Chapter 1: Measurement

Measurement is the cornerstone of physics, enabling precise quantification of physical phenomena. This chapter covers SI units and standards, dimensional analysis, significant figures, and error analysis, essential for JEE Main, JEE Advanced, and NEET. These topics appear in problems on unit conversions, dimensional consistency, and error calculations. With clear explanations, extensive derivations, exam-focused examples, and robust problem-solving strategies, this chapter equips students to master measurement for competitive exams, ensuring precision and conceptual clarity.

1.1 SI Units and Standards

The International System of Units (SI) defines seven base units: meter ($m$, length), kilogram ($kg$, mass), second ($s$, time), ampere ($A$, electric current), kelvin ($K$, temperature), mole ($mol$, amount of substance), and candela ($cd$, luminous intensity). Derived units, such as the newton ($N = kg \cdot m / s^2$), combine base units. Standards, like the meter defined by the speed of light ($c = 299,792,458 , m/s$), ensure global consistency. JEE/NEET problems focus on unit conversions and derived units. Fundamentals of Physics emphasizes SI’s role in standardization.

Derivation: Unit of Force (Newton)
Newton’s second law states: $F = m a$. Mass ($m$) is in $kg$, acceleration ($a$) is in $m / s^2$. Thus:
$$ F = kg \cdot \frac{m}{s^2} = kg \cdot m / s^2 $$
This defines the newton ($N$). To verify, consider work ($W = F \cdot d$):
$$ W = (kg \cdot m / s^2) \cdot m = kg \cdot m^2 / s^2 = J $$
(joule), confirming energy units.

Solved Example: A JEE Main problem requires converting $54 , km/h$ to $m/s$.

• Solution: Speed:
  $$ 54 , km/h = 54 \times \frac{1000 , m}{3600 , s} = 54 \times \frac{5}{18} = 15 , m/s $$
  (2 significant figures, April 14, 2025).
  • JEE Tip: Use conversion factor $5/18$ for $km/h$ to $m/s$. Common error: Incorrect factor or rounding.

Solved Example: A NEET problem asks the SI unit of power.

• Solution: Power ($P = W / t$): work ($W = kg \cdot m^2 / s^2$), time ($t = s$). Thus:
  $$ P = \frac{kg \cdot m^2 / s^2}{s} = kg \cdot m^2 / s^3 = W $$
  (watt). Option (a) $kg \cdot m^2 / s^3$ is correct.
  • NEET Tip: Derive from work/time; memorize watt. Common error: Confusing with joule.

Solved Example: A JEE Main problem involves a mass of $2.5 , kg$ and acceleration $3.2 , m/s^2$. Calculate the force in newtons.

• Solution: Force:
  $$ F = m a = 2.5 \cdot 3.2 = 8.0 , N $$
  (2 significant figures).
  • JEE Tip: Multiply with least significant figures. Common error: Retaining extra digits.

Application: SI units enable precise measurements in GPS (meter-based distances) and medical diagnostics (kelvin for body temperature).

1.2 Dimensional Analysis

Dimensional analysis verifies equations by comparing dimensions: length ($[\text{L}]$), mass ($[\text{M}]$), time ($[\text{T}]$). For example, velocity: $[\text{L} \text{T}^{-1}]$. It derives formulas or checks consistency. JEE/NEET problems involve finding dimensions or validating equations. University Physics uses dimensional analysis for problem-solving.

Derivation: Dimensions of Kinetic Energy
Kinetic energy: $KE = \frac{1}{2} m v^2$. Mass ($m$): $[\text{M}]$, velocity ($v$): $[\text{L} \text{T}^{-1}]$. Thus:
$$ KE = \lbrack \text{M} \rbrack \cdot \lbrack \text{L} \text{T}^{-1} \rbrack^2 = \lbrack \text{M} \text{L}^2 \text{T}^{-2} \rbrack $$
Compare with work ($W = F \cdot d$): force ($F = m a = \lbrack \text{M} \text{L} \text{T}^{-2} \rbrack$), distance ($d = [\text{L}]$):
$$ W = \lbrack \text{M} \text{L} \text{T}^{-2} \rbrack \cdot [\text{L}] = \lbrack \text{M} \text{L}^2 \text{T}^{-2} \rbrack $$
Both match, confirming energy’s dimensions.

Solved Example: A JEE Advanced problem asks the dimensions of pressure.

• Solution: Pressure: $P = \frac{F}{A}$. Force ($F = m a$): $[\text{M} \text{L} \text{T}^{-2}]$, area ($A$): $[\text{L}^2]$. Thus:
  $$ P = \frac{[\text{M} \text{L} \text{T}^{-2}]}{[\text{L}^2]} = [\text{M} \text{L}^{-1} \text{T}^{-2}] $$
  Answer: $[\text{M} \text{L}^{-1} \text{T}^{-2}]$.
  • JEE Tip: Divide dimensions carefully. Common error: Incorrect area dimension.

Solved Example: A NEET problem checks if $s = ut + \frac{1}{2} at^2$ is dimensionally consistent.

• Solution: Displacement ($s$): $[\text{L}]$. Velocity ($u$): $[\text{L} \text{T}^{-1}]$, time ($t$): $[\text{T}]$, acceleration ($a$): $[\text{L} \text{T}^{-2}]$. Term $ut$: $[\text{L} \text{T}^{-1}] \cdot [\text{T}] = [\text{L}]$. Term $\frac{1}{2} at^2$: $[\text{L} \text{T}^{-2}] \cdot [\text{T}^2] = [\text{L}]$. Right side: $[\text{L}] + [\text{L}] = [\text{L}]$. Consistent. Option (a) “Consistent” is correct.
  • NEET Tip: Match dimensions; constants are dimensionless. Common error: Incorrect acceleration dimension.

Solved Example: A JEE Main problem derives the dimensions of work.

• Solution: Work: $W = F \cdot d$. Force: $[\text{M} \text{L} \text{T}^{-2}]$, distance: $[\text{L}]$. Thus:
  $$ W = [\text{M} \text{L} \text{T}^{-2}] \cdot [\text{L}] = [\text{M} \text{L}^2 \text{T}^{-2}] $$
  Answer: $[\text{M} \text{L}^2 \text{T}^{-2}]$.
  • JEE Tip: Use force dimensions correctly. Common error: Omitting length.

Application: Dimensional analysis ensures formula accuracy in astrophysics (e.g., Kepler’s laws) and engineering (e.g., fluid dynamics).

1.3 Significant Figures

Significant figures reflect measurement precision (e.g., $2.34 , m$: 3 significant figures). Rules: non-zero digits are significant, zeros between non-zeros are significant, trailing zeros in decimals are significant. Calculations use the least precise input’s significant figures. JEE/NEET problems require reporting results correctly. Fundamentals of Physics stresses precision (April 14, 2025).

Derivation: Significant Figures in Multiplication
For $z = x \cdot y$, relative uncertainty:
$$ \frac{\Delta z}{z} \approx \frac{\Delta x}{x} + \frac{\Delta y}{y} $$
The result’s significant figures match the input with the fewest, preserving precision. Example: $z = 2.5 \cdot 3.24$. $2.5$ (2 significant figures), $3.24$ (3). Compute: $z = 8.1$ (2 significant figures).

Solved Example: A JEE Main problem involves a length $4.8 , m$ and width $2.56 , m$. Calculate the area with correct significant figures.

• Solution: Area:
  $$ A = 4.8 \cdot 2.56 = 12.288 , m^2 $$
  $4.8$ has 2 significant figures, $2.56$ has 3. Use 2:
  $$ A \approx 12 , m^2 $$
  • JEE Tip: Use least significant figures. Common error: Reporting $12.288$.

Solved Example: A NEET problem asks the significant figures in $0.00450$.

• Solution: Leading zeros are not significant; trailing zero in decimal is. $0.00450$ has 3 significant figures (4, 5, 0). Option (a) “3” is correct.
  • NEET Tip: Check trailing zeros in decimals. Common error: Counting leading zeros.

Solved Example: A JEE Main problem adds $7500$ and $2.567$. Report the sum with correct significant figures.

• Solution: Sum:
  $$ 7500 + 2.567 = 7502.567 $$
  $7500$ has 2 significant figures (no decimal), $2.567$ has 4. Use 0 decimal places:
  $$ 7503 $$
  (April 14, 2025).
  • JEE Tip: Use least precise decimal places for addition. Common error: Retaining decimals.

Application: Significant figures ensure accuracy in lab experiments (e.g., pendulum timing) and pharmaceutical dosing (e.g., drug concentrations).

1.4 Error Analysis

Error analysis quantifies measurement uncertainty. Absolute error: $\Delta x = |x_{\text{measured}} - x_{\text{true}}|$. Relative error: $\frac{\Delta x}{x}$. Percentage error: $\frac{\Delta x}{x} \times 100%$. For functions, error propagates (e.g., $z = x + y$, $\Delta z = \Delta x + \Delta y$). JEE/NEET problems involve error calculations. University Physics connects errors to experimental design.

Derivation: Error Propagation in Product
For $z = x \cdot y$, relative error:
$$ \frac{\Delta z}{z} = \sqrt{\left( \frac{\Delta x}{x} \right)^2 + \left( \frac{\Delta y}{y} \right)^2} $$
Example: $z = 2.0 \cdot 3.0$, $\Delta x = 0.1$, $\Delta y = 0.1$. Relative errors: $\frac{0.1}{2.0} = 0.05$, $\frac{0.1}{3.0} \approx 0.033$. Total: $\frac{\Delta z}{z} = \sqrt{0.05^2 + 0.033^2} \approx 0.06$. If $z = 6.0$, then $\Delta z \approx 0.36$.

Solved Example: A JEE Advanced problem involves a length $12.5 \pm 0.2 , cm$. Calculate the percentage error.

• Solution: Percentage error:
  $$ \frac{\Delta x}{x} \times 100% = \frac{0.2}{12.5} \times 100 = 1.6% $$
  (2 significant figures).
  • JEE Tip: Use measured value for denominator. Common error: Incorrect percentage.

Solved Example: A NEET problem involves $z = x - y$, with $x = 8.0 \pm 0.3 , m$, $y = 5.0 \pm 0.2 , m$. Calculate the absolute error in $z$.

• Solution: Absolute error:
  $$ \Delta z = \Delta x + \Delta y = 0.3 + 0.2 = 0.5 , m $$
  Option (a) $0.5 , m$ is correct.
  • NEET Tip: Sum absolute errors for subtraction. Common error: Subtracting errors.

Solved Example: A JEE Main problem measures $v = 20.0 \pm 0.5 , m/s$, $t = 4.00 \pm 0.05 , s$. Calculate the percentage error in $s = v \cdot t$.

• Solution: Relative error:
  $$ \frac{\Delta s}{s} = \frac{\Delta v}{v} + \frac{\Delta t}{t} = \frac{0.5}{20.0} + \frac{0.05}{4.00} = 0.025 + 0.0125 = 0.0375 $$
  Percentage error:
  $$ 0.0375 \times 100 = 3.75% \approx 3.8% $$
  (2 significant figures).
  • JEE Tip: Sum relative errors for multiplication. Common error: Using absolute errors.

Application: Error analysis ensures reliability in particle physics (e.g., CERN) and structural engineering (e.g., stress tests).

Summary and Quick Revision

• SI Units: Base units ($m$, $kg$, $s$, $A$, $K$, $mol$, $cd$); derived units (e.g., $N = kg \cdot m / s^2$). Convert: $1 , km/h = \frac{5}{18} , m/s$.
• Dimensional Analysis: Dimensions (e.g., velocity: $[\text{L} \text{T}^{-1}]$). Energy: $[\text{M} \text{L}^2 \text{T}^{-2}]$. Checks equation consistency.
• Significant Figures: Precision (e.g., $2.34$: 3 significant figures). Use least significant figures in calculations (e.g., $4.8 \cdot 2.56 = 12 , m^2$).
• Error Analysis: Absolute error: $\Delta x$. Relative error: $\frac{\Delta x}{x}$. Propagation: $\Delta z = \Delta x + \Delta y$ for $z = x + y$.
• SI Units: Length ($m$), mass ($kg$), time ($s$), etc.
• JEE/NEET Tips: Verify units, use dimensional analysis, report significant figures correctly (April 14, 2025), sum absolute errors for addition/subtraction.
• Applications: GPS, engineering, labs, CERN.

Practice Problems

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Note: Content regularly updated to align with current JEE/NEET syllabi.