

The slope of an x-t graph represents speed. B has a steeper graph.

Answer: B walks faster than A.

(d) A and B reach home at the (same/different) time.

Both graphs terminate at the same value of time.

Answer: Same time.

(e) (A/B) overtakes (B/A) on the road (once/twice).

The graphs intersect only once and after that B remains ahead.

Answer: B overtakes A once.

Question 2.3

A woman starts from her home at 9.00 am, walks with a speed of 5 km h⁻¹ on a straight road up to her office 2.5 km away, stays at the office up to 5.00 pm, and returns home by an auto with a speed of 25 km h⁻¹. Plot the x-t graph of her motion.

Given:

  • Distance = 2.5 km
  • Walking speed = 5 km h⁻¹
  • Auto speed = 25 km h⁻¹

Time taken to reach office:

t = Distance / Speed = 2.5 / 5 = 0.5 hour = 30 minutes

She reaches the office at 9:30 am.

Time taken to return:

t = 2.5 / 25 = 0.1 hour = 6 minutes

She reaches home at 5:06 pm.

Time Position (km)
9:00 am 0
9:30 am 2.5
5:00 pm 2.5
5:06 pm 0

The x-t graph consists of an increasing straight line, a horizontal line, and a steep decreasing straight line.

Question 2.4

A drunkard walking in a narrow lane takes 5 steps forward and 3 steps backward repeatedly. Each step is 1 m long and takes 1 s. Determine how long he takes to fall in a pit 13 m away.

Net displacement in one cycle = 5 − 3 = 2 m

Time taken in one cycle = 8 s

After 4 cycles:

Displacement = 4 × 2 = 8 m

Time = 4 × 8 = 32 s

Then he moves forward:

Time (s) Position (m)
33 9
34 10
35 11
36 12
37 13

Therefore, the drunkard falls into the pit after 37 seconds.

Question 2.5

A car moving along a straight highway with speed 126 km h⁻¹ is brought to a stop within a distance of 200 m. What is the retardation and how long does it take to stop?

Initial speed, u = 126 km h⁻¹ = 35 m s⁻¹

Final speed, v = 0

Distance, s = 200 m

Using:

v² = u² + 2as

0 = (35)² + 2(a)(200)

a = −3.06 m s⁻²

Retardation = 3.06 m s⁻²

Using:

v = u + at

0 = 35 − 3.06t

t = 11.4 s

Retardation = 3.06 m s⁻²
Time taken = 11.4 s

Question 2.6

A player throws a ball upwards with an initial speed of 29.4 m s⁻¹.

(a) What is the direction of acceleration during upward motion?

Acceleration is always vertically downward due to gravity.

(b) What are the velocity and acceleration at the highest point?

Velocity = 0

Acceleration = 9.8 m s⁻² downward

v = 0, a = 9.8 m s⁻² downward

(c) Sign convention:

  • During upward motion: Position positive, Velocity negative, Acceleration positive.
  • During downward motion: Position positive, Velocity positive, Acceleration positive.

(d) Maximum height and total time of flight

Using:

v² = u² − 2gh

0 = (29.4)² − 2(9.8)h

h = 44.1 m

Time to rise:

t = u/g = 29.4/9.8 = 3 s

Total time:

T = 2 × 3 = 6 s

Maximum height = 44.1 m
Total time = 6 s

Question 2.7

State whether the following statements are true or false with reasons.

(a) A particle with zero speed may have non-zero acceleration.

True. Example: Ball at highest point of vertical motion.

(b) A particle with zero speed may have non-zero velocity.

False. Velocity is zero when speed is zero.

(c) A particle moving with constant speed must have zero acceleration.

False. Uniform circular motion has non-zero acceleration.

(d) A particle with positive acceleration must be speeding up.

False. Positive acceleration can reduce speed if velocity is negative.

Saturday, June 20, 2026

CBSE Class 11 Physics: Point Object MCQ with Topper Answer

 When Can a Body Be Considered a Point Object? CBSE Class 11 Solution

Q1. In which of the following examples of motion can the body be considered approximately a point object? 

Educational diagram explaining the point object concept with examples of a railway carriage, cyclist's cap, cricket ball, and falling beaker in CBSE Class 11 Physics.
Examples showing when a body can and cannot be considered a point object in Class 11 Physics.
 
- Dr. Sanjay Kumar Pawar 

Principle: A body can be considered a point object if the distance travelled by it is much greater than its own size (dimensions).

(a) A railway carriage moving without jerks between two stations.

Answer: Yes, the railway carriage can be considered a point object.

Explanation: The distance between two stations is very large compared to the length of the carriage. Hence, the size of the carriage is negligible in comparison to the distance travelled. Therefore, it can be treated as a point object.

(b) A cap on top of a man cycling smoothly on a circular track.

Answer: Yes, the cap can be considered a point object.

Explanation: The size of the cap is very small compared to the circumference of the circular track covered during motion. Therefore, its dimensions can be neglected and it may be treated as a point object.

(c) A spinning cricket ball that turns sharply on hitting the ground.

Answer: No, the cricket ball cannot be considered a point object.

Explanation: The sharp turn of the spinning ball involves rotational motion, and the distance over which the change in direction occurs is comparable to the size of the ball. Hence, its dimensions cannot be neglected.

(d) A tumbling beaker that has slipped off the edge of a table.

Answer: No, the beaker cannot be considered a point object.

Explanation: The size of the beaker is comparable to the height through which it falls. Moreover, tumbling involves rotational motion. Therefore, the dimensions of the beaker are important and it cannot be treated as a point object.

Conclusion

Hence, the body can be considered approximately a point object in cases (a) and (b) only.

Final Answer: (a) and (b)

Topper's Tip for the Exam:

The Core Principle: Always state the rule first. An object can be considered a point object if the distance it travels during its motion is much greater than its own linear dimensions.

बहुत अच्छा विचार है। NEET/CBSE Physics के किसी भी टॉपिक को याद करने और ट्रिकी प्रश्न हल करने के लिए "4-Layer Topper Memory Structure" सबसे प्रभावी रहता है। आपके दिए हुए "Point Object" वाले प्रश्न को उदाहरण बनाकर यह संरचना तैयार की जा सकती है।

NEET में इस प्रकार उत्तर Explain करें:

  1. पहले Rule लिखें।
  2. फिर Distance vs Size की तुलना करें।
  3. अंत में Yes/No लिखें।
  4. यदि rotation या tumbling हो तो उल्लेख करें कि आकार और orientation महत्वपूर्ण हो जाते हैं।

Practice Questions: Point Object (CBSE Class 11 Physics)

Key Rule

A body can be considered a point object if the distance travelled by it is much greater than its size.


Question 1

Can an airplane flying from Delhi to Mumbai be considered a point object?

Answer

Yes.

The distance between Delhi and Mumbai is much greater than the size of the airplane. Therefore, the airplane can be treated as a point object while studying its motion.


Question 2

Can a football rolling across a playground be considered a point object?

Answer

Yes.

The distance covered by the football across the playground is much larger than its diameter. Hence, it can be considered a point object.


Question 3

Can a spinning top rotating at one place be considered a point object?

Answer

No.

The top is mainly rotating about its own axis. Its size and rotational motion are important. Therefore, it cannot be treated as a point object.


Question 4

Can a student walking around a 400 m circular track be considered a point object?

Answer

Yes.

The size of the student is very small compared to the distance covered on the track. Hence, the student can be considered a point object.


Question 5

Can a coin spinning on a table be considered a point object?

Answer

No.

The motion depends on the coin's size and rotation. Therefore, it cannot be treated as a point object.


Multiple Choice Questions (MCQs)

Question 6

Which of the following can be treated as a point object?

(a) A train travelling between two cities

(b) A rotating ceiling fan

(c) A spinning coin

(d) A tumbling box

Answer

(a) A train travelling between two cities

Reason: Distance travelled is much greater than the train's size.


Question 7

A body can be treated as a point object when:

(a) Its mass is very small

(b) Its shape is circular

(c) Its size is negligible compared to the distance travelled

(d) It is at rest

Answer

(c) Its size is negligible compared to the distance travelled


Question 8

A cricket ball moving from one boundary to another on a cricket field can be treated as:

(a) A point object

(b) A rigid body only

(c) A fluid

(d) None of these

Answer

(a) A point object

Reason: The distance covered is much larger than the size of the ball.


Assertion-Reason Questions

Question 9

Assertion (A): A train moving between two stations can be treated as a point object.

Reason (R): The distance travelled is much greater than the size of the train.

Answer

Both A and R are true, and R is the correct explanation of A.


Question 10

Assertion (A): A spinning beaker falling from a table can be treated as a point object.

Reason (R): The size of the beaker is important in describing its motion.

Answer

Assertion is false, but Reason is true.


Exam Tip

Remember:

Distance ≫ Size → Point Object ✅

Distance ≈ Size → Not a Point Object ❌

Easy Trick:

"Long Journey = Point Object"

"Rotation Important = Not Point Object"


Internal Links
What Is Motion in Physics? Class 11 Notes
Difference Between Distance and Displacement
Scalars and Vectors Explained for Class 11
NCERT Solutions for Motion in a Straight Line
Rest and Motion: Important Concepts
Average Speed and Velocity Numericals
Class 11 Physics Chapter-wise NCERT Solutions
Physical Quantities and Units Explained
Relative Motion Basics for Students
Important CBSE Class 11 Physics Questions

Tuesday, June 16, 2026

CBSE Class 11 Physics Numericals: Speed of Light = 1 Unit Explained

 Distance Between Sun and Earth When c = 1 | Class 11 Physics Solution

 - Dr.Sanjaykumar Pawar 

Sun and Earth connected by light path showing time 500 seconds and speed of light equal to 1 unit per second in physics concept.
When speed of light is taken as 1 unit, distance becomes numerically equal to time in seconds.


🔗  INTERNAL LINKS 

/class-11-physics-units-and-measurement

/class-11-physics-important-numericals

/natural-units-system-physics-explained

/cbse-class-11-physics-short-notes

/physics-tricks-for-exam-preparation

/sun-earth-distance-numericals

CBSE Class 11 Physics - Speed of Light Unit Question

CBSE Class 11 Physics - Topper Notes

Question

A new unit of length is chosen such that the speed of light in vacuum is unity. What is the distance between the Sun and the Earth in terms of the new unit if light takes 8 min 20 s to cover this distance?

Solution

Given:

  • Speed of light, c = 1 (in new unit system)
  • Time = 8 min 20 s

Step 1: Convert time into seconds

8 min 20 s = (8 × 60) + 20 = 480 + 20 = 500 s

Step 2: Use formula

Distance = Speed × Time

Distance = 1 × 500 = 500 units

Final Answer: Distance = 500 units

Short Notes (Exam Revision)

  • When speed of light c = 1, units become simplified natural units.
  • 1 unit of length = distance travelled by light in 1 second.
  • Distance = Speed × Time → here Speed = 1, so Distance = Time.
  • Always convert time into seconds before solving.
  • This concept is useful in relativity and theoretical physics.

Topper Trick

If c = 1, then simply convert time into seconds → that is the answer!

Practice Questions - Speed of Light (c = 1)

Practice Questions & Answers (c = 1 System)

Q1.

If speed of light is 1 unit/sec, how much distance does light cover in 15 seconds?

Answer:

Distance = Speed × Time = 1 × 15 = 15 units

Q2.

Light takes 2 minutes to travel a distance. Find the distance in this system.

Answer:

2 minutes = 120 seconds

Distance = 1 × 120 = 120 units

Q3.

A signal takes 45 seconds to reach Earth. Find the distance.

Answer:

Distance = 1 × 45 = 45 units

Q4.

Light takes 5 minutes 30 seconds to travel from a star to Earth. Find distance.

Answer:

5 min 30 sec = (5 × 60) + 30 = 330 seconds

Distance = 1 × 330 = 330 units

Q5.

If distance is 1000 units in this system, how much time does light take?

Answer:

Time = Distance / Speed = 1000 / 1 = 1000 seconds

Q6.

Why does distance become equal to time when c = 1?

Answer:

Because speed of light c = distance / time = 1

So, distance = time (numerically in this unit system)

Q7.

Light takes 8 min 20 sec to travel from Sun to Earth. Find distance.

Answer:

8 min 20 sec = 500 seconds

Distance = 1 × 500 = 500 units

Q8.

A spaceship signal takes 1.5 hours to reach Earth. Find distance.

Answer:

1.5 hours = 1.5 × 3600 = 5400 seconds

Distance = 1 × 5400 = 5400 units

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...