Mastering Units, Measurement & Precision

Quantity	Unit	Symbol	Definition Basis
Length	meter	m	Light travels 299,792,458 m in 1 s
Mass	kilogram	kg	Fixed by Planck’s constant (h)
Time	second	s	9,192,631,770 cycles of Cs-133 atom
Electric Current	ampere	A	Flow of 1.6 × 10^-19 C/s
Temperature	kelvin	K	Boltzmann constant (k)
Substance Amount	mole	mol	6.022 × 10²³ entities
Light Intensity	candela	cd	Luminous efficacy of light

Supplementary Units

Radian (rad): Angle → arc length/radius
Steradian (sr): Solid angle → surface area/radius²

Pro Tip: Use SI prefixes for extremes:

Nano (10^-9): Atoms (e.g., H-atom = 0.1 nm)
Giga (10⁹): Planets (e.g., Earth mass = 5.97 × 10²⁴ kg)

Significant Figures: Precision Matters!

Rules Simplified

Non-zero digits → Always significant
Example: 123.4 g → 4 sig figs
Zeros between non-zeros → Significant
Example: 2005 s → 4 sig figs
Leading zeros → Never significant
Example: 0.0023 m → 2 sig figs
Trailing zeros:
- With decimal: Significant (50.0 kg → 3 sig figs)
- No decimal: Ambiguous (1200 J → Use 1.20 × 10³ for 3 sig figs)

Arithmetic Rules

Operation	Rule	Example
Multiplication/Division	Result has least sig figs	$\frac{4.237\, \text{g}}{2.51\, \text{cm}^3} = 1.69\, \text{g/cm}^3$
Addition/Subtraction	Result matches least decimal places	$12.11\, \text{g} + 3.1\, \text{g} = 15.2\, \text{g}$

Rounding Rules

Digit < 5 → Keep unchanged (2.34 → 2.3)
Digit ≥ 5 → Round up (2.37 → 2.4)
Exactly 5:
- Preceding digit even → Keep (2.45 → 2.4)
- Preceding digit odd → Round up (2.35 → 2.4)

Exam Hack: In multi-step calculations, keep 1 extra digit → Round only at the end!
NCERT Example: Cube side = 7.203 m → Volume = $(7.203)^3 = 373.7 \, \text{m}^3$ (4 sig figs)

Errors in Measurement: Understanding Uncertainty

No measurement is perfect! Errors quantify the uncertainty in measurements.

Types of Errors

Error Type	Cause	Example	Reduction Strategy
Systematic Errors	Consistent bias in instrument or method	Clock running fast, zero error in vernier	Calibrate instruments
Random Errors	Unpredictable fluctuations	Human reaction time, temperature changes	Take multiple measurements
Gross Errors	Mistakes or accidents	Misreading scale, recording wrong value	Careful observation

Error Quantification

Absolute Error: $\Delta a = |a_{\text{measured}} - a_{\text{true}}|$
Example: True length = 10.0 cm, Measured = 10.2 cm → Δa = 0.2 cm
Relative Error: $\frac{\Delta a}{a_{\text{true}}}$
Example: $\frac{0.2}{10.0} = 0.02$
Percentage Error: $\text{Relative Error} \times 100\%$
Example: 0.02 × 100% = 2%

Combination of Errors

When measurements with errors are combined:

Operation	Error Formula	Example
Addition/Subtraction $Z = A \pm B$	$\Delta Z = \Delta A + \Delta B$	A = 10.0 ± 0.1 cm, B = 5.0 ± 0.2 cm Z = 15.0 cm, ΔZ = 0.1 + 0.2 = 0.3 cm ∴ Z = 15.0 ± 0.3 cm
Multiplication/Division $Z = A \times B$ or $Z = A/B$	$\frac{\Delta Z}{Z} = \frac{\Delta A}{A} + \frac{\Delta B}{B}$	Mass m = 50 ± 1 g, Volume V = 10 ± 0.5 cm³ Density ρ = m/V = 5 g/cm³ Error: Δρ/ρ = (1/50) + (0.5/10) = 0.02 + 0.05 = 0.07 Δρ = 0.07 × 5 = 0.35 g/cm³ ∴ ρ = 5.0 ± 0.4 g/cm³

NCERT Practice Tip: In a rectangular sheet, length l = 16.2 ± 0.1 cm, breadth b = 10.1 ± 0.1 cm
Area A = l × b = 163.62 cm²
Relative error = ΔA/A = (0.1/16.2) + (0.1/10.1) ≈ 0.0062 + 0.0099 = 0.0161
ΔA = 0.0161 × 163.62 ≈ 2.6 cm²
∴ Report area as 164 ± 3 cm²

Order of Magnitude: Think in Powers of 10!

Express as $a × 10^b$ ( $1 \leq a < 10$ )

Order of magnitude = $10^b$ if $a \leq 5$ , else $10^{b+1}$

Quantity	Value	Order of Magnitude
Earth’s diameter	$1.28 \times 10^7 \, \text{m}$	$10^7 \, \text{m}$
Hydrogen atom diameter	$1.06 \times 10^{-10} \, \text{m}$	$10^{-10} \, \text{m}$
Human hair thickness	$\sim 10^{-4} \, \text{m}$	$10^{-4} \, \text{m}$
Gravitational Constant	$6.67 \times 10^{-11} \, \frac{\text{Nm}^2}{\text{kg}^2}$	$10^{-10} \, \frac{\text{Nm}^2}{\text{kg}^2}$

Key Insight: Earth is 17 orders larger than a H-atom: $7 - (-10) = 17$

Dimensional Analysis: Physics Superpower

Dimensions: [M], [L], [T] for mass, length, time

Quantity	Formula	Dimensional Formula
Speed	$\frac{\text{distance}}{\text{time}}$	$[L T^{-1}]$
Force	$m \times a$	$[M L T^{-2}]$
Energy	$\frac{1}{2}mv^2$	$[M L^2 T^{-2}]$
Pressure	$\frac{Force}{Area}$	$[M L^{-1} T^{-2}]$

Applications

1. Check Equation Validity (Homogeneity Principle):

All terms must have identical dimensions
Example:
- $[v^2] = [u^2] = [L^2 T^{-2}]$
- $[2as] = [L T^{-2}] [L] = [L^2 T^{-2}]$ → Valid!

“If an equation is dimensionally correct, it may be correct. But if it is dimensionally wrong, it must be wrong.“

2. Derive Relationships:

Example: Pendulum time period depends on , ,
1. Assume $T = k \cdot l^x g^y m^z$
2. Dimensional equation: $[T] = [L]^x [L T^{-2}]^y [M]^z$
3. Solve: $x + y = 0$ , $-2y = 1$ , $z = 0$ → $x = \frac{1}{2}$ , $y = -\frac{1}{2}$
4. Thus, $T = k \sqrt{\frac{l}{g}}$ → Actual $k = 2\pi$

Limitations:

Cannot find dimensionless constants (e.g., $2\pi$ , $\frac{1}{2}$ )
Cannot distinguish same-dimensional quantities (e.g., Work vs. Torque both = $[M L^2 T^{-2}]$ )

NCERT Practice Essentials

Sig Fig Application:
Problem: Mass = 5.74 g, Volume = 1.2 cm³ → Density = $\frac{5.74}{1.2} = 4.8 \, \text{g/cm}^3$ (2 sig figs)
Dimensional Check:
Problem: Is $K = \frac{1}{2}mv^2$ valid?
LHS: $[K] = [M L^2 T^{-2}]$ , RHS: $[\frac{1}{2}mv^2] = [M][L T^{-1}]^2 = [M L^2 T^{-2}]$ → Yes!
Error Calculation:
Problem: Time for 20 oscillations = 40.0 ± 0.5 s
Period T = 2.00 s
Error ΔT = 0.5/20 = 0.025 s
∴ T = 2.00 ± 0.03 s

Exercises

Q1.1 Fill in the blanks

(a) Volume = (1 cm)³ = (10^–2 m)³ = 10^–6 m³
(b) Surface Area = 2πr(r + h) = 2 × 3.1416 × 20 × 120 = 15080 mm²
(c) 18 km/h = 5 m/s ⇒ Distance in 1 s = 5 m
(d) Relative density = 11.3 ⇒ Density = 11.3 g/cm³ = 11300 kg/m³
Final Answers:
(a) 10^–6 m³
(b) 1.51 × 10⁴ mm²
(c) 5 m
(d) 11.3 g/cm³ or 11300 kg/m³

Q1.2 Fill in the blanks by suitable conversion

(a) 1 kg m² s⁻² = 10⁷ g cm² s⁻²
(b) 1 m = 1 / (9.47 × 10¹⁵) ly ≈ 1.06 × 10⁻¹⁶ ly
(c) 3.0 m/s² = 38880 km/h²
(d) G = 6.67 × 10^–8 cm³ s⁻² g⁻¹
Final Answers:
(a) 10⁷ g cm² s⁻²
(b) 1.06 × 10⁻¹⁶ ly
(c) 38880 km/h²
(d) 6.67 × 10⁻⁸ cm³ s⁻² g⁻¹

Q1.3 Calorie in new system of units

1 calorie = 4.2 J = 4.2 kg·m²/s²
$n_1u_1=n_2u_2$
$4.2 kg.m^2.s^{-2} = n_2 (\alpha kg)(\beta m)^2 (\gamma s)^{-2}$
Convert with new units: mass = α kg, length = β m, time = γ s
→ $4.2 \times \gamma^2 / (\alpha \times \beta^2)$
Final Answer: $4.2 \alpha^{-1}\beta^{-2}\gamma^2$

Q1.4 Rewriting comparative statements

Calling a dimensional quantity “large/small” needs a standard.
Revised Statements:
(a) Atoms are small compared to grains of sand.
(b) Jet plane speed is greater than cars or buses.
(c) Jupiter’s mass is much more than Earth’s.
(d) Air in the room has far more molecules than visible objects.
(e) A proton’s mass is ~1836 times that of an electron.
(f) Speed of sound (~343 m/s) is much less than speed of light (~3×10⁸ m/s).

Q1.5 Distance to Sun in new unit

Time = 8 min 20 s = 500 s
Speed of light = 1 unit
Distance = speed × time = 1 × 500 = 500 units
Final Answer: 500 new units

Q1.6 Most precise length-measuring device

(a) Vernier caliper LC ≈ 0.05 mm
(b) Screw gauge LC = 1 mm / 200 = 0.005 mm
(c) Optical instrument accuracy ≈ wavelength ≈ 0.0005 mm
Final Answer: (c) Optical instrument

Q1.7 Estimate thickness of hair

Magnified width = 3.5 mm
Magnification = 100
Real thickness = 3.5 / 100 = 0.035 mm = 35 µm
Final Answer: 35 µm

Q1.8 Estimation, Accuracy and Reliability

(a) Coil thread on ruler; measure total length, divide by turns.
(b) No. Too many divisions cause mechanical limitations; accuracy won’t increase indefinitely.
(c) More observations reduce random errors, better statistical mean.
Final Answer: (a) Coil method; (b) Not possible beyond limits; (c) 100 is more reliable.

Q1.9 Linear magnification from slide to screen

Slide image area = 1.75 cm² = 1.75 × 10⁻⁴ m²
Screen area = 1.55 m²
Magnification (area) = 1.55 / 1.75 × 10⁻⁴ ≈ 8857.14
Linear magnification = √(area magnification) = √8857.14 ≈ 94.09
Final Answer: 94.1

Q1.10 Number of significant figures

(a) 0.007 m² → 1 sig fig
(b) 2.64 × 10²⁴ kg → 3 sig fig
(c) 0.2370 g/cm³ → 4 sig fig
(d) 6.320 J → 4 sig fig
(e) 6.032 N/m² → 4 sig fig
(f) 0.0006032 m² → 4 sig fig
Final Answers:
(a) 1, (b) 3, (c) 4, (d) 4, (e) 4, (f) 4

Q1.11 Area and volume to correct significant figures

Length = 4.234 m (4 sig fig), Breadth = 1.005 m (4 sig fig), Thickness = 2.01 cm = 0.0201 m (3 sig fig)
Area = L × B = 4.234 × 1.005 = 4.25517 → 4.255 m² (4 sig fig)
Volume = L × B × T = 4.234 × 1.005 × 0.0201 ≈ 0.08553 → 0.0855 m³ (3 sig fig)
Final Answers:
Area = 4.255 m²
Volume = 0.0855 m³

Q1.12 Mass of box and difference in gold pieces

Box = 2.30 kg, Gold pieces = 20.15 g and 20.17 g
Total gold mass = 40.32 g = 0.04032 kg
Total mass = 2.30 + 0.04032 = 2.34 kg (3 sig fig)
Difference = 20.17 – 20.15 = 0.02 g (1 sig fig)
Final Answers:
(a) 2.34 kg
(b) 0.02 g

Q1.13 Correct placement of c in relativistic mass formula

Incorrect: $m = \dfrac{m_0}{\sqrt{1 - v^2}}$
Correct: $m = \dfrac{m_0}{\sqrt{1 - \dfrac{v^2}{c^2}}}$

Final Answer: $m = \dfrac{m_0}{\sqrt{1 - v^2/c^2}}$

Q1.14 Atomic volume of a mole of hydrogen

Size of atom ≈ 0.5 Å = 0.5 × 10⁻¹⁰ m
Volume of 1 atom = (4/3)πr³ = (4/3)π(0.5 × 10⁻¹⁰)³ ≈ 5.24 × 10⁻³¹ m³
1 mole = 6.022 × 10²³ atoms
Total volume = 6.022 × 10²³ × 5.24 × 10⁻³¹ ≈ 3.15 × 10⁻⁷ m³
Final Answer: 3.15 × 10⁻⁷ m³

Q1.15 Ratio of molar volume to atomic volume

Molar volume = 22.4 L = 22.4 × 10⁻³ m³
Atomic volume (from Q1.14) = 3.15 × 10⁻⁷ m³
Ratio = 22.4 × 10⁻³ / 3.15 × 10⁻⁷ ≈ 7.11 × 10⁴
Reason: Most of gas volume is empty space between molecules.
Final Answer: 7.11 × 10⁴

Q1.16 Why nearby objects move fast and distant look stationary

Due to relative motion, nearby objects appear to move faster because the angular displacement across our field of view changes rapidly. Distant objects (stars, hills) have tiny angular shifts and appear nearly stationary.
Final Answer: Distant objects appear stationary due to negligible angular displacement across field of view during motion.

Q1.17 Estimating mass density of the Sun

Mass = 2.0 × 10³⁰ kg, Radius = 7.0 × 10⁸ m
Volume = (4/3)πr³ = (4/3)π(7 × 10⁸)³ ≈ 1.436 × 10²⁷ m³
Density = Mass / Volume = 2.0 × 10³⁰ / 1.436 × 10²⁷ ≈ 1393 kg/m³
Final Answer: 1393 kg/m³ (in the range of liquids)

Exemplar

Q1. Can a quantity have units but still be dimensionless?

Yes. A quantity can have units but be dimensionless. For example, angle in radians has units (rad), but its dimensional formula is [M⁰L⁰T⁰] – hence dimensionless.
Final Answer: Yes

Q2. Write the number of significant figures in 0.00208.

In 0.00208, leading zeroes are not significant. Only 2, 0, and 8 are counted.
Significant figures = 3
Final Answer: 3

Q3. Can a physical quantity be dimensionless and still have a unit?

Yes. For example, plane angle is dimensionless but its unit is radian. Similarly, strain is dimensionless but has no unit.
Final Answer: Yes

Q4. Out of displacement, speed, velocity, and acceleration, which can be negative?

Displacement, velocity, and acceleration are vector quantities and can be negative depending on direction. Speed is scalar and always positive.
Final Answer: Displacement, velocity, and acceleration

Q5. Name a dimensionless quantity that is not unitless.

Angle in radians is a dimensionless quantity (M⁰L⁰T⁰), but it has the unit “radian”.
Final Answer: Radian (angle)

Q6. What is the difference between precision and accuracy? Illustrate with an example.

Precision refers to how closely measured values agree with each other.
Accuracy refers to how close a measured value is to the true value.

Example: If the true value of a length is 5.00 m, and three measurements give:
– Set A: 5.11 m, 5.10 m, 5.12 m → Precise but not accurate
– Set B: 5.00 m, 5.01 m, 4.99 m → Both accurate and precise

Final Answer: Precision is repeatability; accuracy is closeness to true value.

Q7. Name the dimensional quantities among the following: pressure, angle, strain, force, work.

Dimensional quantities: Pressure, Force, Work
Dimensionless quantities: Angle, Strain

Final Answer: Pressure, Force, and Work are dimensional

Q8. Why do we take significant figures in measurements?

Significant figures indicate the degree of precision and reliability in a measurement. They include all certain digits and the first uncertain digit.

Final Answer: To show the precision of measured values

Q9. Write the number of significant figures in (i) 0.0200 (ii) 1.0050 × 10³

(i) 0.0200 → 3 significant figures
(ii) 1.0050 × 10³ → 5 significant figures

Final Answer: (i) 3, (ii) 5

Q10. Why do we prefer the SI system over others?

SI system is internationally accepted, follows decimal-based conversions, and covers all physical quantities. It simplifies calculations and standardizes scientific communication globally.

Final Answer: Because it is rational, universal, and easy to use

Q11. Explain the following: (i) Why measurement is always uncertain (ii) Importance of significant figures (iii) Difference between dimensions and units

(i) Measurement is always uncertain due to instrumental limitations, observer errors, and external factors. No instrument can give a perfectly exact value.

(ii) Significant figures reflect the precision of a measurement. They include certain digits plus the first uncertain digit. They help communicate how reliable the measurement is.

(iii)
– Dimensions: Describe the nature of physical quantities based on base quantities (e.g., [M¹L²T⁻²] for energy)
– Units: Standard quantities used to express measurements (e.g., joule, metre, kilogram)

Final Answer: Measurement has uncertainty; significant figures indicate reliability; dimensions show physical nature, units are standard references.

Q12. What is dimensional analysis? What are its uses and limitations?

Dimensional Analysis: A method of analyzing physical quantities by comparing their dimensions.

Uses:
– To check dimensional consistency of equations
– To derive relationships between physical quantities
– To convert units from one system to another

Limitations:
– Cannot find dimensionless constants (like 1/2, π)
– Only applicable when functional relation is of product type
– Cannot distinguish between physical quantities with same dimensions (e.g., torque vs. energy)

Final Answer: Dimensional analysis is useful for checking and deriving relations but cannot handle constants or distinguish similar dimension quantities.

Q13. Explain with an example how dimensional analysis can help in checking correctness of a physical equation.

A physical equation is correct only if dimensions are the same on both sides (principle of homogeneity).

Example:
Equation: $s = ut + \frac{1}{2}at^2$
Dimensions:
– LHS: [L]
– u·t = [L T⁻¹]·[T] = [L]
– a·t² = [L T⁻²]·[T²] = [L]
All terms have [L], so the equation is dimensionally consistent.

Final Answer: By ensuring dimensions match on both sides, dimensional analysis verifies equation consistency.

Q14. Derive the formula for time period of a simple pendulum using dimensional method.

Let time period $T$ depend on:
– Length $l$ → [L]
– Acceleration due to gravity $g$ → [L T⁻²]

Assume: $T = k \cdot l^a \cdot g^b$
Dimensions:
[LHS] = [T]
RHS = [L]^a · [L T⁻²]^b = [L]^{a + b} [T]^{-2b}

Equating powers:
– a + b = 0 → (1)
– –2b = 1 → b = –1/2
From (1): a = 1/2
→ $T = k \cdot \sqrt{\frac{l}{g}}$
Final Answer: $T \propto \sqrt{\frac{l}{g}}$

Q15. Why is it necessary to express physical quantities with units?

Units provide standard reference for measuring and comparing physical quantities. Without units, numerical values are meaningless. They ensure clarity, consistency, and reproducibility in scientific communication.

Example: Saying “speed = 10” is incomplete unless it’s “10 m/s” or “10 km/h”.

Final Answer: Units standardize physical quantities and give meaning to numerical values.

IIT-JEE

JEE Main 2021: If the length of the pendulum in a pendulum clock increases by 0.1%, what is the error in time per day?

Options: (A) 86.4 s (B) 4.32 s (C) 43.2 s (D) 8.64 s

Solution:
The time period of a pendulum is given by: $T = 2\pi\sqrt{\frac{L}{g}}$
If length L increases by 0.1%, then:
$\frac{\Delta T}{T} = \frac{1}{2}\times \frac{\Delta L}{L} = 0.0005 = 0.05\%$
Time error per day = $0.0005 \times 86400 s = 43.2 s$

Final Answer: $(C) 43.2 s$

JEE Main 2021: In a screw gauge, fifth division of circular scale coincides when closed. With 50 divisions and pitch 0.5 mm, main scale reads 5.00 mm and 20th division coincides. What is true reading?

Options: (A) 5.00 mm (B) 5.25 mm (C) 5.15 mm (D) 5.20 mm

Solution:
Pitch = 0.5 mm; divisions = 50 → least count = 0.5 / 50 = 0.01 mm
Circular reading = 20 × 0.01 = 0.20 mm
Zero error = −5 × 0.01 = −0.05 mm
True reading = 5.00 + 0.20 − 0.05 = 5.15 mm

Final Answer: $(C) 5.15 mm$

JEE Main 2021: The dimensions of a quantity $P = E \times L^2 \times M^{-5} \times G^{-2}$ are?

Options: (A) $M^0L^0T^0$ (B) $M^1L^1T^{-2}$ (C) $M^{-1}L^0T^{-2}$ (D) $M^0L^2T^{-2}$

Solution:
$E = [M L^2 T^{-2}]; L = [M L^2 T^{-1}]; G = [M^{-1} L^3 T^{-2}]$
$P = E \times L^2 \times M^{-5} \times G^{-2}$
$P = [M L^2 T^{-2}] \times [M L^2 T^{-1}]^2 \times [M]^{-5} \times [M^{-1} L^3 T^{-2}]^{-2}$
$P = [M L^2 T^{-2}] \times [M^2 L^4 T^{-2}] \times [M]^{-5} \times [M^2 L^{-6} T^4]$
Combine: $M^{1+2-5+2}} = M^0; L^{2+4-6} = L^0; T^{-2-2+4}= T^0$

Final Answer: $(A) M^0L^0T^0$ (Dimensionless)

JEE Advanced 2020: Let [position] = Xᵅ, [speed] = Xᵝ, [acceleration] = Xᵖ, [momentum] = Xᵠ, [force] = Xʳ. Which of the following are correct?

**Options:** (A) α + p = 2β (B) p + q − r = β (C) p = α − 2β (D) q = α − β

Solution:
Using dimension relations:
Speed = position/time → β = α − 1
Acceleration = speed/time → p = β − 1
Momentum = mass × speed → q = m + β (mass not defined separately)
Force = mass × acceleration → r = m + p
Subtracting: (q − r = β − p) → p + q − r = β ✅
Also: p = β − 1 and β = α − 1 → p = α − 2 ✅ Also: α + p = α + (β − 1) = α + β − 1; ≠ 2β ❌ Also: q = α − β ❌ (Not provable)

Final Answers: (B) and (C)

JEE Advanced 2019: If angular momentum and mass are dimensionless, find the dimensions of force, energy, and momentum.

Given: – [Mass] = M⁰ – [Angular momentum] = M⁰ L⁰ T⁰ = dimensionless We know angular momentum = mvr → L = M L² T⁻¹
So for it to be dimensionless: M × L² × T⁻¹ = 1 → L = T^½
Now:
– Force = mass × acceleration → [M⁰] × [L T⁻²] = [T⁻³] → L⁻³
– Energy = F × d = [L⁻³] × [L] = [L⁻²]
– Momentum = mass × velocity = [M⁰] × [L T⁻¹] = [L⁻¹]

Final Answers: Force → L⁻³, Energy → L⁻², Momentum → L⁻¹

JEE Main 2016: Screw gauge – thickness of aluminium sheet

Question:
A screw gauge has pitch 0.5 mm and 50 divisions on its circular scale. When the two jaws touch each other, the 45th division coincides with the main scale, and the zero of the main scale is just visible. When measuring a thin aluminium sheet, the main scale reading is 0.5 mm and 25th circular scale division coincides. What is the actual thickness of the sheet?

Solution:

Pitch = 0.5 mm
Number of circular divisions = 50
Least count (LC) = $\frac{0.5}{50} = 0.01 \, \text{mm}$
Zero error = $- (50 - 45) \times \text{LC} = -5 \times 0.01 = -0.05 \, \text{mm}$
Measured thickness = $0.5 + 25 \times 0.01 = 0.5 + 0.25 = 0.75 \, \text{mm}$
Corrected thickness = $0.75 + 0.05 = \boxed{0.80 \, \text{mm}}$

Final Answer: 0.80 mm

JEE Main 2011: Percentage error in density

Question:
Using a screw gauge (pitch = 0.5 mm, 50 divisions), a sphere’s diameter is measured as 2.70 mm. If the percentage error in mass is 2%, what is the percentage error in the calculated density?

Solution:
– Least Count (LC) = $\frac{0.5}{50} = 0.01 \, \text{mm}$
– Measured diameter $D = 2.70 \, \text{mm}$
– Relative error in diameter:

$\frac{\Delta D}{D} = \frac{0.01}{2.70} \approx 0.0037 \, (0.37\%)$

– Density $\rho = \frac{m}{V} \propto \frac{m}{D^3}$
– Error propagation:

$\frac{\Delta \rho}{\rho} = \frac{\Delta m}{m} + 3 \cdot \frac{\Delta D}{D} = 0.02 + 3 \cdot 0.0037 = 0.0311$

– Percentage error = $0.0311 \times 100 = \boxed{3.11\%}$ Final Answer: 3.1%

JEE Main 2015: Accuracy in determining $g$

Question:
A simple pendulum of length $l = 20.0 \, \text{cm}$ is used to determine $g$ . Length is measured with an error of ±1 mm, and time for 100 oscillations is 90 s (stopwatch least count 1 s). What is the percentage error in $g$ ?

Solution:
– Length: $l = 20.0 \, \text{cm} = 0.20 \, \text{m}, \quad \Delta l = 0.001 \, \text{m}$
– Time: $T = \frac{90}{100} = 0.9 \, \text{s}, \quad \Delta T = \frac{1}{100} = 0.01 \, \text{s}$
– Formula: $g = \frac{4\pi^2 l}{T^2}$
– Error propagation:

$\frac{\Delta g}{g} = \frac{\Delta l}{l} + 2 \cdot \frac{\Delta T}{T} = \frac{0.001}{0.20} + 2 \cdot \frac{0.01}{0.9}$

$= 0.005 + 0.0222 = 0.0272 \Rightarrow \boxed{2.72\%}$

Final Answer: 3%

Key Takeaways

SI Units are physics’ universal language → Memorize the 7 base units
Sig Figs = Precision → Apply rules in calculations and rounding
Errors quantify uncertainty → Distinguish systematic/random errors
Dimensional Analysis validates equations and derives relationships

Sharpen Your Skills: Practice NCERT Exercises 1.1–1.17 for exam mastery!

Based on NCERT Class 11 Physics, Chapter 1 (2025-26 Edition)

Mastering Units, Measurement & Precision

Why Units Rule Physics