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MERITROOT FREE LEARNING · CLASS 11 / PHYSICS

Units and Measurements

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Understand the chapter

Chapter overview

Physics studies measurable quantities. A physical quantity is expressed as:

Physical quantity=numerical value×unit\text{Physical quantity}=\text{numerical value}\times\text{unit}

Example:

l=5 ml=5\ \text{m}

Here, 55 is the numerical value and m\text{m} is the unit.

1. Fundamental and derived quantities

Fundamental quantities are independent quantities. Derived quantities are obtained from fundamental quantities.

QuantitySI unitSymbol
Lengthmetrem\text{m}
Masskilogramkg\text{kg}
Timeseconds\text{s}
Electric currentampereA\text{A}
TemperaturekelvinK\text{K}
Amount of substancemolemol\text{mol}
Luminous intensitycandelacd\text{cd}

2. Derived units

QuantityFormulaSI unit
Areal2l^2m2\text{m}^2
Volumel3l^3m3\text{m}^3
Velocitydisplacementtime\frac{\text{displacement}}{\text{time}}m s−1\text{m s}^{-1}
Accelerationvelocitytime\frac{\text{velocity}}{\text{time}}m s−2\text{m s}^{-2}
Forcemamakg m s−2\text{kg m s}^{-2} or N\text{N}
WorkFsFskg m2s−2\text{kg m}^2\text{s}^{-2} or J\text{J}
PowerWt\frac{W}{t}kg m2s−3\text{kg m}^2\text{s}^{-3} or W\text{W}

3. Dimensions

Dimensions represent the powers of fundamental quantities in a physical quantity.

QuantityDimensional formula
Length[L][L]
Mass[M][M]
Time[T][T]
Velocity[LT−1][LT^{-1}]
Acceleration[LT−2][LT^{-2}]
Force[MLT−2][MLT^{-2}]
Work/Energy[ML2T−2][ML^2T^{-2}]
Power[ML2T−3][ML^2T^{-3}]
Pressure[ML−1T−2][ML^{-1}T^{-2}]

Example:

F=maF=ma

Dimensional formula:

[F]=[M][LT−2]=[MLT−2][F]=[M][LT^{-2}]=[MLT^{-2}]

4. Dimensional analysis

Principle of homogeneity

Only physical quantities with the same dimensions can be added or subtracted.

If:

A+B=CA+B=C

then:

[A]=[B]=[C][A]=[B]=[C]

Example:

s=ut+12at2s=ut+\frac{1}{2}at^2

Checking dimensions:

[ut]=[LT−1][T]=[L][ut]=[LT^{-1}][T]=[L] [at2]=[LT−2][T2]=[L][at^2]=[LT^{-2}][T^2]=[L]

So the equation is dimensionally correct.

5. Uses of dimensional analysis

To check correctness of equations

Example:

T=2πlgT=2\pi\sqrt{\frac{l}{g}}

Dimension of RHS:

[lg]=[L][LT−2]=[T2]=[T]\left[\sqrt{\frac{l}{g}}\right] = \sqrt{\frac{[L]}{[LT^{-2}]}} = \sqrt{[T^2]} = [T]

So the formula is dimensionally correct.

To derive relations

Suppose time period of a simple pendulum depends on length ll and acceleration due to gravity gg.

T∝lagbT\propto l^a g^b

So:

[T]=[L]a[LT−2]b[T]=[L]^a[LT^{-2}]^b [T]=[La+bT−2b][T]=[L^{a+b}T^{-2b}]

Equating powers:

a+b=0,−2b=1a+b=0,\qquad -2b=1

Therefore:

b=−12,a=12b=-\frac{1}{2},\qquad a=\frac{1}{2}

Hence:

T∝lgT\propto\sqrt{\frac{l}{g}}

6. Limitations of dimensional analysis

  • It cannot determine dimensionless constants like 22, π\pi, or 12\frac12.
  • It cannot distinguish between scalar and vector quantities.
  • It cannot derive equations involving trigonometric, exponential, or logarithmic functions.
  • It cannot verify equations where dimensions are correct but physics is wrong.

7. Significant figures

Significant figures are meaningful digits in a measured quantity.

Examples:

NumberSignificant figures
2.342.343
0.00450.00452
5.0605.0604
100.0100.04

Rules:

  • All non-zero digits are significant.
  • Zeros between non-zero digits are significant.
  • Leading zeros are not significant.
  • Trailing zeros after decimal are significant.

8. Accuracy and precision

Accuracy means closeness to the true value.

Precision means closeness among repeated measurements.

9. Errors in measurement

Measured value:

a1,a2,a3,…,ana_1,a_2,a_3,\ldots,a_n

Mean value:

aˉ=a1+a2+⋯+ann\bar{a}=\frac{a_1+a_2+\cdots+a_n}{n}

Absolute error:

Δai=∣ai−aˉ∣\Delta a_i=|a_i-\bar{a}|

Mean absolute error:

Δamean=Δa1+Δa2+⋯+Δann\Delta a_{\text{mean}} = \frac{\Delta a_1+\Delta a_2+\cdots+\Delta a_n}{n}

Relative error:

Δaa\frac{\Delta a}{a}

Percentage error:

Δaa×100\frac{\Delta a}{a}\times100

10. Combination of errors

Addition and subtraction

If:

Z=A+BZ=A+B

or:

Z=A−BZ=A-B

then:

ΔZ=ΔA+ΔB\Delta Z=\Delta A+\Delta B

Multiplication and division

If:

Z=ABZ=AB

or:

Z=ABZ=\frac{A}{B}

then:

ΔZZ=ΔAA+ΔBB\frac{\Delta Z}{Z} = \frac{\Delta A}{A} + \frac{\Delta B}{B}

Powers

If:

Z=AnZ=A^n

then:

ΔZZ=nΔAA\frac{\Delta Z}{Z} = n\frac{\Delta A}{A}

General rule:

Z=ApBqCrZ=A^pB^qC^r

then:

ΔZZ=pΔAA+qΔBB+rΔCC\frac{\Delta Z}{Z} = p\frac{\Delta A}{A} + q\frac{\Delta B}{B} + r\frac{\Delta C}{C}

11. Vernier calipers and screw gauge

Least count of vernier calipers:

LC=1 MSDnumber of divisions on vernier scaleLC=\frac{1\ \text{MSD}}{\text{number of divisions on vernier scale}}

Measurement:

Reading=main scale reading+vernier coincidence×LC\text{Reading} = \text{main scale reading} + \text{vernier coincidence}\times LC

Screw gauge least count:

LC=pitchnumber of circular scale divisionsLC=\frac{\text{pitch}}{\text{number of circular scale divisions}}

Screw gauge reading:

Reading=pitch scale reading+circular scale reading×LC\text{Reading} = \text{pitch scale reading} + \text{circular scale reading}\times LC

12. Solved examples

Example 1

Check the dimensions of:

v=u+atv=u+at

Solution:

[v]=[LT−1][v]=[LT^{-1}] [u]=[LT−1][u]=[LT^{-1}] [at]=[LT−2][T]=[LT−1][at]=[LT^{-2}][T]=[LT^{-1}]

All terms have the same dimension, so the equation is dimensionally correct.

Example 2

If:

l=2.0±0.1 ml=2.0\pm0.1\ \text{m}

and:

b=3.0±0.2 mb=3.0\pm0.2\ \text{m}

then area:

A=lbA=lb

Relative error:

ΔAA=0.12.0+0.23.0\frac{\Delta A}{A} = \frac{0.1}{2.0} + \frac{0.2}{3.0}

13. Common mistakes

  • Confusing units and dimensions.
  • Writing SI symbols in plural form, like mts\text{mts} instead of m\text{m}.
  • Ignoring significant zeros after decimal.
  • Adding percentage errors for addition/subtraction problems.
  • Forgetting that dimensional analysis cannot determine constants.

14. Quick revision

ConceptFormula
Force dimension[MLT−2][MLT^{-2}]
Work dimension[ML2T−2][ML^2T^{-2}]
Power dimension[ML2T−3][ML^2T^{-3}]
Meanaˉ=∑ain\bar{a}=\frac{\sum a_i}{n}
Relative errorΔaa\frac{\Delta a}{a}
Percentage errorΔaa×100\frac{\Delta a}{a}\times100
Product errorΔZZ=ΔAA+ΔBB\frac{\Delta Z}{Z}=\frac{\Delta A}{A}+\frac{\Delta B}{B}
Power errorΔZZ=nΔAA\frac{\Delta Z}{Z}=n\frac{\Delta A}{A}

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