Showing posts with label Assertion Reason. Show all posts
Showing posts with label Assertion Reason. Show all posts

Tuesday, July 28, 2026

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

Illustration explaining Class 11 Physics vectors including vector addition, triangle law, parallelogram law, equal vectors, resultant vector and important formulas for CBSE and NEET students.
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 - Vectors Notes

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.
A B
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.

A B 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
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 - Vectors Question Bank

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.

Wednesday, July 22, 2026

Conservation of Mechanical Energy Class 11 Notes, MCQs, Questions & Answers | CBSE & NEET

-  Dr Sanjay Kumar Pawar 

Conservation of Mechanical Energy Class 11 Physics Notes PDF | CBSE & NEET 

Educational diagram showing conservation of mechanical energy for a freely falling ball from height H, illustrating the conversion of potential energy (PE = mgh) into kinetic energy (KE = ½mv²) while total mechanical energy remains constant.
Conservation of Mechanical Energy explained with a falling ball showing the conversion of potential energy into kinetic energy.


Internal Links

Link this page to related Class 11 Physics topics to improve SEO and user navigation:

  1. Work, Energy and Power Class 11 Notes
  2. Work-Energy Theorem Explained
  3. Potential Energy Class 11 Notes
  4. Kinetic Energy Formula and Examples
  5. Conservative and Non-Conservative Forces
  6. Gravitational Potential Energy Notes
  7. Free Fall Motion Class 11
  8. Laws of Motion Class 11
  9. Newton's Laws of Motion Notes
  10. Circular Motion Class 11 Notes
  11. System of Particles and Rotational Motion
  12. Gravitation Class 11 Notes
  13. Mechanical Properties of Solids
  14. Complete Class 11 Physics Notes Index 
  15. Class 11 Physics MCQs with Answers
  16. CBSE Class 11 Physics Important Questions
  17. NEET Physics Chapter-wise Notes
  18. NCERT Solutions for Class 11 Physics
  19. Class 11 Physics Formula Sheet
  20. Previous Year CBSE Class 11 Physics Questions
Conservation of Mechanical Energy - NEET Notes

Chapter 5.8
Conservation of Mechanical Energy

Definition:
Mechanical Energy is the sum of Kinetic Energy (KE) and Potential Energy (PE).
Mechanical Energy = KE + PE

1. Work-Energy Theorem

Suppose a body moves through a small distance Δx under the action of force F. According to the Work-Energy Theorem,

ΔK = F(x) Δx

This means the work done by a force changes the kinetic energy of the body.

  • Positive work increases kinetic energy.
  • Negative work decreases kinetic energy.

2. Conservative Force

If the force is conservative, then potential energy can be defined.

−ΔU = F(x) Δx

The negative sign shows that whenever potential energy decreases, kinetic energy increases.

Example:
A falling stone loses potential energy and gains kinetic energy.

3. Combining the Equations

From the two equations:

ΔK = −ΔU

Therefore,

ΔK + ΔU = 0

or

Δ(K + U) = 0

4. Conservation of Mechanical Energy

Since Δ(K+U)=0, the total mechanical energy never changes.

K + U = Constant

This is called the Law of Conservation of Mechanical Energy.

5. Equation Between Two Positions

Ki + Ui = Kf + Uf

The total mechanical energy before motion equals the total mechanical energy after motion.

6. Conservative Force - Important Properties

  • Potential energy can be defined.
  • Work depends only on initial and final positions.
  • Work does not depend on the path.
  • Work done in a closed path is zero.
  • Mechanical energy remains conserved.

7. Example - Falling Ball

A ball of mass m is dropped from height H. Initially the velocity is zero.

At Height H

PE = mgH
KE = 0
EH = mgH

At Height h

PE = mgh
KE = ½mv²h
Eh = mgh + ½mv²h

At Ground Level

PE = 0
KE = ½mv²f
E0 = ½mv²f

8. Conservation of Energy

EH = Eh = E0

Since only gravity acts on the body, mechanical energy remains constant.

mgH = mgh + ½mv²h = ½mv²f

9. Final Velocity

Using conservation of energy,

mgH = ½mv²f

After simplifying,

vf = √(2gH)

10. Velocity at Height h

mgH = mgh + ½mv²h

Therefore,

vh² = 2g(H − h)

11. Energy Conversion

Position Potential Energy Kinetic Energy
Top Maximum Zero
Middle Decreasing Increasing
Ground Zero Maximum

12. Important Points for NEET

  • Mechanical Energy = KE + PE
  • Gravity is a conservative force.
  • Spring force is also conservative.
  • Mechanical energy remains constant if only conservative forces act.
  • Work done by a conservative force depends only on the initial and final positions.
  • Work done in a closed path is zero.
  • At the highest point, PE is maximum and KE is zero.
  • At the ground, KE is maximum and PE is zero.
  • Potential energy converts into kinetic energy during free fall.

13. Formula Sheet

Mechanical Energy = KE + PE
ΔK + ΔU = 0
K + U = Constant
Ki + Ui = Kf + Uf
PE = mgh
KE = ½mv²
vf = √(2gH)
vh² = 2g(H − h)
Work done in a Closed Path = 0
Conservation of Mechanical Energy

Conservation of Mechanical Energy

What is Mechanical Energy?

Mechanical Energy is the sum of Kinetic Energy (KE) and Potential Energy (PE).

Mechanical Energy = KE + PE

Work-Energy Theorem

When a force acts on an object, its kinetic energy changes.

ΔKE = Work Done

For conservative forces, Potential Energy decreases when Kinetic Energy increases.

ΔKE + ΔPE = 0

Energy Conversion During Falling

Top
PE Maximum
KE Zero
Middle
PE ↓
KE ↑
Ground
PE Zero
KE Maximum

Visual Falling Ball

As the ball falls, Potential Energy continuously converts into Kinetic Energy.

Example

Position Potential Energy Kinetic Energy Total Energy
Top 100 J 0 J 100 J
Middle 60 J 40 J 100 J
Ground 0 J 100 J 100 J

Important Formulae

PE = mgh
KE = ½mv²
KE + PE = Constant
vf = √(2gH)
vh² = 2g(H − h)
NEET Remember:
  • Gravity is a conservative force.
  • Total Mechanical Energy remains constant if only conservative forces act.
  • At the highest point: PE is maximum and KE is zero.
  • At the ground: KE is maximum and PE is zero.
  • Potential Energy converts into Kinetic Energy during falling.
Class 11 Physics - Conservation of Mechanical Energy Question Bank

CBSE Class 11 Physics

Chapter 5.8 - Conservation of Mechanical Energy

Question Bank with Answers


Part A - Multiple Choice Questions

1. Mechanical energy is the sum of
  1. Potential energy and Heat energy
  2. Kinetic energy and Potential energy
  3. Heat energy and Electrical energy
  4. Sound energy and Potential energy
Answer: B
2. Mechanical energy remains constant when
  1. Friction acts
  2. Air resistance acts
  3. Only conservative forces act
  4. External force acts
Answer: C
3. Which one is a conservative force?
  1. Friction
  2. Gravity
  3. Air resistance
  4. Viscous force
Answer: B
4. Work done by a conservative force depends on
  1. Path followed
  2. Distance travelled
  3. Initial and final positions only
  4. Speed
Answer: C
5. Work done in a closed path by gravity is
  1. Positive
  2. Negative
  3. Zero
  4. Infinite
Answer: C

Part B - Very Short Answer Questions

1. Define mechanical energy.

Answer: Mechanical energy is the sum of kinetic energy and potential energy.

2. Write the formula of mechanical energy.

Answer: E = KE + PE

3. Name one conservative force.

Answer: Gravitational force.

4. Write the formula of kinetic energy.

Answer: KE = ½mv²

5. Write the formula of potential energy.

Answer: PE = mgh


Part C - Short Answer Questions

1. What is conservation of mechanical energy?

Answer:
When only conservative forces act on a body, the total mechanical energy (kinetic energy + potential energy) remains constant throughout the motion.

2. What is a conservative force?

Answer:
A conservative force is a force whose work depends only on the initial and final positions and not on the path followed. Examples:

  • Gravity
  • Spring force


Part D - Long Answer Questions

1. State the law of conservation of mechanical energy.

Answer:
If only conservative forces act on a body, its total mechanical energy remains constant. Mechanical Energy = Kinetic Energy + Potential Energy Initial Energy Ki + Ui Final Energy Kf + Uf Therefore, Ki + Ui = Kf + Uf Example: A freely falling body loses potential energy and gains kinetic energy. The total mechanical energy remains constant.


Part E - Assertion and Reason

Assertion: Mechanical energy remains constant when only conservative forces act.

Reason: Gravity is a conservative force.

Answer: Both Assertion and Reason are true, and Reason is the correct explanation.


Part F - Fill in the Blanks

  1. Mechanical energy is the sum of ______ and ______.
  2. Gravity is a ______ force.
  3. Potential energy converts into ______ during free fall.
  4. Work done in a closed path is ______.
  5. Mechanical energy remains ______ when only conservative forces act.

Answers:

  1. Kinetic energy, Potential energy
  2. Conservative
  3. Kinetic energy
  4. Zero
  5. Constant


Part G - Match the Columns

Column A Column B
Gravity Conservative force
Friction Non-conservative force
Potential Energy mgh
Kinetic Energy ½mv²

Matching Answers 1 → Conservative force 2 → Non-conservative force 3 → mgh 4 → ½mv²


Part H - Case Study

A ball of mass 2 kg is dropped from a height of 20 m. Ignore air resistance.

Q1. Which energy is maximum at the top?

Answer: Potential Energy

Q2. Which energy is maximum at the ground?

Answer: Kinetic Energy

Q3. Calculate total mechanical energy at the top. Take g = 10 m/s².

PE = mgh = 2 × 10 × 20 = 400 J Answer = 400 Joule

Q4. Is mechanical energy conserved?

Yes. Only gravity acts.


Important Formulae

  • E = KE + PE
  • KE = ½mv²
  • PE = mgh
  • Ki + Ui = Kf + Uf
  • Δ(KE + PE) = 0
  • v = √(2gH)
  • v² = 2g(H − h)
  • Work in closed path = 0

End of CBSE Class 11 Physics Question Bank

Friday, July 10, 2026

Train Crossing Bridge Questions and Answers Class 11 Physics | CBSE & NEET Practice

 - Dr.Sanjaykumar Pawar  

Train Crossing Bridge Numerical Questions with Solutions | Class 11 Physics 

Illustration showing a train crossing a bridge with labeled train length, bridge length, velocity, acceleration, and displacement for Class 11 Physics and NEET preparation.
Train Crossing Bridge Numerical for CBSE Class 11 Physics with Step-by-Step Solution


Internal Links

  1. Motion in a Straight Line Complete Notes
  2. Equations of Motion Explained
  3. Relative Motion Notes
  4. Average Speed and Average Velocity
  5. Instantaneous Velocity Notes
  6. Acceleration Complete Notes
  7. Graphical Analysis of Motion
  8. Kinematics Formula Sheet
  9. Work, Energy and Power Notes
  10. Laws of Motion Complete Notes
  11. NEET Physics Chapter-wise MCQs
  12. CBSE Class 11 Physics Important Questions
  13. Assertion and Reason Questions for Physics
  14. Case Study Questions for Class 11 
  15. NCERT Solutions for Motion in a Straight Line
  16. Previous Year CBSE Physics Questions
  17. NEET Physics Practice Tests
  18. One-Dimensional Motion Numericals
  19. Train Crossing and River Boat Problems
  20. Class 11 Physics Revision Notes
CBSE Class 11 Physics Practice Questions and Answers | Train Crossing Bridge

CBSE Class 11 Physics

Practice Questions and Answers

Topic: Motion in a Straight Line – Train Crossing Bridge with Constant Acceleration


Question 1

A 120 m long train crosses a 480 m long bridge with a constant acceleration of 1 m/s². The train enters the bridge with an initial velocity of 20 m/s.

Find the time taken by the train to completely cross the bridge.

Answer

Given:
  • Length of Train = 120 m
  • Length of Bridge = 480 m
  • Initial Velocity (u) = 20 m/s
  • Acceleration (a) = 1 m/s²
Step 1: Total Distance Covered
Distance = Length of Train + Length of Bridge
s = 120 + 480 = 600 m
Step 2: Apply Equation of Motion
s = ut + ½at²
600 = 20t + ½(1)t²
600 = 20t + 0.5t²
Multiply by 2
1200 = 40t + t²
t² + 40t − 1200 = 0
Factorisation
(t + 60)(t − 20) = 0
Time Taken = 20 s

Question 2

A 150 m long train crosses a 450 m long bridge with constant acceleration 2 m/s². The initial velocity of the train is 10 m/s.

Find the time taken to completely cross the bridge.

Answer

Given:
  • Length of Train = 150 m
  • Length of Bridge = 450 m
  • Initial Velocity = 10 m/s
  • Acceleration = 2 m/s²
Step 1: Calculate Total Distance
s = 150 + 450 = 600 m
Step 2: Apply Equation
600 = 10t + ½(2)t²
600 = 10t + t²
t² + 10t − 600 = 0
Factorisation
(t + 30)(t − 20) = 0
Time Taken = 20 s

Question 3

A 200 m long train crosses a 300 m long bridge. It enters the bridge with a speed of 15 m/s and accelerates uniformly at 1 m/s².

Find the time taken by the train to completely cross the bridge.

Answer

Given:
  • Length of Train = 200 m
  • Length of Bridge = 300 m
  • Initial Velocity = 15 m/s
  • Acceleration = 1 m/s²
Step 1: Total Distance
s = 200 + 300 = 500 m
Step 2: Equation of Motion
500 = 15t + ½(1)t²
500 = 15t + 0.5t²
Multiply by 2
1000 = 30t + t²
t² + 30t − 1000 = 0
Using quadratic formula,
t = 20 s
Time Taken = 20 s

Question 4

A 100 m long train completely crosses a 400 m long bridge. The train enters the bridge with an initial speed of 25 m/s and moves with a constant acceleration of 2 m/s².

Find the time taken by the train to completely cross the bridge.

Answer

Given:
  • Length of Train = 100 m
  • Length of Bridge = 400 m
  • Initial Velocity (u) = 25 m/s
  • Acceleration (a) = 2 m/s²
Step 1: Calculate Total Distance
Distance = Length of Train + Length of Bridge
s = 100 + 400 = 500 m
Step 2: Apply Equation of Motion
s = ut + ½at²
500 = 25t + ½(2)t²
500 = 25t + t²
Rearranging,
t² + 25t − 500 = 0
Using the quadratic formula,
t = 13.9 s (approximately)
Time Taken = 13.9 s

Question 5

A 250 m long train crosses a 350 m long bridge. It enters the bridge with an initial speed of 20 m/s and accelerates uniformly at 2 m/s².

Find the time taken by the train to completely cross the bridge.

Answer

Given:
  • Length of Train = 250 m
  • Length of Bridge = 350 m
  • Initial Velocity = 20 m/s
  • Acceleration = 2 m/s²
Step 1: Total Distance
s = 250 + 350 = 600 m
Step 2: Apply Equation of Motion
600 = 20t + ½(2)t²
600 = 20t + t²
Rearranging,
t² + 20t − 600 = 0
Using the quadratic formula,
t = 15.6 s (approximately)
Time Taken = 15.6 s

Question 6 (Assertion & Reason)

Assertion (A):

To completely cross a bridge, a train travels a distance equal to the sum of the length of the train and the length of the bridge.


Reason (R):

The rear end of the train leaves the bridge only after the front end has already crossed the bridge.

Choose the correct option.

  1. Both A and R are true, and R is the correct explanation of A.
  2. Both A and R are true, but R is not the correct explanation of A.
  3. A is true, but R is false.
  4. A is false, but R is true.

Answer

Correct Option: A. Both Assertion and Reason are true, and Reason is the correct explanation of Assertion.

Explanation:

When a train completely crosses a bridge, its front end first reaches the other end of the bridge. However, the train is considered completely out of the bridge only when its rear end also leaves the bridge. Therefore, the train travels a distance equal to the sum of its own length and the length of the bridge.

Correct Answer: Option A

Question 7 (Very Short Answer)

Why do we add the length of the train and the length of the bridge while solving train crossing problems?

Answer

The train completely crosses the bridge only when its rear end leaves the bridge. Therefore, the front end of the train travels a distance equal to the sum of the train's length and the bridge's length.

Distance = Length of Train + Length of Bridge

Question 8 (Short Answer)

Which equation of motion is generally used when displacement, initial velocity, acceleration, and time are involved in train crossing problems?

Answer

The required equation of motion is:

s = ut + ½at²

Where,

  • s = Displacement
  • u = Initial Velocity
  • a = Acceleration
  • t = Time Taken
Equation Used: s = ut + ½at²

Question 9 (Multiple Choice Question)

A 100 m long train crosses a 500 m long bridge with an initial velocity of 10 m/s and an acceleration of 2 m/s².

The time taken is:

  1. 10 s
  2. 15 s
  3. 20 s
  4. 25 s

Answer

Total Distance
100 + 500 = 600 m
Using
600 = 10t + t²
t² + 10t − 600 = 0
(t + 30)(t − 20) = 0
Correct Option: C (20 s)

Question 10 (Concept-Based Question)

A train takes 20 seconds to completely cross a 500 m long bridge. The train is 100 m long.

Find the total displacement of the front end of the train during crossing.

Answer

The front end of the train travels:
Distance = Length of Train + Length of Bridge
Distance = 100 + 500 = 600 m
Total Displacement = 600 m

CBSE Exam Tips

  • Always calculate the total distance as the length of the train + length of the bridge.
  • Choose the correct equation of motion according to the given data.
  • Write every step clearly to score full marks in CBSE board examinations.
  • Include SI units (m, m/s, s) in every calculation.
  • Highlight the final answer inside a box.


Question 7 (Very Short Answer)

Why do we add the length of the train and the length of the bridge while solving train crossing problems?

Answer

The train completely crosses the bridge only when its rear end leaves the bridge. Therefore, the front end of the train travels a distance equal to the sum of the train's length and the bridge's length.

Distance = Length of Train + Length of Bridge

Question 8 (Short Answer)

Which equation of motion is generally used when displacement, initial velocity, acceleration, and time are involved in train crossing problems?

Answer

The required equation of motion is:

s = ut + ½at²

Where,

  • s = Displacement
  • u = Initial Velocity
  • a = Acceleration
  • t = Time Taken
Equation Used: s = ut + ½at²

Question 9 (Multiple Choice Question)

A 100 m long train crosses a 500 m long bridge with an initial velocity of 10 m/s and an acceleration of 2 m/s².

The time taken is:

  1. 10 s
  2. 15 s
  3. 20 s
  4. 25 s

Answer

Total Distance
100 + 500 = 600 m
Using
600 = 10t + t²
t² + 10t − 600 = 0
(t + 30)(t − 20) = 0
Correct Option: C (20 s)

Question 10 (Concept-Based Question)

A train takes 20 seconds to completely cross a 500 m long bridge. The train is 100 m long.

Find the total displacement of the front end of the train during crossing.

Answer

The front end of the train travels:
Distance = Length of Train + Length of Bridge
Distance = 100 + 500 = 600 m
Total Displacement = 600 m

CBSE Exam Tips

  • Always calculate the total distance as the length of the train + length of the bridge.
  • Choose the correct equation of motion according to the given data.
  • Write every step clearly to score full marks in CBSE board examinations.
  • Include SI units (m, m/s, s) in every calculation.
  • Highlight the final answer inside a box.

Monday, June 29, 2026

NEET Physics: Precision & Least Count Practice Questions with Answers

 - Dr.Sanjaykumar Pawar

Educational diagram comparing metre scale, vernier callipers, screw gauge, and optical instrument showing least count and precision levels.
Comparison of measuring instruments showing how least count affects precision in Physics for NEET preparation.

  CBSE Class 11 Physics Notes - Precision of Measuring Instruments

CBSE Class 11 Physics Notes

Chapter: Units and Measurements


Question

Which of the following is the most precise device for measuring length?

(a) A vernier callipers with 20 divisions on the sliding scale.

(b) A screw gauge of pitch 1 mm and 100 divisions on the circular scale.

(c) An optical instrument that can measure length to within a wavelength of light.


Step-by-Step Solution

Step 1: Find the Least Count of Vernier Callipers

20 Vernier Scale Divisions = 19 Main Scale Divisions

Least Count = 1 MSD − 1 VSD

LC = 1/20 mm

Least Count = 0.05 mm


Step 2: Find the Least Count of Screw Gauge

Given:

  • Pitch = 1 mm
  • Circular Scale Divisions = 100
Least Count = Pitch ÷ Number of Circular Scale Divisions

LC = 1 ÷ 100

Least Count = 0.01 mm


Step 3: Optical Instrument

Optical instruments measure length up to one wavelength of light.

Average wavelength of visible light

λ ≈ 5 × 10-7 m
≈ 0.0005 mm

Therefore, Least Count ≈ 0.0005 mm


Step 4: Comparison

Instrument Least Count
Vernier Callipers 0.05 mm
Screw Gauge 0.01 mm
Optical Instrument 0.0005 mm
Smaller Least Count = Greater Precision

0.0005 mm is the smallest.

Correct Answer: (c) Optical Instrument

Complete Notes

1. Measurement

Measurement means comparing an unknown quantity with a standard quantity.

Examples

  • Measuring length using a ruler.
  • Measuring wire diameter using a screw gauge.
  • Measuring pipe diameter using a vernier callipers.

2. Least Count

Least Count is the smallest value that an instrument can measure accurately.

Smaller Least Count = Higher Precision

3. Accuracy

Accuracy means how close a measured value is to the true value.

Example: Actual length = 10.00 cm Measured length = 10.01 cm This is highly accurate.


4. Precision

Precision means how close repeated measurements are to each other.

Example

  • 10.02 cm
  • 10.01 cm
  • 10.02 cm
  • 10.01 cm

These readings are highly precise.


5. Difference Between Accuracy and Precision

Accuracy Precision
Close to true value Close to repeated values
Correctness Repeatability

6. Vernier Callipers

Uses

  • External diameter
  • Internal diameter
  • Depth
Least Count = 1 MSD − 1 VSD

For 20 divisions, LC = 0.05 mm


7. Screw Gauge

Uses

  • Diameter of wire
  • Thickness of paper
  • Metal sheet thickness

Pitch = Distance moved in one complete rotation.

Least Count = Pitch ÷ Circular Scale Divisions

Example
Pitch = 1 mm
Divisions = 100
LC = 0.01 mm


8. Optical Instrument

Uses light waves to measure extremely small distances.

Approximate precision

0.0005 mm

This is the most precise among the given instruments.


9. Comparison of Measuring Instruments

Instrument Least Count Precision
Metre Scale 1 mm Low
Vernier Callipers 0.05 mm Better
Screw Gauge 0.01 mm Very Good
Optical Instrument 0.0005 mm Highest

Important Formulas

Least Count (Screw Gauge)

LC = Pitch ÷ Number of Circular Scale Divisions

Least Count (Vernier Callipers)

LC = 1 MSD − 1 VSD

Quick Revision

  • Measurement compares an unknown quantity with a standard.
  • Least Count = Smallest measurable value.
  • Smaller Least Count means Higher Precision.
  • Accuracy = Closeness to true value.
  • Precision = Closeness of repeated readings.
  • Vernier Callipers LC = 0.05 mm.
  • Screw Gauge LC = 0.01 mm.
  • Optical Instrument LC ≈ 0.0005 mm.
  • Most Precise Instrument = Optical Instrument.
Exam Tip:
Whenever you are asked to compare measuring instruments, always compare their least counts. The instrument with the smallest least count is the most precise.
NEET Smart Learning Structure

NEET Smart Learning Structure

Topic: Precision of Measuring Instruments


1. Exam Snapshot (30 Seconds Rule)

✔ Least Count जितना कम होगा → Precision उतनी अधिक होगी

Order:
Metre Scale → Vernier Callipers → Screw Gauge → Optical Instrument

2. Visual Memory Table

Instrument Least Count Precision Uses NEET Keyword
Metre Scale 1 mm Low Basic length Easy
Vernier Callipers 0.05 mm Better Diameter, Depth Medium
Screw Gauge 0.01 mm High Wire thickness Important
Optical Instrument 0.0005 mm Highest Atomic level Advanced

3. Memory Trick (Mnemonic)

Mnemonic: "मेरा वीर शेर ऑपरेशन करेगा"

म = Metre Scale
वीर = Vernier Callipers
शेर = Screw Gauge
ऑपरेशन = Optical Instrument
✔ Top to Bottom: Least Count ↓, Precision ↑

4. Golden Rule

⭐ सबसे छोटा Least Count = सबसे अधिक Precision

5. Formula Box

Vernier Callipers:
LC = 1 MSD − 1 VSD

Screw Gauge:
LC = Pitch / Circular Scale Division

6. Concept Tree

Measurement → Instrument → Least Count → Precision → Comparison → MCQ → NEET Trick

7. Flow Chart

Measurement → Least Count → Smaller LC → Higher Precision → Correct Answer

8. Common Mistakes

❌ Accuracy और Precision को एक समझ लेना
❌ Larger Least Count को ज्यादा Precision समझ लेना
❌ Formula भूल जाना

9. NEET Trick

Question में सभी instruments की Least Count compare करो

✔ सबसे छोटा LC → Correct Answer

10. PYQ Strategy

Step 1: Instrument पहचानो
Step 2: Least Count निकालो
Step 3: Compare करो
Step 4: Smallest LC चुनो

11. One Page Revision

Measurement → Least Count → Accuracy → Precision → Vernier → Screw Gauge → Optical → Smallest LC = Best

12. 15 Second Revision

Least Count छोटा → Precision बड़ा → Optical Best → Answer (C)

13. Sample MCQ

Q: सबसे अधिक precise instrument कौन है?

A) Metre Scale
B) Vernier Callipers
C) Screw Gauge
D) Optical Instrument

Answer: D

14. Assertion Reason

Assertion: Optical instrument सबसे precise है
Reason: इसका least count सबसे कम है

Answer: Both true and correct explanation

15. Final Memory Map

Measurement → Least Count → Precision → Comparison → Trick → MCQ → Revision

Golden Formula

Smallest Least Count = Highest Precision = Correct Answer
NEET Practice Questions - Measurement & Precision

NEET Practice Questions

Topic: Precision, Least Count, Measuring Instruments


1. Direct MCQs (Single Correct Option)

Q1 (Easy)
Which instrument has the smallest least count?

A) Metre Scale
B) Vernier Callipers
C) Screw Gauge
D) Optical Instrument
Answer: D (Optical Instrument)
Q2 (Easy)
Least count is defined as:

A) Maximum measurement
B) Smallest measurable value
C) Average value
D) Error in measurement
Answer: B
Q3 (Moderate)
A screw gauge has pitch 1 mm and 100 divisions. Its least count is:

A) 1 mm
B) 0.1 mm
C) 0.01 mm
D) 0.001 mm
Answer: C
Q4 (Hard)
Vernier callipers has 20 VSD = 19 MSD. If 1 MSD = 1 mm, least count is:

A) 0.1 mm
B) 0.05 mm
C) 0.02 mm
D) 0.2 mm
Answer: B

2. Statement-Based Questions

Q5
Statement I: Smaller least count means higher precision.
Statement II: Screw gauge is more precise than vernier callipers.

A) Both true and II explains I
B) Both true but II does not explain I
C) I true II false
D) I false II true
Answer: A
Q6
Statement I: Optical instruments can measure up to wavelength of light.
Statement II: They are less precise than screw gauge.

A) Both true
B) I true II false
C) I false II true
D) Both false
Answer: B

3. Assertion and Reason (A & R)

Q7
Assertion (A): Screw gauge is more precise than vernier callipers.
Reason (R): It has smaller least count.

A) Both true and R explains A
B) Both true but R not explanation
C) A true R false
D) A false R true
Answer: A
Q8
Assertion: Optical instrument is most precise measuring device.
Reason: Its least count is equal to wavelength of light.

A) Both true and correct explanation
B) Both true but not explanation
C) A true R false
D) A false R true
Answer: A

4. Match the Columns

Q9
Match Column I with Column II:

Column I:
1. Metre Scale
2. Vernier Callipers
3. Screw Gauge
4. Optical Instrument

Column II:
A. 0.0005 mm
B. 1 mm
C. 0.05 mm
D. 0.01 mm
Answer:
1 → B
2 → C
3 → D
4 → A

5. Diagram / Graph-Based Questions

Q10
A graph shows comparison of least count of instruments:

Metre Scale → ██████████ (1 mm)
Vernier Callipers → ████ (0.05 mm)
Screw Gauge → ██ (0.01 mm)
Optical Instrument → █ (0.0005 mm)

Which conclusion is correct? A) Metre scale most precise
B) Screw gauge least precise
C) Optical instrument most precise
D) Vernier is most precise
Answer: C
Q11
If least count decreases in a graph, precision: A) Decreases
B) Increases
C) Remains same
D) Becomes zero
Answer: B

Final NEET Shortcut Rule

✔ Least Count ↓ → Precision ↑
✔ Smallest LC = Correct Answer
✔ Optical Instrument = Highest Precision
NEET Practice Questions - Measurement & Precision

NEET Practice Questions

Topic: Precision, Least Count, Measuring Instruments


1. Direct MCQs (Single Correct Option)

Q1 (Easy)
Which instrument has the smallest least count?

A) Metre Scale
B) Vernier Callipers
C) Screw Gauge
D) Optical Instrument
Answer: D (Optical Instrument)
Q2 (Easy)
Least count is defined as:

A) Maximum measurement
B) Smallest measurable value
C) Average value
D) Error in measurement
Answer: B
Q3 (Moderate)
A screw gauge has pitch 1 mm and 100 divisions. Its least count is:

A) 1 mm
B) 0.1 mm
C) 0.01 mm
D) 0.001 mm
Answer: C
Q4 (Hard)
Vernier callipers has 20 VSD = 19 MSD. If 1 MSD = 1 mm, least count is:

A) 0.1 mm
B) 0.05 mm
C) 0.02 mm
D) 0.2 mm
Answer: B

2. Statement-Based Questions

Q5
Statement I: Smaller least count means higher precision.
Statement II: Screw gauge is more precise than vernier callipers.

A) Both true and II explains I
B) Both true but II does not explain I
C) I true II false
D) I false II true
Answer: A
Q6
Statement I: Optical instruments can measure up to wavelength of light.
Statement II: They are less precise than screw gauge.

A) Both true
B) I true II false
C) I false II true
D) Both false
Answer: B

3. Assertion and Reason (A & R)

Q7
Assertion (A): Screw gauge is more precise than vernier callipers.
Reason (R): It has smaller least count.

A) Both true and R explains A
B) Both true but R not explanation
C) A true R false
D) A false R true
Answer: A
Q8
Assertion: Optical instrument is most precise measuring device.
Reason: Its least count is equal to wavelength of light.

A) Both true and correct explanation
B) Both true but not explanation
C) A true R false
D) A false R true
Answer: A

4. Match the Columns

Q9
Match Column I with Column II:

Column I:
1. Metre Scale
2. Vernier Callipers
3. Screw Gauge
4. Optical Instrument

Column II:
A. 0.0005 mm
B. 1 mm
C. 0.05 mm
D. 0.01 mm
Answer:
1 → B
2 → C
3 → D
4 → A

5. Diagram / Graph-Based Questions

Q10
A graph shows comparison of least count of instruments:

Metre Scale → ██████████ (1 mm)
Vernier Callipers → ████ (0.05 mm)
Screw Gauge → ██ (0.01 mm)
Optical Instrument → █ (0.0005 mm)

Which conclusion is correct? A) Metre scale most precise
B) Screw gauge least precise
C) Optical instrument most precise
D) Vernier is most precise
Answer: C
Q11
If least count decreases in a graph, precision: A) Decreases
B) Increases
C) Remains same
D) Becomes zero
Answer: B

Final NEET Shortcut Rule

✔ Least Count ↓ → Precision ↑
✔ Smallest LC = Correct Answer
✔ Optical Instrument = Highest Precision
🔗 11. Internal Links

/neet-physics-measurement-basics
/least-count-vernier-callipers-explained
/screw-gauge-formula-and-examples
/optical-instruments-physics-neet
/physics-mcqs-class-11-chapter-1
/assertion-reason-questions-physics
/neet-short-tricks-physics

Saturday, May 30, 2026

The Law of Inertia Class 11 Physics Notes, MCQs, Questions & Answers

  

3. THE LAW OF INERTIA (NEET Notes – Easy Line-by-Line Explanation)

1. Galileo's Study of Motion on an Inclined Plane

Original: Galileo studied motion of objects on an inclined plane.

Easy Note: Galileo performed experiments using sloping surfaces (inclined planes) to understand how objects move.


Original: Objects moving down an inclined plane accelerate.

Easy Note: When an object rolls down a slope, its speed increases continuously. This increase in speed is called acceleration.


Original: Objects moving up retard.

Easy Note: When an object moves upward on a slope, its speed decreases gradually. This decrease in speed is called retardation (deceleration).


Original: Motion on a horizontal plane is an intermediate situation.

Easy Note: A flat (horizontal) surface is between the above two situations because the object neither gains nor loses speed due to the slope.


Original: An object moving on a frictionless horizontal plane should move with constant velocity.

Easy Note: If there is no friction, an object on a horizontal surface will keep moving forever with the same speed and direction (constant velocity).

NEET Point

  • Frictionless surface → No acceleration
  • Velocity remains constant

2. Galileo's Double Inclined Plane Experiment

Original: A ball released from rest on one plane rolls down and climbs up the other.

Easy Note: Galileo placed two slopes facing each other. A ball released from one side rolled down and then moved up the opposite slope.


Original: If the planes are smooth, the final height is nearly the same as the initial height.

Easy Note: When friction is very small, the ball reaches almost the same height from which it was released.


Original: In the ideal situation, when friction is absent, the final height equals the initial height.

Easy Note: Without friction, the ball would rise exactly to its original height.

NEET Point

  • No friction → Initial height = Final height

3. Effect of Reducing the Slope

Original: If the slope of the second plane is decreased, the ball still reaches the same height.

Easy Note: Making the second slope less steep does not change the height reached by the ball.


Original: The ball travels a longer distance to reach that height.

Easy Note: A gentler slope means the ball must travel farther to reach the same height.


Original: When the second plane becomes horizontal, the ball travels an infinite distance.

Easy Note: If the second plane is completely flat and frictionless, the ball will never stop moving.

NEET Point

  • Horizontal surface + No friction → Motion continues forever.

4. Real Situation

Original: In practice, the ball comes to a stop because of friction.

Easy Note: In real life, friction opposes motion and gradually stops the ball.


Original: Friction can never be totally eliminated.

Easy Note: Some friction is always present in practical situations.


Original: Without friction, the ball would continue with constant velocity.

Easy Note: If friction were absent, the ball would move forever with the same speed and direction.


5. Galileo's Important Conclusion

Original: State of rest and state of uniform linear motion are equivalent.

Easy Note: A body at rest and a body moving with constant velocity are both in similar conditions because neither experiences a net force.

NEET Point

Both situations have:

Net Force = 0

  • Body at rest → remains at rest
  • Body moving uniformly → continues moving uniformly

6. Common Misconception Corrected

Original: It is incorrect to assume that a net force is needed to keep a body in uniform motion.

Easy Note: A force is not required to keep an object moving with constant velocity.

Important NEET Fact

Force is needed only to:

  • Start motion
  • Stop motion
  • Change speed
  • Change direction

7. Why Do We Apply Force in Daily Life?

Original: We apply force to counter friction.

Easy Note: In daily life, friction slows objects down. Therefore, we apply force to balance friction and maintain constant speed.

Example:

  • A cyclist keeps pedaling to overcome friction and air resistance.

8. Inertia

Original: This property of the body is called inertia.

Easy Note: The tendency of a body to resist any change in its current state is called inertia.

Definition

Inertia = Resistance to change in the state of rest or uniform motion.


9. Law of Inertia (First Law of Motion)

Statement

A body continues to remain at rest or continues to move with uniform velocity in a straight line unless acted upon by an external unbalanced force.

NEET Shortcut

No Net Force ⇒ No Change in Motion


10. Key Points for NEET Revision

🔹 Galileo proposed the concept of inertia.

🔹 Friction is the force that stops moving objects in real life.

🔹 In the absence of friction, an object continues moving forever with constant velocity.

🔹 Rest and uniform motion are equivalent states.

🔹 Inertia means resistance to change.

🔹 Greater mass ⇒ Greater inertia.

🔹 Net external force = 0 ⇒ Velocity remains constant.


One-Line NEET Summary

Law of Inertia: A body remains at rest or continues to move with constant velocity unless an external unbalanced force acts on it. ✔️  

Educational diagram showing Galileo's double inclined plane experiment explaining the law of inertia, constant velocity, friction, and Newton's First Law for Class 11 Physics students.
Galileo's double inclined plane experiment demonstrating the Law of Inertia and uniform motion.


CBSE Class 11 Physics – The Law of Inertia

Question Bank with Answers


A. MCQs (1 Mark Each)

1. Who first inferred the law of inertia?

a) Newton b) Aristotle c) Galileo d) Einstein

Answer: c) Galileo


2. In the absence of friction, a moving body on a horizontal surface will:

a) Stop immediately b) Accelerate c) Move with constant velocity d) Move in a circle

Answer: c) Move with constant velocity


3. Inertia is the property of a body to:

a) Change its state b) Resist change in its state c) Increase velocity d) Decrease velocity

Answer: b) Resist change in its state


4. The SI unit of inertia is:

a) Newton b) Joule c) No unit d) kg

Answer: c) No unit


5. Greater the mass of a body:

a) Smaller the inertia b) Greater the inertia c) No inertia d) Constant inertia

Answer: b) Greater the inertia


6. A body remains at rest or in uniform motion when:

a) Net force is maximum b) Friction is present c) Net external force is zero d) Acceleration is maximum

Answer: c) Net external force is zero


7. Which force opposes motion?

a) Gravitational force b) Frictional force c) Magnetic force d) Electrostatic force

Answer: b) Frictional force


8. Uniform motion means:

a) Constant speed in a straight line b) Changing speed c) Circular motion d) Accelerated motion

Answer: a) Constant speed in a straight line


B. Very Short Answer Questions (1 Mark)

1. What is inertia?

Answer: Inertia is the property of a body to resist any change in its state of rest or uniform motion.


2. Who proposed the concept of inertia?

Answer: Galileo.


3. Which quantity measures inertia?

Answer: Mass.


4. State Newton's First Law.

Answer: A body remains at rest or in uniform motion unless acted upon by an external unbalanced force.


5. What is meant by net force?

Answer: The vector sum of all forces acting on a body.


C. Short Answer Questions (2–3 Marks)

1. Why does a moving ball stop after some time on a horizontal surface?

Answer: The ball stops because friction acts opposite to its motion. Friction gradually reduces its speed until it comes to rest.


2. Why is force not required to maintain uniform motion?

Answer: According to the law of inertia, a body moving with constant velocity continues moving unless an external force acts on it. Therefore, no force is required to maintain uniform motion.


3. Define inertia of rest with an example.

Answer: Inertia of rest is the tendency of a body to remain at rest.

Example: Passengers fall backward when a bus starts suddenly.


4. Define inertia of motion with an example.

Answer: Inertia of motion is the tendency of a moving body to continue moving.

Example: Passengers fall forward when a moving bus stops suddenly.


D. Long Answer Questions (4–5 Marks)

1. Explain Galileo's double inclined plane experiment.

Answer:

  1. Galileo used two inclined planes facing each other.
  2. A ball released from one plane rolled down and climbed the opposite plane.
  3. The ball reached nearly the same height from which it was released.
  4. As the slope of the second plane was reduced, the ball travelled a longer distance to reach the same height.
  5. When the second plane became horizontal, the ball would continue moving indefinitely in the absence of friction.
  6. Galileo concluded that a body continues in its state of motion if no external force acts on it.

2. Explain the law of inertia.

Answer:

The law of inertia states that a body remains at rest or continues to move with uniform velocity in a straight line unless acted upon by an external unbalanced force.

Key points:

  • Inertia means resistance to change.
  • Rest and uniform motion are equivalent states.
  • No net force is needed to maintain motion.
  • Mass is the measure of inertia.

E. Assertion and Reason Questions

1.

Assertion (A): A force is necessary to keep a body moving with constant velocity.

Reason (R): Uniform motion is possible only when net external force is zero.

Answer: Assertion is False, Reason is True.


2.

Assertion (A): A body at rest remains at rest if no external force acts on it.

Reason (R): This property is called inertia.

Answer: Both A and R are True and R is the correct explanation.


3.

Assertion (A): Greater mass means greater inertia.

Reason (R): Mass measures the resistance of a body to change in motion.

Answer: Both A and R are True and R is the correct explanation.


4.

Assertion (A): Friction helps a moving body continue forever.

Reason (R): Friction opposes motion.

Answer: Assertion is False, Reason is True.


F. Fill in the Blanks

  1. The law of inertia was first inferred by Galileo.

  2. Inertia means resistance to change.

  3. A body moving with constant velocity has zero acceleration.

  4. Friction acts opposite to motion.

  5. Mass is a measure of inertia.

  6. In the absence of friction, a body moves with constant velocity.

  7. Rest and uniform motion are equivalent states.

  8. The net force on a body in uniform motion is zero.


G. True / False

  1. Galileo studied motion using inclined planes. True

  2. Friction increases the speed of a body. False

  3. Inertia depends on mass. True

  4. A body in motion always requires force to keep moving. False

  5. Uniform motion means constant velocity. True


H. Case Study Questions

Case Study 1

Galileo released a ball from one inclined plane. The ball rolled down and climbed another inclined plane. When the second plane's slope was reduced, the ball travelled a longer distance but still reached the same height.

Questions

1. Who performed this experiment?

Answer: Galileo

2. What conclusion did Galileo draw?

Answer: A body continues in motion if no external force acts on it.

3. Which force prevents infinite motion in real life?

Answer: Friction

4. What happens in the absence of friction?

Answer: The body continues moving with constant velocity.


Case Study 2

A cyclist stops pedaling but the bicycle continues moving for some distance before stopping.

Questions

1. Which property keeps the bicycle moving?

Answer: Inertia of motion

2. Why does the bicycle eventually stop?

Answer: Due to friction and air resistance.

3. What would happen if friction were absent?

Answer: The bicycle would continue moving with constant velocity.

4. Which law explains this behavior?

Answer: Law of Inertia (Newton's First Law).


I. Statement-Based Questions

1. Consider the following statements:

I. Inertia is the resistance to change in state.

II. Mass is a measure of inertia.

III. Friction helps maintain motion.

Choose the correct option:

a) I only b) I and II only c) II and III only d) I, II and III

Answer: b) I and II only


2. Consider the statements:

I. Uniform motion requires zero net force.

II. Rest and uniform motion are equivalent states.

III. Galileo inferred the law of inertia.

Choose the correct answer:

a) I only b) II only c) I and III only d) I, II and III

Answer: d) I, II and III


J. Match the Columns

Column A Column B
1. Galileo a. Resistance to change
2. Inertia b. Opposes motion
3. Friction c. Double inclined plane
4. Mass d. Measure of inertia

Answer

1 → c

2 → a

3 → b

4 → d


K. Competency-Based Questions

1. Why do passengers fall backward when a bus starts suddenly?

Answer: Due to inertia of rest, the lower part of the body moves with the bus while the upper part tends to remain at rest.


2. Why do passengers fall forward when a moving bus stops suddenly?

Answer: Due to inertia of motion, the upper body continues moving forward even after the bus stops.


One-Mark CBSE Revision Questions

  1. Define inertia.
  2. Name the scientist who inferred the law of inertia.
  3. What is the measure of inertia?
  4. State Newton's First Law.
  5. What is uniform motion?
  6. What is meant by net force?
  7. Name the force opposing motion.
  8. Can a body move without force? Explain briefly.

Answers: Inertia, Galileo, Mass, Newton's First Law, Constant velocity, Resultant force, Friction, Yes—if net external force is zero.

LAW OF INERTIA

├── Galileo's Observations

│   │

│   ├── Inclined Plane

│   │   ├── Moving Down → Acceleration

│   │   ├── Moving Up → Retardation

│   │   └── Horizontal Surface → Constant Velocity

│   │

│   └── Conclusion

│       └── Frictionless Horizontal Surface

│           └── Motion Continues Forever

├── Double Inclined Plane Experiment

│   │

│   ├── Ball Released from One Side

│   ├── Rolls Down First Plane

│   ├── Climbs Second Plane

│   │

│   ├── Smooth Planes

│   │   └── Final Height ≈ Initial Height

│   │

│   ├── No Friction (Ideal Case)

│   │   └── Final Height = Initial Height

│   │

│   └── Slope of Second Plane Reduced

│       ├── Same Height Reached

│       ├── Longer Distance Travelled

│       └── Horizontal Plane

│           └── Infinite Motion

├── Role of Friction

│   │

│   ├── Opposes Motion

│   ├── Causes Ball to Stop

│   └── Cannot Be Completely Eliminated

├── Galileo's Insight

│   │

│   ├── State of Rest

│   ├── State of Uniform Motion

│   └── Both Are Equivalent

│       └── Net Force = 0

├── Force and Motion

│   │

│   ├── Force Not Needed

│   │   └── To Maintain Uniform Motion

│   │

│   └── Force Needed

│       ├── Change Speed

│       ├── Change Direction

│       ├── Start Motion

│       └── Stop Motion

├── Inertia

│   │

│   ├── Meaning

│   │   └── Resistance to Change

│   │

│   ├── Rest Inertia

│   │   └── Resists Change from Rest

│   │

│   └── Motion Inertia

│       └── Resists Change in Motion

└── Law of Inertia (Newton's First Law)

    │

    ├── Net External Force = 0

    │   ├── Body at Rest → Remains at Rest

    │   └── Body in Motion → Moves with Constant Velocity

    │

    └── External Unbalanced Force

        └── Changes State of Motion


NEET KEYWORDS:

Inertia • Friction • Constant Velocity • Net Force = 0

Rest State • Uniform Motion • Galileo • Newton's First Law



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...