Significant Figures NEET Notes: Rules, Tricks, Examples & Practice Questions

 - Dr.Sanjaykumar Pawar 

Significant Figures and Measurement Errors | NEET Smart Notes

NEET Smart Notes: Significant Figures Rules, Calculation Tricks and Quick Revision Guide

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NEET Smart Notes: Significant Figures & Mass Calculation

Part 1: Question and Answer

Question:

The mass of a box measured by a grocer’s balance is 2.30 kg. Two gold pieces of masses 20.15 g and 20.17 g are added to the box. Find:

(a) The total mass of the box.
(b) The difference in the masses of the pieces to correct significant figures.

Solution:

Given:

• Mass of box = 2.30 kg
• Mass of gold piece 1 = 20.15 g
• Mass of gold piece 2 = 20.17 g

Step 1: Convert kg into grams

1 kg = 1000 g

2.30 kg = 2.30 × 1000 = 2300 g

(a) Total Mass of Box

Mass of gold pieces:

20.15 g + 20.17 g = 40.32 g

Total mass:

2300 g + 40.32 g = 2340.32 g

For addition, answer should have the same decimal places as the least precise value.

2340.32 g ≈ 2340 g

Final Answer (a): 2.34 kg

(b) Difference in Masses

20.17 g - 20.15 g = 0.02 g

Final Answer (b): 0.02 g

Part 2: Short Notes

Significant Figures

Significant figures are the meaningful digits present in a measurement. They indicate the accuracy and precision of a measurement.

Example:
2.30 kg has 3 significant figures.

Rules for Addition and Subtraction

In addition and subtraction, the number of decimal places decides the final answer.

Addition/Subtraction → Check Decimal Places

Example:
20.15 + 20.17 = 40.32

Unit Conversion

Before calculation, all quantities must be converted into the same unit.

1 kg = 1000 g

Steps to Solve Questions

1. Convert all quantities into the same unit.
2. Perform addition or subtraction.
3. Apply significant figure rules.
4. Write final answer with correct unit.

Part 3: Concept Table

Concept	Meaning	Example
Measurement	Finding the value of a physical quantity	Mass = 2.30 kg
Significant Figures	Meaningful digits in measurement	2.30 has 3 SF
Addition/Subtraction Rule	Decimal places decide answer	20.15 + 20.17
Unit Conversion	Change units before calculation	1 kg = 1000 g

Part 4: Memory Mnemonics

Mnemonic 1:

"ADD-SUB → Decimal Dekho"

Addition and subtraction में हमेशा decimal places count करो.

Mnemonic 2:

"MULTI-DIV → Significant Gino"

Multiplication and division में significant figures count करो.

Mnemonic 3: Zero Rule

"Between and Behind = Important, Before = Ignore"

• Between zeros → Significant
• Decimal के बाद ending zeros → Significant
• Starting zeros → Not significant

30-Second Revision Box

SIGNIFICANT FIGURES

Addition/Subtraction
        ↓
Decimal Places

Multiplication/Division
        ↓
Significant Figures

Zero Rules:
Between digits → Significant
After decimal → Significant
Before first digit → Not Significant

1 kg = 1000 g

Convert units before calculation.

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

 - Dr.Sanjaykumar Pawar  

Rectangular Sheet Area and Volume Calculation | Significant Figures NEET Guide

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

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 Complete NEET Physics Formula Sheet

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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।

Significant Figures Rules for NEET | Short Notes, Tricks & Solved Questions

 - Dr.Sanjaykumar Pawar  

Significant Figures in Physics & Chemistry | NEET Preparation Notes 

Significant Figures Rules and Tricks for NEET: Easy Chart to Remember Counting Rules

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NEET Previous Year Questions

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Significant Figures (महत्वपूर्ण अंक) - NEET Smart Notes

Part 1: Questions and Answers

Question: State the number of significant figures in the following:

1. 0.007 m²
2. 2.64 × 10²⁴ kg
3. 0.2370 g cm⁻³
4. 6.320 J
5. 6.032 N m⁻²
6. 0.0006032 m²

Solution:

Question	Explanation	Answer
0.007 m²	Leading zeros are not significant. Only 7 is counted.	1 Significant Figure
2.64 × 10²⁴ kg	All non-zero digits are significant.	3 Significant Figures
0.2370 g cm⁻³	Leading zero ignored. 2,3,7 and final zero after decimal counted.	4 Significant Figures
6.320 J	Final zero after decimal is significant.	4 Significant Figures
6.032 N m⁻²	Zero between non-zero digits is significant.	4 Significant Figures
0.0006032 m²	Starting zeros ignored. Zero between 6 and 3 is significant.	4 Significant Figures

Part 2: Short Notes for Beginners

What are Significant Figures?

Significant figures are the meaningful digits in a measurement. They show the accuracy or precision of a measured value.

Rules of Significant Figures

Rule	Description	Example
1	All non-zero digits are significant.	245 = 3 S.F.
2	Zeros between two non-zero digits are significant.	1002 = 4 S.F.
3	Zeros before the first non-zero digit are not significant.	0.0045 = 2 S.F.
4	Trailing zeros after decimal are significant.	4.500 = 4 S.F.
5	Scientific notation: only digits before power of 10 are counted.	3.20 × 10⁵ = 3 S.F.

Part 3: Mnemonics (याद करने की ट्रिक)

N-Z-B-D Rule

N = Non-zero digits

1 to 9 are always significant.

Z = Zero in between

Zero between two non-zero digits is significant.
Example: 506 = 3 S.F.

B = Before zero

Zeros before the first non-zero digit are ignored.
Example: 0.005 = 1 S.F.

D = Decimal ending zero

Zeros at the end after decimal are significant.
Example: 2.500 = 4 S.F.

Easy Hindi Memory Line:

"शुरुआत का Zero छोड़ो, बीच का Zero पकड़ो, Decimal के बाद का Zero जोड़ो।"

One Line Revision Formula:

START ZERO ❌
MIDDLE ZERO ✅
END ZERO WITH DECIMAL ✅
NON ZERO ALWAYS ✅

NEET Question Solving Strategy

1. सबसे पहले देखें कि Decimal है या नहीं।
2. शुरुआती zeros को ignore करें।
3. Non-zero digits को हमेशा count करें।
4. दो digits के बीच वाले zeros को count करें।
5. Decimal के बाद अंतिम zeros को count करें।

Answer Explanation Format:

Given Number: 0.006030

Leading zeros ignored.

Significant digits: 6, 0, 3, 0

Final Answer = 4 Significant Figures

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

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

 INTERNAL LINKS

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

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

3. Class 11 Physics Important Formulas PDF

4. CBSE Class 11 Physics Sample Questions

5. Significant Figures Explained with Examples

6. SI Units and Derived Units Guide

7. Class 11 Physics MCQs with Answers

8. NCERT Class 11 Physics Complete Solutions

9. Measurement and Error Analysis Notes

10. CBSE Class 11 Study Material Hub

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 - 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

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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