Uniformly Accelerated Motion Class 11 Physics Notes | NEET & JEE MCQs

 - Dr.Sanjaykumar Pawar  

Uniformly Accelerated Motion (1-D) Physics Notes, Formulas & NEET Questions 

Uniformly Accelerated Motion (1-D): Equations of Motion, Graphs and NEET Physics Concepts

Internal Links

1. Motion in One Dimension Foundation

 Motion in One Dimension Class 11 Physics Notes

Before introducing uniformly accelerated motion, link readers to the basic concepts of displacement, velocity and speed.

2. Instantaneous Velocity & Acceleration

Instantaneous Velocity and Acceleration Explained

NCERT Physics Class 11 Chapter 2: Instantaneous Velocity & Acceleration

Under "Important Terms" section after explaining acceleration.

3. Vectors in Physics

Vectors Class 11 Physics Notes

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

  explaining displacement as a vector quantity.

4. Newton's Laws of Motion

Newton's Laws of Motion Class 11 Physics

Block and Trolley System NEET Solution | Acceleration & Tension Explained

acceleration concepts because force causes acceleration.

5. Free Fall Motion

Free Fall and Acceleration Due to Gravity

NEET Tips section where free fall problems are mentioned.

6. Graphical Motion Analysis

Velocity-Time and Acceleration-Time Graphs

 diagram-based questions.

Uniform acceleration formula

Equations of motion derivation

Motion in one dimension notes

Class 11 Physics chapter 2 notes

NEET physics motion questions

JEE physics kinematics notes

Velocity time graph questions

Acceleration numericals with solutions

CBSE Physics important questions

Physics revision notes for NEET

FAQ Schema Questions

Q1. What is uniformly accelerated motion?

Uniformly accelerated motion is motion in which acceleration remains constant with time.

Q2. What are the three equations of motion?

The three equations are v = u + at, s = ut + ½at² and v² = u² + 2as.

Q3. What does the slope of a velocity-time graph represent?

The slope of a velocity-time graph represents acceleration.

Q4. What does the area under a velocity-time graph represent?

The area represents displacement.

Q5. Is uniformly accelerated motion important for NEET and JEE?

Yes, it is a fundamental topic used in many mechanics problems.

Uniformly Accelerated Motion (1-D)

1. Meaning of Uniformly Accelerated Motion

Uniformly accelerated motion means an object is moving in a straight line and its acceleration remains constant with time.

• The velocity changes by the same amount in equal intervals of time.
• The motion takes place in one dimension.

Example: A bike increases its speed by 5 m/s every second.

Important Terms

1. Initial Velocity (u)

Initial velocity is the velocity of an object at the starting time. It is represented by u.

At starting time:
t = 0, velocity = u

2. Final Velocity (v)

Final velocity is the velocity of an object after a certain time. It is represented by v.

3. Acceleration (a)

Acceleration is the rate of change of velocity.

a = (v - u) / t

Unit of acceleration = m/s²

4. Displacement (s)

Displacement is the distance travelled by an object in a particular direction.

Unit = metre (m)

First Equation of Motion

v = u + at

Derivation:

Acceleration:

a = dv/dt

Rearranging:

a dt = dv

Integrating:

∫a dt = ∫dv

a(t - 0) = v - u

at = v - u

v = u + at

Final velocity = Initial velocity + Change in velocity

Second Equation of Motion

s = ut + 1/2 at²

Derivation:

Velocity:

v = ds/dt

Therefore:

ds = v dt

Using:
v = u + at

ds = (u + at)dt

After integration:

s = ut + 1/2 at²

Displacement = Distance due to initial velocity + Distance due to acceleration

Third Equation of Motion

v² = u² + 2as

Derivation:

From first equation:

v = u + at

Rearranging:

t = (v - u)/a

Using second equation:

s = ut + 1/2 at²

After simplification:

v² = u² + 2as

This equation is useful when time is not given.

Three Equations of Motion Summary

1. Velocity Equation

v = u + at

Used to find final velocity when time is given.

2. Displacement Equation

s = ut + 1/2 at²

Used to find displacement when time is given.

3. Time Independent Equation

v² = u² + 2as

Used when time is not given.

Easy Memory Trick

• V-U-AT: v = u + at
• S-U-T-A-T: s = ut + 1/2 at²
• V-U-AS: v² = u² + 2as

Symbols at a Glance

Symbol	Meaning	Unit
u	Initial Velocity	m/s
v	Final Velocity	m/s
a	Acceleration	m/s²
t	Time	second
s	Displacement	metre

Conclusion

The three equations of motion are used to solve problems involving constant acceleration in one-dimensional motion.

NEET Physics Practice Questions

Chapter: Uniformly Accelerated Motion (1-D)

NEET Question Types

• Direct MCQs (Single Correct Option)
• Statement Based Questions
• Assertion and Reason
• Match the Columns
• Diagram Based / Graphical Questions

PART 1: Direct MCQs (Single Correct Option)

Q1. A car starts from rest and accelerates uniformly at 4 m/s². Its velocity after 5 seconds will be:

A) 10 m/s
B) 20 m/s
C) 25 m/s
D) 40 m/s

Solution:
u = 0
a = 4 m/s²
t = 5 s

v = u + at
v = 0 + 4 × 5
v = 20 m/s

Answer: B) 20 m/s

Q2. The SI unit of acceleration is:

A) m/s
B) m²/s
C) m/s²
D) km/h

Answer: C) m/s²

Q3. A body moving with velocity 20 m/s is brought to rest in 5 seconds. The acceleration is:

A) +4 m/s²
B) -4 m/s²
C) +5 m/s²
D) -5 m/s²

a = (v-u)/t
a = (0-20)/5
a = -4 m/s²

Answer: B) -4 m/s²

Q4. The velocity-time graph for uniformly accelerated motion is:

A) Straight line
B) Circle
C) Parabola
D) Hyperbola

Answer: A) Straight line

Q5. A particle starts from rest and travels 100 m in 5 seconds with uniform acceleration. Find acceleration.

A) 4 m/s²
B) 6 m/s²
C) 8 m/s²
D) 10 m/s²

s = ut + 1/2 at²
100 = 0 + 1/2 × a × 25
a = 8 m/s²

Answer: C) 8 m/s²

PART 2: Statement Based Questions

Options:
A) Both statements true and II explains I
B) Both true but II does not explain I
C) I true, II false
D) I false, II true

Q6.

Statement I: Velocity-time graph for uniformly accelerated motion is a straight line.

Statement II: Acceleration is constant in uniformly accelerated motion.

Answer: A

Q7.

Statement I: Displacement can be zero even if distance travelled is not zero.

Statement II: Displacement depends only on initial and final position.

Answer: A

Q8.

Statement I: Acceleration due to gravity is constant near Earth's surface.

Statement II: Value of g is approximately 9.8 m/s².

Answer: A

PART 3: Assertion and Reason

Options:
A) Both true and R explains A
B) Both true but R does not explain A
C) A true, R false
D) A false, R true

Q9.

Assertion: A body moving with constant velocity has zero acceleration.

Reason: Acceleration is the rate of change of velocity.

Answer: A

Q10.

Assertion: Area under velocity-time graph gives displacement.

Reason: Velocity is displacement divided by time.

Answer: A

Q11.

Assertion: A body can have zero velocity and non-zero acceleration.

Reason: At highest point of upward motion, velocity is zero but acceleration acts downward.

Answer: A

PART 4: Match the Columns

Column I	Column II
Velocity	m/s
Acceleration	m/s²
Displacement	m
Time	second

Answer: Velocity-m/s, Acceleration-m/s², Displacement-m, Time-second

PART 5: Diagram Based / Graphical Questions

Q14. The slope of velocity-time graph represents:

A) Distance
B) Velocity
C) Acceleration
D) Displacement

Answer: C) Acceleration

Q15. Area under velocity-time graph represents:

A) Acceleration
B) Displacement
C) Force
D) Momentum

Answer: B) Displacement

Q16. Velocity-time graph:

v
|
|        /
|       /
|      /
|_____/________ t

The particle has:

A) Constant velocity
B) Constant acceleration
C) Zero acceleration
D) Variable acceleration

Answer: B) Constant acceleration

Hard Numerical Practice

Q17. A train moving at 72 km/h stops in 10 seconds. Find retardation.

72 km/h = 20 m/s a = (0-20)/10 a = -2 m/s² Retardation = 2 m/s²

Q18. A particle has initial velocity 5 m/s and acceleration 2 m/s². Find distance in 10 seconds.

s = ut + 1/2at² s = 5×10 + 1/2×2×100 s = 150 m Answer: 150 m

NEET Formula Revision

v = u + at

s = ut + 1/2 at²

v² = u² + 2as

s = ((u+v)/2)t

NEET Tips

• Practice velocity-time graphs.
• Remember sign convention.
• Understand distance and displacement difference.
• Use correct equation according to given data.
• Practice free fall problems.

NEET Physics: Precision & Least Count Practice Questions with Answers

 - Dr.Sanjaykumar Pawar

Comparison of measuring instruments showing how least count affects precision in Physics for NEET preparation.

 

CBSE Class 11 Physics Notes

Chapter: Units and Measurements

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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.

CBSE Class 11 Physics | Units and Measurements | Complete Beginner Notes

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

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

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

NCERT Solutions Class 11 Physics Chapter 2 Exercise 2.15 to 2.18 with Answers

 - Dr.Sanjaykumar Pawar  

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NCERT Solutions

Class 11 Physics

Chapter 2: Motion in a Straight Line

Exercise 2.15 to 2.18

Question 2.15

Suggest a suitable physical situation for each of the following graphs (Fig. 2.12).

(a) Position-Time Graph (x–t)

Observation:

• The particle remains at rest initially.
• Then moves away from the origin with constant velocity.
• Returns towards the origin with constant velocity.
• Finally comes to rest again.

Suitable Physical Situation:

A person standing on a straight road starts walking uniformly away from the origin, reaches a point A, turns back, crosses the origin at B and finally stops at another point.

The graph represents a person moving away from the origin, returning back and finally coming to rest.

---

(b) Velocity-Time Graph (v–t)

Observation:

• Velocity decreases linearly.
• Changes sign repeatedly.
• Magnitude decreases after every interval.

Suitable Physical Situation:

A ball thrown vertically upward repeatedly strikes the ground and rebounds. Each collision reduces its speed due to loss of energy.

The graph represents the motion of a bouncing ball with decreasing speed after each rebound.

---

(c) Acceleration-Time Graph (a–t)

Observation:

• Acceleration is zero for most of the time.
• A sudden positive acceleration occurs for a short interval.

Suitable Physical Situation:

A vehicle moving uniformly receives a sudden push or the accelerator is pressed for a short duration.

The graph represents a body receiving acceleration for a short time and then moving uniformly again.

Question 2.16

Figure 2.13 gives the x–t plot of a particle executing one-dimensional simple harmonic motion. Give the signs of position (x), velocity (v) and acceleration (a) at: (i) t = 0.3 s (ii) t = 1.2 s (iii) t = –1.2 s

Principle Used:

Acceleration in SHM: a = -ω²x Velocity is given by the slope of the x-t graph.

(i) At t = 0.3 s

• x is negative.
• Slope is negative.
• Acceleration is positive.

x < 0, v < 0, a > 0

(ii) At t = 1.2 s

• x is positive.
• Slope is positive.
• Acceleration is negative.

x > 0, v > 0, a < 0

(iii) At t = –1.2 s

• x is positive.
• Slope is positive.
• Acceleration is negative.

x > 0, v > 0, a < 0

Time	Position (x)	Velocity (v)	Acceleration (a)
0.3 s	Negative	Negative	Positive
1.2 s	Positive	Positive	Negative
–1.2 s	Positive	Positive	Negative

Question 2.17

Figure 2.14 gives the x–t plot of a particle in one-dimensional motion. Three different equal intervals of time are shown. In which interval is the average speed greatest and in which is it least? Give the sign of average velocity for each interval.

Concept:

Average Velocity = Change in Position / Time Average Speed = Distance Travelled / Time

Since all intervals have equal duration, the interval with the greatest change in position will have the greatest average speed.

Interval 1

• Position increases slowly.
• Average velocity is positive.

Interval 2

• Position increases rapidly.
• Average velocity is positive.

Interval 3

• Position decreases rapidly.
• Average velocity is negative.

Greatest Average Speed = Interval 3

Least Average Speed = Interval 1

Interval	Sign of Average Velocity
1	Positive
2	Positive
3	Negative

Question 2.18

Figure 2.15 gives a speed-time graph of a particle moving in a constant direction. Three equal intervals of time are shown. (a) In which interval is the average acceleration greatest in magnitude? (b) In which interval is the average speed greatest? (c) Choosing the positive direction as direction of motion, give signs of velocity and acceleration in the three intervals. (d) What are the accelerations at points A, B, C and D?

(a) Average Acceleration

Average Acceleration = Change in Speed / Time

Since all intervals have equal duration, the third interval shows the largest change in speed.

Greatest Average Acceleration occurs in Interval 3.

---

(b) Average Speed

The average speed is greatest in the interval where the speed values are highest.

Average Speed is greatest in Interval 3.

---

(c) Signs of Velocity and Acceleration

Interval	Velocity	Acceleration
1	Positive	Positive
2	Positive	Negative
3	Positive	Positive

---

(d) Accelerations at A, B, C and D

Acceleration is equal to the slope of the speed-time graph.

At A → Positive slope → aA > 0 At B → Slope = 0 → aB = 0 At C → Slope = 0 → aC = 0 At D → Slope = 0 → aD = 0

aA > 0, aB = 0, aC = 0, aD = 0

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NCERT Class 11 Physics Chapter 2 Motion in a Straight Line Exercise 2.15 to 2.18 solved with graphs, SHM concepts, velocity and acceleration analysis. 

Internal Links

Class 11 Physics Motion in a Straight Line Notes

Instantaneous Velocity and Acceleration Notes

NCERT Physics Chapter 2 Important Questions

Numerical Problems on Motion in a Straight Line

Kinematics MCQs for NEET and JEE

Previous Year Questions on Motion in One Dimension

Difference Between Speed and Velocity

Graphs in Physics Explained

Simple Harmonic Motion Basics

Class 11 Physics Revision Notes

NCERT Solutions Chapter 2 Complete Series

Physics Formula Sheet for Class 11

NEET Physics Kinematics Practice Questions

JEE Motion in a Straight Line Problems

CBSE Class 11 Physics Sample Questions

NCERT Class 11 Physics Chapter 2 Exercise 2.8 to 2.14 Solutions (CBSE 2026)

-  Dr.Sanjaykumar Pawar  

 

Internal Links

1. NCERT Class 11 Physics Chapter 2 Motion in a Straight Line Notes

2. NCERT Class 11 Physics Exercise 2.1 to 2.7 Solutions

3. NCERT Class 11 Physics Chapter 3 Motion in a Plane Solutions

4. Important Class 11 Physics Formulas PDF

5. CBSE Class 11 Physics Previous Year Questions

6. Motion in a Straight Line MCQs with Answers

7. Speed, Velocity and Acceleration Notes

8. Class 11 Physics Revision Notes

9. NCERT Exemplar Class 11 Physics Solutions

10. Complete Class 11 Physics Study Material

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NCERT Solutions Class 11 Physics

Chapter 2: Motion in a Straight Line

Exercise 2.8 to 2.14

Question 2.8

A ball is dropped from a height of 90 m on a floor. At each collision with the floor, the ball loses one tenth of its speed. Plot the speed-time graph of its motion between t = 0 to 12 s.

Given:

• Height = 90 m
• g = 9.8 m/s²
• Loss of speed after every collision = 10%

Step 1: Time taken to reach the floor

h = ½gt²

90 = ½ × 9.8 × t²

t² = 180/9.8

t = 4.29 s

Step 2: Speed just before collision

v = gt

v = 9.8 × 4.29

v = 42 m/s

Step 3: Speed after collision

v' = 0.9 × 42

v' = 37.8 m/s

Step 4: Time taken to move upward

t = v'/g

t = 37.8/9.8

t = 3.86 s

Graph Description:

• Speed increases uniformly from 0 to 42 m/s during first 4.29 s.
• At collision speed suddenly decreases to 37.8 m/s.
• Speed decreases linearly to zero while moving upward.
• Then increases again while falling downward.

Speed-Time graph consists of straight line segments with sudden vertical drops at collisions.

Question 2.9

Explain clearly, with examples, the distinction between: (a) Magnitude of displacement and total path length (b) Magnitude of average velocity and average speed.

(a) Magnitude of Displacement and Total Path Length

Magnitude of Displacement	Total Path Length
Shortest distance between initial and final positions.	Actual distance travelled.
Depends only on initial and final positions.	Depends on actual path followed.
Always less than or equal to path length.	Always greater than or equal to displacement.

Example:

Particle moves 3 m east and 4 m north. Displacement = √(3² + 4²) = 5 m Path Length = 3 + 4 = 7 m

(b) Magnitude of Average Velocity and Average Speed

Average Velocity = Displacement / Time

Average Speed = Total Path Length / Time

Since total path length ≥ displacement,

Average Speed ≥ Magnitude of Average Velocity

Equality occurs only when motion is along a straight line without changing direction.

Question 2.10

A man walks on a straight road from his home to a market 2.5 km away with a speed of 5 km h⁻¹. Finding the market closed, he instantly turns and walks back home with a speed of 7.5 km h⁻¹. Find: (a) Magnitude of average velocity (b) Average speed

(a) Magnitude of Average Velocity

Net displacement = 0 Average Velocity = 0 / Total Time = 0

Magnitude of Average Velocity = 0 km h⁻¹

(b) Average Speed

Time to market = 2.5/5 = 0.5 h

Time to return = 2.5/7.5 = 0.333 h

Total distance = 5 km Total time = 0.5 + 0.333 = 0.833 h

Average Speed = 5 / 0.833 = 6 km h⁻¹

Average Speed = 6 km h⁻¹

Question 2.11

Why is there no distinction between instantaneous speed and magnitude of instantaneous velocity?

Instantaneous Speed = Magnitude of Instantaneous Velocity

Velocity has both magnitude and direction, whereas speed is only magnitude. At any instant, speed is simply the magnitude of velocity.

Instantaneous Speed = |Instantaneous Velocity|

Question 2.12

Look at the graphs (a) to (d) carefully and state which of these cannot represent one-dimensional motion of a particle.

• Graph (a): Impossible because one instant corresponds to more than one position.
• Graph (b): Impossible because one instant corresponds to more than one velocity.
• Graph (c): Impossible because speed cannot be negative.
• Graph (d): Impossible because total path length can never decrease.

All graphs (a), (b), (c) and (d) cannot represent one-dimensional motion.

Question 2.13

Figure 2.11 shows the x-t plot of one-dimensional motion of a particle. Is it correct to say from the graph that the particle moves in a straight line for t < 0 and on a parabolic path for t > 0?

No. The graph is an x-t graph and not the actual trajectory of the particle.

For t < 0, x remains constant, showing that the particle is at rest.

For t > 0, x increases with time and velocity increases continuously.

The graph represents variation of position with time and not the actual path of motion.

Question 2.14

A policeman moving at 30 km h⁻¹ fires a bullet at a thief's car speeding away in the same direction with a speed of 192 km h⁻¹. If the muzzle speed of the bullet is 150 m s⁻¹, with what speed does the bullet hit the thief's car?

Step 1: Convert speeds into m/s

Policeman = 30 × 5/18 = 8.33 m/s

Thief = 192 × 5/18 = 53.33 m/s

Step 2: Bullet speed relative to ground

Bullet speed = 150 + 8.33 = 158.33 m/s

Step 3: Relative speed of bullet with respect to thief

158.33 − 53.33 = 105 m/s

Speed of bullet relative to thief's car = 105 m/s

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NCERT Physics Class 11 Chapter 2 Instantaneous Acceleration Notes with Graphs

NCERT Physics Class 11 Chapter 2 Easy Line-by-Line Notes

Instantaneous Acceleration

“Instantaneous acceleration is defined in the same way as instantaneous velocity.”

• Just like instantaneous velocity gives velocity at one instant,

• Instantaneous acceleration gives acceleration at one instant.

Formula of Instantaneous Acceleration

a = lim(Δt → 0) [ Δv / Δt ]

• Take very small time interval.

• Find change in velocity.

• Divide by tiny time interval.

Calculus Form

a = dv / dt

Acceleration = rate of change of velocity with time.

“The acceleration at an instant is the slope of the tangent to the v–t curve.”

Important Graph Rule:

On velocity-time graph:

Slope of tangent = instantaneous acceleration

Velocity and Acceleration

“Velocity has both magnitude and direction.”

Velocity is a vector quantity.

It can change by:

• Changing speed
• Changing direction
• Changing both

“Acceleration may result from change in speed, direction or both.”

Situation	Type of Change
Car speeding up	Speed changes
Circular motion	Direction changes
Curved fast motion	Both change

Positive, Negative and Zero Acceleration

Positive Acceleration

• Velocity increases with time.
• Object speeds up.

Example: Bike accelerating forward.

Negative Acceleration (Retardation)

• Velocity decreases with time.
• Object slows down.

Example: Applying brakes.

Zero Acceleration

• Velocity remains constant.
• No change in motion.

Example: Car moving at constant speed.

Position-Time Graphs

“Graph curves upward for positive acceleration.”

• Slope keeps increasing.

• Object moves faster and faster.

“Graph curves downward for negative acceleration.”

• Slope decreases with time.

• Object slows down.

“Straight line for zero acceleration.”

• Constant slope.

• Constant velocity.

Constant Acceleration

“Our study will be restricted to constant acceleration.”

Acceleration remains same throughout motion.

Example: Free fall near Earth.

Equation for Constant Acceleration

a = (v - v₀) / t

Rearranging Equation

v = v₀ + at

First Equation of Motion

Meaning of Symbols

Symbol	Meaning
v₀	Initial velocity
v	Final velocity
a	Acceleration
t	Time

Velocity-Time Graph Cases

Case (a)

Object moving in positive direction with positive acceleration.

• Velocity positive
• Acceleration positive

Object moves forward faster and faster.

Graph: Upward sloping line above time axis.

Case (b)

Object moving in positive direction with negative acceleration.

• Velocity positive
• Acceleration negative

Object still moves forward but slows down.

Example: Car braking while moving forward.

Case (c)

Object moving in negative direction with negative acceleration.

• Velocity negative
• Acceleration negative

Object moves backward and speeds up backward.

Case (d)

Object moving in positive direction till t₁ and then turns back.

• Initially moving forward
• Velocity becomes zero at t₁
• Then object reverses direction

v = 0

Important Point: At turning point velocity becomes zero.

Area Under Velocity-Time Graph

“Area under the curve represents displacement.”

Very Important Concept

On velocity-time graph:

Displacement = Area under v-t graph

Why Area Gives Displacement

v = Δx / Δt

Δx = v Δt

Area = velocity × time.

Constant Velocity Case

Velocity-time graph becomes:

• Straight horizontal line

Area rectangle under graph gives displacement.

Important Graph Rules Summary

Graph	Slope Gives	Area Gives
Position-Time	Velocity	—
Velocity-Time	Acceleration	Displacement

Real-Life Examples

Situation	Acceleration Type
Bike speeding up	Positive
Car braking	Negative
Cruise control	Zero
Falling object	Constant acceleration

One-Line Summary:

Acceleration describes how velocity changes with time, and on a velocity-time graph, slope gives acceleration while area under the graph gives displacement.

- Dr.Sanjaykumar Pawar   

Instantaneous acceleration on a velocity-time graph showing tangent slope and displacement area for Class 11 Physics students.

Internal Links

Motion in a Straight Line Notes

Difference Between Speed and Velocity

Average Velocity and Instantaneous Velocity

Position-Time Graph Explained

Velocity-Time Graph Explained

Equations of Motion Derivation

Uniform and Non-Uniform Motion

Numerical Problems on Acceleration

Class 11 Physics Chapter 3 Exercise Solutions

Important NEET Kinematics Questions

Exam Preparation Links

CBSE Class 11 Physics Important Questions

NEET Physics Motion Chapter MCQs

JEE Main Kinematics Practice Problems

Previous Year Questions on Acceleration

Physics Revision Notes for Class 11 

CBSE Class 11 Physics Chapter 2 Important Questions and Answers – Instantaneous Acceleration

Very Short Answer Questions

What is instantaneous acceleration?

Instantaneous acceleration is the rate of change of velocity at a particular instant of time.

Write the formula for instantaneous acceleration.

a = lim (Δt → 0) (Δv / Δt)

What is the SI unit of acceleration?

The SI unit of acceleration is metre per second squared (m/s²).

Is acceleration a scalar or vector quantity?

Acceleration is a vector quantity because it has both magnitude and direction.

What does the slope of a velocity-time graph represent?

The slope of a velocity-time graph represents acceleration.

What does the area under a velocity-time graph represent?

The area under a velocity-time graph represents displacement.

Short Answer Questions

Define instantaneous acceleration.

Instantaneous acceleration is the acceleration of an object at a particular instant of time. Mathematically, it is expressed as:

a = dv/dt

It measures how quickly velocity changes at a given moment.

How can velocity change?

Velocity can change in three ways:

• Change in speed
• Change in direction
• Change in both speed and direction

Any of these changes produces acceleration.

Differentiate between positive and negative acceleration.

Positive Acceleration	Negative Acceleration
Velocity increases with time.	Velocity decreases with time.
Object speeds up.	Object slows down.
Slope of v-t graph is positive.	Slope of v-t graph is negative.

What is zero acceleration?

Zero acceleration occurs when velocity remains constant. There is no change in speed or direction.

Example: A car moving at constant speed on a straight road.

Long Answer Questions

Explain instantaneous acceleration using a velocity-time graph.

Instantaneous acceleration is the rate of change of velocity at a particular instant.

a = dv/dt

On a velocity-time graph, the instantaneous acceleration at any point is obtained by drawing a tangent at that point. The slope of the tangent gives the instantaneous acceleration.

Therefore:

Instantaneous Acceleration = Slope of Tangent to the v-t Graph

Explain positive, negative and zero acceleration with examples.

Positive Acceleration: Velocity increases with time. Example: A bike speeding up.

Negative Acceleration: Velocity decreases with time. Example: A car slowing down after brakes are applied.

Zero Acceleration: Velocity remains constant. Example: A train moving uniformly on a straight track.

Derive the first equation of motion.

For constant acceleration:

a = (v − v₀)/t

Multiplying both sides by t:

at = v − v₀

Rearranging:

v = v₀ + at

This is known as the first equation of motion.

Multiple Choice Questions (MCQs)

Instantaneous acceleration is represented by:

A. Δx/Δt
B. Δv/Δt
C. dv/dt
D. Both B and C

Answer: D. Both B and C

Area under a velocity-time graph gives:

A. Velocity
B. Acceleration
C. Displacement
D. Speed

Answer: C. Displacement

Slope of a velocity-time graph gives:

A. Velocity
B. Distance
C. Acceleration
D. Time

Answer: C. Acceleration

Assertion and Reason Questions

Assertion:

Acceleration can occur even when speed remains constant.

Reason:

Velocity changes when direction changes.

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

Assertion:

The area under a velocity-time graph gives displacement.

Reason:

Velocity is displacement per unit time.

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

Fill in the Blanks

Instantaneous acceleration is represented by dv/dt.

The slope of a velocity-time graph gives acceleration.

The area under a velocity-time graph gives displacement.

Negative acceleration is also called retardation.

At the turning point, velocity becomes zero.

Match the Following

Column A	Column B
Slope of v-t graph	Acceleration
Area under v-t graph	Displacement
Constant velocity	Zero acceleration
Retardation	Negative acceleration

Case Study Questions

A car starts moving with an initial velocity of 10 m/s and accelerates uniformly at 2 m/s² for 5 seconds.

What is the initial velocity of the car?

Answer: 10 m/s

What is the acceleration of the car?

Answer: 2 m/s²

Calculate the final velocity.

Using:

v = v₀ + at

v = 10 + (2 × 5) = 20 m/s

Answer: 20 m/s

Is the acceleration positive or negative?

Answer: Positive acceleration.

Which equation of motion is used?

Answer: v = v₀ + at

Instantaneous Acceleration

Definition

• Acceleration at a particular instant.
• Rate of change of velocity with time.

a = dv/dt

Formula

• Very small time interval considered.
• Change in velocity divided by time interval.

a = lim(Δt→0) (Δv/Δt)

Velocity-Time Graph

• Slope of tangent gives acceleration.
• Positive slope → Positive acceleration.
• Negative slope → Negative acceleration.

Types of Acceleration

• Positive Acceleration
• Negative Acceleration (Retardation)
• Zero Acceleration

Velocity Changes By

• Changing Speed
• Changing Direction
• Changing Both

Position-Time Graph

• Upward Curve → Positive Acceleration
• Downward Curve → Negative Acceleration
• Straight Line → Zero Acceleration

Constant Acceleration

• Acceleration remains constant.
• Example: Free Fall.

v = v₀ + at

Area Under v-t Graph

• Area under curve = Displacement
• Rectangle area for constant velocity.

Real-Life Examples

• Bike speeding up
• Car braking
• Cruise control
• Falling object

NCERT Class 11 Physics Chapter 2.2 Instantaneous Velocity and Speed Notes

NCERT Physics Class 11 Chapter 2

2.2 Instantaneous Velocity and Speed

“The average velocity tells us how fast an object has been moving over a given time interval...”

Easy Meaning

• Average velocity gives motion over a long interval of time.
• It does NOT tell the exact velocity at a particular moment.

Example

If a car travels:

• 100 m in 10 s

v_avg = Δx / Δt

= 100 / 10 = 10 m/s

But the car may move:

• slowly at first
• faster later

So average velocity hides instant changes.

“...but does not tell us how fast it moves at different instants of time...”

Easy Meaning

• We need velocity at one exact moment.
• Example:
  • velocity exactly at 4 s
  • not from 0 to 10 s

Instantaneous Velocity

“For this, we define instantaneous velocity or simply velocity v at an instant t.”

Definition

Instantaneous velocity = velocity at a particular instant of time.

Example

• Speedometer of a bike shows instantaneous speed.

Mathematical Definition

“The velocity at an instant is defined as the limit of average velocity...”

Idea

Take smaller and smaller time intervals.

• When interval becomes extremely tiny:
• average velocity becomes instantaneous velocity.

Formula of Instantaneous Velocity

v = lim_Δt→0 (Δx / Δt)

Meaning of Symbols

• (v) → instantaneous velocity
• (Δx) → small displacement
• (Δt) → small time interval

Calculus Form

v = dx / dt

Meaning

Velocity = rate of change of position with time.

“...the quantity on the right hand side is the differential coefficient of x with respect to t...”

Easy Meaning

In calculus:

dx / dt

means how quickly position changes with time.

“It is the rate of change of position with respect to time...”

Important Point

• Velocity tells:
• how fast position changes
• and in which direction

Finding Velocity Graphically

“We can use Eq. (2.1a) for obtaining the value of velocity graphically or numerically.”

Two Methods

1. Graph method
2. Numerical/table method

“Suppose we want to obtain graphically the value of velocity at t = 4 s...”

Meaning

We want exact velocity at 4 seconds.

“Let us take Δt = 2 s centered at t = 4 s.”

Meaning

Choose time interval around 4 s:

• from 3 s to 5 s

Average Velocity from Graph

Slope of Line

Slope = Δx / Δt

Meaning

Slope of position-time graph gives velocity.

“The slope of line P₁P₂ gives average velocity over interval 3 s to 5 s.”

Easy Meaning

• Draw line between two points.
• Measure rise/change in position.
• Divide by time interval.

“Now, we decrease the value of Δt from 2 s to 1 s.”

Meaning

Take smaller intervals:

• 3.5 to 4.5
• 3.75 to 4.25
• etc.

“In the limit Δt → 0, the line P₁P₂ becomes tangent to the curve...”

Very Important Concept

• When interval becomes extremely small:
• secant line changes into tangent.

Tangent Slope = Instantaneous Velocity

Tangent Concept

“Velocity at t = 4 s is given by the slope of tangent at point P.”

Final Conclusion

Instantaneous velocity at a point = slope of tangent to position-time graph at that point.

Numerical Method

“It is difficult to show this process graphically...”

Meaning

Exact tangent drawing is hard. So tables and calculations are easier.

“Table 2.1 gives values of Δx/Δt...”

What Table Shows

As:

Δt → 0

The value of:

Δx / Δt

approaches a fixed number.

Observation from Table

Δt gets smaller	Velocity value
2.0	3.92
1.0	3.86
0.5	3.845
0.1	3.8402
0.01	3.8400

Final Instantaneous Velocity

v ≈ 3.84 m/s

Important Concepts Summary

Average Velocity

v_avg = Δx / Δt

• For large interval

Instantaneous Velocity

v = dx / dt

• Velocity at exact instant

Graph Rule

Position-Time Graph

• Slope of secant → average velocity
• Slope of tangent → instantaneous velocity

Easy Real-Life Examples

Situation	Type
Average speed of trip	Average velocity
Car speedometer reading	Instantaneous velocity
Straight line slope	Velocity from graph

One-Line Summary: Instantaneous velocity is the velocity of an object at one exact moment and is equal to the slope of the tangent to the position-time graph at that instant.

Learn Instantaneous Velocity and Speed Class 11 Physics with formulas, graphs, examples, numerical methods, and NCERT explanations. 

Internal Links

Add these naturally throughout the article:

Motion & Kinematics

Motion in a Straight Line Notes

Distance and Displacement Explained

Speed vs Velocity Difference

Average Velocity Formula and Examples

Acceleration in One Dimension

Position-Time Graph Explained

Velocity-Time Graph Explained

Equations of Motion Class 11

Calculus & Concepts

Introduction to Differentiation in Physics

Graphical Interpretation of Motion

Slope of a Graph in Physics

Numerical Methods in Kinematics

Exam Preparation

Class 11 Physics Important Questions

NEET Kinematics Questions

JEE Motion in a Straight Line Problems

NCERT Solutions Motion in a Straight Line

Previous Year Physics Questions 

Instantaneous velocity is equal to the slope of the tangent to the position-time graph at a given instant.

Calorie in New Unit System | Class 11 Physics Dimensional Analysis Solution

 NCERT Class 11 Physics: Calorie Conversion Using α β γ Unit System

-  Dr.Sanjaykumar Pawar 

Step-by-step dimensional analysis showing how calorie is converted into a new unit system using α, β, and γ base units.

🔗  Internal Links

/class-11-physics-units-and-dimensions

/dimensional-analysis-notes

/energy-and-work-physics-notes

/ncert-class-11-physics-solutions

/neet-physics-important-questions

/jee-main-physics-unit-conversion

/mcq-on-units-and-dimensions

/previous-year-questions-physics-class-11

Class 11 Physics (CBSE)

NCERT Exercise 1.3

Question

A calorie is a unit of heat (energy in transit) and it equals about 4.2 J, where

1 J = 1 kg m² s⁻²

Suppose we employ a system of units in which the unit of mass equals α kg, the unit of length equals β m, and the unit of time equals γ s. Show that a calorie has a magnitude 4.2 α⁻¹ β⁻² γ² in terms of the new units.

Topper's Answer

Given:

1 calorie = 4.2 J

1 J = 1 kg m² s⁻²

New units are:

• Unit of mass = α kg
• Unit of length = β m
• Unit of time = γ s

Step 1: Dimensions of Energy

The dimensional formula of energy is:

[M L² T⁻²]

Therefore, one new unit of energy is:

(α kg)(β m)²(γ s)⁻²

= αβ²γ⁻² kg m² s⁻²

= αβ²γ⁻² J

Hence,

1 New Energy Unit = αβ²γ⁻² J

Step 2: Express 1 Joule in New Units

1 J = α⁻¹β⁻²γ² (New Energy Unit)

Step 3: Express 1 Calorie in New Units

1 calorie = 4.2 J

= 4.2(α⁻¹β⁻²γ²)(New Energy Unit)

Therefore,

1 calorie = 4.2 α⁻¹ β⁻² γ²

Hence Proved.

Short Notes (Exam Revision)

• Energy has dimensions [ML²T⁻²].
• New energy unit = αβ²γ⁻² J.
• 1 J = α⁻¹β⁻²γ² (new energy units).
• 1 calorie = 4.2 J.
• Therefore, 1 calorie = 4.2 α⁻¹ β⁻² γ² in the new system of units.

Final Answer: 4.2 α⁻¹ β⁻² γ²

Class 11 Physics

Unit Conversion Practice Questions (Easy to High Level)

Q1 (Easy)

In a system, unit of mass = α kg, unit of length = β m, unit of time = γ s. Find the magnitude of 1 Joule in the new system.

1 J = kg m² s⁻²

Answer:
New unit of energy = αβ²γ⁻² J
So,
1 J = α⁻¹β⁻²γ²

Q2 (Easy)

Find 1 Newton in a system where base units are α kg, β m, γ s.

1 N = kg m s⁻²

Answer:
New form = αβγ⁻² N unit
So,
1 N = α⁻¹β⁻¹γ²

Q3 (Moderate)

Find the magnitude of 1 Pascal in the new system.

1 Pa = kg m⁻¹ s⁻²

Answer:
Substitute new units:
= αβ⁻¹γ⁻²

So,
1 Pa = α⁻¹β⁻³γ²

Q4 (Moderate)

If mass = 2 kg, length = 3 m, time = 5 s, find 1 Joule in new system.

1 J = kg m² s⁻²

Answer:
= 2⁻¹ × 3⁻² × 5²
= 25 / (2 × 9)
= 25/18

Q5 (High)

Show that gravitational constant G has magnitude α⁻¹β³γ⁻² in new system.

G = kg⁻¹ m³ s⁻²

Answer:
Substitute units:
= α⁻¹β³γ⁻²

Q6 (High Concept)

If mass = 4 kg, length = 2 m, time = 1 s, find ratio of new energy unit to Joule.

Energy = M L² T⁻²

Answer:
= 4 × 2² × 1⁻²
= 4 × 4 = 16

Final Answer: 16