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Physics · Chapter

Units and Measurements

This chapter establishes the language of physics: measurement. It introduces the concepts of fundamental and derived physical quantities, and the standardized systems of units used to measure them, with a focus on the International System (SI). You will learn how to express any physical quantity in terms of its fundamental dimensions of mass (M), length (L), and time (T). This skill is the basis for the powerful technique of dimensional analysis.

Dimensional analysis is the core analytical tool of this chapter. It is used to check the consistency of physical equations, a principle known as homogeneity of dimensions. More advanced applications involve deriving the relationship between different physical quantities in a phenomenon, and converting values from one system of units to another. Mastery of this section is essential for problem-solving across all of physics.

The chapter then moves to the practical aspects of measurement, covering accuracy, precision, and the analysis of errors. You will learn to calculate absolute, relative, and percentage errors, and how to combine them when measured quantities are used in calculations. This topic, along with problems involving measuring instruments like the Vernier caliper and screw gauge, can be challenging. The chapter relies on basic algebra but is otherwise self-contained, forming the foundation for all subsequent physics topics.

Formulas for Units and Measurements

Conversion of a physical quantity between unit systems
n2=n1(M1M2)a(L1L2)b(T1T2)cn_2 = n_1 \left( \frac{M_1}{M_2} \right)^a \left( \frac{L_1}{L_2} \right)^b \left( \frac{T_1}{T_2} \right)^c n₁ and n₂ are numerical values in two systems with base units (M₁, L₁, T₁) and (M₂, L₂, T₂).
Relative Error
δa=Δameanamean\delta a = \frac{\Delta a_{mean}}{a_{mean}} Δa_mean is the mean absolute error and a_mean is the true or mean value of the quantity.
Error in a sum or difference
ΔZ=ΔA+ΔB\Delta Z = \Delta A + \Delta B For Z = A ± B. The absolute errors are always added.
Error propagation in a general relation
ΔZZ=pΔAA+qΔBB+rΔCC\frac{\Delta Z}{Z} = p \frac{\Delta A}{A} + q \frac{\Delta B}{B} + r \frac{\Delta C}{C} For a quantity Z = AᵖB۹/Cʳ. The fractional errors are multiplied by the power and added.
Least Count of a Vernier Caliper
LC=1 MSD−1 VSD\text{LC} = 1 \text{ MSD} - 1 \text{ VSD} MSD is Main Scale Division and VSD is Vernier Scale Division.
Vernier Caliper Reading
Reading=MSR+(VSC×LC)\text{Reading} = \text{MSR} + (\text{VSC} \times \text{LC}) MSR is Main Scale Reading, VSC is Vernier Scale Coincidence, and LC is Least Count.
Screw Gauge Reading
Reading=PSR+(CSD×LC)\text{Reading} = \text{PSR} + (\text{CSD} \times \text{LC}) PSR is Pitch Scale Reading, CSD is Circular Scale Division, and LC is Least Count.

One of these questions, worked through

JEE Advanced 2016

A length-scale (ll) depends on the permittivity (ε\varepsilon) of a dielectric material, Boltzmann constant (kBk_{\text{B}}), the absolute temperature (TT), the number per unit volume (nn) of certain charged particles, and the charge (qq) carried by each of the particles. Which of the following expression(s) for ll is(are) dimensionally correct?
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Step 1: Identify the dimensions of the given physical quantities.

The dimensions of the length-scale ll, permittivity ε\varepsilon, Boltzmann constant kBk_{\text{B}}, absolute temperature TT, number density nn, and charge qq are determined from their physical definitions.

[l]=L[l] = L [ε]=[q2][F][r2]=M−1L−3T4I2[\varepsilon] = \frac{[q^2]}{[F][r^2]} = M^{-1}L^{-3}T^4 I^2 [kB]=[energy][T]=ML2T−2Θ−1[k_{\text{B}}] = \frac{[\text{energy}]}{[T]} = ML^2 T^{-2} \Theta^{-1} [T]=Θ[T] = \Theta [n]=L−3[n] = L^{-3} [q]=IT[q] = IT

Step 2: Find the dimensions of key combinations.

Determine the dimension of the combination εkBT\varepsilon k_{\text{B}} T.

[εkBT]=(M−1L−3T4I2)(ML2T−2Θ−1)(Θ)[\varepsilon k_{\text{B}} T] = (M^{-1}L^{-3}T^4 I^2)(ML^2 T^{-2} \Theta^{-1})(\Theta) [εkBT]=L−1T2I2[\varepsilon k_{\text{B}} T] = L^{-1}T^2 I^2

Determine the dimension of the combination q2εkBT\dfrac{q^2}{\varepsilon k_{\text{B}} T}.

[q2εkBT]=I2T2L−1T2I2\left[\frac{q^2}{\varepsilon k_{\text{B}} T}\right] = \frac{I^2 T^2}{L^{-1}T^2 I^2} [q2εkBT]=L\left[\frac{q^2}{\varepsilon k_{\text{B}} T}\right] = L

Step 3: Test the dimensional correctness of Option 1.

Check the dimension of the expression nq2εkBT\sqrt{\dfrac{nq^{2}}{\varepsilon k_{\text{B}} T}}.

[nq2εkBT]=[n]⋅[q2εkBT]\left[\sqrt{\frac{nq^{2}}{\varepsilon k_{\text{B}} T}}\right] = \sqrt{[n] \cdot \left[\frac{q^2}{\varepsilon k_{\text{B}} T}\right]} [nq2εkBT]=L−3⋅L\left[\sqrt{\frac{nq^{2}}{\varepsilon k_{\text{B}} T}}\right] = \sqrt{L^{-3} \cdot L} [nq2εkBT]=L−1≠L\left[\sqrt{\frac{nq^{2}}{\varepsilon k_{\text{B}} T}}\right] = L^{-1} \neq L

Step 4: Test the dimensional correctness of Option 2.

Check the dimension of the expression εkBTnq2\sqrt{\dfrac{\varepsilon k_{\text{B}} T}{nq^{2}}}.

[εkBTnq2]=1[n]⋅[εkBTq2]\left[\sqrt{\frac{\varepsilon k_{\text{B}} T}{nq^{2}}}\right] = \sqrt{\frac{1}{[n]} \cdot \left[\frac{\varepsilon k_{\text{B}} T}{q^2}\right]} [εkBTnq2]=L3⋅L−1\left[\sqrt{\frac{\varepsilon k_{\text{B}} T}{nq^{2}}}\right] = \sqrt{L^3 \cdot L^{-1}} [εkBTnq2]=L\left[\sqrt{\frac{\varepsilon k_{\text{B}} T}{nq^{2}}}\right] = L

Step 5: Test the dimensional correctness of Option 3.

Check the dimension of the expression q2εn2/3kBT\sqrt{\dfrac{q^{2}}{\varepsilon n^{2/3} k_{\text{B}} T}}.

[q2εn2/3kBT]=[n−2/3]⋅[q2εkBT]\left[\sqrt{\frac{q^{2}}{\varepsilon n^{2/3} k_{\text{B}} T}}\right] = \sqrt{[n^{-2/3}] \cdot \left[\frac{q^2}{\varepsilon k_{\text{B}} T}\right]} [q2εn2/3kBT]=L2⋅L\left[\sqrt{\frac{q^{2}}{\varepsilon n^{2/3} k_{\text{B}} T}}\right] = \sqrt{L^2 \cdot L} [q2εn2/3kBT]=L3/2≠L\left[\sqrt{\frac{q^{2}}{\varepsilon n^{2/3} k_{\text{B}} T}}\right] = L^{3/2} \neq L

Step 6: Test the dimensional correctness of Option 4.

Check the dimension of the expression q2εn1/3kBT\sqrt{\dfrac{q^{2}}{\varepsilon n^{1/3} k_{\text{B}} T}}.

[q2εn1/3kBT]=[n−1/3]⋅[q2εkBT]\left[\sqrt{\frac{q^{2}}{\varepsilon n^{1/3} k_{\text{B}} T}}\right] = \sqrt{[n^{-1/3}] \cdot \left[\frac{q^2}{\varepsilon k_{\text{B}} T}\right]} [q2εn1/3kBT]=L⋅L\left[\sqrt{\frac{q^{2}}{\varepsilon n^{1/3} k_{\text{B}} T}}\right] = \sqrt{L \cdot L} [q2εn1/3kBT]=L\left[\sqrt{\frac{q^{2}}{\varepsilon n^{1/3} k_{\text{B}} T}}\right] = L

Answer: The dimensionally correct expressions are (2) εkBTnq2\sqrt{\dfrac{\varepsilon k_{\text{B}} T}{nq^{2}}} and (4) q2εn1/3kBT\sqrt{\dfrac{q^{2}}{\varepsilon n^{1/3} k_{\text{B}} T}}.

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Past-paper questions on Units and Measurements