Physics Teacher

Question

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

Answer 1

Step 1: Take a pencil or a cylindrical rod.

Step 2: Wind the thread closely around the pencil without leaving any gap.

Step 3: Make 20–50 turns of the thread.

Step 4: Measure the total length (L) of all turns using the metre scale.

Step 5: Count the total number of turns (N).

Diameter of Thread = Total Length (L) / Number of Turns (N)

Answer: Divide the total length of all turns by the number of turns to obtain the diameter of the thread.

Answer 2

Step 1: Use the formula

Least Count = Pitch / Circular Scale Divisions

Step 2:

Least Count = 1.0 / 200 = 0.005 mm

Step 3: Increasing the number of divisions decreases the least count.

Step 4: However, accuracy cannot increase forever because of:

Human error
Instrument error
Manufacturing limitations
Backlash error

Final Answer:

No. Accuracy cannot be increased arbitrarily by increasing the number of divisions.

Answer 3

Step 1: Every measurement contains small random errors.

Step 2: With only 5 readings, random errors affect the average more.

Step 3: With 100 readings, positive and negative errors cancel each other.

Step 4: Therefore, the mean value becomes closer to the true value.

More Measurements → Less Random Error → Better Mean Value

Answer 4

(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.

Answer 5

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

Answer 6

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

Answer 7

(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

Answer 8

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.

Answer 9

(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.

Answer 10

(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

Answer 11

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|

Answer 12

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.

Answer 13

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.

Answer 14

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

Answer 15

A body can be treated as a point object if its size is negligible compared to the distance travelled.

(a) Yes. The distance between stations is much greater than the size of the carriage.
(b) Yes. The radius of the circular track is much larger than the dimensions of the monkey-man system.
(c) No. Rotation and spin are important.
(d) No. The tumbling motion depends on the size and shape of the beaker.

Final Answer: (a) and (b)

Answer 16

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

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

Answer: A lives closer to the school than B.

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

A starts at t = 0 while B starts later.

Answer: A starts earlier than B.

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

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

Answer: B walks faster than A.

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

Both graphs terminate at the same value of time.

Answer: Same time.

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

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

Answer: B overtakes A once.

Answer 17

Given:

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

Time taken to reach office:

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

She reaches the office at 9:30 am.

Time taken to return:

t = 2.5 / 25 = 0.1 hour = 6 minutes

She reaches home at 5:06 pm.

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

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

Answer 18

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

Time taken in one cycle = 8 s

After 4 cycles:

Displacement = 4 × 2 = 8 m

Time = 4 × 8 = 32 s

Then he moves forward:

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

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

Answer 19

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

Final speed, v = 0

Distance, s = 200 m

Using:

v² = u² + 2as

0 = (35)² + 2(a)(200)

a = −3.06 m s⁻²

Retardation = 3.06 m s⁻²

Using:

v = u + at

0 = 35 − 3.06t

t = 11.4 s

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

Answer 20

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

Acceleration is always vertically downward due to gravity.

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

Velocity = 0

Acceleration = 9.8 m s⁻² downward

v = 0, a = 9.8 m s⁻² downward

(c) Sign convention:

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

(d) Maximum height and total time of flight

Using:

v² = u² − 2gh

0 = (29.4)² − 2(9.8)h

h = 44.1 m

Time to rise:

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

Total time:

T = 2 × 3 = 6 s

Maximum height = 44.1 m
Total time = 6 s

Answer 21

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

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

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

False. Velocity is zero when speed is zero.

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

False. Uniform circular motion has non-zero acceleration.

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

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

Answer 22

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.

Answer 23

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

Answer 24

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

Answer 25

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

So,
1 Pa = α⁻¹β⁻³γ²

Answer 26

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

Answer 27

Answer:
Substitute units:
= α⁻¹β³γ⁻²

Answer 28

Answer:
= 4 × 2² × 1⁻²
= 4 × 4 = 16

Final Answer: 16

Answer 29

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

Answer 30

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

Answer 31

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

Final Answer: 5 m

Answer 32

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

Answer 33

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

Answer 34

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

Answer 35

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

Answer 36

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

Answer 37

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

Answer 38

500 ÷ 100 = 5 m
Answer: 5 m

Answer 39

1 km = 1000 m
Answer: 1000 m

Answer 40

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

Answer 41

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

Answer 42

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

Answer 43

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

Answer 44

1 litre = 1000 cm³
Answer: 1000 cm³

Answer 45

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

Answer 46

Answer: Newton (N)

Answer 47

Answer: Joule (J)

Answer 48

3 × 1000 = 3000 g
Answer: 3000 g

Answer 49

2500 ÷ 1000 = 2.5 kg
Answer: 2.5 kg

Answer 50

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

Answer 51

Answer: Second (s)

Answer 52

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

Answer 53

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

Answer 54

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

Answer 55

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

Answer 56

Conclusion:

In both cases, force on the wall is perpendicular (normal) to the wall.

The force is NOT inclined at 30°.

Answer 57

(i) Direction of Force on Wall

Case (a): Normal to wall, positive x-direction.
Case (b): Normal to wall, positive x-direction.

(ii) Ratio of Impulses

Ja : Jb
= 2mu : 2mu cos30°
= 1 : cos30°
= 2/√3 : 1

or

Ja/Jb = 2/√3 ≈ 1.15 ≈ 1.2

Topic	Key Point
Thread Diameter	Wind the thread around a pencil and use Length ÷ Turns.
Screw Gauge	Measures very small thickness like wire diameter.
Least Count	Least Count = Pitch ÷ Circular Divisions.
Vernier Calipers	Measures external diameter, internal diameter and depth.
Repeated Measurements	More readings reduce random errors.

Word	Meaning
My	Meter Scale
Very	Vernier Calipers
Smart	Screw Gauge

Instrument	Accuracy
Meter Scale	Low
Vernier Calipers	Medium
Screw Gauge	Highest

If Question Says	Remember
More Readings	Better Mean Value
Increase Circular Divisions	Accuracy improves only up to practical limits.
Least Count	Smaller LC → Better Precision

Statement	Comparison Given?	Correct / Incorrect
Atom is small	❌ No	Incorrect
Atom is smaller than a grain of sand	✅ Yes	Correct
Jet is fast	❌ No	Incorrect
Jet is faster than a car	✅ Yes	Correct
Proton is heavier than electron	✅ Yes	Correct
Speed of sound is less than speed of light	✅ Yes	Correct

Quantity	Compared With	Final Statement
Atom	Everyday objects	Very small
Jet Plane	Car	Very fast
Jupiter	Earth	Very massive
Air Molecules	Humans	Very large number
Proton	Electron	Much heavier
Sound	Light	Much slower

Monday, July 6, 2026

UNITS AND MEASUREMENTS

Question (a)

Step-by-Step Answer

Question (b)

Step-by-Step Answer

Question (c)

Step-by-Step Answer

Short Notes (Quick Revision)

Mnemonics (Easy Memory Tricks)

1. Instrument Order

Meter → Vernier → Screw

2. Accuracy Order

Meter < Vernier < Screw

3. Formula Memory Trick

4. NEET MCQ Trick

Monday, June 29, 2026

Q. 1.4 Explain this statement clearly:

Answer (CBSE Topper Style)

Reframing the given statements

(a) Atoms are very small objects.

(b) A jet plane moves with great speed.

(c) The mass of Jupiter is very large.

(d) The air inside this room contains a large number of molecules.

(e) A proton is much more massive than an electron.

(f) The speed of sound is much smaller than the speed of light.

Final Answer (Exam Points)

NEET Master Learning Structure (NMLS)

हर टॉपिक के लिए एक यूनिवर्सल फॉर्मेट

1. Topic Name

2. One-Line Definition (Exam Definition)

3. Why? (Concept Building)

याद रखने वाला नियम

4. Visual Flow Chart

5. Memory Trick (Mnemonic)

Mnemonic

"LSC"

दूसरा आसान Mnemonic

"No Comparison = No Conclusion"

6. Quick Table

7. Comparison Table

8. Mind Map

9. Common Mistakes

10. Examiner's Favourite Questions

Type-1

Type-2

Type-3

Type-4

Type-5

11. NEET Trick

"Compared to What?"

12. PYQ Thinking Method

13. 10-Second Revision

14. Golden Rule (100% Exam Point)

15. One-Line Formula

NEET Practice Question Bank

Chapter: Units and Measurements

Topic: Standard of Comparison (Large & Small Quantities)

Section A: Direct MCQs (Single Correct Option)

Easy Level

Q1.

Q2.

Moderate Level

Q3.

Q4.

Hard Level

Q5.

Section B: Statement-Based Questions

Q6.

Q7.

Section C: Assertion & Reason

Q8.

Q9.

Section D: Match the Columns

Column I

Column II

Section E: Diagram-Based Question

Section F: Higher Order Thinking (HOTS)

Q10.

Q11.

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