Dimension of Area

Introduction

Dimension of area place an important role in the measurement process. Measurement is the most valuable part of science. Also, it is most important for living too. Measurement plays a crucial role in our lives. It is the foundation for all technological development. The accuracy of measurement is more important in science. Metrology is the comparison of any physical quantity with its standard quantity. The measurement method is essentially to measure a quantity, it is always compared to a standard quantity. A unit is defined as a physical quantity having a definite numerical value accepted by a rule or conversion. In earlier times, different unit systems were used by different nations. But at the end of World War two, there was a need for a global unit system.

Dimensional Analysis

• All physical quantities are described by some set of seven basic quantities. These seven fundamental dimensions are the dimensions of the physical world.

• Dimensions are indicated by [ ](square brackets). In mechanics, there are three dimensions: the dimension of length [L], the dimension of mass [M] and the dimension of time [T].

• In electrochemistry, the dimension of current is [A], in thermodynamics the dimension of temperature is [K], in optics the dimension of light intensity is [cd] or [Φ], and the dimension of the volume of matter is [mol].

Applications of dimensional analysis

• This method is used to convert a physical quantity from one unit of measurement to another unit of measurement.

• It is used to check if the given equation is dimensionally correct.

• This method is based on the principle that the nume rical value (n) of a quantity multiplied by its unit (u) is a constant.

$$\mathrm{n[u]=constant}$$

$$\mathrm{or}$$

$$\mathrm{n_1[u_1]=n_2[u_2]}$$

• Let us assume that a physical quantity has dimension 'a' of mass, dimension 'b' of length and dimension 'c' of time.

• If the basic units of one unit of measurement are M₁,L₁ and T₁, the base units of another unit aisM₂,L₂ and T₂ respectively, then

$$\mathrm{n_1 [M_1^a L_1^b T_1^c]=n_2 [M_2^a L_2^b T_2^c]}$$

• From this, the numerical value of a physical quantity can be converted from one unit of measurement to another.

Fundamental Quantities and their dimensions

• The dimensions of a physical quantity are the enhanced steps of the dimensions of the fundamental quantities to obtain the dimensions of the physical quantity.

  $$\mathrm{For\: example\:, velocity\:=\frac{displacement}{Time}=\frac{[L]}{[T]}=[M^0 LT^{-1}]}$$

• So, the dimension of velocity has zero dimension of mass, 1 dimension of length and -1 dimension of time.

• A dimensional term is an equation that tells us which basic dimensions are used to specify a physical quantity and how.

  $$\mathrm{For\: example\:, [M^0 LT^{-2}]\: is\: a\: dimensional\: expression\: of\: acceleration.}$$

• An equation describing the dimensional formula of a physical quantity in the form of an equation is called a dimensional equation.

• Depending on the dimensions, physical quantities can be classified into four categories.

Dimensional variables

• Any physical quantity having dimension and different values are called a dimensional variable.

  Ex:- Area, Volume, Velocity etc.

Dimensionless variables

• Any physical quantities which are dimensionless but have different values are called dimensionless variables.

  Ex: - Contrast, strain, refractive index etc.

Dimensional constants

• • Any physical quantities that have a constant value with respect to dimension are called dimensionless constants.

  Ex: - Planck's constant and gravitational constant etc.

Dimensionless constants

• If a constant is dimensionless, they are called a dimensionless constant.

  Ex: π,e (Euler number) numbers and so on

Dimension of Area

• The area is a measure of how much two-dimensional surfaces or shapes spread over a surface.

$$\mathrm{Area (Rectangle)=length\times breadth}$$

$$\mathrm{∴ A=[M^0 L^1 T^0 ]×[M^0 L^1 T^0 ]=[M^0 L^2 T^0 ]}$$

$$\mathrm{A=[L^2 ]}$$

Solved examples

1.Convert pressure at 76 cm of mercury to Nm^-2 in the dimensional method.

Solution −

76cm mercury pressure in the CGS method

$$\mathrm{P_1= 76 \times 13.6\times 980\: dyne \:\:cm^{-2}}$$

$$\mathrm{Dimensional\: formula\: of\: pressure\: [ML^{-1} T^{-2}]}$$

$$\mathrm{P_1 [M_1^a L_1^b T_1^c]=P_2 [M_2^a L_2^b T_2^c]}$$

$$\mathrm{P_2=P_1 [\frac{M_1}{M_2} ]^a [\frac{L_1}{L_2} ]^b [\frac{T_1}{T_2} ]^c}$$

$$\mathrm{M_1=1g; M_2=1kg}$$

$$\mathrm{L_1=1cm\:\:;\: L_2=1m}$$

$$\mathrm{T_1=1s\:\:\:; T_2=1s}$$

Therefore a=1,b=-1 and c=-2

$$\mathrm{P_2=76 \times 13.6\times 980 [\frac{1g}{1kg}]^1 [\frac{1cm}{1m}]^{-1} [\frac{1s}{1s}]^{-2}}$$

$$\mathrm{P_2=76 × 13.6× 980 [\frac{10^{-3} kg}{1kg}]^1 [\frac{10^{-2} m}{1m}]^{-1} [\frac{1s}{1s}]^{-2}}$$

$$\mathrm{P_2=76 × 13.6× 980 ×[10^{-3} ]\times 10^2}$$

$$\mathrm{P_2=1.01\times 10^5\: Nm^{-2}}$$

2.Verify the equation $\mathrm{\frac{1}{2} mv^2=mgh}$ is dimensionally correct by the dimensional analysis method

$$\mathrm{Dimensional\: formula\: of\: \frac{1}{2} mv^2=mgh=[M] [LT^{-1} ]^2=[ML^2 T^{-2}] }$$

$$\mathrm{Dimensional\: formula\: of\: mgh=[M][LT^{-2} ][L]=[ML^2 T^{-2}]}$$

$$\mathrm{∴[ML^2 T^{-2}]=[ML^2 T^{-2}]}$$

Dimensions are equal on both sides. Hence the equation $\mathrm{\frac{1}{2} mv^2=mgh}$ is dimensionally correct.

Conclusion

Metrology is the comparison of any physical quantity with its standard quantity. Measurement is the most valuable part of science. Also, it is most important for living too. Measurement plays a crucial role in our lives. A unit is defined as a physical quantity having a definite numerical value accepted by a rule or conversion. All physical quantities are described by some set of seven basic quantities. These seven fundamental dimensions are the dimensions of the physical world. It is used to check if the given equation is dimensionally correct. The area is a measure of how much two-dimensional surfaces or shapes spread over a surface.

FAQs

1. Define the principle of homogeneity of dimensions

According to this principle, the dimensions of each element in an equal same equal. For example, in the equation v²=u²+2as the magnitudes of v², u² and 2as are the same and equal to[L² T^-2].

2. What are the limitations of dimensional analysis?

• This method cannot determine whether a given quantity is a vector quantity or a scalar quantity.

• It is not possible to find correlations of equations involving trigonometry, exponential, and logarithms.

• This method cannot be used for equations involving more than three physical quantities.

• In this method it is possible to verify that an equation is dimensionally correct, but not to find its true equation.

3. What is unit conversion?

Unit conversion refers to the process of changing a unit of a physical quantity from one unit to another unit. To simplify the calculations, unit conversion is necessary to express the results necessarily.

4. What are Screw Gauge and Vernier Calliper?

• A screw gauge is a measuring instrument accurate to one-hundredth of a millimeter (0.01 mm). This instrument can measure the diameter of a thin wire, the thickness of a thin metal plate, etc.

• Generally, the dimensions of various objects can be calculated using a vernier calliper. After calculating the length, width, and height of objects, the volume of the object is.

5. Explain the steps for unit conversion

• First Pick out the unit which has to be converted. The selected unit to be converted is called starting unit.

• Choose the desired unit, which is converted by the selected unit. This unit is known as the desired unit.

• Find the suitable unit conversion factors.

• After finding the factors, apply the numerical calculations to find the same value of the selected unit. As result, the desired unit value and selected unit value must be equal.

6. What are fundamental and derived units?

The units used to measure basic quantities are called fundamental units and are units obtained by appropriate multiplication or division of the fundamental units to measure other physical quantities also called derived units.