This MCQ module is based on: Dimensions and Analysis
TOPIC 3 OF 33
Dimensions and Analysis
🎓 Class 11
Physics
CBSE
Theory
Ch 1 – Units and Measurements
⏱ ~14 min
🌐 Language: [gtranslate]
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Dimensions and Analysis
1.4 Dimensions of Physical Quantities
Every physical quantity can be expressed in terms of the fundamental (base) quantities. The dimensional formula expresses which base quantities are involved and with what powers (exponents).
We use the following symbols for dimensions of the seven base quantities:
| Base Quantity | Dimension Symbol |
|---|---|
| Mass | [M] |
| Length | [L] |
| Time | [T] |
| Electric Current | [A] |
| Temperature | [K] |
| Amount of Substance | [mol] |
| Luminous Intensity | [cd] |
In mechanics, most quantities involve only [M], [L], and [T].
Dimensional Formulas of Common Quantities
| Physical Quantity | Expression | Dimensional Formula |
|---|---|---|
| Velocity | displacement / time | \([M^0\,L\,T^{-1}]\) |
| Acceleration | velocity / time | \([M^0\,L\,T^{-2}]\) |
| Force | mass × acceleration | \([M\,L\,T^{-2}]\) |
| Work / Energy | force × displacement | \([M\,L^2\,T^{-2}]\) |
| Power | work / time | \([M\,L^2\,T^{-3}]\) |
| Pressure | force / area | \([M\,L^{-1}\,T^{-2}]\) |
| Momentum | mass × velocity | \([M\,L\,T^{-1}]\) |
| Gravitational constant G | \(F = \frac{Gm_1 m_2}{r^2}\) | \([M^{-1}\,L^3\,T^{-2}]\) |
| Planck's constant h | \(E = h\nu\) | \([M\,L^2\,T^{-1}]\) |
Dimensionless Quantities: Some physical quantities have no dimensions — all base-quantity exponents are zero. Examples include angle (radian), refractive index, strain, relative density, and mathematical constants like \(\pi\).
1.6 Dimensional Analysis and Its Applications
Dimensional analysis is a powerful method with three main applications:
Application 1: Checking Dimensional Consistency
An equation is dimensionally correct if the dimensions of every term on both sides match. Note that a dimensionally correct equation is not necessarily physically correct, but a dimensionally inconsistent equation is always wrong.
Example 1: Check dimensional consistency of \(v = u + at\)
Solution:
LHS: \([v] = [M^0\,L\,T^{-1}]\)
RHS term 1: \([u] = [M^0\,L\,T^{-1}]\)
RHS term 2: \([at] = [M^0\,L\,T^{-2}] \times [T] = [M^0\,L\,T^{-1}]\)
All terms have the same dimensions \([M^0\,L\,T^{-1}]\).
\(\boxed{\text{The equation is dimensionally consistent.}}\)
LHS: \([v] = [M^0\,L\,T^{-1}]\)
RHS term 1: \([u] = [M^0\,L\,T^{-1}]\)
RHS term 2: \([at] = [M^0\,L\,T^{-2}] \times [T] = [M^0\,L\,T^{-1}]\)
All terms have the same dimensions \([M^0\,L\,T^{-1}]\).
\(\boxed{\text{The equation is dimensionally consistent.}}\)
Example 2: Check dimensional consistency of \(s = ut + \frac{1}{2}at^2\)
Solution:
LHS: \([s] = [M^0\,L\,T^0] = [L]\)
RHS term 1: \([ut] = [L\,T^{-1}][T] = [L]\)
RHS term 2: \([\frac{1}{2}at^2] = [L\,T^{-2}][T^2] = [L]\)
(Note: \(\frac{1}{2}\) is dimensionless)
All terms have dimensions \([L]\). \(\boxed{\text{Dimensionally consistent.}}\)
LHS: \([s] = [M^0\,L\,T^0] = [L]\)
RHS term 1: \([ut] = [L\,T^{-1}][T] = [L]\)
RHS term 2: \([\frac{1}{2}at^2] = [L\,T^{-2}][T^2] = [L]\)
(Note: \(\frac{1}{2}\) is dimensionless)
All terms have dimensions \([L]\). \(\boxed{\text{Dimensionally consistent.}}\)
Application 2: Deriving Relations (Method of Dimensions)
Example 3: Derive the expression for the time period of a simple pendulum
Assume the time period \(T\) of a simple pendulum depends on: mass \(m\), length \(l\), and acceleration due to gravity \(g\). Find the relation using dimensional analysis.
Solution:
Let \(T = k\, m^a\, l^b\, g^c\), where \(k\) is a dimensionless constant.
Writing dimensions:
\[[M^0\,L^0\,T^1] = [M]^a \cdot [L]^b \cdot [L\,T^{-2}]^c\] \[[M^0\,L^0\,T^1] = [M^a\,L^{b+c}\,T^{-2c}]\] Equating exponents:
Mass: \(a = 0\)
Length: \(b + c = 0 \Rightarrow b = -c\)
Time: \(-2c = 1 \Rightarrow c = -\frac{1}{2}\)
Therefore: \(b = \frac{1}{2}\), \(a = 0\), \(c = -\frac{1}{2}\)
\[T = k\, m^0\, l^{1/2}\, g^{-1/2} = k\sqrt{\frac{l}{g}}\] \[\boxed{T = k\sqrt{\frac{l}{g}}}\] The dimensionless constant \(k\) turns out to be \(2\pi\) (determined experimentally), so \(T = 2\pi\sqrt{l/g}\).
Key observation: The time period does not depend on mass!
Let \(T = k\, m^a\, l^b\, g^c\), where \(k\) is a dimensionless constant.
Writing dimensions:
\[[M^0\,L^0\,T^1] = [M]^a \cdot [L]^b \cdot [L\,T^{-2}]^c\] \[[M^0\,L^0\,T^1] = [M^a\,L^{b+c}\,T^{-2c}]\] Equating exponents:
Mass: \(a = 0\)
Length: \(b + c = 0 \Rightarrow b = -c\)
Time: \(-2c = 1 \Rightarrow c = -\frac{1}{2}\)
Therefore: \(b = \frac{1}{2}\), \(a = 0\), \(c = -\frac{1}{2}\)
\[T = k\, m^0\, l^{1/2}\, g^{-1/2} = k\sqrt{\frac{l}{g}}\] \[\boxed{T = k\sqrt{\frac{l}{g}}}\] The dimensionless constant \(k\) turns out to be \(2\pi\) (determined experimentally), so \(T = 2\pi\sqrt{l/g}\).
Key observation: The time period does not depend on mass!
Application 3: Converting Units Between Systems
Example 4: Convert 1 newton to CGS (dyne)
Solution:
Force has dimensions \([M\,L\,T^{-2}]\).
In SI: \(1\text{ N} = 1\text{ kg}\cdot\text{m}\cdot\text{s}^{-2}\)
Converting each base unit to CGS:
\(1\text{ kg} = 1000\text{ g} = 10^3\text{ g}\)
\(1\text{ m} = 100\text{ cm} = 10^2\text{ cm}\)
\(1\text{ s} = 1\text{ s}\)
\[1\text{ N} = 1 \times (10^3\text{ g}) \times (10^2\text{ cm}) \times (1\text{ s})^{-2}\] \[= 10^3 \times 10^2 \times 1 = 10^5 \text{ dyne}\] \[\boxed{1\text{ N} = 10^5\text{ dyne}}\]
Force has dimensions \([M\,L\,T^{-2}]\).
In SI: \(1\text{ N} = 1\text{ kg}\cdot\text{m}\cdot\text{s}^{-2}\)
Converting each base unit to CGS:
\(1\text{ kg} = 1000\text{ g} = 10^3\text{ g}\)
\(1\text{ m} = 100\text{ cm} = 10^2\text{ cm}\)
\(1\text{ s} = 1\text{ s}\)
\[1\text{ N} = 1 \times (10^3\text{ g}) \times (10^2\text{ cm}) \times (1\text{ s})^{-2}\] \[= 10^3 \times 10^2 \times 1 = 10^5 \text{ dyne}\] \[\boxed{1\text{ N} = 10^5\text{ dyne}}\]
Example 5: Convert 1 joule to erg
Solution:
Energy has dimensions \([M\,L^2\,T^{-2}]\).
\[1\text{ J} = 1\text{ kg}\cdot\text{m}^2\cdot\text{s}^{-2}\] \[= (10^3\text{ g})(10^2\text{ cm})^2(1\text{ s})^{-2}\] \[= 10^3 \times 10^4 \times 1 = 10^7\text{ erg}\] \[\boxed{1\text{ J} = 10^7\text{ erg}}\]
Energy has dimensions \([M\,L^2\,T^{-2}]\).
\[1\text{ J} = 1\text{ kg}\cdot\text{m}^2\cdot\text{s}^{-2}\] \[= (10^3\text{ g})(10^2\text{ cm})^2(1\text{ s})^{-2}\] \[= 10^3 \times 10^4 \times 1 = 10^7\text{ erg}\] \[\boxed{1\text{ J} = 10^7\text{ erg}}\]
Example 6: Find the dimensions of the gravitational constant G
From Newton's law of gravitation: \(F = \frac{Gm_1 m_2}{r^2}\)
Solution:
Rearranging: \(G = \frac{Fr^2}{m_1 m_2}\)
\[[G] = \frac{[M\,L\,T^{-2}] \cdot [L^2]}{[M] \cdot [M]}\] \[= \frac{[M\,L^3\,T^{-2}]}{[M^2]}\] \[= [M^{-1}\,L^3\,T^{-2}]\] \[\boxed{[G] = [M^{-1}\,L^3\,T^{-2}]}\] SI unit of G: \(\text{N}\cdot\text{m}^2\cdot\text{kg}^{-2}\) or equivalently \(\text{m}^3\,\text{kg}^{-1}\,\text{s}^{-2}\).
Rearranging: \(G = \frac{Fr^2}{m_1 m_2}\)
\[[G] = \frac{[M\,L\,T^{-2}] \cdot [L^2]}{[M] \cdot [M]}\] \[= \frac{[M\,L^3\,T^{-2}]}{[M^2]}\] \[= [M^{-1}\,L^3\,T^{-2}]\] \[\boxed{[G] = [M^{-1}\,L^3\,T^{-2}]}\] SI unit of G: \(\text{N}\cdot\text{m}^2\cdot\text{kg}^{-2}\) or equivalently \(\text{m}^3\,\text{kg}^{-1}\,\text{s}^{-2}\).
Limitations of Dimensional Analysis
Important Limitations:
(i) Cannot determine dimensionless constants (like \(2\pi\) in the pendulum formula, or \(\frac{1}{2}\) in \(s = \frac{1}{2}at^2\)).
(ii) Cannot derive equations involving trigonometric, exponential, or logarithmic functions (since these are dimensionless).
(iii) Cannot distinguish between two quantities having the same dimensions (e.g., work and torque both have \([M\,L^2\,T^{-2}]\)).
(iv) If a quantity depends on more than three unknowns (in mechanics), the method fails because we have only three equations (from M, L, T).
(i) Cannot determine dimensionless constants (like \(2\pi\) in the pendulum formula, or \(\frac{1}{2}\) in \(s = \frac{1}{2}at^2\)).
(ii) Cannot derive equations involving trigonometric, exponential, or logarithmic functions (since these are dimensionless).
(iii) Cannot distinguish between two quantities having the same dimensions (e.g., work and torque both have \([M\,L^2\,T^{-2}]\)).
(iv) If a quantity depends on more than three unknowns (in mechanics), the method fails because we have only three equations (from M, L, T).
Activity — Dimensional Detective
L4 Analyse
Predict first: Can you tell whether the equation \(v^2 = u^2 + 2as\) is dimensionally consistent just by checking the dimensions of each term?
- Write down the dimensional formula of each term in the equation \(v^2 = u^2 + 2as\).
- Check whether all terms have the same dimensions.
- Now consider a made-up equation: \(v = u + at^2\). Check its dimensional consistency.
- What does this tell you about the validity of each equation?
Analysis:
For \(v^2 = u^2 + 2as\):
\([v^2] = [L^2\,T^{-2}]\), \([u^2] = [L^2\,T^{-2}]\), \([2as] = [L\,T^{-2}]\times[L] = [L^2\,T^{-2}]\)
All terms match — dimensionally consistent.
For \(v = u + at^2\):
\([v] = [L\,T^{-1}]\), \([u] = [L\,T^{-1}]\), \([at^2] = [L\,T^{-2}][T^2] = [L]\)
The third term has dimensions [L], not [L T¹] — dimensionally inconsistent. This equation is definitely wrong!
For \(v^2 = u^2 + 2as\):
\([v^2] = [L^2\,T^{-2}]\), \([u^2] = [L^2\,T^{-2}]\), \([2as] = [L\,T^{-2}]\times[L] = [L^2\,T^{-2}]\)
All terms match — dimensionally consistent.
For \(v = u + at^2\):
\([v] = [L\,T^{-1}]\), \([u] = [L\,T^{-1}]\), \([at^2] = [L\,T^{-2}][T^2] = [L]\)
The third term has dimensions [L], not [L T¹] — dimensionally inconsistent. This equation is definitely wrong!
Interactive: Dimension Checker L3 Apply
Select a physical quantity to see its dimensional formula:
Competency-Based Questions
A physics teacher gives students a formula derived by a student: \(F = m v^2 / r\), claiming it represents the centripetal force on an object moving in a circle of radius \(r\) with speed \(v\) and mass \(m\). The teacher asks the class to verify this using dimensional analysis and further explore the method.
Q1. L2 Understand What is the dimensional formula of centripetal force \(F = mv^2/r\)?
Answer: B. \([F] = \frac{[M][L\,T^{-1}]^2}{[L]} = \frac{[M\,L^2\,T^{-2}]}{[L]} = [M\,L\,T^{-2}]\), which is the dimension of force.
Q2. L3 Apply Using dimensional analysis, derive the relationship between the speed \(v\) of transverse waves on a stretched string, the tension \(T\) (force), and the linear mass density \(\mu\) (mass per unit length). (3 marks)
Answer:
Let \(v = k\,T^a\,\mu^b\) where \(k\) is dimensionless.
\([L\,T^{-1}] = [M\,L\,T^{-2}]^a \cdot [M\,L^{-1}]^b\)
\([L\,T^{-1}] = [M^{a+b}\,L^{a-b}\,T^{-2a}]\)
Equating: M: \(a+b = 0\), L: \(a-b = 1\), T: \(-2a = -1\)
From T: \(a = \frac{1}{2}\), then \(b = -\frac{1}{2}\)
Check L: \(\frac{1}{2}-(-\frac{1}{2}) = 1\) ✓
\[\boxed{v = k\sqrt{\frac{T}{\mu}}}\]
Let \(v = k\,T^a\,\mu^b\) where \(k\) is dimensionless.
\([L\,T^{-1}] = [M\,L\,T^{-2}]^a \cdot [M\,L^{-1}]^b\)
\([L\,T^{-1}] = [M^{a+b}\,L^{a-b}\,T^{-2a}]\)
Equating: M: \(a+b = 0\), L: \(a-b = 1\), T: \(-2a = -1\)
From T: \(a = \frac{1}{2}\), then \(b = -\frac{1}{2}\)
Check L: \(\frac{1}{2}-(-\frac{1}{2}) = 1\) ✓
\[\boxed{v = k\sqrt{\frac{T}{\mu}}}\]
Q3. L3 Apply Convert the gravitational constant \(G = 6.67 \times 10^{-11}\) N m² kg² to CGS units. (3 marks)
Answer:
\([G] = [M^{-1}\,L^3\,T^{-2}]\)
\[G_{\text{CGS}} = 6.67 \times 10^{-11} \times \frac{(10^3)^{-1} \times (10^2)^3}{(1)^2}\] \[= 6.67 \times 10^{-11} \times 10^{-3} \times 10^6 \times 1\] \[= 6.67 \times 10^{-11} \times 10^3\] \[\boxed{G = 6.67 \times 10^{-8}\;\text{dyne}\cdot\text{cm}^2\cdot\text{g}^{-2}}\]
\([G] = [M^{-1}\,L^3\,T^{-2}]\)
\[G_{\text{CGS}} = 6.67 \times 10^{-11} \times \frac{(10^3)^{-1} \times (10^2)^3}{(1)^2}\] \[= 6.67 \times 10^{-11} \times 10^{-3} \times 10^6 \times 1\] \[= 6.67 \times 10^{-11} \times 10^3\] \[\boxed{G = 6.67 \times 10^{-8}\;\text{dyne}\cdot\text{cm}^2\cdot\text{g}^{-2}}\]
Q4. L4 Analyse A student uses dimensional analysis to derive the formula for kinetic energy and gets \(E = k\,m\,v^2\). Why can't dimensional analysis determine whether \(k = \frac{1}{2}\), \(k = 1\), or \(k = 2\)? (2 marks)
Answer: Dimensional analysis equates the dimensions on both sides of an equation, but dimensionless constants (pure numbers like \(\frac{1}{2}\), \(2\pi\), etc.) have no dimensions and therefore do not appear in the dimensional equation at all. The value of such constants can only be determined through actual derivation using physical laws or through experiments.
Q5. L5 Evaluate The equation \(v = \sqrt{\frac{2GM}{R}}\) gives the escape velocity from a planet of mass \(M\) and radius \(R\). Verify that this equation is dimensionally correct. (3 marks)
Answer:
LHS: \([v] = [L\,T^{-1}]\)
RHS: \(\left[\sqrt{\frac{GM}{R}}\right] = \left[\frac{[M^{-1}\,L^3\,T^{-2}][M]}{[L]}\right]^{1/2}\)
\(= \left[\frac{[L^3\,T^{-2}]}{[L]}\right]^{1/2} = [L^2\,T^{-2}]^{1/2} = [L\,T^{-1}]\)
LHS = RHS = \([L\,T^{-1}]\). The equation is dimensionally consistent. ✓
LHS: \([v] = [L\,T^{-1}]\)
RHS: \(\left[\sqrt{\frac{GM}{R}}\right] = \left[\frac{[M^{-1}\,L^3\,T^{-2}][M]}{[L]}\right]^{1/2}\)
\(= \left[\frac{[L^3\,T^{-2}]}{[L]}\right]^{1/2} = [L^2\,T^{-2}]^{1/2} = [L\,T^{-1}]\)
LHS = RHS = \([L\,T^{-1}]\). The equation is dimensionally consistent. ✓
Assertion-Reason Questions
Assertion (A): A dimensionally correct equation is not necessarily a physically correct equation.
Reason (R): Dimensional analysis cannot account for dimensionless constants and functions like sine, cosine, or exponential.
Answer: A. A dimensionally correct equation could still be wrong because dimensional analysis cannot detect pure numbers or trigonometric/exponential terms. For example, \(s = ut + at^2\) is dimensionally correct but physically wrong (missing the factor \(\frac{1}{2}\)). The Reason correctly explains this limitation.
Assertion (A): Work and torque have the same dimensional formula \([M\,L^2\,T^{-2}]\).
Reason (R): They are physically the same quantity.
Answer: C. The Assertion is true: both work (\(W = F \cdot d\)) and torque (\(\tau = F \times r\)) have dimensions \([M\,L^2\,T^{-2}]\). However, the Reason is false: they are physically different quantities. Work is a scalar (dot product of force and displacement), while torque is a vector (cross product of position vector and force). This illustrates a limitation of dimensional analysis.
Assertion (A): 1 joule = 10&sup7; erg.
Reason (R): Energy has dimensions \([M\,L^2\,T^{-2}]\), and 1 kg = 10³ g, 1 m = 10² cm.
Answer: A. Using the dimensional formula and unit conversion: \(1\text{ J} = 1\text{ kg}\cdot\text{m}^2\cdot\text{s}^{-2} = 10^3\text{ g}\cdot(10^2)^2\text{ cm}^2\cdot\text{s}^{-2} = 10^3\times 10^4 = 10^7\text{ erg}\). The Reason provides the correct conversion factors.
Frequently Asked Questions - Dimensions and Analysis
What is the main concept covered in Dimensions and Analysis?
In NCERT Class 11 Physics Chapter 1 (Units and Measurements), "Dimensions and Analysis" covers core principles and equations needed for board exam success. The MyAiSchool lesson explains the topic with definitions, derivations, worked examples, and interactive simulations. Key formulas and dimensional analysis are included to build conceptual depth and problem-solving skills aligned with the CBSE 2025-26 syllabus.
How is Dimensions and Analysis useful in real-life applications?
Real-life applications of Dimensions and Analysis from NCERT Class 11 Physics Chapter 1 include engineering design, satellite mechanics, sports biomechanics, transportation safety, and electrical/electronic devices. The MyAiSchool lesson links every concept to a tangible example so students see physics as a problem-solving framework for the physical world, not as abstract formulas.
What are the key formulas in Dimensions and Analysis?
Key formulas in Dimensions and Analysis (NCERT Class 11 Physics Chapter 1 Units and Measurements) are derived step-by-step in the MyAiSchool lesson. Students should memorize the final formula AND understand its derivation for full board marks. Each formula is listed with its dimensional formula, SI unit, applicability range, and common pitfalls. The Summary section at the end of each part includes a quick-reference formula card.
How does this part connect to other parts of Chapter 1?
NCERT Class 11 Physics Chapter 1 (Units and Measurements) is structured so each part builds on the previous one. Dimensions and Analysis connects directly to neighbouring parts via shared definitions, units, and methodology. The MyAiSchool lesson cross-references related concepts with internal links so students can navigate the whole chapter as one connected story rather than disconnected fragments.
What types of CBSE board questions come from Dimensions and Analysis?
CBSE board questions from Dimensions and Analysis typically include: (1) 1-mark MCQs on definitions and formulas, (2) 2-mark short-answer derivations or applications, (3) 3-mark numerical problems with units, (4) 5-mark long-answer derivations followed by application. The MyAiSchool lesson tags each Competency-Based Question (CBQ) with Bloom level (L1-L6) so students know how to study for each weight.
How can students use the interactive simulation effectively?
The interactive simulation in the Dimensions and Analysis lesson allows students to adjust input parameters (sliders or selectors) and see physical quantities update in real time. To use it effectively: (1) try extreme values to understand limiting cases, (2) compare with the analytical formula, (3) check unit consistency, (4) test special configurations from worked examples. The simulation reinforces conceptual intuition that pure formula manipulation cannot.
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