Measurement and Units: NMAT Physics Review

Why Measurement Matters in NMAT Physics

Physics is built on measurement. Every law, formula, graph, and experiment depends on describing the real world with numbers and units. In the NMAT, many “hard-looking” problems become simple once you handle units correctly. If you can convert units quickly, identify the right base quantities, and apply dimensional analysis, you can solve questions faster and avoid common traps.

This review focuses on the essential ideas you need: the SI system, prefixes, unit conversion strategies, dimensional analysis, significant figures, and common measurement instruments. You’ll also learn practical exam shortcuts that help you check answers without doing long computations.

Physical Quantities: Base vs Derived

A physical quantity is something measurable with a magnitude and a unit. In physics, quantities fall into two main types:

• Base (fundamental) quantities: defined independently and form the foundation of the unit system.
• Derived quantities: defined using base quantities through relationships (like speed = distance/time).

SI Base Quantities and Units

• Length — meter (m)
• Mass — kilogram (kg)
• Time — second (s)
• Electric current — ampere (A)
• Temperature — kelvin (K)
• Amount of substance — mole (mol)
• Luminous intensity — candela (cd)

Most NMAT questions use the first three heavily (m, kg, s), plus derived units such as Newtons, Joules, and Pascals. You should be able to express any derived unit in terms of base units when needed.

SI Derived Units You Must Know

Derived units combine base units. Here are the most important ones for NMAT-level physics:

• Area: m²
• Volume: m³
• Speed/Velocity: m/s
• Acceleration: m/s²
• Force (Newton, N): kg·m/s²
• Work/Energy (Joule, J): N·m = kg·m²/s²
• Power (Watt, W): J/s = kg·m²/s³
• Pressure (Pascal, Pa): N/m² = kg/(m·s²)
• Density: kg/m³
• Charge (Coulomb, C): A·s
• Voltage (Volt, V): J/C = kg·m²/(s³·A)
• Resistance (Ohm, Ω): V/A

Even if you don’t memorize every derived unit, you should memorize at least Newton, Joule, and Pascal in base-unit form. These show up often in dimensional analysis and unit-check questions.

Metric Prefixes and Powers of Ten

NMAT physics often tests comfort with scientific notation and metric prefixes. Learn these common prefixes:

• Giga (G) = 10⁹
• Mega (M) = 10⁶
• Kilo (k) = 10³
• Deci (d) = 10^-1
• Centi (c) = 10^-2
• Milli (m) = 10^-3
• Micro (μ) = 10^-6
• Nano (n) = 10^-9

Fast mental rule: moving from kilo → base → milli is a 10⁶ change (because 10³ to 10^-3).

Also remember: the symbol “m” can mean meter (unit) or milli- (prefix). Context matters. “mm” is millimeter; “m” alone is meter.

Unit Conversion: The Factor-Label Method

The safest conversion strategy is the factor-label (dimensional) method. You multiply by conversion factors written as fractions equal to 1.

Example 1: Convert 72 km/h to m/s

• Use 1 km = 1000 m
• Use 1 h = 3600 s

72 km/h × (1000 m / 1 km) × (1 h / 3600 s) = 72 × 1000 / 3600 m/s = 20 m/s

Example 2: Convert 2500 cm³ to m³

First convert cm to m: 1 cm = 10^-2 m. For volume, cube the factor:

1 cm³ = (10^-2 m)³ = 10^-6 m³

So 2500 cm³ = 2500 × 10^-6 m³ = 2.5 × 10^-3 m³

Key NMAT trap: when converting squared or cubed units, students forget to square/cube the conversion factor.

Dimensional Analysis: Your Shortcut for Checking Formulas

Dimensional analysis checks whether an equation makes sense by comparing dimensions (not units). The dimension symbols commonly used are:

• [L] for length
• [M] for mass
• [T] for time

A valid physical equation must be dimensionally consistent: both sides must have the same dimensions.

Example: Is v = at dimensionally correct?

• Velocity v has dimensions [L][T]^-1
• Acceleration a has [L][T]^-2
• Multiply by time t ([T]): a·t = [L][T]^-2·[T] = [L][T]^-1

So v = at is dimensionally consistent.

Example: Deriving a relationship quickly

Suppose you forget the formula for period T of a pendulum but recall it depends on length L and gravitational acceleration g.
Assume T ∝ L^a g^b.

• [T] = [L]^a ([L][T]^-2)^b = [L]^a+b[T]^-2b
• Match powers: -2b = 1 ⇒ b = -1/2
• a + b = 0 ⇒ a = 1/2

So T ∝ √(L/g). This technique is powerful for NMAT conceptual questions.

Scientific Notation and Estimation Skills

NMAT exams reward speed. Scientific notation helps you multiply and divide large/small numbers quickly while tracking significant digits.

Rules to remember

• (a × 10^m) (b × 10ⁿ) = (ab) × 10^m+n
• (a × 10^m) / (b × 10ⁿ) = (a/b) × 10^m-n
• Keep the coefficient between 1 and 10 in final standard form.

Quick estimation tip: Round numbers to 1–2 significant digits before computing. Many multiple-choice options are separated enough that an estimate is sufficient.

Example: Estimate (3.2 × 10⁵)(4.8 × 10^-3)

≈ (3 × 5) × 10^5-3 = 15 × 10² = 1.5 × 10³

Accuracy vs Precision and Types of Error

These terms often appear in NMAT conceptual questions:

• Accuracy: how close a measurement is to the true value.
• Precision: how consistent repeated measurements are with each other.

You can be precise but not accurate (tight clustering around the wrong value) or accurate but not precise (average is close but values scatter).

Common sources of error

• Systematic error: consistent bias (e.g., miscalibrated instrument). Affects accuracy.
• Random error: unpredictable fluctuations (e.g., reaction time). Affects precision.
• Zero error: instrument does not read zero when it should (common in vernier calipers and micrometers).
• Parallax error: reading a scale from an angle instead of straight-on.

Significant Figures and Rounding Rules

Significant figures communicate measurement certainty. In NMAT problems, you may need to round final answers correctly.

Counting significant figures

• Non-zero digits are significant (123 has 3 s.f.).
• Zeros between non-zero digits are significant (1002 has 4 s.f.).
• Leading zeros are not significant (0.0032 has 2 s.f.).
• Trailing zeros are significant only if a decimal point is shown (2.500 has 4 s.f.; 2500 is ambiguous without notation).

Rules in calculations

• Multiplication/division: final answer has the same number of significant figures as the factor with the fewest s.f.
• Addition/subtraction: final answer has the same number of decimal places as the term with the fewest decimal places.

Example (Multiplication): (2.3)(4.56) = 10.488 → 2 s.f. → 10

Example (Addition): 12.11 + 0.3 = 12.41 → 1 decimal place → 12.4

Common Measurements and Instruments

NMAT physics may include basic instrument questions. Know what each instrument measures and typical precision.

• Meter stick / ruler: length; least count often 1 mm (0.1 cm).
• Vernier caliper: more precise length/diameter; least count depends on vernier scale (often 0.01 cm or 0.02 mm depending on type).
• Micrometer screw gauge: very small thickness/diameter; can measure to 0.01 mm (varies by instrument).
• Stopwatch: time measurement; human reaction time can introduce random error.
• Triple-beam balance / digital balance: mass measurement.
• Graduated cylinder: volume measurement; better than beaker for accuracy.
• Thermometer: temperature measurement; note scale (°C vs K).

For many NMAT problems, you won’t need the instrument’s exact least count, but you should know the idea: a smaller least count means higher resolution (more precision).

Temperature Scales and Conversion

Temperature commonly appears in physics and chemistry overlaps. The NMAT may ask about Celsius and Kelvin.

• Kelvin is the SI unit for temperature.
• Conversion: T(K) = T(°C) + 273.15 (often approximated as +273)

Example: 27°C = 27 + 273 = 300 K (approx.)

Be careful: temperature differences in Celsius and Kelvin have the same numerical value (a change of 1°C equals a change of 1 K), but absolute values differ by 273.

Density, Specific Gravity, and Unit Awareness

Density is mass per unit volume: ρ = m/V. The SI unit is kg/m³, but g/cm³ is commonly used in lab-style contexts.

Useful conversion:

• 1 g/cm³ = 1000 kg/m³

Specific gravity is the ratio of a substance’s density to the density of water (at a reference condition). It has no units because it’s a ratio.

Exam Tactics: Unit-Based Shortcuts

Here are fast, high-value tactics for NMAT multiple-choice problems:

• Check units early: If an answer choice has the wrong unit, eliminate it immediately.
• Use dimensional reasoning: If the question asks for energy, your expression must reduce to kg·m²/s².
• Convert only when needed: Sometimes you can keep everything in cm and g as long as you stay consistent, then convert at the end (if required).
• Watch squared/cubed conversions: This is one of the most common error sources.
• Estimate first: If your computed result is 10× bigger than all options, you likely made a power-of-ten mistake.
• Use “unit cancellation”: Write conversions so units cancel neatly; it prevents inverted factors.

Mini Practice Set (Concept-Focused)

Use these as quick self-checks. Try answering without a calculator, focusing on units and powers of ten.

1. Convert 5.0 m/s to km/h. (Hint: multiply by 3.6)
2. Express 1 N in base units.
3. If pressure is measured in Pa, what are the base units of Pa?
4. Which is larger: 0.45 mm or 4.5 × 10^-5 m?
5. A cube has side 2 cm. What is its volume in m³?

If you can do these quickly, you’re in strong shape for measurement and units questions.

Summary: What to Master for NMAT

To score well on NMAT Physics measurement questions, you should be able to:

• Recognize SI base units and common derived units (N, J, Pa).
• Use metric prefixes confidently and move between powers of ten.
• Convert units using factor-label method (including squared/cubed units).
• Apply dimensional analysis to check or infer formulas.
• Handle scientific notation and quick estimation.
• Understand accuracy vs precision and basic measurement errors.
• Use significant figures and rounding rules properly.

These skills don’t just help in a single topic—they make nearly every physics problem easier. With consistent practice, unit logic becomes automatic, and your speed and accuracy improve across the entire NMAT Physics section.

Measurement and Units: Problem Sets with Answer Keys (NMAT Physics)

How to Use These Problem Sets

These questions focus on the most tested skills in Measurement and Units for NMAT-level physics: SI base/derived units,
metric prefixes, unit conversion (including squared and cubed units), dimensional analysis, significant figures, and estimation.
Try solving first without looking at the answer keys. Then check your work and review the notes included in some solutions.

Problem Set 1: SI Units and Derived Units

1. Identify the SI base unit for each quantity:
   1. Length
   2. Mass
   3. Time
   4. Temperature
   5. Electric current
2. Express the Newton (N) in SI base units.
3. Express the Joule (J) in SI base units.
4. Express the Pascal (Pa) in SI base units.
5. A quantity has units kg·m²/s³. What is its common SI unit name?
6. Which of the following are dimensionless (no units)?
   1. Strain
   2. Density
   3. Coefficient of friction
   4. Specific gravity
   5. Velocity

Answer Key: Problem Set 1

1. 1. meter (m)
   2. kilogram (kg)
   3. second (s)
   4. kelvin (K)
   5. ampere (A)
2. 1 N = 1 kg·m/s²
3. 1 J = 1 N·m = (kg·m/s²)·m = kg·m²/s²
4. 1 Pa = 1 N/m² = (kg·m/s²)/m² = kg/(m·s²)
5. Watt (W), since 1 W = 1 J/s = kg·m²/s³
6. Dimensionless: (a) Strain, (c) Coefficient of friction, (d) Specific gravity

Problem Set 2: Metric Prefixes and Scientific Notation

1. Convert 7.2 km to meters.
2. Convert 0.045 m to millimeters.
3. Convert 3.6 × 10⁵ μs to seconds.
4. Write 0.00000082 in scientific notation.
5. Write 5.07 × 10⁶ in standard (decimal) form.
6. Arrange from smallest to largest:
   1. 2.5 nm
   2. 2.5 μm
   3. 2.5 mm
   4. 2.5 m

Answer Key: Problem Set 2

1. 7.2 km = 7.2 × 10³ m = 7200 m
2. 0.045 m = 0.045 × 1000 mm = 45 mm
3. 1 μs = 10^-6 s
   3.6 × 10⁵ μs = 3.6 × 10⁵ × 10^-6 s = 3.6 × 10^-1 s = 0.36 s
4. 0.00000082 = 8.2 × 10^-7
5. 5.07 × 10⁶ = 5,070,000
6. Smallest → largest: 2.5 nm < 2.5 μm < 2.5 mm < 2.5 m

Problem Set 3: Unit Conversions (Including Squared and Cubed)

1. Convert 90 km/h to m/s.
2. Convert 15 m/s to km/h.
3. Convert 250 cm to meters.
4. Convert 2.5 m² to cm².
5. Convert 1.2 × 10⁶ mm³ to cm³.
6. Convert 3.0 L to m³.
7. Convert 4500 cm³ to liters (L).
8. Convert 1.8 g/cm³ to kg/m³.

Answer Key: Problem Set 3

1. 90 km/h = 90 × (1000/3600) m/s = 25 m/s
2. 15 m/s × 3.6 = 54 km/h
3. 250 cm = 2.50 m
4. 1 m = 100 cm ⇒ 1 m² = (100)² cm² = 10,000 cm²
   2.5 m² = 2.5 × 10,000 = 25,000 cm²
5. 1 cm = 10 mm ⇒ 1 cm³ = (10 mm)³ = 1000 mm³
   1.2 × 10⁶ mm³ ÷ 1000 = 1.2 × 10³ cm³ = 1200 cm³
6. 1 L = 10^-3 m³ ⇒ 3.0 L = 3.0 × 10^-3 m³
7. 1000 cm³ = 1 L ⇒ 4500 cm³ = 4.5 L
8. 1 g/cm³ = 1000 kg/m³ ⇒ 1.8 g/cm³ = 1800 kg/m³

Problem Set 4: Dimensional Analysis and Formula Checking

1. Show that the unit of acceleration is m/s² using the definition of acceleration.
2. A student proposes the formula for distance: d = vt². Is it dimensionally correct? Explain briefly.
3. Using dimensional analysis, determine the dimensions of pressure in terms of M, L, and T.
4. The period T of a simple pendulum depends on length L and gravitational acceleration g.
   Use dimensional analysis to find how T depends on L and g (ignore constants).
5. If energy E has dimensions ML²T^-2, what are the dimensions of power P = E/t?

Answer Key: Problem Set 4

1. Acceleration a = Δv/Δt. Velocity has units m/s. Dividing by s gives (m/s)/s = m/s².
2. Check dimensions: left side d is [L]. Right side v t² is [L/T]·[T²] = [L·T]. Not equal to [L]. So it is NOT dimensionally correct.
3. Pressure = Force/Area. Force is [M L T^-2]. Area is [L²].
   Pressure dimensions = [M L T^-2]/[L²] = [M L^-1 T^-2].
4. Assume T ∝ L^a g^b. Dimensions: [T] = [L]^a ([L T^-2])^b = [L]^a+b[T]^-2b.
   Match exponents: -2b = 1 ⇒ b = -1/2. Then a + b = 0 ⇒ a = 1/2.
   Therefore T ∝ √(L/g).
5. Power P = E/t ⇒ [P] = [M L² T^-2]/[T] = [M L² T^-3].

Problem Set 5: Significant Figures, Rounding, and Measurement Concepts

1. How many significant figures are in 0.00450?
2. How many significant figures are in 2050 (no decimal shown)?
3. Round 12.349 to 3 significant figures.
4. Multiply and report with correct significant figures: (2.4)(3.56)
5. Add and report with correct decimal places: 18.2 + 0.045 + 1.1
6. Explain briefly: what is the difference between accuracy and precision?
7. A miscalibrated balance always reads 2 g heavier than the true mass. Is this systematic or random error?

Answer Key: Problem Set 5

1. 0.00450 has 3 significant figures (4, 5, and the trailing 0 after the decimal).
2. 2050 (no decimal shown) is ambiguous, but typically counted as 3 significant figures (2, 0, 5). The last 0 may or may not be significant without a decimal point.
3. 12.349 → 3 significant figures = 12.3 (since the next digit is 4).
4. (2.4)(3.56) = 8.544. Fewest s.f. is 2 (from 2.4). Final = 8.5
5. 18.2 + 0.045 + 1.1 = 19.345. Least decimal places is 1 (from 18.2 and 1.1). Final = 19.3
6. Accuracy is closeness to the true value; precision is closeness of repeated measurements to each other.
7. Systematic error (consistent bias of +2 g).

Problem Set 6: Mixed NMAT-Style Questions (Speed + Unit Sense)

1. A runner covers 400 m in 50 s. What is the average speed in m/s?
2. The density of a material is 2.7 g/cm³. What is it in kg/m³?
3. Convert 5.0 × 10^-2 m to micrometers (μm).
4. A force of 12 N acts over a distance of 0.50 m. How much work is done?
5. A pressure of 2.0 × 10⁵ Pa is applied over an area of 0.10 m². What is the force?
6. Which has the larger magnitude: 3.6 × 10⁴ mm or 3.6 × 10¹ m?
7. A cube has side length 5.0 cm. Find its volume in m³.
8. If v has units m/s and p has units kg·m/s, what quantity could p represent?

Answer Key: Problem Set 6

1. v = d/t = 400/50 = 8 m/s
2. 2.7 g/cm³ = 2700 kg/m³
3. 5.0 × 10^-2 m = 0.05 m. Since 1 m = 10⁶ μm, 0.05 m = 0.05 × 10⁶ μm = 5.0 × 10⁴ μm
4. W = Fd = 12 × 0.50 = 6.0 J
5. F = PA = (2.0 × 10⁵)(0.10) = 2.0 × 10⁴ N
6. 3.6 × 10⁴ mm = 3.6 × 10⁴ × 10^-3 m = 3.6 × 10¹ m = 36 m.
   They are equal.
7. Side = 5.0 cm = 5.0 × 10^-2 m. Volume = s³ = (5.0 × 10^-2)³ m³ = 125 × 10^-6 = 1.25 × 10^-4 m³
8. p = kg·m/s is momentum (p = mv).

Bonus: Quick Conversion & Unit Facts (Must-Memorize)

• 1 hour = 3600 s
• 1 km = 1000 m
• 1 m/s = 3.6 km/h
• 1 L = 1000 cm³ = 10^-3 m³
• 1 g/cm³ = 1000 kg/m³
• 1 N = kg·m/s²
• 1 J = N·m = kg·m²/s²
• 1 Pa = N/m² = kg/(m·s²)

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