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Saturday, 7 June 2025

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: 4.237g4.237 \, \text{g} (4 sig. figs)

  • Volume: 2.51cm32.51 \, \text{cm}^3 (3 sig. figs)

Raw calculation:

Density=4.2372.51=1.68804780876g/cm3\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:

Density=1.69g/cm3\text{Density} = \boxed{1.69 \, \text{g/cm}^3}

Example: Light Year Calculation

Given:

  • Speed of light: 3.00×108m/s3.00 \times 10^8 \, \text{m/s} (3 sig. figs)

  • 1 year = 3.1557×107s3.1557 \times 10^7 \, \text{s} (5 sig. figs)

Light year=(3.00×108)×(3.1557×107)=9.4671×1015m\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:

Light year=9.47×1015m\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:

  • 436.32g436.32 \, \text{g} (2 decimal places)

  • 227.2g227.2 \, \text{g} (1 decimal place)

  • 0.301g0.301 \, \text{g} (3 decimal places)

Raw sum:

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

Apply rule (least decimal places = 1):

Final Answer:

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

Example: Length Subtraction

Measured:

  • 0.307m0.307 \, \text{m} (3 decimal places)

  • 0.304m0.304 \, \text{m} (3 decimal places)

Raw difference:

ΔL=0.3070.304=0.003m\Delta L = 0.307 - 0.304 = 0.003 \, \text{m}

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

Final Answer:

ΔL=3×103m\Delta L = \boxed{3 \times 10^{-3} \, \text{m}}

Incorrect to write: 3.00×103m3.00 \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.


Part A Counting Significant Figures

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

  1. 0.08040 kg0.08040\ \text{kg}

  2. 2.307×103 m2.307 \times 10^{3}\ \text{m}

  3. 6.0700 cm6.0700\ \text{cm}

  4. 5200 N5200\ \text{N}

  5. 4.700×102 mol4.700 \times 10^{-2}\ \text{mol}


Part B Multiplication & Division

Give each result with the correct significant‐figure limit.

  1. Mass =  3.42 g=\;3.42\ \text{g}; Volume =  1.1 cm3=\;1.1\ \text{cm}^{3}.
    ρ=mV\displaystyle \rho = \frac{m}{V}

  2. Charge q=1.602×1019 Cq = 1.602 \times 10^{-19}\ \text{C}; Time t=0.235 st = 0.235\ \text{s}.
    I=qt\displaystyle I = \frac{q}{t}

  3. Area A=2.45 m2A = 2.45\ \text{m}^{2}; Pressure P=1.01325×105 PaP = 1.01325 \times 10^{5}\ \text{Pa}.
    F=PA\displaystyle F = P \, A

  4. Frequency f=3.0×102 Hzf = 3.0 \times 10^{2}\ \text{Hz}; Period T=1/fT = 1/f.

  5. Wavelength λ=6.626 nm\lambda = 6.626\ \text{nm}; Convert λ\lambda to metres. (Remember: 1 nm=109 m1\ \text{nm}=10^{-9}\ \text{m}.)


Part C Addition & Subtraction

Express each result with the correct decimal‐place limit.

  1. 0.073 m+2.1 m+0.0085 m0.073\ \text{m} + 2.1\ \text{m} + 0.0085\ \text{m}

  2. 438.6 g37.18 g438.6\ \text{g} - 37.18\ \text{g}

  3. (6.021 cm)+(0.31 cm)(4.263 cm)(6.021\ \text{cm}) + (0.31\ \text{cm}) - (4.263\ \text{cm})

  4. Temperature rise: 36.582C36.582^{\circ}\text{C}98.6C98.6^{\circ}\text{C}.
    Calculate ΔT\Delta T.

  5. Potential difference:
    12.0 V+0.354 V+0.007 V12.0\ \text{V} + 0.354\ \text{V} + 0.007\ \text{V}.


Part D Mixed Operations

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

  1. (4.37 m1.25 s)+2.1 m s1\displaystyle \Big(\tfrac{4.37\ \text{m}}{1.25\ \text{s}}\Big) + 2.1\ \text{m s}^{-1}

  2. (6.022×1023)×(0.0023)1.5×1021\displaystyle (6.022 \times 10^{23}) \times (0.0023) - 1.5 \times 10^{21}

  3. A block’s density is ρ=2.65 g cm3\rho = 2.65\ \text{g cm}^{-3}.
    Its measured volume is 12.3 cm312.3\ \text{cm}^{3}.
    (a) Find the mass in grams.
    (b) Convert that mass to kilograms. (Give each answer with proper sig. figs.)


Part E Concept Questions

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

  1. Trailing zeros in 3.050 m are not significant.

  2. In 7.00×1037.00 \times 10^{3}, the power of ten affects the count of significant figures.

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

  4. 22 in the formula r=d/2r = d/2 has infinite significant figures.

  5. Converting 1500 m1500\ \text{m} to 1.500×103 m1.500 \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:


Part A – Counting Significant Figures

  1. 0.08040 kg0.08040\ \text{kg}4 sig. figs

  2. 2.307×103 m2.307 \times 10^3\ \text{m}4 sig. figs

  3. 6.0700 cm6.0700\ \text{cm}5 sig. figs

  4. 5200 N5200\ \text{N}2 sig. figs (no decimal point)

  5. 4.700×102 mol4.700 \times 10^{-2}\ \text{mol}4 sig. figs


Part B – Multiplication & Division

  1. 3.421.1=3.109\frac{3.42}{1.1} = 3.1093.1 g/cm³ (2 sig. figs)

  2. 1.602×10190.235=6.82×1019 A\frac{1.602 \times 10^{-19}}{0.235} = 6.82 \times 10^{-19}\ \text{A}6.82 × 10⁻¹⁹ A (3 sig. figs)

  3. 1.01325×105×2.45=2.4824625×105 N1.01325 \times 10^5 \times 2.45 = 2.4824625 \times 10^5\ \text{N}2.48 × 10⁵ N (3 sig. figs)

  4. T=13.0×102=3.33×103 sT = \frac{1}{3.0 \times 10^2} = 3.33 \times 10^{-3}\ \text{s}3.3 × 10⁻³ s (2 sig. figs)

  5. 6.626 nm=6.626×109 m6.626\ \text{nm} = 6.626 \times 10^{-9}\ \text{m}6.626 × 10⁻⁹ m (4 sig. figs)


Part C – Addition & Subtraction

  1. 0.073+2.1+0.0085=2.18150.073 + 2.1 + 0.0085 = 2.18152.2 m (least 1 decimal place)

  2. 438.637.18=401.42438.6 - 37.18 = 401.42401.4 g (1 decimal place)

  3. 6.021+0.314.263=2.0686.021 + 0.31 - 4.263 = 2.0682.07 cm (2 decimal places)

  4. 98.636.582=62.01898.6 - 36.582 = 62.01862.0 °C (1 decimal place)

  5. 12.0+0.354+0.007=12.36112.0 + 0.354 + 0.007 = 12.36112.4 V (1 decimal place)


Part D – Mixed Operations

  1. 4.371.25=3.496\frac{4.37}{1.25} = 3.496, then +2.1=5.596+2.1 = 5.5965.6 m/s (2 sig. figs)

  2. (6.022×1023)×0.0023=1.384×1021(6.022 \times 10^{23}) \times 0.0023 = 1.384 \times 10^{21},
     Then 1.5×1021=0.116×1021=1.2×1020- 1.5 \times 10^{21} = -0.116 \times 10^{21} = -1.2 \times 10^{20} (2 sig. figs)

(a) 2.65×12.3=32.595 g2.65 \times 12.3 = 32.595\ \text{g}32.6 g (3 sig. figs)
(b) 32.6 g=0.0326 kg32.6\ \text{g} = 0.0326\ \text{kg}0.0326 kg (3 sig. figs)


Part E – Concept Questions

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

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

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

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

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


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