📘 Chapter 1: Physical Quantities and Measurements – Numerical Problems
Prepared by Muhammad Tayyab, Subject Specialist Physics, Govt Christian High School Daska. Based on PECTAA 2026 syllabus (National Curriculum 2023).
📖 What's Inside: This section covers numerical problems from Chapter 1 Physical Quantities and Measurements including time conversions using SI prefixes, scientific notation, addition and subtraction in scientific notation, multiplication and division in scientific notation, significant figures, prefix conversions, light year calculation, and density unit conversion. Each problem is presented with Given Data, To Find, and step-by-step Solution as per the official PECTAA 2026 Physics curriculum. Perfect for Punjab Boards (Lahore, Gujranwala, Multan, etc.) and all BISE boards across Pakistan.
📚 Related Resources – Chapter 1: Physical Quantities and Measurements
📋 Exercise Short Qs
📝 Short Qs
📖 Comprehensive Qs
🔢 Numerical Problems
✅ MCQs
✏️ Constructed Response
Physical Quantities and Measurements covers fundamental concepts of measurement, SI units, and measuring instruments.
📐 Numerical Problems & Solutions (PECTAA 2026)
1.1
Calculate the number of seconds in (a) a day (b) a week (c) a month and state your answers using SI prefixes.
Given Data
(a)\(t = 1\) day
(b)\(t = 1\) week
(c)\(t = 1\) month
To Find
Number of seconds in SI prefixes
Solution
(a) 1 day
\[
t = 1 \text{ day} = 1 \times 24 \times 60 \times 60 \text{ s} = 86400 \text{ s}
\]
\[
t = 8.64 \times 10^4 \text{ s} = 86.4 \times 10^{-1} \times 10^4 \text{ s} = 86.4 \times 10^3 \text{ s}
\]
\[
t = 86.4 \text{ ks} \quad (\because k = 10^3)
\]
(b) 1 week
\[
t = 1 \times 7 \times 24 \times 60 \times 60 \text{ s} = 604800 \text{ s}
\]
\[
t = 6.048 \times 10^5 \text{ s} = 604.8 \times 10^{-2} \times 10^5 \text{ s} = 604.8 \times 10^3 \text{ s}
\]
\[
t = 604.8 \text{ ks}
\]
(c) 1 month
\[
t = 1 \times 30 \times 24 \times 60 \times 60 \text{ s} = 2592000 \text{ s}
\]
\[
t = 2.592 \times 10^6 \text{ s} = 2.592 \text{ Ms} \quad (\because M = 10^6)
\]
✅ (a) 86.4 ks (b) 604.8 ks (c) 2.592 Ms
1.2
State the answers of problem 1.1 in scientific notation.
Given Data
(a)86.4 ks
(b)604.8 ks
(c)2.592 Ms
To Find
Answers in scientific notation
Solution
(a) 86.4 ks
\[
86.4 \text{ ks} = 86.4 \times 10^3 \text{ s} = 8.64 \times 10^1 \times 10^3 \text{ s} = 8.64 \times 10^4 \text{ s}
\]
(b) 604.8 ks
\[
604.8 \text{ ks} = 604.8 \times 10^3 \text{ s} = 6.048 \times 10^2 \times 10^3 \text{ s} = 6.048 \times 10^5 \text{ s}
\]
(c) 2.592 Ms
\[
2.592 \text{ Ms} = 2.592 \times 10^6 \text{ s}
\]
✅ (a) \(8.64 \times 10^4 \text{ s}\) (b) \(6.048 \times 10^5 \text{ s}\) (c) \(2.592 \times 10^6 \text{ s}\)
1.3
Solve the following addition or subtraction. State your answers in scientific notation. (a) \(4 \times 10^{-4} \text{ kg} + 3 \times 10^{-5} \text{ kg}\) (b) \(5.4 \times 10^{-6} \text{ m} - 3.2 \times 10^{-5} \text{ m}\)
Given Data
(a)\(4 \times 10^{-4} \text{ kg} + 3 \times 10^{-5} \text{ kg}\)
(b)\(5.4 \times 10^{-6} \text{ m} - 3.2 \times 10^{-5} \text{ m}\)
To Find
Answers in scientific notation
Solution
(a)
\[
4 \times 10^{-4} \text{ kg} + 3 \times 10^{-5} \text{ kg}
\]
\[
= 4 \times 10^{-4} \text{ kg} + 0.3 \times 10^{-4} \text{ kg}
\]
\[
= (4 + 0.3) \times 10^{-4} \text{ kg} = 4.03 \times 10^{-4} \text{ kg}
\]
(b)
\[
5.4 \times 10^{-6} \text{ m} - 3.2 \times 10^{-5} \text{ m}
\]
\[
= 0.54 \times 10^{-5} \text{ m} - 3.2 \times 10^{-5} \text{ m}
\]
\[
= (0.54 - 3.2) \times 10^{-5} \text{ m} = -2.66 \times 10^{-5} \text{ m}
\]
✅ (a) \(4.03 \times 10^{-4} \text{ kg}\) (b) \(-2.66 \times 10^{-5} \text{ m}\)
1.4
Solve the following multiplication or division. State your answers in scientific notation. (a) \((5 \times 10^4 \text{ m}) \times (3 \times 10^{-2} \text{ m})\) (b) \(\frac{6 \times 10^8 \text{ kg}}{3 \times 10^4 \text{ m}^3}\)
Given Data
(a)\((5 \times 10^4 \text{ m}) \times (3 \times 10^{-2} \text{ m})\)
(b)\(\frac{6 \times 10^8 \text{ kg}}{3 \times 10^4 \text{ m}^3}\)
To Find
Answers in scientific notation
Solution
(a)
\[
(5 \times 10^4 \text{ m}) \times (3 \times 10^{-2} \text{ m})
\]
\[
= 15 \times 10^{4-2} \text{ m}^2 = 15 \times 10^2 \text{ m}^2
\]
\[
= 1.5 \times 10^3 \text{ m}^2
\]
(b)
\[
\frac{6 \times 10^8 \text{ kg}}{3 \times 10^4 \text{ m}^3} = \frac{6}{3} \times 10^{8-4} \text{ kg m}^{-3} = 2.0 \times 10^4 \text{ kg m}^{-3}
\]
✅ (a) \(1.5 \times 10^3 \text{ m}^2\) (b) \(2.0 \times 10^4 \text{ kg m}^{-3}\)
1.5
Calculate the following and state your answer in scientific notation. \(\frac{(3 \times 10^2 \text{ kg}) \times (4.0 \text{ km})}{5 \times 10^2 \text{ s}^2}\)
Given Data
Expression\(\frac{(3 \times 10^2 \text{ kg}) \times (4.0 \text{ km})}{5 \times 10^2 \text{ s}^2}\)
To Find
Answer in scientific notation
Solution
\[
\frac{(3 \times 10^2 \text{ kg}) \times (4.0 \text{ km})}{5 \times 10^2 \text{ s}^2}
\]
\[
= \frac{(3 \times 10^2 \text{ kg}) \times (4.0 \times 10^3 \text{ m})}{5 \times 10^2 \text{ s}^2}
\]
\[
= \frac{3 \times 4.0}{5} \times 10^{2+3-2} \text{ kg m s}^{-2}
\]
\[
= 2.4 \times 10^3 \text{ kg m s}^{-2}
\]
✅ \(2.4 \times 10^3 \text{ kg m s}^{-2}\)
1.6
State the number of significant digits in each measurement. (a) \(0.0045 \text{ m}\) (b) \(2.047 \text{ m}\) (c) \(3.40 \text{ m}\) (d) \(3.420 \times 10^4 \text{ m}\)
Given Data
(a)\(0.0045 \text{ m}\)
(b)\(2.047 \text{ m}\)
(c)\(3.40 \text{ m}\)
(d)\(3.420 \times 10^4 \text{ m}\)
To Find
Number of significant digits
Solution
(a) \(0.0045 \text{ m}\) → 2 significant digits (4 and 5)
(b) \(2.047 \text{ m}\) → 4 significant digits (2, 0, 4, 7)
(c) \(3.40 \text{ m}\) → 3 significant digits (3, 4, 0)
(d) \(3.420 \times 10^4 \text{ m}\) → 4 significant digits (3, 4, 2, 0)
✅ (a) 2 (b) 4 (c) 3 (d) 4
1.7
Write in scientific notation: (a) \(0.0035 \text{ m}\) (b) \(206.4 \times 10^2 \text{ m}\)
Given Data
(a)\(0.0035 \text{ m}\)
(b)\(206.4 \times 10^2 \text{ m}\)
To Find
Answers in scientific notation
Solution
(a)
\[
0.0035 \text{ m} = 3.5 \times 10^{-3} \text{ m}
\]
(b)
\[
206.4 \times 10^2 \text{ m} = 2.064 \times 10^2 \times 10^2 \text{ m} = 2.064 \times 10^4 \text{ m}
\]
✅ (a) \(3.5 \times 10^{-3} \text{ m}\) (b) \(2.064 \times 10^4 \text{ m}\)
1.8
Write using correct prefixes: (a) \(5.0 \times 10^4 \text{ cm}\) (b) \(580 \times 10^2 \text{ g}\) (c) \(45 \times 10^{-4} \text{ s}\)
Given Data
(a)\(5.0 \times 10^4 \text{ cm}\)
(b)\(580 \times 10^2 \text{ g}\)
(c)\(45 \times 10^{-4} \text{ s}\)
To Find
Correct prefixes
Solution
(a)
\[
5.0 \times 10^4 \text{ cm} = 5.0 \times 10^4 \times 10^{-2} \text{ m} = 5.0 \times 10^2 \text{ m} = 0.5 \text{ km}
\]
(b)
\[
580 \times 10^2 \text{ g} = 58.0 \times 10^3 \text{ g} = 58.0 \text{ kg}
\]
(c)
\[
45 \times 10^{-4} \text{ s} = 4.5 \times 10^{-3} \text{ s} = 4.5 \text{ ms}
\]
✅ (a) 0.5 km (b) 58.0 kg (c) 4.5 ms
1.9
Light year is a unit of distance used in Astronomy. It is the distance covered by light in one year. Taking the speed of light as \(3.0 \times 10^8 \text{ ms}^{-1}\), calculate the distance.
Given Data
Speed of light\(c = 3.0 \times 10^8 \text{ ms}^{-1}\)
Time\(t = 1\) year
To Find
Distance covered = \(S = ?\)
Solution
\[
t = 1 \times 365 \times 24 \times 60 \times 60 \text{ s} = 31536000 \text{ s} = 3.1536 \times 10^7 \text{ s}
\]
By using formula of distance:
\[
S = vt = ct
\]
\[
S = (3.0 \times 10^8) \times (3.1536 \times 10^7)
\]
\[
S = 9.5 \times 10^{15} \text{ m}
\]
✅ \(9.5 \times 10^{15} \text{ m}\)
1.10
Express the density of mercury given as \(13.6 \text{ gcm}^{-3}\) in \(\text{kgm}^{-3}\).
Given Data
Density of mercury\(13.6 \text{ gcm}^{-3}\)
To Find
Density in \(\text{kgm}^{-3}\)
Solution
\[
13.6 \text{ gcm}^{-3} = 13.6 \times \frac{1 \text{ g}}{1 \text{ cm}^3}
\]
\[
= 13.6 \times \frac{10^{-3} \text{ kg}}{10^{-6} \text{ m}^3}
\]
\[
= 13.6 \times 10^3 \text{ kgm}^{-3}
\]
\[
= 1.36 \times 10^4 \text{ kgm}^{-3}
\]
✅ \(1.36 \times 10^4 \text{ kgm}^{-3}\)
📘 Solved Examples (PECTAA 2026)
Example 1.1
Solve the following: (a) \(5.123 \times 10^4 \text{ m} + 3.28 \times 10^5 \text{ m}\) (b) \(2.57 \times 10^{-2} \text{ mm} - 3.43 \times 10^{-3} \text{ mm}\)
Given Data
(a)\(5.123 \times 10^4 \text{ m} + 3.28 \times 10^5 \text{ m}\)
(b)\(2.57 \times 10^{-2} \text{ mm} - 3.43 \times 10^{-3} \text{ mm}\)
To Find
Answers in scientific notation
Solution
(a)
\[
5.123 \times 10^4 \text{ m} + 3.28 \times 10^5 \text{ m}
\]
\[
= 5.123 \times 10^4 \text{ m} + 32.8 \times 10^4 \text{ m}
\]
\[
= (5.123 + 32.8) \times 10^4 \text{ m}
\]
\[
= 37.923 \times 10^4 \text{ m} = 3.7923 \times 10^5 \text{ m}
\]
(b)
\[
2.57 \times 10^{-2} \text{ mm} - 3.43 \times 10^{-3} \text{ mm}
\]
\[
= 2.57 \times 10^{-2} \text{ mm} - 0.343 \times 10^{-2} \text{ mm}
\]
\[
= (2.57 - 0.343) \times 10^{-2} \text{ mm}
\]
\[
= 2.227 \times 10^{-2} \text{ mm} = 2.227 \times 10^{-5} \text{ m}
\]
✅ (a) \(3.7923 \times 10^5 \text{ m}\) (b) \(2.227 \times 10^{-5} \text{ m}\)
Example 1.2
Find the value of each of the following quantities: (a) \((4 \times 10^3 \text{ kg})(6 \times 10^6 \text{ m})\) (b) \(\frac{6 \times 10^6 \text{ m}^3}{2 \times 10^{-2} \text{ m}^2}\)
Given Data
(a)\((4 \times 10^3 \text{ kg})(6 \times 10^6 \text{ m})\)
(b)\(\frac{6 \times 10^6 \text{ m}^3}{2 \times 10^{-2} \text{ m}^2}\)
To Find
Answers in scientific notation
Solution
(a)
\[
(4 \times 10^3 \text{ kg})(6 \times 10^6 \text{ m})
\]
\[
= 24 \times 10^9 \text{ kg m} = 2.4 \times 10^{10} \text{ kg m}
\]
(b)
\[
\frac{6 \times 10^6 \text{ m}^3}{2 \times 10^{-2} \text{ m}^2}
\]
\[
= \frac{6}{2} \times 10^{6+2} \text{ m}^{3-2} = 3 \times 10^8 \text{ m}
\]
✅ (a) \(2.4 \times 10^{10} \text{ kg m}\) (b) \(3 \times 10^8 \text{ m}\)
📐 Key Formulas – Numerical Problems
Time Conversion: \( t = \text{days} \times 24 \times 60 \times 60 \text{ s} \)
Scientific Notation Addition/Subtraction: Convert to same power of 10, then add/subtract coefficients
Scientific Notation Multiplication: \( (a \times 10^m)(b \times 10^n) = (a \times b) \times 10^{m+n} \)
Scientific Notation Division: \( \frac{a \times 10^m}{b \times 10^n} = \frac{a}{b} \times 10^{m-n} \)
Distance: \( S = vt \)
Unit Conversion: \( 1 \text{ g} = 10^{-3} \text{ kg}, \quad 1 \text{ cm}^3 = 10^{-6} \text{ m}^3 \)
💡 Exam Tip:
For numerical problems, always write the given data, the required quantity, and then show all steps of the solution clearly. Pay attention to scientific notation rules and significant figures. Use SI prefixes correctly (k, M, m, μ, n, etc.). These problems follow the PECTAA 2026 pattern and are prepared by Subject Specialist Muhammad Tayyab.
Chapter 1✨ Physical Quantities and Measurements
Chapter 2Kinematics
Chapter 3Dynamics
Chapter 4Turning Effect of Forces
Chapter 5Work and Energy
Chapter 6Mechanical Properties of Matter
Chapter 7Thermal Properties of Matter
Chapter 8Magnetism
Chapter 9Nature of Science
📖 Complete syllabus coverage for Class 9 Physics (PECTAA 2026) – Chapters 1 to 9
Created by Hira Science Academy | Aligned with PECTAA 2026 Syllabus