Units And MeasurementClass 11 Physics Notes

Introduction

In physics, measurement is the process of comparing a physical quantity with a basic, internationally accepted reference standard called a unit. The result of any measurement is always a number followed by a unit.

For example, if you say the length of a table is "2 metres," the number is "2" and the unit is "metres."

Even though there are many physical quantities, we only need a few fundamental units to describe them all.

• Fundamental or Base Units are the units for fundamental quantities like length, mass, and time.
• Derived Units are units for all other physical quantities, which are created by combining the base units. For example, the unit for speed (metres per second) is derived from the base units for length (metre) and time (second).

A complete set of both base and derived units is called a system of units.

The International System of Units

In the past, scientists used different systems of units, which could be confusing. The three most common systems were:

• CGS system: Based on centimetre, gram, and second.
• FPS (or British) system: Based on foot, pound, and second.
• MKS system: Based on metre, kilogram, and second.

To create a single, global standard, the Système Internationale d' Unites, or SI, was developed. This is now the internationally accepted system for all scientific and commercial work. The SI system is convenient because it's a decimal system, making conversions between units very simple.

The SI system is built on seven base units:

Besides these seven base units, there are two more supplementary units:

1. Plane Angle: Measured in radians (rad). It's the ratio of the length of an arc to its radius.
2. Solid Angle: Measured in steradians (sr). It's the ratio of an intercepted surface area of a sphere to the square of its radius.

Both radians and steradians are dimensionless quantities because they are ratios of similar quantities (length/length or area/area).

Significant Figures

Every measurement has some uncertainty or error. Significant figures (or significant digits) are the digits in a measured value that are known reliably, plus the first digit that is uncertain. They indicate the precision of a measurement.

Reporting a result with more digits than are significant is misleading because it suggests a higher precision than was actually achieved.

Rules for Determining Significant Figures

1. All non-zero digits are significant.
   • Example: 287.5 cm has 4 significant figures.
2. All zeros between two non-zero digits are significant.
   • Example: 2.308 cm has 4 significant figures.
3. For numbers less than 1, zeros to the right of the decimal point but to the left of the first non-zero digit are NOT significant. These are just placeholders.
   • Example: 0.002308 m has 4 significant figures (the leading zeros are not significant).
4. Trailing zeros in a number WITHOUT a decimal point are NOT significant.
   • Example: 12300 cm has 3 significant figures.
5. Trailing zeros in a number WITH a decimal point ARE significant. They indicate the precision of the measurement.
   • Example: 3.500 has 4 significant figures. 0.06900 also has 4 significant figures.

Using Scientific Notation to Avoid Confusion

Trailing zeros can be ambiguous. For example, does 4700 mm have 2, 3, or 4 significant figures? To remove this doubt, we use scientific notation, which expresses a number as $a \times 10^b$, where a is a number between 1 and 10.

In scientific notation, all digits in the base number a are significant.

• $4.700 \times 10^2 \text{ cm}$
• $4.700 \times 10^3 \text{ mm}$
• $4.700 \times 10^{-3} \text{ km}$

In all these cases, it is clear that the measurement has four significant figures.

Rules for Arithmetic Operations with Significant Figures

When you calculate with measured values, the result cannot be more precise than the least precise measurement used.

1. For Multiplication or Division: The final result should have the same number of significant figures as the original number with the least number of significant figures.

   Example

   To find the density of an object with mass $4.237 \text{ g}$ (4 significant figures) and volume $2.51 \text{ cm}^3$ (3 significant figures): $\text{Density} = \frac{4.237 \text{ g}}{2.51 \text{ cm}^3} = 1.68804... \text{ g cm}^{-3}$ Since the least precise value (volume) has 3 significant figures, the answer must be rounded to 3 significant figures. Final Answer: $1.69 \text{ g cm}^{-3}$

2. For Addition or Subtraction: The final result should have the same number of decimal places as the number with the least number of decimal places.

   Example

   To add three masses: $436.32 \text{ g}$, $227.2 \text{ g}$, and $0.301 \text{ g}$. $436.32 + 227.2 + 0.301 = 663.821 \text{ g}$ The least precise measurement, $227.2 \text{ g}$, is only accurate to one decimal place. Therefore, the result must be rounded to one decimal place. Final Answer: $663.8 \text{ g}$

Rounding off the Uncertain Digits

When rounding, follow these conventions:

• If the digit to be dropped is greater than 5, increase the preceding digit by 1.
  • 2.746 rounded to three figures is 2.75.
• If the digit to be dropped is less than 5, leave the preceding digit as it is.
  • 1.743 rounded to three figures is 1.74.
• If the digit to be dropped is exactly 5:
  • If the preceding digit is even, leave it unchanged. (2.745 becomes 2.74).
  • If the preceding digit is odd, increase it by 1. (2.735 becomes 2.74).

Rules for Determining Uncertainty

• Combining Data: When you multiply or divide experimental values, their percentage errors add up. For example, if you measure length as $16.2 \pm 0.6\%$ and breadth as $10.1 \pm 1\%$, the area will have an uncertainty of $0.6\% + 1\% = 1.6\%$.
• Subtraction: Subtracting two similar numbers can reduce the number of significant figures. For example, $12.9 \text{ g} - 7.06 \text{ g} = 5.84 \text{ g}$. Following the rule for subtraction (least decimal places), the answer should be reported as $5.8 \text{ g}$.
• Relative Error: The relative error depends on the number itself, not just the number of significant figures. A measurement of $1.02 \text{ g}$ (with an uncertainty of $\pm 0.01 \text{ g}$) has a relative error of about $1\%$, while a measurement of $9.89 \text{ g}$ (also $\pm 0.01 \text{ g}$) has a relative error of only $0.1\%$.

Given

• Length of side, $l = 7.203 \text{ m}$ (4 significant figures)

To Find

• Total surface area of the cube
• Volume of the cube

Formula

$\text{Surface area} = 6 \times l^2$ $\text{Volume} = l^3$

Solution

The measured length has 4 significant figures, so the calculated area and volume must be rounded to 4 significant figures.

Surface Area: $\text{Area} = 6 \times (7.203 \text{ m})^2 = 311.299254 \text{ m}^2$ Rounding to 4 significant figures gives $311.3 \text{ m}^2$.

Volume: $\text{Volume} = (7.203 \text{ m})^3 = 373.714754 \text{ m}^3$ Rounding to 4 significant figures gives $373.7 \text{ m}^3$.

Final Answer The surface area is $311.3 \text{ m}^2$ and the volume is $373.7 \text{ m}^3$.

Given

• Mass, $m = 5.74 \text{ g}$ (3 significant figures)
• Volume, $V = 1.2 \text{ cm}^3$ (2 significant figures)

To Find

Density of the substance

Formula

$\text{Density} = \frac{\text{Mass}}{\text{Volume}}$

Solution

The number of significant figures in the result of a division is determined by the measurement with the least number of significant figures. Here, the volume has only 2 significant figures.

$\text{Density} = \frac{5.74 \text{ g}}{1.2 \text{ cm}^3} = 4.7833... \text{ g cm}^{-3}$ Rounding the result to 2 significant figures gives $4.8 \text{ g cm}^{-3}$.

Final Answer The density is $4.8 \text{ g cm}^{-3}$.

Dimensions of Physical Quantities

The dimensions of a physical quantity describe its fundamental nature. All physical quantities can be expressed in terms of the seven base quantities, which we call the seven dimensions of the physical world. These are denoted with square brackets:

• Length: [L]
• Mass: [M]
• Time: [T]
• Electric Current: [A]
• Thermodynamic Temperature: [K]
• Luminous Intensity: [cd]
• Amount of Substance: [mol]

The dimensions of a quantity are the powers to which these base quantities are raised. For example:

• Volume: It is length × breadth × height, so its dimensions are $[L] \times [L] \times [L] = [L^3]$.
• Force: It is mass × acceleration, or mass × (length/time²). Its dimensions are $[M] \times \frac{[L]}{[T]^2} = [M L T^{-2}]$.

Dimensional Formulae and Dimensional Equations

A dimensional formula is an expression that shows how a physical quantity is related to the base quantities.

• Dimensional formula for volume: $[M^0 L^3 T^0]$
• Dimensional formula for velocity: $[M^0 L T^{-1}]$
• Dimensional formula for acceleration: $[M^0 L T^{-2}]$

A dimensional equation is an equation formed by setting a physical quantity equal to its dimensional formula.

• $[V] = [M^0 L^3 T^0]$
• $[v] = [M^0 L T^{-1}]$
• $[F] = [M L T^{-2}]$

Dimensional Analysis and its Applications

Dimensional analysis is a powerful tool used to check equations and deduce relationships between physical quantities. It is based on a simple rule called the principle of homogeneity of dimensions:

You can only add or subtract quantities that have the same dimensions.

This means that in any valid physical equation, the dimensions of every term on both sides of the equation must be identical.

Checking the Dimensional Consistency of Equations

We can use the principle of homogeneity to test if an equation is dimensionally correct. If the dimensions don't match, the equation is wrong.

• Dimension of $x$ (distance): $[L]$
• Dimension of $x_0$ (initial position): $[L]$
• Dimension of $v_0 t$ (velocity × time): $[L T^{-1}] \times [T] = [L]$
• Dimension of $\frac{1}{2} a t^2$ (acceleration × time²): $[L T^{-2}] \times [T^2] = [L]$ (Note: the number $\frac{1}{2}$ is dimensionless)

Since every term in the equation has the dimension of length [L], the equation is dimensionally correct.

To Check

The dimensional consistency of the equation $\frac{1}{2} m v^2 = mgh$.

Solution

Dimensions of the Left-Hand Side (LHS): The term is $\frac{1}{2} m v^2$. The constant $\frac{1}{2}$ has no dimensions.

• Dimension of mass, $[m] = [M]$
• Dimension of velocity, $[v] = [L T^{-1}]$
• So, $[v^2] = [L T^{-1}]^2 = [L^2 T^{-2}]$
• Dimensions of LHS = $[M] \times [L^2 T^{-2}] = [M L^2 T^{-2}]$

Dimensions of the Right-Hand Side (RHS): The term is $mgh$.

• Dimension of mass, $[m] = [M]$
• Dimension of acceleration due to gravity, $[g] = [L T^{-2}]$
• Dimension of height, $[h] = [L]$
• Dimensions of RHS = $[M] \times [L T^{-2}] \times [L] = [M L^2 T^{-2}]$

Conclusion: Since the dimensions of the LHS ($[M L^2 T^{-2}]$) are the same as the dimensions of the RHS ($[M L^2 T^{-2}]$), the equation is dimensionally correct.

Given

• Dimensions of kinetic energy, $[K] = [M L^2 T^{-2}]$
• Dimensions of mass, $[m] = [M]$
• Dimensions of speed, $[v] = [L T^{-1}]$
• Dimensions of acceleration, $[a] = [L T^{-2}]$

To Find

Which formulas are dimensionally incorrect.

Solution

We will check the dimensions of the right side of each formula.

(a) $[m^2 v^3] = [M]^2 [L T^{-1}]^3 = [M^2 L^3 T^{-3}]$. This does not match $[K]$. Ruled out.

(b) $[(1/2) m v^2] = [M] [L T^{-1}]^2 = [M L^2 T^{-2}]$. This matches $[K]$. Dimensionally correct.

(c) $[ma] = [M] [L T^{-2}] = [M L T^{-2}]$. This does not match $[K]$. Ruled out.

(d) $[(3/16) m v^2] = [M] [L T^{-1}]^2 = [M L^2 T^{-2}]$. This matches $[K]$. Dimensionally correct.

(e) In $K = (1/2) m v^2 + ma$, two quantities with different dimensions ($[M L^2 T^{-2}]$ and $[M L T^{-2}]$) are being added. This violates the principle of homogeneity. Ruled out.

Final Answer Formulas (a), (c), and (e) can be ruled out on the basis of dimensional arguments.

Deducing Relations Among Physical Quantities

Dimensional analysis can also be used to derive a formula if you know which physical quantities it depends on.

Given

• Time period $T$ is a function of length ($l$), mass ($m$), and acceleration due to gravity ($g$).

To Find

An expression for the time period $T$.

Solution

1. Assume a relationship. We assume the relationship is a product of the quantities raised to some powers $x, y,$ and $z$. $T = k l^x g^y m^z$ where $k$ is a dimensionless constant.

2. Write the dimensional equation. We write the dimensions for each quantity in the equation.

   • $[T] = [T^1] = [M^0 L^0 T^1]$
   • $[l] = [L]$
   • $[g] = [L T^{-2}]$
   • $[m] = [M]$

   Substituting these into the equation: $[M^0 L^0 T^1] = [L]^x [L T^{-2}]^y [M]^z$ $[M^0 L^0 T^1] = L^{x+y} T^{-2y} M^z$

3. Equate the exponents. For the equation to be dimensionally consistent, the exponents of M, L, and T must be equal on both sides.

   • For M: $z = 0$
   • For T: $-2y = 1 \implies y = -1/2$
   • For L: $x + y = 0 \implies x + (-1/2) = 0 \implies x = 1/2$

4. Substitute the exponents back into the formula. $T = k l^{1/2} g^{-1/2} m^0$ $T = k \sqrt{\frac{l}{g}}$

Final Answer The expression for the time period is $T = k \sqrt{\frac{l}{g}}$. Dimensional analysis cannot determine the value of the constant $k$ (which is actually $2\pi$).