Showing posts with label Exam Preparation. Show all posts

Wednesday, July 22, 2026

Potential Energy of a Spring Notes for NEET 2026 | Hooke's Law Explained

- Dr.Sanjaykumar pawar

Potential Energy of a Spring Class 11 Physics Notes for NEET

Illustration explaining the potential energy of a spring, Hooke's Law, spring force, force-displacement graph, work done, and energy conversion for NEET Physics students.

Potential Energy of a Spring explained with Hooke's Law, formulas, graphs, and NEET exam shortcuts.

NEET Notes - Potential Energy of a Spring

NEET Physics Notes

Potential Energy of a Spring (Hooke's Law)

1. Introduction

A spring is an elastic object that returns to its original shape after stretching or compressing.

Examples

Pen Spring
Vehicle Shock Absorber
Spring Balance
Toy Spring

NEET Point: Spring force is a restoring force.

2. Hooke's Law

The restoring force of a spring is directly proportional to the displacement from its equilibrium position.

F = -kx

Symbol	Meaning	SI Unit
F	Restoring Force	Newton (N)
k	Spring Constant	N/m
x	Displacement	m

Negative sign shows that the spring force always acts opposite to displacement.

3. Spring Constant (k)

The spring constant measures the stiffness of a spring.

Large k	Small k
Hard Spring	Soft Spring

SI Unit

N/m

Dimension

M T^-2

4. Force-Displacement Graph

The graph between force and displacement is a straight line passing through the origin.

Slope = -k

Area under the Force-Displacement graph gives the work done.

5. Work Done by External Force

To stretch the spring slowly from 0 to x, the external force acts in the same direction as displacement.

F = kx

Work done

W = ∫F dx

W = ½kx²

This energy gets stored as Potential Energy inside the spring.

6. Work Done by Spring

Since spring force acts opposite to displacement, its work is negative.

W = -½kx²

External Force → Positive Work

Spring Force → Negative Work

16. NEET Practice MCQs

According to Hooke's law, spring force is
- A. Constant
- B. Proportional to displacement ✔
- C. Inversely proportional to displacement
- D. Zero
Answer: B
The SI unit of spring constant is
- A. N
- B. J
- C. N/m ✔
- D. Nm
Answer: C
Potential energy stored in a spring is
U = ½kx²
At equilibrium position,
- A. KE Maximum ✔
- B. PE Maximum
- C. Both Zero
- D. Speed Zero
Total mechanical energy is
- A. Variable
- B. Constant ✔
- C. Infinite
- D. Zero

17. Assertion – Reason Questions

Q1.

Assertion: Spring force is a restoring force.

Reason: Spring force acts opposite to displacement.

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

Q2.

Assertion: Potential energy is zero at equilibrium.

Reason: Displacement is zero.

Both are true.

18. Previous Year NEET Questions

Question 1

A spring is stretched by x. Potential energy becomes

A. kx
B. kx²
C. ½kx² ✔
D. 2kx²

Question 2

The slope of Force-Displacement graph equals

A. k
B. -k ✔
C. 1/k
D. Zero

19. One Minute Revision

Concept	Formula
Hooke's Law	F = -kx
Potential Energy	½kx²
Work by Spring	-½kx²
Mechanical Energy	K + U
Maximum Speed	xₘ√(k/m)
Maximum Compression	v√(m/k)

20. Quick NEET Tips

Remember the negative sign in Hooke's law.
Potential energy depends on x².
Spring force always opposes displacement.
KE is maximum at mean position.
PE is maximum at extreme positions.
Total mechanical energy remains constant.
Hooke's law is valid only within the elastic limit.
Area under F-x graph represents work done.

16. NEET Practice MCQs

CBSE Class 11 Physics Question Bank

CBSE Class 11 Physics

Chapter: Work, Energy and Power

Topic: Potential Energy of a Spring (Hooke's Law)

1. Multiple Choice Questions (MCQs)

The restoring force of a spring is
A) kx
B) -kx
C) x/k
D) k/x
Answer: B
SI unit of spring constant is
A) N
B) J
C) N/m
D) Nm
Answer: C
Potential energy stored in a spring is
A) kx
B) k/x
C) ½kx²
D) k²x
Answer: C
At equilibrium position, spring potential energy is
A) Maximum
B) Minimum
C) Infinite
D) Negative
Answer: B
Hooke's law is valid only within
A) Elastic limit
B) Plastic limit
C) Breaking point
D) Melting point
Answer: A

2. Very Short Answer Questions (1 Mark)

State Hooke's Law.
Within elastic limit, restoring force is directly proportional to displacement.
Write the formula of spring force.
F = -kx
What is the SI unit of spring constant?
Newton per metre (N/m)
Write the formula of spring potential energy.
U = ½kx²
At which position is kinetic energy maximum?
At the equilibrium position.

3. Short Answer Questions (2–3 Marks)

Why is the spring force negative?
The negative sign shows that the restoring force acts opposite to the displacement and always tries to bring the spring back to equilibrium.
Derive the expression for work done by stretching a spring.
W = ∫Fdx = ∫kx dx = ½kx²
Define spring constant.
Spring constant is the force required to produce unit displacement in a spring.

4. Long Answer Questions (5 Marks)

Derive the expression for the potential energy stored in a spring.
Given, F = kx Work done, W = ∫Fdx = ∫kx dx = ½kx² Hence, Potential Energy, U = ½kx²
State and explain conservation of mechanical energy in a spring block system.
Total Energy E = K + U = ½mv² + ½kx² The total mechanical energy remains constant.

5. Assertion and Reason

Q1.

Assertion (A): Spring force is a restoring force.

Reason (R): Spring force always acts opposite to displacement.

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

Q2.

Assertion: Potential energy of spring is maximum at equilibrium.

Reason: Velocity is maximum at equilibrium.

Answer: Assertion is False, Reason is True.

6. Fill in the Blanks

Spring force is ______ proportional to displacement.
Directly
Hooke's law is F = ______
-kx
Potential energy stored in spring is ______
½kx²
The SI unit of spring constant is ______
N/m
Mechanical energy is the sum of kinetic and ______ energy.
Potential

7. True / False

Spring force always acts opposite to displacement.
True
Potential energy is maximum at equilibrium.
False
Hooke's law is valid within elastic limit.
True
Spring constant is measured in joule.
False

8. Match the Columns

Column A	Column B
Hooke's Law	F = -kx
Spring Potential Energy	½kx²
SI Unit of k	N/m
Equilibrium Position	PE = 0

9. Case Study Questions

A block is attached to a spring fixed at one end. The block is pulled by 20 cm and released. The spring constant is 200 N/m.

Q1. Write Hooke's law.

F = -kx

Q2. Calculate potential energy stored.

x = 0.20 m U = ½kx² = ½ × 200 × (0.20)² = 4 J

Q3. At which position is kinetic energy maximum?

At equilibrium position.

Q4. At which position is potential energy maximum?

At maximum extension or compression.

10. Important Formula Sheet

Hooke's Law: F = -kx
Work Done: W = ½kx²
Potential Energy: U = ½kx²
Total Mechanical Energy: E = K + U
Maximum Speed: v = xm√(k/m)
Maximum Compression: x = v√(m/k)

End of Question Bank

Tuesday, July 7, 2026

Significant Figures Explained | NEET Physics Notes, Rules, Tricks & Numericals

- Dr.Sanjaykumar Pawar

Rectangular Sheet Area and Volume Calculation | Significant Figures NEET Guide

NEET Physics educational infographic explaining significant figures, rules of zeros, rectangular sheet area and volume calculation, unit conversion, formulas, and memory tricks for students.

Significant Figures NEET Notes: Learn measurement rules, area-volume formulas, unit conversion, and numerical solving tricks with easy memory techniques.

Internal Links

Units and Measurements for NEET

Complete Units and Measurements Notes for NEET

Dimensional Analysis

Learn Dimensional Formula and Applications

Errors in Measurement

Physics Errors and Error Calculation Methods

Scientific Notation

Scientific Notation Rules and Examples

NEET Physics Formula Revision

Complete NEET Physics Formula Sheet

Numerical Problem Solving Strategy

How to Solve Physics Numericals Faster

SI Units and Conversion

SI Units List and Unit Conversion Tricks

Significant Figures - NEET Concept Notes

NEET Concept Notes: Significant Figures, Area and Volume Calculation

Question

The length, breadth and thickness of a rectangular sheet of metal are 4.234 m, 1.005 m and 2.01 cm respectively. Give the area and volume of the sheet to correct significant figures.

Solution (Step by Step)

Given Data:

Quantity	Value	Significant Figures
Length	4.234 m	4
Breadth	1.005 m	4
Thickness	2.01 cm = 0.0201 m	3

Step 1: Convert Unit

Thickness = 2.01 cm

1 cm = 10^-2 m

Thickness = 2.01 × 10^-2 m = 0.0201 m

Step 2: Calculate Area

Area = Length × Breadth

A = 4.234 × 1.005

A = 4.25517 m²

Least significant figures = 4

Area = 4.255 m²

Step 3: Calculate Volume

Volume = Length × Breadth × Thickness

V = 4.234 × 1.005 × 0.0201

V = 0.085527917 m³

Least significant figures = 3

Volume = 0.0855 m³

Short Notes: Significant Figures

Significant figures show the accuracy of a measurement.
Final answer should contain the same number of significant figures as the value having the least significant figures.
All non-zero digits are always significant.
Zeros between non-zero digits are significant.
Starting zeros are not significant.

Number	Significant Figures
4.234	4
1.005	4
0.0201	3

Formula Notes

Rectangle Area:
A = l × b

Rectangle Volume:
V = l × b × t

Remember:

Area uses two dimensions.
Volume uses three dimensions.
Always convert units before calculation.

Mnemonics (Easy Memory Tricks)

1. Significant Figures Rule

"Non-zero digits are HERO"

HERO:

H - Hidden zeros (between numbers) are significant
E - Ending zeros after decimal can be significant
R - Real non-zero digits are always significant
O - Opening zeros are not significant

2. Zero Rule

"शुरू के Zero सोते हैं, बीच के Zero काम करते हैं।"

3. Calculation Method

D-U-F-C-A Method

Letter	Meaning
D	Data पहचानो
U	Unit सही करो
F	Formula लगाओ
C	Calculate करो
A	Accuracy (Significant Figures) check करो

NEET Exam Trick

"पहले Unit, फिर Formula, फिर Calculation, अंत में Significant Figure"

इस तरीके से विद्यार्थी numerical questions को जल्दी और सही तरीके से solve कर सकता है।

Tuesday, June 30, 2026

Microscope Magnification Formula | NEET & CBSE Class 10–12 Physics Notes with Solved Numericals

- Dr.Sanjaykumar Pawar

Physics diagram showing microscope magnification concept with labeled image size, actual size, and formula M = I/A used for NEET CBSE notes.

Microscope Magnification Explained – Image Size, Actual Size & Formula (NEET/CBSE Physics)

🔗 Internal Links

/neet-physics-microscopy-and-measurement

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

/cbse-class-10-physics-light-and-optics

/neet-important-formulas-physics

/physics-mcqs-class-11-chapter-2

/measurement-errors-and-significant-figures

/physics-short-notes-revision-pdf

Microscope Magnification Notes (NEET/CBSE)

📘 Microscope Magnification - NEET/CBSE Notes

🌟 Basic Concept

Very small objects like human hair, cells, bacteria cannot be measured directly. A microscope is used to magnify them.

🔬 Magnification Formula

Magnification (M) = Image size / Actual size

👉 A = I / M (Actual size = Image size ÷ Magnification)

📏 Units Conversion

1 mm = 1000 µm
1 µm = 10⁻⁶ m
1 nm = 10⁻⁹ m

📝 Solved Question (CBSE Style)

Q: A student measures the thickness of a human hair using a microscope of magnification 100. He finds the average width of the hair image is 3.5 mm. Find the actual thickness of the hair.

Given:

Magnification (M) = 100
Image size (I) = 3.5 mm

Step 1: Formula

A = I / M

Step 2: Substitute values

A = 3.5 / 100

Step 3: Calculation

A = 0.035 mm

Step 4: Convert to micrometer

0.035 mm = 35 µm

Final Answer: 0.035 mm = 35 µm

⚡ Important NEET Points

Always use A = I / M
Convert units before final answer
Magnification increases image size
Actual size is always very small

🔥 Quick Revision Formula

M = I / A

A = I / M

NEET Notes - Magnification & Microscope

🧠 NEET Notes: Measurement Using Microscope & Magnification

🌟 1. Introduction

Some objects like human hair, bacteria, and cells are very small and cannot be measured directly using a ruler. So, we use a microscope to observe and measure them.

🔬 2. Magnification

Definition: Magnification tells how many times an object appears larger than its actual size.

M = Image size / Actual size

🧾 3. Important Formula

Actual size = Image size / Magnification

🧪 4. Important Terms

Image size: Size seen under microscope
Actual size: Real size of object
Magnification: How many times object is enlarged

📏 5. Units Conversion

1 mm = 1000 µm
1 µm = 10⁻⁶ m
1 nm = 10⁻⁹ m

🧠 6. Step-by-Step Method

Write formula M = I / A
Rearrange A = I / M
Substitute values
Solve carefully
Convert units if needed

📘 7. Example

Given: Magnification = 100, Image size = 3.5 mm

Step 1: A = 3.5 / 100

Step 2: A = 0.035 mm

Step 3: Convert to µm = 35 µm

✅ Final Answer

Thickness of hair = 0.035 mm = 35 µm

⚡ 8. Important Points for NEET

Always check units before solving
Use M = I / A formula correctly
Divide image size by magnification
Convert mm to µm if required

🔥 9. Quick Revision

M = I / A
A = I / M

NEET Smart Study Structure - Microscope

🧠 NEET SMART STUDY STRUCTURE

🔬 Topic: Microscope Magnification & Hair Thickness

📊 1. BASIC DATA TABLE

Concept	Symbol	Formula	Meaning
Magnification	M	M = I / A	Object कितना बड़ा दिख रहा है
Image size	I	—	Microscope में दिखने वाला size
Actual size	A	A = I / M	Real object का size

⚡ 2. ONE-LINE FORMULA BOX

M = I / A
A = I / M
I = M × A

🧠 3. MNEMONIC (MIA RULE)

MIA Rule:

M = Magnification
I = Image (जो दिखता है)
A = Actual (असल आकार)

Trick: Image हमेशा बड़ा, Actual हमेशा छोटा

🧩 4. NEET SOLVING TRICK (MIA METHOD)

👉 Use triangle logic:

      I
   --------
   M |  A

A चाहिए → I ÷ M
I चाहिए → M × A
M चाहिए → I ÷ A

📘 5. EXAM ANSWER STRUCTURE

Given data लिखो
Formula लिखो
Rearrange करो
Substitution करो
Final answer लिखो

🧪 6. EXAMPLE

Given: M = 100, I = 3.5 mm

Step 1: A = I / M

Step 2: A = 3.5 / 100

Step 3: A = 0.035 mm

Step 4: 0.035 mm = 35 µm

Final Answer: 35 µm

⚡ 7. UNIT TRICK

MM → Micro Magic Rule

1 mm = 1000 µm
1 µm = 1000 nm

🔥 8. FINAL REVISION

M = I / A
A = I / M
MIA Rule → Image ÷ Magnification = Actual

NEET Practice Questions - Microscope

🧠🔬 NEET Practice Questions

Topic: Microscope & Magnification

🟢 1. DIRECT MCQs (Single Correct Option)

Q1 (Easy): A microscope has magnification 100. Image size is 2 mm. Actual size is:

A) 2 mm B) 0.2 mm C) 0.02 mm D) 20 mm

✔ Answer: C (0.02 mm)

Q2 (Moderate): If magnification increases, actual size:

A) Increases B) Decreases C) Remains same D) Becomes zero

✔ Answer: C

Q3 (Hard): A hair appears 5 mm under 200× microscope. Real thickness is:

A) 0.25 mm B) 25 mm C) 0.025 mm D) 2.5 mm

✔ Answer: C (0.025 mm)

🟡 2. STATEMENT-BASED QUESTIONS

Q4: Statement I: Magnification is ratio of image size to actual size. Statement II: Magnification has no unit.

A) Both true B) Both false C) I true, II false D) I false, II true

✔ Answer: A

Q5: Statement I: Actual size increases with magnification. Statement II: Image size increases with magnification.

✔ Answer: D (I false, II true)

🔵 3. ASSERTION & REASON (A & R)

Q6: Assertion: Microscopes help measure small objects. Reason: Magnification reduces actual size.

✔ Answer: C (A true, R false)

Q7: Assertion: Hair thickness can be found using microscope. Reason: M = I / A.

✔ Answer: A

🟣 4. MATCH THE COLUMNS

Match:

A. Magnification → 2. I / A

B. Image size → 3. Seen size

C. Actual size → 1. Real size

✔ Answer: A-2, B-3, C-1

🔷 5. DIAGRAM / GRAPHICAL QUESTIONS

Q9: Image = 10 mm, Magnification = 100×. Find real size.

A = I / M = 10 / 100 = 0.1 mm

✔ Answer: 0.1 mm

Q10: If magnification increases from 50× to 200×, image size:

A) Decreases B) Increases C) Same D) Zero

✔ Answer: B

🔥 QUICK REVISION

M = I / A
A = I / M
I = M × A

✔ Image increases with magnification

✔ Actual size remains constant

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:

पहले Rule लिखें।
फिर Distance vs Size की तुलना करें।
अंत में Yes/No लिखें।
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

(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

(d) It is at rest

Answer

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

(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

Monday, June 15, 2026

Class 11 Physics Units and Measurements Solutions | NCERT Exercise 1.1 & 1.2

- Dr.Sanjaykumar Pawar

NCERT Class 11 Physics Chapter 1 Questions and Answers PDF

Student studying Class 11 Physics Units and Measurements chapter with SI units, measurements, formulas, and solved NCERT exercises.

Class 11 Physics Chapter 1 Units and Measurements with NCERT solutions and practice questions.

Class 11 Physics - Exercise 1.1 and 1.2

NCERT Class 11 Physics

Chapter 1: Units and Measurements

Exercise 1.1

Q1(a). The volume of a cube of side 1 cm is equal to ____ m³.

Answer:
Volume = (1 cm)³ = 1 cm³
1 cm = 10⁻² m
1 cm³ = (10⁻²)³ m³
= 10⁻⁶ m³

Final Answer: 1 × 10⁻⁶ m³

Q1(b). The surface area of a solid cylinder of radius 2.0 cm and height 10.0 cm is equal to ____ (mm)².

Answer:
Surface Area = 2πr(r+h)
= 2 × 3.14 × 2.0 × (2.0 + 10.0)
= 150.72 cm²
1 cm² = 100 mm²
150.72 × 100 = 15072 mm²

Final Answer: 1.5 × 10⁴ mm²

Q1(c). A vehicle moving with a speed of 18 km h⁻¹ covers ____ m in 1 s.

Answer:
18 km h⁻¹ = 18 × (1000/3600)
= 5 m s⁻¹
Distance = Speed × Time
= 5 × 1
= 5 m

Final Answer: 5 m

Q1(d). The relative density of lead is 11.3. Its density is ____ g cm⁻³ or ____ kg m⁻³.

Answer:
Density = Relative Density × Density of Water
= 11.3 × 1
= 11.3 g cm⁻³

1 g cm⁻³ = 1000 kg m⁻³
11.3 × 1000 = 1.13 × 10⁴ kg m⁻³

Final Answer:
11.3 g cm⁻³
1.13 × 10⁴ kg m⁻³

Exercise 1.2

Q2(a). 1 kg m² s⁻² = ____ g cm² s⁻²

Answer:
1 kg = 10³ g
1 m² = 10⁴ cm²

1 kg m² s⁻²
= 10³ × 10⁴ g cm² s⁻²
= 10⁷ g cm² s⁻²

Final Answer: 10⁷ g cm² s⁻²

Q2(b). 1 m = ____ ly

Answer:
1 light year = 9.46 × 10¹⁵ m

1 m = 1 / (9.46 × 10¹⁵)
= 1.06 × 10⁻¹⁶ ly

Final Answer: 1.06 × 10⁻¹⁶ ly

Q2(c). 3.0 m s⁻² = ____ km h⁻²

Answer:
1 m = 10⁻³ km
1 s = 1/3600 h

3.0 × 10⁻³ × (3600)²
= 38880 km h⁻²

Final Answer: 3.9 × 10⁴ km h⁻²

Q2(d). G = 6.67 × 10⁻¹¹ N m² (kg)⁻² = ____ cm³ s⁻² g⁻¹

Answer:
1 N = kg m s⁻²

G = 6.67 × 10⁻¹¹ m³ kg⁻¹ s⁻²

1 m³ = 10⁶ cm³
1 kg⁻¹ = 10⁻³ g⁻¹

G = 6.67 × 10⁻¹¹ × 10⁶ × 10⁻³
= 6.67 × 10⁻⁸ cm³ s⁻² g⁻¹

Final Answer: 6.67 × 10⁻⁸ cm³ s⁻² g⁻¹

Class 11 Physics Practice Questions

Class 11 Physics - Units and Measurements

20 Practice Questions with Answers

Q1. A cube has a side of 2 cm. Find its volume.

Volume = a³
= 2³
= 8 cm³
Answer: 8 cm³

Q2. Convert 500 cm into metres.

500 ÷ 100 = 5 m
Answer: 5 m

Q3. Convert 1 km into metres.

1 km = 1000 m
Answer: 1000 m

Q4. Convert 72 km h⁻¹ into m s⁻¹.

72 × (1000/3600)
= 20 m s⁻¹
Answer: 20 m s⁻¹

Q5. A body moves with a speed of 15 m s⁻¹ for 4 s. Find the distance covered.

Distance = Speed × Time
= 15 × 4
= 60 m
Answer: 60 m

Q6. Convert 1 m² into cm².

1 m = 100 cm
1 m² = (100)² cm²
= 10000 cm²
Answer: 10000 cm²

Q7. Find the area of a square of side 5 m.

Area = side²
= 5²
= 25 m²
Answer: 25 m²

Q8. Convert 1 litre into cm³.

1 litre = 1000 cm³
Answer: 1000 cm³

Q9. A body has mass 200 g and volume 50 cm³. Find density.

Density = Mass / Volume
= 200 / 50
= 4 g cm⁻³
Answer: 4 g cm⁻³

Q10. What is the SI unit of force?

Answer: Newton (N)

Q11. What is the SI unit of energy?

Answer: Joule (J)

Q12. Convert 3 kg into grams.

3 × 1000 = 3000 g
Answer: 3000 g

Q13. Convert 2500 g into kilograms.

2500 ÷ 1000 = 2.5 kg
Answer: 2.5 kg

Q14. Find the total surface area of a cube of side 4 cm.

TSA = 6a²
= 6 × 4²
= 6 × 16
= 96 cm²
Answer: 96 cm²

Q15. What is the SI unit of time?

Answer: Second (s)

Q16. A cyclist covers 120 m in 20 s. Find speed.

Speed = Distance / Time
= 120 / 20
= 6 m s⁻¹
Answer: 6 m s⁻¹

Q17. Convert 36 km h⁻¹ into m s⁻¹.

36 × (1000 / 3600)
= 10 m s⁻¹
Answer: 10 m s⁻¹

Q18. What is the relative density of water?

Answer: 1

Q19. Find the volume of a cylinder of radius 3 cm and height 7 cm.

Volume = πr²h
= (22/7) × 3² × 7
= 198 cm³
Answer: 198 cm³

Q20. Express 1 day in seconds.

1 day = 24 × 60 × 60
= 86400 s
Answer: 86400 s

Quick Revision

SI unit of length = metre (m)
SI unit of mass = kilogram (kg)
SI unit of time = second (s)
SI unit of force = newton (N)
SI unit of energy = joule (J)
1 km = 1000 m
1 m = 100 cm
1 litre = 1000 cm³
Density = Mass / Volume
Speed = Distance / Time

Thursday, June 4, 2026

Block and Trolley System NEET Solution | Acceleration & Tension Explained

NEET Physics Example 4.9: Block and Trolley System (Step-by-Step)

- Dr.Sanjaykumar Pawar

INTERNAL LINKS

/neet-physics-newtons-laws

/friction-notes-class-11

/tension-in-string-problems

/pulley-system-problems-neet

/mechanics-important-questions

/class-11-physics-motion-in-a-line

/neet-important-derivations

Diagram of a block and trolley system showing a 3 kg hanging mass connected to a 20 kg trolley with forces, tension, friction, and acceleration vectors.

Physics NEET diagram showing a block and trolley system with forces, tension, friction, and acceleration clearly labeled.

Example 4.9 – Block and Trolley System

Example 4.9 – Block and Trolley System (Easy NEET Notes)

Given:

Mass of hanging block, \( m_1 = 3\,kg \)
Mass of trolley, \( m_2 = 20\,kg \)
Coefficient of kinetic friction, \( \mu_k = 0.04 \)
Acceleration due to gravity, \( g = 10\,m\,s^{-2} \)
String is light (massless) and inextensible
Pulley is smooth (frictionless)

To Find:

Acceleration of the system \( a \)
Tension in the string \( T \)

Step 1: Understand the Motion

The 3 kg block hangs vertically and moves downward.
The 20 kg trolley moves horizontally.
Both bodies have the same acceleration due to inextensible string.

NEET Point: Connected bodies with a light inextensible string have the same acceleration.

Step 2: Forces on 3 kg Block

Weight:

\[ W = mg = 3 \times 10 = 30\,N \]

Applying Newton's Second Law:

\[ 30 - T = 3a \quad \cdots (1) \]

Step 3: Forces on 20 kg Trolley

Newton's Second Law:

\[ T - f_k = 20a \quad \cdots (2) \]

Step 4: Friction Calculation

\[ N = mg = 20 \times 10 = 200\,N \]

\[ f_k = \mu_k N = 0.04 \times 200 = 8\,N \]

Step 5: Substitute Friction

\[ T - 8 = 20a \quad \cdots (3) \]

Step 6: Solve Equations

From (1):

\[ T = 30 - 3a \]

Substitute into (3):

\[ (30 - 3a) - 8 = 20a \]

\[ 22 = 23a \]

\[ a = \frac{22}{23} = 0.96\,m\,s^{-2} \]

Acceleration: \( a = 0.96\,m\,s^{-2} \)

Step 7: Tension

\[ T = 30 - 3a = 30 - 2.88 = 27.12\,N \]

Tension: \( T = 27.1\,N \)

Final Answer

Acceleration: \( 0.96\,m\,s^{-2} \)

Tension: \( 27.1\,N \)

NEET Quick Revision Points

Smooth pulley ⇒ same tension throughout string
Inextensible string ⇒ same acceleration
\( f_k = \mu_k N \)
\( N = mg \) (horizontal surface)
\( F = ma \)
Always draw FBD and form equations separately

Tuesday, June 2, 2026

Newton's Third Law of Motion Explained with Examples for Class 11

Newton's Third Law: CBSE Class 11 Physics Complete Guide

Newton’s Third Law of Motion – Easy NEET Notes (Line-by-Line Explanation)

Educational infographic explaining Newton's Third Law of Motion with examples of walking, swimming, rocket propulsion, gun recoil and wall pushing, showing equal and opposite action-reaction force pairs.

Newton's Third Law of Motion: Every action force has an equal and opposite reaction force acting simultaneously on another body.

- Dr.Sanjaykumar pawar

Introduction

Newton’s Second Law tells us how an external force causes acceleration.
But a question arises: Where does this external force come from?
In Newtonian mechanics, an external force always comes from another body or object.

Understanding Force Between Two Bodies

Consider two bodies: A and B.
If body B exerts a force on A, we may ask:
- Does A also exert a force on B?
Newton's answer is Yes.

Example 1: Spring and Hand

When you press a spring, your hand exerts force on the spring.
The spring gets compressed.
At the same time, the spring pushes back on your hand.
You can feel this force.

Example 2: Earth and Stone

Earth pulls a stone downward due to gravity.
We may not notice it, but the stone also pulls the Earth upward.
The force exerted by the stone on Earth is equal in magnitude.
Since Earth is extremely massive, its acceleration is very small and difficult to observe.

Newton's Third Law of Motion

Statement

"To every action, there is always an equal and opposite reaction."

"For every force, there is an equal and opposite force acting on another body."

Mathematical Form

Where:

= Force on A by B
= Force on B by A
Negative sign (−) shows opposite direction.

Main Idea of Third Law

Force never exists alone.
Forces always occur in pairs.
Whenever one body exerts a force on another body, the second body exerts an equal and opposite force on the first body.
These two forces are called an action-reaction pair.

Important Points for NEET

Point 1: Action and Reaction Are Both Forces

The words action and reaction simply mean forces.
There is no special difference between them.
A better statement is:

"Force on A by B is equal and opposite to force on B by A."

Point 2: Action Does Not Come Before Reaction

Common Misconception

Many students think action happens first and reaction happens later.

Correct Concept

Action and reaction occur simultaneously.
They act at the same instant.
Neither force is the cause nor the effect of the other.

Example

When you hit a wall:
- Your hand exerts force on the wall.
- The wall exerts force on your hand at exactly the same moment.

Point 3: Action and Reaction Act on Different Bodies

Very Important NEET Point

Action and reaction never act on the same body.
They always act on two different bodies.

Example

Person pushes wall:

Force	Acts On
Person pushes wall	Wall
Wall pushes person	Person

Since they act on different bodies, they cannot cancel each other.

Why Action and Reaction Do Not Cancel?

Example

Suppose:

You push a wall with 100 N force.
Wall pushes you back with 100 N force.

Students often say:

Net force = 100 N − 100 N = 0

This is wrong.

Reason

100 N force acts on the wall.
100 N reaction force acts on you.
Since they act on different bodies, they cannot be added together.

Internal Forces in a System

Consider System (A + B)

Here, force on A by B and force on B by A are internal forces.
Internal forces always occur in equal and opposite pairs.
Therefore, they cancel each other when considering the whole system.

Result

Internal forces do not affect the motion of the entire system.
Only external forces can change the motion of a system.

Real-Life Examples of Newton’s Third Law

1. Walking

Foot pushes the ground backward.
Ground pushes the foot forward.
This forward reaction helps us walk.

2. Swimming

Swimmer pushes water backward.
Water pushes swimmer forward.

3. Rocket Launch

Rocket throws gases downward.
Gases push rocket upward.

4. Gun Recoil

Bullet moves forward.
Gun experiences backward recoil.

5. Jumping

Person pushes ground downward.
Ground pushes person upward.

NEET Quick Revision Points

✅ Force always occurs in pairs.

✅ Action and reaction are equal in magnitude.

✅ Action and reaction are opposite in direction.

✅ They act simultaneously.

✅ They act on different bodies.

✅ They never cancel each other on a single body.

✅ Internal forces cancel within a system.

✅ Only external forces can change the motion of a system.

One-Line NEET Definition

Newton’s Third Law states that whenever one body exerts a force on another body, the second body simultaneously exerts an equal and opposite force on the first body.

NEWTON'S THIRD LAW OF MOTION – CBSE CLASS 11 QUESTION BANK

A. Multiple Choice Questions (MCQs)

1. Newton's Third Law states that:

(a) Force is proportional to acceleration (b) Every object remains at rest (c) To every action there is an equal and opposite reaction (d) Momentum is conserved

Answer: (c)

2. Action and reaction forces:

(a) Act on the same body (b) Act on different bodies (c) Are unequal (d) Act at different times

Answer: (b)

3. Which of the following is an example of Newton's Third Law?

(a) Falling of an apple (b) Walking on the ground (c) Inertia of rest (d) Motion of planets

Answer: (b)

4. The force exerted by Earth on a stone and by the stone on Earth are:

(a) Unequal (b) Equal and opposite (c) Same direction (d) Zero

Answer: (b)

5. Action and reaction forces:

(a) Cancel each other (b) Act simultaneously (c) Are internal forces only (d) Produce no motion

Answer: (b)

B. Very Short Answer Questions (1 Mark)

1. State Newton's Third Law of Motion.

Answer: To every action, there is always an equal and opposite reaction.

2. Do action and reaction act on the same body?

Answer: No, they act on different bodies.

3. What is the direction of reaction force?

Answer: Opposite to the action force.

4. Can action and reaction occur separately?

Answer: No, they always occur together.

5. Give one example of Newton's Third Law.

Answer: Walking on the ground.

C. Short Answer Questions (2–3 Marks)

1. Why can a person walk on the ground?

Answer: The foot pushes the ground backward. The ground exerts an equal and opposite force on the foot in the forward direction. This reaction force enables the person to walk.

2. Why does a gun recoil when fired?

Answer: When the bullet moves forward, it exerts an equal and opposite force on the gun. As a result, the gun moves backward, called recoil.

3. Explain why action and reaction forces do not cancel each other.

Answer: Action and reaction forces act on different bodies. Since they act on different objects, they cannot be added together and therefore do not cancel each other.

4. What are internal forces?

Answer: Internal forces are forces acting between particles of the same system. They occur in equal and opposite pairs and cancel each other.

D. Long Answer Questions (5 Marks)

1. State and explain Newton's Third Law of Motion with examples.

Answer: Newton's Third Law states that to every action there is always an equal and opposite reaction.

Characteristics:

Forces always occur in pairs.
Action and reaction are equal in magnitude.
They are opposite in direction.
They act simultaneously.
They act on different bodies.

Examples: • Walking • Swimming • Rocket propulsion • Gun recoil • Jumping from a boat

Thus, every force in nature has an equal and opposite counterpart.

2. Explain rocket propulsion using Newton's Third Law.

Answer: A rocket expels hot gases downward at high speed. The gases exert a downward force. According to Newton's Third Law, the gases exert an equal and opposite force on the rocket. This upward reaction force propels the rocket upward.

E. Assertion and Reason Questions

1.

Assertion (A): Action and reaction forces are equal and opposite.

Reason (R): They act on the same body.

Answer: Assertion is true but Reason is false.

2.

Assertion (A): A swimmer moves forward in water.

Reason (R): Water pushes the swimmer forward when the swimmer pushes water backward.

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

3.

Assertion (A): Action and reaction occur simultaneously.

Reason (R): There is no cause-effect relationship between them.

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

F. Fill in the Blanks

Newton's Third Law states that every action has an equal and ______ reaction.

Answer: opposite

Action and reaction forces act on ______ bodies.

Answer: different

Walking is possible because of the ______ force of the ground.

Answer: reaction

Forces always occur in ______.

Answer: pairs

Internal forces of a system ______ each other.

Answer: cancel

G. Statement-Based Questions

1. Identify True or False:

(a) Action and reaction act on different bodies. Answer: True

(b) Action occurs before reaction. Answer: False

(d) Action and reaction can cancel each other. Answer: False

(e) Rocket propulsion is based on Newton's Third Law. Answer: True

H. Match the Columns

Column A

A. Walking

B. Gun

C. Rocket

D. Swimming

Column B

Water pushes swimmer
Ground pushes person
Gases push rocket upward
Recoil

Answers

A → 2

B → 4

C → 3

D → 1

I. Case Study Questions

Case Study

A student pushes a wall with a force of 50 N. The wall exerts a force of 50 N on the student. The student notices that the wall does not move.

Questions

1. Which law explains this situation?

Answer: Newton's Third Law of Motion.

2. What is the reaction force?

Answer: Force exerted by the wall on the student.

3. Are action and reaction equal?

Answer: Yes.

4. Why does the wall not move?

Answer: Because the wall is firmly supported and its acceleration is negligible.

5. Do action and reaction act on the same body?

Answer: No.

J. Important One-Mark CBSE Questions

Define action-reaction pair.
Give one example of Newton's Third Law.
Why does a boat move backward when a person jumps forward?
Why does a rocket move upward?
Can action exist without reaction?

Answers

Pair of equal and opposite forces acting on different bodies.
Walking.
Due to equal and opposite reaction force.
Due to reaction force of escaping gases.
No.

NEWTON'S THIRD LAW OF MOTION
│
├── Definition
│   │
│   └── To every action,
│       there is an equal
│       and opposite reaction.
│
├── Main Idea
│   │
│   ├── Force never exists alone
│   ├── Forces occur in pairs
│   └── Mutual interaction between bodies
│
├── Mathematical Form
│   │
│   └── FAB = -FBA
│
├── Action & Reaction
│   │
│   ├── Equal magnitude
│   ├── Opposite direction
│   ├── Act simultaneously
│   └── Act on different bodies
│
├── Important Points
│   │
│   ├── Action = Force
│   ├── Reaction = Force
│   ├── No cause-effect relation
│   ├── No time gap between them
│   └── Cannot cancel each other
│       on a single body
│
├── Internal Forces
│   │
│   ├── Present within a system
│   ├── Equal and opposite
│   └── Cancel each other
│
├── Examples
│   │
│   ├── Walking
│   │   ├── Foot pushes ground
│   │   └── Ground pushes foot
│   │
│   ├── Swimming
│   │   ├── Swimmer pushes water
│   │   └── Water pushes swimmer
│   │
│   ├── Rocket
│   │   ├── Gases move downward
│   │   └── Rocket moves upward
│   │
│   ├── Gun Recoil
│   │   ├── Bullet forward
│   │   └── Gun backward
│   │
│   └── Jumping
│       ├── Person pushes ground
│       └── Ground pushes person
│
└── NEET Revision
    │
    ├── Forces occur in pairs
    ├── Equal magnitude
    ├── Opposite direction
    ├── Simultaneous action
    ├── Different bodies
    ├── Do not cancel on one body
    └── Internal forces cancel 

INTERNAL LINKS
Newton's First Law of Motion Notes
Newton's Second Law of Motion Explained
Laws of Motion Complete Chapter Notes
Force and Inertia Class 11 Physics
Momentum and Impulse Notes
Conservation of Momentum
Friction Class 11 Physics
Work, Energy and Power Notes
Circular Motion Class 11 Physics
NEET Physics Important Questions
CBSE Class 11 Physics MCQ Bank
Physics Formula Sheet for Class 11
Motion in a Straight Line Notes
Motion in a Plane Notes
Gravitation Complete Notes

Newton's Second Law Notes for Class 11 Physics | CBSE & NEET Guide

Newton's Second Law Explained: Projectile Motion and Force Components

Easy Notes (NEET Level) – Important Points from Newton’s Second Law

- Dr.Sanjaykumar pawar

1. Component of Velocity Normal (Perpendicular) to the Force Remains Unchanged

A force can only change the velocity component in its own direction.
The velocity component perpendicular (normal) to the force does not change.
This is because there is no acceleration in the perpendicular direction.

Example: Projectile Motion

In projectile motion, only gravitational force acts on the particle.
Gravity acts vertically downward.
Therefore, acceleration is only in the vertical direction.
There is no horizontal force acting on the projectile (neglecting air resistance).
Hence, the horizontal component of velocity remains constant throughout the motion.
Only the vertical component of velocity changes due to gravity.

Key NEET Point:
Force changes velocity only along its direction; perpendicular velocity remains unchanged.

2. Newton's Second Law for a Single Particle

Newton's second law is written as:

Here, F is the net external force acting on the particle.
a is the acceleration produced in the particle.
This form is directly applicable to a single point particle.

Key NEET Point:
Always use the net external force, not individual forces separately.

3. Newton's Second Law for a System of Particles

The same law can also be applied to:
- A rigid body
- A group of particles (system)
In such cases:
- F = total external force on the system.
- a = acceleration of the centre of mass of the system.
Internal forces between particles of the system are not included.
Internal forces cancel each other and do not affect the motion of the whole system.

Example

Two blocks connected by a string.
Tension between the blocks is an internal force.
Only external forces like gravity or an applied pull are considered.

Key NEET Point:
For a system, consider only external forces; ignore internal forces.

4. Acceleration Depends on Present Force Only

Acceleration at any instant is determined by the force acting at that same instant.
Past motion does not affect present acceleration.
A particle has no "memory" of its previous motion.

Example: Stone Dropped from an Accelerated Train

A train is moving with acceleration.
A stone is dropped from the train.
Immediately after release:
- The stone is no longer connected to the train.
- No horizontal force acts on the stone (ignoring air resistance).
- Therefore, horizontal acceleration becomes zero.
The stone keeps its horizontal velocity but does not keep the train's acceleration.

Key NEET Point:
Velocity may continue, but acceleration changes instantly according to the force acting at that moment.

5. Newton's Second Law is a Local Relation

Newton's second law is called a local relation.
It relates force and acceleration at the same place and same time.

Meaning

Force acting here and now determines acceleration here and now.
Previous positions, velocities, or accelerations do not directly determine present acceleration.
Only the current force matters.

Key NEET Point:
Present acceleration depends only on present force, not on the history of motion.

Quick Revision for NEET

Force changes velocity only in its own direction.
Velocity perpendicular to force remains unchanged.
In projectile motion, horizontal velocity remains constant.
Newton's second law: .
For a system, use total external force only.
Internal forces are ignored.
Acceleration depends on the force acting at that instant.
A body has no memory of past acceleration.
Newton's second law is a local relation.
Present force determines present acceleration.

Educational diagram showing Newton's Second Law, force and acceleration relationship, projectile motion, centre of mass and external forces for Class 11 Physics students.

Newton's Second Law explains how force produces acceleration and why horizontal velocity remains constant in projectile motion.

CBSE Class 11 Physics – Newton's Second Law (Important Questions with Answers)

A. MCQs (1 Mark Each)

1. In projectile motion, the horizontal component of velocity remains constant because:

a) Gravity acts horizontally
b) No horizontal force acts on the projectile
c) Air resistance is maximum
d) Vertical velocity is constant

Answer: (b) No horizontal force acts on the projectile

2. The force in Newton's second law represents:

a) Internal force
b) External force
c) Gravitational force only
d) Friction only

Answer: (b) External force

3. For a system of particles, acceleration refers to the acceleration of:

a) Any particle
b) Largest particle
c) Centre of mass
d) Geometric centre

Answer: (c) Centre of mass

4. Internal forces are:

a) Included in net external force
b) Ignored while applying Newton's second law to a system
c) Greater than external forces
d) Always zero

Answer: (b)

5. A body remembers:

a) Past acceleration
b) Past force
c) Present force only affects acceleration
d) Future force

Answer: (c)

B. Very Short Answer Questions (1 Mark)

1. What is meant by the net force on a particle?

Answer: The vector sum of all external forces acting on the particle.

2. Which velocity component remains unchanged in projectile motion?

Answer: Horizontal component of velocity.

3. What is the SI unit of force?

Answer: Newton (N).

4. Which force is not included while applying Newton's second law to a system?

Answer: Internal force.

5. What is meant by a local relation?

Answer: A relation that connects physical quantities at the same place and same instant.

C. Short Answer Questions (2–3 Marks)

1. Why does the horizontal velocity of a projectile remain constant?

Answer:

Gravity acts vertically downward.
No horizontal force acts on the projectile.
Therefore, horizontal acceleration is zero.
Hence, horizontal velocity remains constant.

2. What is the role of internal forces in a system?

Answer:

Internal forces act between particles of the same system.
They cancel each other in pairs.
Therefore, they do not affect the motion of the system as a whole.

3. Explain why a stone dropped from an accelerating train has no horizontal acceleration.

Answer:

After release, the stone is no longer connected to the train.
No horizontal force acts on it.
According to Newton's second law, acceleration depends on force.
Hence horizontal acceleration becomes zero.

D. Long Answer Questions (3–5 Marks)

1. Explain Newton's second law for a system of particles.

Answer: Newton's second law states that the net external force acting on a body equals the product of its mass and acceleration.

For a system of particles:

F represents the total external force on the system.
a represents the acceleration of the centre of mass.
Internal forces are not included because they cancel each other.
The motion of the entire system depends only on external forces.

2. Explain why Newton's second law is called a local relation.

Answer:

Force and acceleration are related at the same place and same instant.
Present acceleration depends only on present force.
Past motion does not influence present acceleration directly.
A body has no memory of previous forces or accelerations.
Therefore Newton's second law is called a local relation.

E. Assertion and Reason Questions

1.

Assertion (A): Horizontal velocity remains constant in projectile motion.

Reason (R): No horizontal force acts on the projectile.

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

2.

Assertion (A): Internal forces are included in the net force acting on a system.

Reason (R): Internal forces cancel each other.

Answer: Assertion is false, Reason is true.

3.

Assertion (A): A body remembers its previous acceleration.

Reason (R): Present acceleration depends only on present force.

Answer: Assertion is false, Reason is true.

F. Fill in the Blanks

In projectile motion, the ________ component of velocity remains constant. Answer: horizontal
Newton's second law relates force and ________. Answer: acceleration
For a system, acceleration refers to the acceleration of the ________. Answer: centre of mass
Internal forces are ________ while applying Newton's second law to a system. Answer: ignored
Present acceleration depends on ________ force. Answer: present

G. True or False

Gravity changes the horizontal velocity of a projectile. Answer: False
Internal forces affect the motion of the centre of mass. Answer: False
External force determines acceleration. Answer: True
Newton's second law is a local relation. Answer: True
A body remembers its past acceleration. Answer: False

H. Match the Columns

Column A	Column B
1. Projectile motion	(a) Centre of mass
2. System acceleration	(b) Horizontal velocity constant
3. Internal forces	(c) Present force
4. Local relation	(d) Ignored
5. Acceleration depends on	(e) Same place and time

Answers

1 → (b)
2 → (a)
3 → (d)
4 → (e)
5 → (c)

I. Statement-Based Questions

Statement 1:

A projectile is moving in air with negligible air resistance.

Statement 2:

Its horizontal velocity remains constant.

a) Both statements are true and Statement 2 explains Statement 1.
b) Both statements are true but Statement 2 does not explain Statement 1.
c) Statement 1 is true, Statement 2 is false.
d) Statement 1 is false, Statement 2 is true.

Answer: (a)

J. Case Study Questions (4 Marks)

Case Study

A train is moving with acceleration. A stone is dropped from the train. Immediately after release, the stone continues moving forward but experiences only gravitational force.

Questions

1. Which force acts on the stone after release?

Answer: Gravitational force.

2. What is the horizontal acceleration of the stone?

Answer: Zero.

3. Why does the stone continue moving horizontally?

Answer: Due to its existing horizontal velocity.

4. What does this example prove about Newton's second law?

Answer: Present acceleration depends only on present force and not on past motion.

Exam-Oriented One-Line Revision

Force changes velocity only in its own direction.
Velocity perpendicular to force remains unchanged.
Horizontal velocity remains constant in projectile motion.
Only external forces are considered for a system.
Internal forces cancel each other.
Acceleration of a system is acceleration of its centre of mass.
Present force determines present acceleration.
Newton's second law is a local relation.

NEWTON'S SECOND LAW – IMPORTANT POINTS

│

├── 1. Velocity Component Perpendicular to Force

│ │

│ ├── Force changes velocity only along its direction

│ ├── No acceleration perpendicular to force

│ ├── Perpendicular (normal) velocity remains constant

│ │

│ └── Example: Projectile Motion

│ │

│ ├── Gravity acts vertically downward

│ ├── Vertical velocity changes

│ ├── No horizontal force (air resistance neglected)

│ └── Horizontal velocity remains constant

│

├── 2. Newton's Second Law for a Particle

│ │

│ ├── F = ma

│ ├── F = Net external force

│ ├── a = Acceleration produced

│ └── Applicable to a single particle

│

├── 3. Newton's Second Law for a System

│ │

│ ├── Applicable to

│ │ ├── Rigid body

│ │ └── System of particles

│ │

│ ├── F = Total external force

│ ├── a = Acceleration of centre of mass

│ │

│ └── Internal Forces

│ │

│ ├── Not included in F

│ ├── Cancel each other

│ └── Do not affect system motion

│

├── 4. Acceleration Depends on Present Force

│ │

│ ├── Present force → Present acceleration

│ ├── Past motion does not matter

│ └── Body has no memory of past acceleration

│

├── 5. Example: Stone Dropped from Accelerated Train

│ │

│ ├── Train is accelerating

│ ├── Stone is released

│ ├── Connection with train ends

│ ├── No horizontal force on stone

│ ├── Horizontal acceleration = 0

│ └── Horizontal velocity continues unchanged

│

├── 6. Local Nature of Newton's Second Law

│ │

│ ├── Local Relation

│ │

│ ├── Force at a point

│ │ ↓

│ ├── Determines acceleration

│ │ ↓

│ └── At the same place and same time

│

└── NEET QUICK FACTS

│

├── Force changes velocity only in its direction

├── Perpendicular velocity remains unchanged

├── Horizontal velocity is constant in projectile motion

├── F = ma

├── Use only external forces for a system

├── Ignore internal forces

├── Present force determines present acceleration

├── No memory of previous acceleration

└── Newton's second law is a local relation