Dimensional Formulae and Dimensional Equations

Section 1.5 – Dimensional Formulae and Dimensional Equations from Class 11 Physics (Units and Measurements)

📘 1.5 – Dimensional Formulae and Dimensional Equations

🧠 What is a Dimensional Formula?

A dimensional formula is an expression that shows how and which of the base physical quantities (Mass $[M]$, Length $[L]$, Time $[T]$, etc.) are used to represent a given physical quantity.

$$\text{Dimensional Formula} = [M^a L^b T^c A^d K^e \text{mol}^f cd^g]$$

Here,

• $a, b, c, d, e, f, g$ are integers (positive, negative, or zero)

• Each represents the power of the respective base quantity

📌 Examples of Dimensional Formulae

Physical Quantity	Dimensional Formula
Volume $V$	$[M^0 L^3 T^0]$
Speed $v$	$[M^0 L T^{-1}]$
Acceleration $a$	$[M^0 L T^{-2}]$
Density $ρ$	$[M^1 L^{-3} T^0]$
Force $F$	$[M^1 L^1 T^{-2}]$
Energy $E$	$[M^1 L^2 T^{-2}]$

🧮 What is a Dimensional Equation?

A dimensional equation is formed when a physical quantity is equated to its dimensional formula.

For example:

• Volume:          $[V] = [M^0 L^3 T^0]$

• Velocity:         $[v] = [M^0 L T^{-1}]$

• Force:           $[F] = [M L T^{-2}]$

• Density:         $[ρ] = [M L^{-3} T^0]$

🧩 Deriving Dimensional Equations from Known Relations

You can derive the dimensional equation of a quantity from the physical law or formula it follows.

📌 Example 1: Force

We know:

$$F = m \cdot a$$

Where:

• Mass $m = [M]$

• Acceleration $a = [L T^{-2}]$

$$\Rightarrow [F] = [M] \cdot [L T^{-2}] = [M L T^{-2}]$$

So the dimensional equation of force is:

$$[F] = [M L T^{-2}]$$

📌 Example 2: Density

We know:

$$ρ = \frac{m}{V}$$

Where:

• Mass $m = [M]$

• Volume $V = [L^3]$

$$\Rightarrow [ρ] = \frac{[M]}{[L^3]} = [M L^{-3}]$$

So the dimensional equation of density is:

$$[ρ] = [M L^{-3} T^0]$$

✅ Why Dimensional Equations Are Useful

• 🔎 Check dimensional consistency of physical equations

• 🧩 Derive new relations between physical quantities

• 🌐 Convert units from one system to another

• 🧪 Help identify hidden physical relations in complex formulas

⚠️ Limitations

• Dimensional analysis cannot determine constants (like $\frac{1}{2}, 2\pi$, etc.)

• It cannot distinguish between scalar and vector quantities

• It fails if a physical quantity is expressed as a sum/difference of different dimensional terms

🧾 Summary Table

Quantity	Formula	Dimensional Equation
Volume $V$	$L \times B \times H$	$[V] = [M^0 L^3 T^0]$
Velocity $v$	$\frac{d}{t}$	$[v] = [M^0 L T^{-1}]$
Force $F$	$m \cdot a$	$[F] = [M L T^{-2}]$
Pressure $P$	$\frac{F}{A}$	$[P] = [M L^{-1} T^{-2}]$
Energy $E$	$F \cdot d$	$[E] = [M L^2 T^{-2}]$
Density $ρ$	$\frac{m}{V}$	$[ρ] = [M L^{-3} T^0]$

🔚 Conclusion

Dimensional formulae and dimensional equations form the backbone of unit analysis and physical reasoning in physics. They allow us to:

• Express physical quantities independent of any unit system

• Test equations for correctness

• Develop insights into physical laws

They act like a "grammar" in the language of physics, helping ensure that all expressions and equations are physically and mathematically valid.

Dimensions of Physical Quantities

Dimensions of Physical Quantities from Section 1.4 – CBSE Class 11 Physics (Units and Measurements)

📘 1.4 – Dimensions of Physical Quantities

🔍 What are Dimensions?

The dimension of a physical quantity refers to the power (or exponent) to which a base quantity must be raised to represent that physical quantity.

For any derived quantity, we express it in terms of fundamental dimensions like:

$$\text{[Length]} = [L], \quad \text{[Mass]} = [M], \quad \text{[Time]} = [T], \quad \text{[Electric Current]} = [A], \\ \text{[Temperature]} = [K], \quad \text{[Luminous Intensity]} = [cd], \quad \text{[Amount of Substance]} = [mol]$$

These seven are the fundamental or base dimensions in physics.

🔢 Representing Dimensions

The notation [Q] refers to the dimensions of a physical quantity Q.

For example:

• If $Q = \text{Force}$, then $[Q] = [M L T^{-2}]$

• If $Q = \text{Speed}$, then $[Q] = [L T^{-1}]$

📐 Examples in Mechanics

In mechanics, all quantities can be expressed using only [M], [L], [T].

📦 Example 1: Volume

Volume = Length × Breadth × Height
Each is of dimension $[L]$, so:

$$[\text{Volume}] = [L] \cdot [L] \cdot [L] = [L^3]$$

It has:

• 3 dimensions in length

• 0 in mass $[M^0]$

• 0 in time $[T^0]$

🧲 Example 2: Force

$$\text{Force} = \text{Mass} \times \text{Acceleration}$$ $$[\text{Mass}] = [M], \quad [\text{Acceleration}] = \frac{[L]}{[T^2]} = [L T^{-2}]$$ $$[\text{Force}] = [M] \cdot [L T^{-2}] = [M L T^{-2}]$$

Force has:

• 1 dimension in mass

• 1 in length

• –2 in time

🚗 Example 3: Velocity

Velocity = Displacement / Time

$$[\text{Velocity}] = \frac{[L]}{[T]} = [L T^{-1}]$$

Same applies to:

• Initial velocity

• Final velocity

• Average velocity

• Speed

Even though magnitudes vary, their dimensions remain the same.

✅ Important Characteristics of Dimensional Representation

• Only the type or nature of quantity is described.

• It does not involve magnitude or unit.

• Different quantities with same dimensional formula are called dimensionally similar.

• It helps:

  • Check dimensional consistency of equations

  • Derive relations between quantities

  • Convert units between systems

🧮 Zero Dimensions in a Base Quantity

A quantity can have zero power in a particular base quantity, meaning it doesn't depend on it.

For example:

• Volume: $[M^0 L^3 T^0]$

• Speed: $[M^0 L^1 T^{-1}]$

🚫 Not All Physical Properties Are Dimensioned

• Quantities like refractive index, strain, angle (in radians) are dimensionless.

• Still physical, but have no associated base dimension.

🧠 Summary Table

Physical Quantity	Dimensional Formula
Speed/Velocity	$[L T^{-1}]$
Acceleration	$[L T^{-2}]$
Force	$[M L T^{-2}]$
Work/Energy	$[M L^2 T^{-2}]$
Power	$[M L^2 T^{-3}]$
Pressure	$[M L^{-1} T^{-2}]$
Density	$[M L^{-3}]$
Momentum	$[M L T^{-1}]$

🧾 Conclusion

Understanding dimensions is essential for:

• Analyzing physical relationships

• Verifying equations

• Deriving new formulas

• Performing conversions across unit systems

By using dimensional analysis, we gain powerful insight into the structure of physical laws, without the need for detailed experiments at every step.

Rules for Determining the Uncertainty in the Results of Arithmetic Calculations

Section 1.3.3 – Rules for Determining the Uncertainty in the Results of Arithmetic Calculations from CBSE Class 11 Physics Chapter 1: Units and Measurements.

📏 Section 1.3.3: Rules for Determining the Uncertainty in the Results of Arithmetic Calculations

In experimental physics, every measured quantity carries a certain degree of uncertainty. When we perform calculations using these quantities, it's essential to propagate or combine the uncertainties in a systematic way to ensure that the final result reflects the limitations of the measurements.

This section outlines the rules and reasoning for how uncertainties (or errors) should be managed during addition, subtraction, multiplication, division, and multi-step calculations.

🔢 1. Error Propagation in Multiplication and Division

When quantities are multiplied or divided, the relative errors (percentage errors) are added to estimate the total uncertainty in the result.

📌 Example:

Let’s say:

• Length $l = 16.2 \pm 0.1\ \text{cm} \Rightarrow \text{Relative error} = \frac{0.1}{16.2} \times 100 \approx 0.6\%$

• Breadth $b = 10.1 \pm 0.1\ \text{cm} \Rightarrow \text{Relative error} = \frac{0.1}{10.1} \times 100 \approx 1.0\%$

Then:

$$\text{Area} = l \times b = 16.2 \times 10.1 = 163.62\ \text{cm}^2$$ $$\text{Total relative error} = 0.6\% + 1.0\% = 1.6\%$$ $$\text{Uncertainty in area} = \frac{1.6}{100} \times 163.62 \approx 2.6\ \text{cm}^2$$

✅ Final answer:

$$A = (164 \pm 3)\ \text{cm}^2$$

➕ 2. Error Propagation in Addition and Subtraction

For addition or subtraction, the absolute uncertainties (not percentage) are combined. The result must be reported with the least number of decimal places among the given quantities.

📌 Example:

$$12.9\ \text{g} - 7.06\ \text{g} = 5.84\ \text{g}$$

But:

• 12.9 has 1 decimal place

• 7.06 has 2 decimal places

Thus, the result must be rounded to 1 decimal place:

$$\boxed{5.8\ \text{g}}$$

⚖️ 3. Dependence of Relative Error on Magnitude

The same absolute error has different impacts on measurements of different magnitudes.

📌 Example:

Measurement	Absolute Error	Relative Error
$1.02\ \text{g}$	$\pm 0.01 \text{g}$	$\frac{0.01}{1.02} \times 100 \approx 1\%$
$9.89\ \text{g}$	$\pm 0.01 \text{g}$	$\frac{0.01}{9.89} \times 100 \approx 0.1\%$

🔎 Inference: Smaller measurements suffer more from the same level of absolute uncertainty.

🧮 4. Multi-Step Calculations and Intermediate Rounding

In multi-step calculations, if we round off after every step, it can lead to compounding of errors and loss of accuracy.

🔑 Rule: Retain one extra significant figure in intermediate results than the least precise measurement. Round off only in the final step.

📌 Example:

• Reciprocal of $9.58$:

$$1/9.58 = 0.1044 \ (\text{Retain one extra digit})$$

• If we round off this as 0.104 (3 sig. figs),
  Then:

$$1/0.104 = 9.62 \ne 9.58$$

🧠 This small discrepancy emphasizes why we should avoid premature rounding and retain extra precision temporarily.

✅ Final Summary (Quick Rules):

Operation	Uncertainty Rule	Result Rule
Multiplication / Division	Add relative errors	Keep result to least sig. figs
Addition / Subtraction	Add absolute errors	Keep result to least decimal places
Multi-step Calculation	Retain extra digit in steps	Round at the end

📝 Important Note:

Exact numbers (like 2, π, etc.) in physical formulae are considered to have infinite significant figures and do not contribute to uncertainty.

Formulas for Combination of Errors (Uncertainty Propagation), which is an essential part of experimental physics and measurement analysis in CBSE Class 11 Physics (Chapter 1: Units and Measurements).

📘 Combination of Errors: A Comprehensive Note

In experimental sciences, measured quantities are often used to calculate other physical quantities through mathematical relationships (e.g., products, quotients, powers). Since each measurement comes with its own uncertainty, we need rules to combine these uncertainties to find the final uncertainty in the result.

These rules are known as error propagation formulas or combinations of errors.

⚙️ Let’s Define Some Terms First:

• $\Delta x$: Absolute error in measurement $x$

• $\Delta y$: Absolute error in measurement $y$

• $\delta x = \frac{\Delta x}{x}$: Relative error in $x$

• $\% \text{error} = \delta x \times 100$: Percentage error in $x$

🧮 Rules for Combination of Errors

📌 1. Addition and Subtraction

If:

$$Q = A + B \quad \text{or} \quad Q = A - B$$

Then:

$$\Delta Q = \Delta A + \Delta B$$

• Absolute errors are added directly.

• This is a conservative estimate to account for the worst-case uncertainty.

✅ Example:

$$A = 12.3 \pm 0.1,\quad B = 5.6 \pm 0.2 \Rightarrow Q = A + B = 17.9 \pm 0.3$$

📌 2. Multiplication and Division

If:

$$Q = A \times B \quad \text{or} \quad Q = \frac{A}{B}$$

Then:

$$\frac{\Delta Q}{Q} = \frac{\Delta A}{A} + \frac{\Delta B}{B} \quad \text{or} \quad \delta Q = \delta A + \delta B$$

• Relative errors are added.

✅ Example:

$$A = 3.4 \pm 0.1,\quad B = 2.0 \pm 0.1$$ $$Q = A \times B = 6.8,\quad \frac{\Delta Q}{Q} = \frac{0.1}{3.4} + \frac{0.1}{2.0} \approx 0.029 + 0.05 = 0.079 \Rightarrow \Delta Q \approx 6.8 \times 0.079 \approx 0.537 \Rightarrow Q = 6.8 \pm 0.54$$

📌 3. Error in Powers (Exponents)

If:

$$Q = A^n$$

Then:

$$\frac{\Delta Q}{Q} = |n| \cdot \frac{\Delta A}{A} \quad \text{or} \quad \delta Q = |n| \cdot \delta A$$

• The relative error is multiplied by the absolute value of the exponent.

✅ Example:

$$A = 2.0 \pm 0.1,\quad Q = A^3 \Rightarrow Q = 8.0$$ $$\frac{\Delta Q}{Q} = 3 \cdot \frac{0.1}{2.0} = 0.15 \Rightarrow \Delta Q = 8.0 \times 0.15 = 1.2 \Rightarrow Q = 8.0 \pm 1.2$$

📌 4. Error in a Function of Multiple Powers

If:

$$Q = A^p \cdot B^q / C^r$$

Then:

$$\frac{\Delta Q}{Q} = |p| \cdot \frac{\Delta A}{A} + |q| \cdot \frac{\Delta B}{B} + |r| \cdot \frac{\Delta C}{C}$$

• Relative errors of all variables are added, each multiplied by the magnitude of its exponent.

✅ Example:

$$Q = \frac{A^2 \cdot B}{C^3} \Rightarrow \frac{\Delta Q}{Q} = 2 \cdot \frac{\Delta A}{A} + 1 \cdot \frac{\Delta B}{B} + 3 \cdot \frac{\Delta C}{C}$$

🧠 Best Practices:

1. Always carry extra digits during intermediate calculations. Round off only in the final result.

2. Use appropriate significant figures based on the least precise measured value.

3. Remember that errors should reflect the precision of the instruments used and should not be overstated or understated.

4. Exact numbers (like π, 2, etc.) are error-free and are not included in error propagation.

Rounding off the Uncertain Digits

 Section 1.3.2: Rounding off the Uncertain Digits, from CBSE Class 11 Physics Chapter 1: Units and Measurements:

📘 Rounding Off Uncertain Digits: Precision and Convention

In experimental physics and calculations involving measured quantities, significant figures express the precision of measurements. Often, results derived from measurements include uncertain digits—values that go beyond the precision of the original data. To preserve meaningful accuracy and avoid misleading results, it is essential to round off such numbers correctly.

🔢 General Rule of Rounding

Given a number, when rounding to a required number of significant figures:

• If the digit to be dropped (i.e., the first insignificant digit) is less than 5, the preceding digit remains unchanged.

  • Example: $1.743 \rightarrow 1.74$ (dropping 3)

• If the digit to be dropped is more than 5, the preceding digit is increased by 1.

  • Example: $2.746 \rightarrow 2.75$ (dropping 6)

🔁 Special Case: Digit to Be Dropped = 5

When the digit to be dropped is exactly 5, the rule depends on the digit preceding the 5:

• If the preceding digit is even, leave it unchanged.

  • Example: $2.745 \rightarrow 2.74$

• If the preceding digit is odd, raise it by 1.

  • Example: $2.735 \rightarrow 2.74$

This approach is called "round to even" or "bankers' rounding", and it helps reduce rounding bias over large datasets.

🧮 Multi-step Calculations: Rule of Thumb

In complex or multi-step calculations:

• Do not round off at intermediate steps.

• Instead, carry one extra digit than required.

• Perform rounding only at the final step to maintain overall accuracy.

Example:
If you need 3 significant figures in the final result, use 4 in the intermediate steps.

⚖️ Examples

🔷 Example 1.1: Surface Area and Volume of a Cube

Given: Side of cube = $7.203\ \text{m}$ (4 significant figures)

• Surface Area =

  $$6 \times (7.203)^2 = 6 \times 51.849 \approx 311.299\ \text{m}^2 \Rightarrow \boxed{311.3\ \text{m}^2}$$

• Volume =

  $$(7.203)^3 = 373.7147\ \text{m}^3 \Rightarrow \boxed{373.7\ \text{m}^3}$$

Result is rounded to 4 significant figures, matching the original measurement.

🔷 Example 1.2: Density of a Substance

Given:

• Mass = $5.74\ \text{g}$ (3 significant figures)

• Volume = $1.2\ \text{cm}^3$ (2 significant figures)

$$\text{Density} = \frac{5.74}{1.2} = 4.783\ \text{g/cm}^3 \Rightarrow \boxed{4.8\ \text{g/cm}^3}$$

Since volume has only 2 significant figures, the final answer must be limited to 2 significant figures.

🔁 Use of Constants and Infinite Precision

• Constants like $2$, $\pi$, or $10$ (used in formulas) are considered exact numbers, having infinite significant figures.

• For practical calculations:

  • Use $\pi \approx 3.14$ (2 sig. figs) or $\pi \approx 3.142$ (4 sig. figs) as required.

  • Speed of light $c = 2.99792458 \times 10^8\ \text{m/s} \approx 3.00 \times 10^8\ \text{m/s}$ for simplicity.

✅ Conclusion

• Rounding ensures consistency with the precision of measured quantities.

• Always round the final answer, not intermediates.

• Apply standard rounding rules, especially for digits like 5, using the even–odd rule to prevent cumulative bias.

This approach ensures that results are scientifically meaningful and not overstated in precision.

Worksheet on Rounding Off Uncertain Digits based on CBSE Class 11 Physics, Chapter 1 – Units and Measurement (Section 1.3.2):

🔬 Worksheet: Rounding Off Uncertain Digits

Subject: Physics | Class: 11 | Chapter: Units and Measurements
Topic: Rounding Off Uncertain Digits | Time: 25 mins
Name: _____________________   Date: ____________
 

🧠 Part A: Conceptual Understanding

Q1. State the rule for rounding off when the digit to be dropped is:
a) Less than 5 → __________________________
b) Greater than 5 → _______________________
c) Equal to 5, and the preceding digit is even → _____________________
d) Equal to 5, and the preceding digit is odd → ______________________

Q2. Why should rounding be done only at the final step in a multi-step calculation?

🧮 Part B: Rounding Practice

Round the following to 3 significant figures:

Q3. $2.748$ → __________
Q4. $5.646$ → __________
Q5. $0.03567$ → __________
Q6. $19.4505$ → __________
Q7. $7.500$ → __________
Q8. $4.735$ → __________

🔢 Part C: Application-Based Problems

Q9. The radius of a sphere is measured as $6.378\ \text{m}$. Calculate:
a) Surface Area $A = 4\pi r^2$ → _______________
b) Volume $V = \frac{4}{3}\pi r^3$ → _______________
(Use $\pi = 3.14$; give your answers to 4 significant figures.)

Q10. A student measured the mass of a metal block to be $8.527\ \text{g}$ and its volume as $2.1\ \text{cm}^3$. Calculate the density, expressing the final answer with correct significant figures.

$$\text{Density} = \frac{\text{Mass}}{\text{Volume}} = ?$$

Answer: ___________________________

Q11. If the speed of light is $2.99792458 \times 10^8\ \text{m/s}$, express it rounded off to:
a) 3 significant figures → ________________
b) 2 significant figures → ________________

✍️ Part D: True or False

Q12. ( ) Trailing zeros in a number like 4.500 are not significant.
Q13. ( ) The digit zero placed before the decimal point in 0.0487 is not significant.
Q14. ( ) The number 5.745 rounded to 3 significant figures is 5.74.
Q15. ( ) $\pi$ is a constant with infinite significant figures.

Worksheet on Rounding Off Uncertain Digits (Class 11 Physics – CBSE Chapter 1: Units and Measurement, Section 1.3.2):

✅ Answer Key

🧠 Part A: Conceptual Understanding

Q1.
a) Preceding digit remains unchanged
b) Preceding digit is raised by 1
c) Preceding digit remains unchanged
d) Preceding digit is raised by 1

Q2.
Because rounding at intermediate steps may lead to accumulated rounding errors. Keeping one extra digit during intermediate steps maintains precision, and final rounding ensures the correct number of significant figures in the final result.

🧮 Part B: Rounding Practice

To 3 significant figures:

Q3. $2.748$ → 2.75
Q4. $5.646$ → 5.65
Q5. $0.03567$ → 0.0357
Q6. $19.4505$ → 19.5
Q7. $7.500$ → 7.50
Q8. $4.735$ → 4.74

🔢 Part C: Application-Based Problems

Q9. Given:
$r = 6.378\ \text{m}$, $\pi = 3.14$

a)

$$A = 4\pi r^2 = 4 \times 3.14 \times (6.378)^2 \approx 4 \times 3.14 \times 40.684 \approx 510.99\ \text{m}^2 \Rightarrow \boxed{511.0\ \text{m}^2}$$

(because 4 significant figures)

b)

$$V = \frac{4}{3}\pi r^3 \approx \frac{4}{3} \times 3.14 \times (6.378)^3 \approx \frac{4}{3} \times 3.14 \times 259.2 \approx 1085.35 \Rightarrow \boxed{1085\ \text{m}^3}$$

(Rounded to 4 significant figures)

Q10.

$$\text{Density} = \frac{8.527\ \text{g}}{2.1\ \text{cm}^3} = 4.060\ \text{g/cm}^3$$

Volume has 2 significant figures, so final answer = $\boxed{4.1\ \text{g/cm}^3}$

Q11.
a) $2.99792458 \times 10^8$ → $\boxed{3.00 \times 10^8\ \text{m/s}}$
b) → $\boxed{3.0 \times 10^8\ \text{m/s}}$

✍️ Part D: True or False

Q12. ❌ False (Trailing zeros in 4.500 are significant because of the decimal point)
Q13. ✅ True
Q14. ✅ True (5.745 → 5.74; digit before 5 is even)
Q15. ✅ True (π is an irrational number with infinite significant digits)

Arithmetic Operations with Significant Figures

📘 Comprehensive Note: 

In physics and other sciences, we often perform arithmetic operations (like addition, subtraction, multiplication, and division) on measured values. These measured values contain uncertainties, and therefore the resulting quantity should not have greater precision than the least precise measurement used in the calculation.

To maintain scientific accuracy, we follow specific rules when performing arithmetic operations with significant figures (sig. figs).

✴️ General Principle

The result of any arithmetic operation (addition, subtraction, multiplication, or division) must reflect the precision of the least precise value used in the calculation.

🔢 Rule 1: Multiplication and Division

Rule:

In multiplication or division, the final result should retain as many significant figures as the quantity with the least number of significant figures.

✅ Example: Density Calculation

Measured:

• Mass: $4.237 \, \text{g}$ (4 sig. figs)

• Volume: $2.51 \, \text{cm}^3$ (3 sig. figs)

Raw calculation:

$$\text{Density} = \frac{4.237}{2.51} = 1.68804780876 \, \text{g/cm}^3$$

Applying rule:

• Least sig. figs = 3 → Round result to 3 significant figures.

Final Answer:

$$\text{Density} = \boxed{1.69 \, \text{g/cm}^3}$$

✅ Example: Light Year Calculation

Given:

• Speed of light: $3.00 \times 10^8 \, \text{m/s}$ (3 sig. figs)

• 1 year = $3.1557 \times 10^7 \, \text{s}$ (5 sig. figs)

$$\text{Light year} = (3.00 \times 10^8) \times (3.1557 \times 10^7) = 9.4671 \times 10^{15} \, \text{m}$$

Apply rule (least sig. figs = 3):

Final Answer:

$$\text{Light year} = \boxed{9.47 \times 10^{15} \, \text{m}}$$

➕ Rule 2: Addition and Subtraction

Rule:

In addition or subtraction, the final result should retain as many decimal places as the quantity with the least number of decimal places.

✅ Example: Mass Addition

Measured:

• $436.32 \, \text{g}$ (2 decimal places)

• $227.2 \, \text{g}$ (1 decimal place)

• $0.301 \, \text{g}$ (3 decimal places)

Raw sum:

$$\text{Total mass} = 436.32 + 227.2 + 0.301 = 663.821 \, \text{g}$$

Apply rule (least decimal places = 1):

Final Answer:

$$\text{Total mass} = \boxed{663.8 \, \text{g}}$$

✅ Example: Length Subtraction

Measured:

• $0.307 \, \text{m}$ (3 decimal places)

• $0.304 \, \text{m}$ (3 decimal places)

Raw difference:

$$\Delta L = 0.307 - 0.304 = 0.003 \, \text{m}$$

No change in decimal places; answer has 3 decimal places.

Final Answer:

$$\Delta L = \boxed{3 \times 10^{-3} \, \text{m}}$$

❌ Incorrect to write: $3.00 \times 10^{-3} \, \text{m}$ (suggests more precision than original data)

🚫 Common Mistake

Do not apply multiplication/division rules (significant figures) to addition/subtraction problems.
Use decimal place rules instead.

🧠 Summary Table

Operation	Rule	What to Count
Multiplication / Division	Result must have same number of significant figures as the input with least sig. figs	Count total sig. figs
Addition / Subtraction	Result must have same number of decimal places as the input with least decimal places	Count decimal places

Worksheet

Class 11 – Physics (CBSE)
Chapter 1: Units & Measurement — Section 1.2: Significant Figures & Arithmetic Operations

Instructions

1. Show all working.

2. Write the final answer with the correct number of significant figures or decimal places, as required by the rules you have learned.

3. Calculator use is allowed, but round only at the final step for each part.

4. Do not write the answer key on this sheet.

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Part A Counting Significant Figures

State the number of significant figures in each of the following measurements.

1. $0.08040\ \text{kg}$

2. $2.307 \times 10^{3}\ \text{m}$

3. $6.0700\ \text{cm}$

4. $5200\ \text{N}$

5. $4.700 \times 10^{-2}\ \text{mol}$

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Part B Multiplication & Division

Give each result with the correct significant‐figure limit.

6. Mass $=\;3.42\ \text{g}$; Volume $=\;1.1\ \text{cm}^{3}$.
   $\displaystyle \rho = \frac{m}{V}$

7. Charge $q = 1.602 \times 10^{-19}\ \text{C}$; Time $t = 0.235\ \text{s}$.
   $\displaystyle I = \frac{q}{t}$

8. Area $A = 2.45\ \text{m}^{2}$; Pressure $P = 1.01325 \times 10^{5}\ \text{Pa}$.
   $\displaystyle F = P \, A$

9. Frequency $f = 3.0 \times 10^{2}\ \text{Hz}$; Period $T = 1/f$.

10. Wavelength $\lambda = 6.626\ \text{nm}$; Convert $\lambda$ to metres. (Remember: $1\ \text{nm}=10^{-9}\ \text{m}$.)

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Part C Addition & Subtraction

Express each result with the correct decimal‐place limit.

11. $0.073\ \text{m} + 2.1\ \text{m} + 0.0085\ \text{m}$

12. $438.6\ \text{g} - 37.18\ \text{g}$

13. $(6.021\ \text{cm}) + (0.31\ \text{cm}) - (4.263\ \text{cm})$

14. Temperature rise: $36.582^{\circ}\text{C}$ → $98.6^{\circ}\text{C}$.
    Calculate $\Delta T$.

15. Potential difference:
    $12.0\ \text{V} + 0.354\ \text{V} + 0.007\ \text{V}$.

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Part D Mixed Operations

Apply each rule where appropriate (do multiplication/division first, then addition/subtraction).

16. $\displaystyle \Big(\tfrac{4.37\ \text{m}}{1.25\ \text{s}}\Big) + 2.1\ \text{m s}^{-1}$

17. $\displaystyle (6.022 \times 10^{23}) \times (0.0023) - 1.5 \times 10^{21}$

18. A block’s density is $\rho = 2.65\ \text{g cm}^{-3}$.
    Its measured volume is $12.3\ \text{cm}^{3}$.
    (a) Find the mass in grams.
    (b) Convert that mass to kilograms. (Give each answer with proper sig. figs.)

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Part E Concept Questions

Write T (true) or F (false). Each is worth 1 mark.

19. Trailing zeros in 3.050 m are not significant.

20. In $7.00 \times 10^{3}$, the power of ten affects the count of significant figures.

21. While adding, the number of significant figures of the result equals the least significant figure count of the addends.

22. $2$ in the formula $r = d/2$ has infinite significant figures.

23. Converting $1500\ \text{m}$ to $1.500 \times 10^{3}\ \text{m}$ changes its number of significant figures.

Here’s the Answer Key for the Worksheet on Significant Figures & Arithmetic Operations from CBSE Class 11 Physics – Chapter 1:

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✅ Part A – Counting Significant Figures

1. $0.08040\ \text{kg}$ → 4 sig. figs

2. $2.307 \times 10^3\ \text{m}$ → 4 sig. figs

3. $6.0700\ \text{cm}$ → 5 sig. figs

4. $5200\ \text{N}$ → 2 sig. figs (no decimal point)

5. $4.700 \times 10^{-2}\ \text{mol}$ → 4 sig. figs

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✅ Part B – Multiplication & Division

6. $\frac{3.42}{1.1} = 3.109$ → 3.1 g/cm³ (2 sig. figs)

7. $\frac{1.602 \times 10^{-19}}{0.235} = 6.82 \times 10^{-19}\ \text{A}$ → 6.82 × 10⁻¹⁹ A (3 sig. figs)

8. $1.01325 \times 10^5 \times 2.45 = 2.4824625 \times 10^5\ \text{N}$ → 2.48 × 10⁵ N (3 sig. figs)

9. $T = \frac{1}{3.0 \times 10^2} = 3.33 \times 10^{-3}\ \text{s}$ → 3.3 × 10⁻³ s (2 sig. figs)

10. $6.626\ \text{nm} = 6.626 \times 10^{-9}\ \text{m}$ → 6.626 × 10⁻⁹ m (4 sig. figs)

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✅ Part C – Addition & Subtraction

11. $0.073 + 2.1 + 0.0085 = 2.1815$ → 2.2 m (least 1 decimal place)

12. $438.6 - 37.18 = 401.42$ → 401.4 g (1 decimal place)

13. $6.021 + 0.31 - 4.263 = 2.068$ → 2.07 cm (2 decimal places)

14. $98.6 - 36.582 = 62.018$ → 62.0 °C (1 decimal place)

15. $12.0 + 0.354 + 0.007 = 12.361$ → 12.4 V (1 decimal place)

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✅ Part D – Mixed Operations

16. $\frac{4.37}{1.25} = 3.496$, then $+2.1 = 5.596$ → 5.6 m/s (2 sig. figs)

17. $(6.022 \times 10^{23}) \times 0.0023 = 1.384 \times 10^{21}$,
     Then $- 1.5 \times 10^{21} = -0.116 \times 10^{21} = -1.2 \times 10^{20}$ (2 sig. figs)

(a) $2.65 \times 12.3 = 32.595\ \text{g}$ → 32.6 g (3 sig. figs)
(b) $32.6\ \text{g} = 0.0326\ \text{kg}$ → 0.0326 kg (3 sig. figs)

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✅ Part E – Concept Questions

19. F (Trailing zeros in decimal numbers are significant)

20. F (Power of 10 does not affect sig. figs)

21. F (Use decimal places in addition/subtraction, not sig. figs)

22. T (2 is an exact number, infinite sig. figs)

23. T (Scientific notation defines sig. figs explicitly)