Physical quantities & SI units

Physical quantities & SI units

Physical Quantities & SI Units

Physics is the science of measurement. To describe the world around us quantitatively, we need a standardized system of units. This lesson will build your foundation in physical quantities, the universally accepted SI system, and the powerful technique of dimensional analysis.

What is a Physical Quantity?

A physical quantity is any property of a material or system that can be quantified by measurement. In simple terms, if you can put a number to it, it's a physical quantity. For example, the length of a desk, the mass of a cricket ball, and the temperature of a room are all physical quantities.

Every measurement of a physical quantity has two parts:

1. A numerical value (the magnitude).
2. A unit (the standard of measurement).

For example, if we say a rod is 5 metres long, 5 is the numerical value and metres is the unit. Without the unit, the number is meaningless.

Physical quantities are broadly classified into two categories:

• Fundamental (or Base) Quantities: These are the basic building blocks. They are independent of each other and cannot be expressed in terms of other physical quantities. The table above lists the seven fundamental quantities recognised by the SI system.
• Derived Quantities: These are quantities that are expressed as combinations of fundamental quantities. For example, speed is derived from length and time (distance/time).

{{KEY: type=definition | title=The SI System (Le Système International d'Unités) | text=The SI system is the modern form of the metric system and is the most widely used system of measurement. It is built upon the seven base units listed above, from which all other units are derived. Its coherence and decimal-based nature make scientific and engineering calculations much simpler.}}

Derived Quantities and Their Units

Derived units are created by multiplying, dividing, or raising the base units to certain powers. They can be simple or complex, but they always trace back to the seven fundamental building blocks.

Let's see how this works:

• Area: It's length × width. Both are lengths.
  • Unit: metre × metre = metre² (m²)
  • Dimension: [L] × [L] = [L²]
• Velocity: It's displacement ÷ time.
  • Unit: metre ÷ second = metre/second (m/s or ms⁻¹)
  • Dimension: [L] ÷ [T] = [LT⁻¹]
• Force: From Newton's second law, Force = mass × acceleration. Acceleration itself is velocity ÷ time.
  • Unit: kg × (m/s²) = kg⋅m/s²
  • This combination is given a special name: the newton (N).
  • Dimension: [M] × ([L]/[T²]) = [MLT⁻²]

Here is a table of some common derived quantities. Notice how each one's dimensional formula is composed solely of M, L, T, etc.

{{VISUAL: diagram: A flowchart illustrating how base quantities like Length [L], Mass [M], and Time [T] combine to form derived quantities like Velocity [LT⁻¹], which in turn combines with Mass to form Force [MLT⁻²].}}

Dimensional Analysis: The Physicist's Superpower

Dimensional analysis is the study of the relationship between physical quantities with respect to their dimensions. Dimensions are the fundamental quantities ([M], [L], [T], etc.) that make up a physical quantity.

This powerful tool has three main uses:

1. To check the dimensional consistency of an equation (Principle of Homogeneity).
2. To convert a quantity from one unit system to another (e.g., SI to CGS).
3. To derive a relationship between different physical quantities.

The Principle of Homogeneity

This principle is the cornerstone of dimensional analysis.

{{KEY: type=concept | title=Principle of Homogeneity | text=An equation is dimensionally correct if and only if the dimensions of all the terms on both sides of the equation are the same. You can only add or subtract quantities that have the same dimensions. You can't add apples and oranges; you can't add force to velocity.}}

Let's test this principle on the first equation of motion: v = u + at.

• Dimension of final velocity v: [LT⁻¹]
• Dimension of initial velocity u: [LT⁻¹]
• Dimension of at: [a] × [t] = [LT⁻²] × [T] = [LT⁻¹]

Since all three terms (v, u, and at) have the same dimension [LT⁻¹], the equation is dimensionally correct.

Solved Example 1: Checking Dimensional Correctness (Easy)

Question: Check the dimensional correctness of the equation T = 2π√(L/g), where T is the time period of a simple pendulum, L is its length, and g is the acceleration due to gravity.

Given:

• Equation: T = 2π√(L/g)
• T = Time period, Dimension = [T]
• L = Length, Dimension = [L]
• g = Acceleration due to gravity, Dimension = [LT⁻²]

To Find: Whether the dimensions on the Left Hand Side (LHS) equal the dimensions on the Right Hand Side (RHS).

Solution:

1. Analyze the LHS: The dimension of the time period T is simply [T].

   LHS = [T]
   

2. Analyze the RHS: We need the dimension of 2π√(L/g). The term 2π is a dimensionless constant, so we can ignore it for dimensional analysis.

   RHS = √([L]/[g])
   

3. Substitute the dimensions of L and g into the RHS expression.

   RHS = √([L] / [LT⁻²])
   

4. Simplify the expression inside the square root. The [L] terms cancel out.

   RHS = √(1 / [T⁻²]) = √([T²])
   

5. Take the square root to find the final dimension of the RHS.

   RHS = [T]
   

Conclusion: Since LHS = RHS = [T], the equation is dimensionally correct.

Solved Example 2: Finding Dimensions of a Constant (Medium)

Question: The gravitational force F between two masses m₁ and m₂ separated by a distance r is given by F = G(m₁m₂/r²). Find the dimensional formula for the universal gravitational constant, G.

Given:

• Equation: F = G(m₁m₂/r²)
• Dimension of Force, [F] = [MLT⁻²]
• Dimension of mass, [m₁] = [m₂] = [M]
• Dimension of distance, [r] = [L]

To Find: The dimensional formula for G, [G].

Solution:

1. Rearrange the formula to make G the subject.

   G = F × r² / (m₁m₂)
   

2. Write the dimensional equation by replacing each quantity with its dimensional formula.

   [G] = [F] × [r²] / ([m₁] × [m₂])
   

3. Substitute the known dimensions into the equation.

   [G] = [MLT⁻²] × [L²] / ([M] × [M])
   

4. Simplify the expression by combining the powers of M, L, and T.

   [G] = [ML³T⁻²] / [M²]
   

   [G] = [M¹⁻² L³ T⁻²]
   

Final Answer: The dimensional formula for the universal gravitational constant G is [M⁻¹L³T⁻²].

{{CALLOUT: type=warning | text=Limitations of Dimensional Analysis: While powerful, this method has limits. It cannot determine dimensionless constants (like 2π in the pendulum formula). It fails for equations involving trigonometric, logarithmic, or exponential functions, as their arguments must be dimensionless. It also cannot be used if a quantity depends on more than three fundamental quantities (M, L, T) or if the formula involves a sum or difference of terms.}}

SI Prefixes

To handle very large or very small quantities, the SI system uses a standard set of prefixes. Each prefix represents a power of 10. Knowing these is essential for numerical problems.

{{VISUAL: diagram: A horizontal logarithmic scale showing SI prefixes. The center is 10⁰ (base unit). To the right are positive powers like kilo (10³), mega (10⁶), giga (10⁹). To the left are negative powers like milli (10⁻³), micro (10⁻⁶), nano (10⁻⁹).}}

Solved Example 3: Unit Conversion using Dimensions (Medium)

Question: Convert 1 Newton (the SI unit of force) into its CGS equivalent, the dyne. The CGS system uses centimetre (cm), gram (g), and second (s).

Approach: We will use the dimensional formula of force, [MLT⁻²]. Let the subscript 1 denote the SI system and 2 denote the CGS system. The conversion formula is: n₂ = n₁ [M₁/M₂]ᵃ [L₁/L₂]ᵇ [T₁/T₂]ᶜ. Here, a=1, b=1, c=-2.

Given:

• n₁ = 1 (Newton)
• SI System (1): M₁ = 1 kg, L₁ = 1 m, T₁ = 1 s
• CGS System (2): M₂ = 1 g, L₂ = 1 cm, T₂ = 1 s

To Find: The value of n₂ in dynes.

Solution:

1. Set up the conversion equation using the dimensional powers.

   n₂ = 1 × [1 kg / 1 g]¹ × [1 m / 1 cm]¹ × [1 s / 1 s]⁻²
   

2. Convert the base units in the numerator to match the denominator's units.

   • 1 kg = 1000 g
   • 1 m = 100 cm

   n₂ = 1 × [1000 g / 1 g]¹ × [100 cm / 1 cm]¹ × [1 s / 1 s]⁻²
   

3. Cancel the units inside the brackets.

   n₂ = 1 × [1000]¹ × [100]¹ × [1]⁻²
   

4. Calculate the final numerical value.

   n₂ = 1000 × 100 × 1 = 100000 = 10⁵
   

Final Answer: 1 Newton = 10⁵ dyne.

Solved Example 4: Deriving a Physical Relation (Hard)

Question: The frequency (f) of a stretched string depends on its length (L), its mass per unit length (μ), and the tension (F) in the string. Using dimensional analysis, derive the formula for frequency.

Given:

• Frequency f is proportional to Lᵃ μᵇ Fᶜ
• Dimension of frequency [f] = [T⁻¹] (since f = 1/T)
• Dimension of length [L] = [L]
• Dimension of mass per unit length [μ] = [M/L] = [ML⁻¹]
• Dimension of tension [F] (a type of force) = [MLT⁻²]

To Find: The values of a, b, and c to establish the formula.

Solution:

1. Write the proportionality equation in terms of dimensions.

   [f] = k [L]ᵃ [μ]ᵇ [F]ᶜ
   

   (where k is a dimensionless constant)

2. Substitute the dimensional formulas for each quantity.

   [T⁻¹] = [L]ᵃ [ML⁻¹]ᵇ [MLT⁻²]ᶜ
   

3. Group the powers of M, L, and T on the RHS.

   [M⁰L⁰T⁻¹] = [Mᵇ⁺ᶜ] [Lᵃ⁻ᵇ⁺ᶜ] [T⁻²ᶜ]
   

   (We write LHS dimensions as M⁰L⁰T⁻¹ for easier comparison).

4. Equate the powers of M, L, and T on both sides (Principle of Homogeneity).

   • For M: 0 = b + c (Equation i)
   • For L: 0 = a - b + c (Equation ii)
   • For T: -1 = -2c (Equation iii)

5. Solve the system of equations.

   • From (iii): c = ½
   • Substitute c in (i): b + ½ = 0 → b = -½
   • Substitute b and c in (ii): a - (-½) + ½ = 0 → a + 1 = 0 → a = -1

6. Substitute the powers a, b, and c back into the original relation f = k Lᵃ μᵇ Fᶜ.

   f = k L⁻¹ μ⁻½ F½
   

Final Answer: The derived relation is f = k/L × √(F/μ). Experiments show that the constant k is ½.

Common Numerical Traps

Students often lose marks due to simple oversights. Here are some common traps to avoid.

Multiple Choice Question Bank

Test your understanding with these exam-style questions.

1. Which of the following is NOT a unit of energy? (a) joule (b) N⋅m (c) kg⋅m²/s² (d) kg⋅m/s

💡 Answer: (d) Energy (Work) = Force × Distance. Its dimensions are [ML²T⁻²]. (a) Joule is the SI unit. (b) N⋅m is Force × distance. (c) kg⋅m²/s² is the base unit representation of Joule. (d) kg⋅m/s is the unit of momentum ([MLT⁻¹]).

2. The physical quantity having the dimensional formula [ML⁻¹T⁻²] is: (a) Work (b) Power (c) Pressure (d) Momentum

💡 Answer: (c) Pressure = Force / Area = [MLT⁻²] / [L²] = [ML⁻¹T⁻²]. Work is [ML²T⁻²], Power is [ML²T⁻³], Momentum is [MLT⁻¹].

3. If the units of length and force are doubled, the unit of energy will: (a) be halved (b) be doubled (c) become four times (d) remain unchanged

💡 Answer: (c) Energy = Force × Length. Let the original units be E = F × L. The new units are E' = (2F) × (2L) = 4 (F × L) = 4E. So, the unit of energy becomes four times larger.

4. In the Vander Waals equation, (P + a/V²)(V - b) = RT, what are the dimensions of the constant 'a'? (a) [ML⁵T⁻²] (b) [ML⁻¹T⁻²] (c) [M⁰L³T⁰] (d) [M⁰L⁶T⁰]

💡 Answer: (a) According to the principle of homogeneity, quantities being added must have the same dimensions. Therefore, the dimensions of P must be equal to the dimensions of a/V². [P] = [a/V²] → [a] = [P][V²] [a] = [ML⁻¹T⁻²] × ([L³])² = [ML⁻¹T⁻²] × [L⁶] = [ML⁵T⁻²].

5. The velocity v of a particle depends upon time t as v = At² + Bt + C. The unit of B will be: (a) m (b) m/s (c) m/s² (d) m/s³

💡 Answer: (c) Each term being added must have the dimension of velocity, [LT⁻¹]. So, [v] = [Bt]. [LT⁻¹] = [B] × [T] [B] = [LT⁻¹] / [T] = [LT⁻²]. This corresponds to the unit m/s².

Practice Problem Set

Solve these problems to solidify your skills.

1. Find the dimensional formula for Planck's constant (h) from the equation E = hf, where E is energy and f is frequency.

   💡 Answer: [ML²T⁻¹]

2. The centripetal force F acting on a particle moving in a circle depends on its mass (m), velocity (v), and the radius of the circle (r). Derive the formula for F using dimensional analysis.

   💡 Answer: F = k(mv²/r)

3. A calorie is a unit of heat or energy and it equals about 4.2 J, where 1 J = 1 kg m²/s². Suppose we employ a system of units in which the unit of mass equals α kg, the unit of length equals β m, the unit of time is γ s. Show that a calorie has a magnitude of 4.2 α⁻¹β⁻²γ² in terms of the new units.

   💡 Answer: Follows the conversion method shown in Example 3. [E] = [ML²T⁻²], so n₂ = n₁ (M₁/M₂)¹(L₁/L₂)₂(T₁/T₂)⁻². Substitute the values.

4. The speed of sound v in a medium depends on its modulus of elasticity E and its density ρ. Derive the relationship between v, E, and ρ. (Hint: E has the same dimensions as pressure).

   💡 Answer: v = k√(E/ρ)

5. The position of a particle at time t is given by x = (v₀/α)(1 - e⁻ᵅᵗ), where v₀ is a constant and α > 0. Find the dimensions of v₀ and α.

   💡 Answer: [v₀] = [LT⁻¹], [α] = [T⁻¹]

Summary & Cheatsheet

{{FLASHCARD: q=What is the Principle of Homogeneity? | a=It states that a physical equation is dimensionally correct only if the dimensions of all the terms on both sides of the equation are identical. This is the fundamental rule for checking equations and deriving new ones.}}