Courses for Kids
Free study material
Offline Centres
Store Icon

Dimensions of Area

Last updated date: 28th May 2024
Total views: 74.1k
Views today: 1.74k
hightlight icon
highlight icon
highlight icon
share icon
copy icon

We can obtain the dimension of any physical quantity from the fundamental units. By fundamental units, we mean the units of mass, length, and time. Here, our focus is on finding the dimension of the area.

In dimensional analysis, for finding out the dimension of any physical quantity, whether it is a force, acceleration, or many more; we represent mass by [M], length by [L], and time by [T].

We know that area of any geometric shape, let’s say, the square is the square of its side. The side is the length, so the dimension for length is [L], and the square of this length will be [L2].

So, the dimensional formula of the area of the square is [M0L2T0].

Two Methods to Write a Dimensional Formula

There are two tricks to write the dimensional formula. They are:

  1. Using formula

  2. Using units 

Let’s understand by an example:

We know that the surface area of the sphere is 4 * π * r2 . . .(1) and the unit of r is meter.

So, r*r becomes m2.

We used the formula and got the unit m2.

Now, we know that m (meter) is the unit of length and the dimension of length is [L], and

The dimensions of constant values, 4 and π are \[[M^{0}L^{0}T^{0}]\].

We know that when two similar quantities with different powers are multiplied, then the two powers are added. Let’s see an example of the same:

\[a^{b} \times a^{c} = a^{b + c}\]

Let’s apply this concept in equation (1):\[[M^{0}L^{0}T^{0}] \times [M^{0}L^{0}T^{0}] \times [M^{0}L^{2}T^{0}]\]

Now adding the powers on common terms:

\[[M^{0 + 0 + 0} L^{0 + 0 + 2} T^{0 + 0 + 0}] = [M^{0} L^{2} T^{0}]\] is the dimensional formula for the surface area of the sphere.

Let’s do the derivation of dimensions of area for different geometric shapes:

Dimensional Formula of the Area for Different Shapes

Let’s find the dimensional formula of area for different geometrical shapes:

  1. Rectangle

We know that area of the rectangle = L x B

⇒ The dimension of L and B  = [L]

So, the dimensional formula for the area of the rectangle = \[[M^{0} L^{2} T^{0}]\]

Total surface area of the rectangle = 2 * (l*b + b*h + h*l) 

The dimension of l = b = h = [L]

So, \[2 \times ([L^{2}] + [L^{2}] + [L^{2}]) = 6 \times [L^{2}]\]

The dimension of 6 (a constant value) = \[[M^{0} L^{0} T^{0}]\]

= \[6 \times [L^{2}] = [M^{0} L^{0} T^{0}] \times [M^{0}L^{0} T^{0}]\] . . .(2)

Now, adding the powers of the term, as shown below:

= \[[M^{0 + 0}L^{0 + 2} T^{0 + 0}]\]

We get the dimensional formula for the total surface area of the rectangle is:

\[[M^{0} L^{2} T^{0}]\]

  1. Circle

(image will be uploaded soon)

We know that area of the circle is πr2

Here, the dimension of radius r is [L]

The dimension formula of \[r^{2} = [M^{0} L^{2} T^{0}]\]

π is constant, so its dimension is \[[M^{0} L^{0} T^{0}]\]

Now, using the formula = \[\pi r^{2} = [M^{0} L^{0} T^{0}] \times [M^{0} L^{0} T^{0}]\]

We know that in multiplication, we add the powers in similar terms.

Let’s add the powers now:

\[[M^{0 + 0} L^{0 + 2} T^{0 + 0}]\]

We get the dimension formula for the area of the circle as \[[M^{0}L^{2}T^{0}]\]. 

  1. Sector

(image will be uploaded soon)

The area of a sector is \[\frac{\theta}{360^{\circ}} \pi r^{2}\] ...(3)

The unit of θ is radian, whose dimension is \[[M^{0}L^{0}T^{0}]\] ….(4)

Similarly, for the constant values, i.e., 360° and π, the dimension is:

\[[M^{0} L^{0} T^{0}]\]….(5)

The dimension of \[r^{2} = [M^{0} L^{2} T^{0}]\]...(6)

On putting the value of (4), (5), and  (6) in equation (3), we get

= \[\frac{[M^{0}L^{0}T^{0}]}{[M^{0}L^{0}T^{0}]} [M^{0}L^{0}T^{0}] \times [M^{0} L^{2} T^{0}]\] …(7)

We know that the powers in the common terms in the division form are subtracted, let’s use the formula mentioned below:

= Using the formula \[\frac{a^{b}}{a^{c}} = a^{b - c}\] and \[a^{b} \times a^{c} = a^{(b + c)}\] we get:

= \[[M^{0 - 0}L^{0 - 0} T^{0 - 0}] \times [M^{0 + 0} L^{2 + 0} T^{0 + 0}] = 1 \times [M^{0} L^{2} T^{0}]\]

So, the dimensional formula for the area of a sector is \[[M^{0} L^{2} T^{0}]\].

  1. Ellipse

(image will be uploaded soon)

The area of the ellipse is π *a * b.

Here, a and b are the lengths of the major and the minor axis, respectively.  The dimensional formula for each of these is [L].

So, for a * b, the dimensional formula will be \[[M^{0} L^{2} T^{0}]\].

We know for π, the formula is \[[M^{0}L^{0}T^{0}]\].

π *a * b = \[[M^{0}L^{0}T^{0}] \times [M^{0}L^{2}T^{0}]\]

So, the dimensional formula for the area of an ellipse = \[[M^{0}L^{2}T^{0}]\].

FAQs on Dimensions of Area

Q1: Which Physical Quantity has a Dimension, but no Unit?

Ans: There is not any physical quantity that has a dimension, but no unit. It’s because the dimensional formula is based on the fundamental units. If there is no unit, then there is no dimension of the physical quantity.

Q2: Write the Seven Basic Fundamental Units.

Ans: Seven fundamental units are as follows:

  1. Mass

  2. Length

  3. Time

  4. Temperature

  5. Electric current

  6. Amount of light

  7. Amount of substance (mole).

Q3: What does Dimensional Formula Mean? Why do we Find Dimensional Formulas?

Ans: Dimensional formula is an expression for the unit of a physical quantity in terms of the fundamental quantities.

The fundamental quantities are mass, length, and time. The dimensional formula of these quantities is M, L, and T, respectively.

The reason why we find the dimensional formulas are:

  1. To check the accuracy of physical relations.

  2. Conversion of the value of a physical quantity from one system of the unit to another system.

  3. To derive a relationship between various physical quantities.

Q4: Write Some Examples of Dimensional Constants.

Ans: The quantities whose values are constant and they possess dimensions.

For example, the velocity of light in vacuum, permeability of free space, gravitational constant, universal gas constant, Boltzmann’s constant, Planck’s constant, etc.