Note on Applications of Dimensional Method

  • Note
  • Things to remember

The dimensional equations have the following applications:

  1. To check the correctness of the equation
  2. To find the relation between the physical quantities involved in the process.
  3. To convert the unit of one physical quantity from one system to another.
  4. To find the relation between the physical quantities involved in the process.

1. To check the correctness of the equation

Example 1:Check the correctness of the following equation.

$$s = ut + \frac {1}{2 }a^2$$

$$s =[M^0L^1T^0 ], u = [M^0L^1T^{-1}], t=[M^0L^0T^1], a = [M^0L^1T^2]$$

In dimensional form, this equation can be written as

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

All the terms of this equation have the same dimensions, then, according to the principle of homogeneity, this equation is dimensionally correct.

This shows that a dimensionally correct equation is not necessarily correct mathematically but a dimensionally wrong equation must be wrong.

Example 2: Check the correctness of the equation \(f=\frac {mv^2}{r}\) where the symbols carry their usual meanings.

Here,

$$f=\frac {mv^2}{r}$$

In dimensional form, this equation can be written as

$$[M^1L^1 T^{-2}] =\frac {[M^1][L^1T^{-1}]}{[L^1]}$$

$$= [M^1L^1T^{-2}]$$

Dimensions of the terms on both sides of the equation are the same. Hence, the equation is dimensionally correct.To find the relation between the physical quantities involved in the process.

2. To find the relation between the physical quantities involved in the process.

Example: Find the time period of a simple pendulum.

The period of a simple pendulum is the time taken to complete one oscillation. The time period ‘t’ is found to depend on length ‘l’ of the pendulum and acceleration due to gravity ‘g’ at that place.

$$\text{i.e. }t\propto l^xg^y$$

$$\text{or, } t =K. l^xg^y \dots (i)$$

where ‘k’ is a dimensionless constant and ‘x’ and ‘y’ are the constants to be determined by the dimensional method.

In dimensional form, equation (i) becomes,

$$[M^0L^0T^1] =[L^x][L^1T^{-2}]$$

$$[M^0L^0T^1] =[M^0L^xL^yT^{-2y}]$$

$$[M^0L^0T^1] =[M^0L^{x + y}T^{-2y}]$$

Equating powers of similar quantities on both sides of the equation,

$$x +y = 0 \dots(ii)$$

$$-2y =1$$

$$y= -\frac12 \dots(iii)$$

From equation (ii) and (iii)

$$x - \frac12 = 0$$

$$x = \frac12$$

Substituting for x and y in equation (i) we obtain

$$\text{or,} t=k.l^\frac12.g^\frac12$$

$$\text{or,} t=k.l^\frac12.\frac{1}.{g^\frac12}$$

$$\text{or,} t=k\sqrt(l/g)$$

Since ‘k’ is a dimensionless constant, we cannot determine its value by this method. However, analytical method shows that k = 2π.

Example: Find the expression for centripetal force.

The centripetal force on a body moving on a circular path depends on the mass ‘m’ and velocity ‘v’ of the body and radius ‘r’ of the circular path.

$$\text(i.e.) F \propto m^x.y^y. r^z$$

$$\text(or) F = m^x.y^y. r^z\dots(i)$$

where ‘k’ is a dimensionless constant and x, y and z are the constants to be determined by the dimensional method.

In dimensional form, equation (i) becomes,

$$[M^1l^1t^-2]=[M]^x. [LT^-1]^y. [L]^z$$

$$[M^1l^1t^-2]=[M]^x. [L^y+z]. [T]^-y$$

Equating powers of similar quantities on both sides of the equation,

$$x = 1\dots(ii)$$

$$x + y = 1\dots(iii)$$

$$ -2 = -y$$

$$ \text{or,}-2 = -y$$

Putting value of y in equation (iii)

$$2 + z =1$$

$$z =-1$$

Substituting for x, y and z in equation (i)

$$F = K m^1v^2.r^1$$

$$F = K.\frac{mv^2}{r}$$

3. To convert the unit of one physical quantity from one system to another.

From principle of homogeneity

$$n_1[M_1]^x[L_1]^y[T_1]^z = n_2[M_2]^2[L_2]^y[T_2]^z$$

where the subscript 1 and 2 represent two different systems of units and n1 and n2 are the numerical values of the physical quantity in these systems.

Example: Convert 1 N into dyne.

In SI (MKS)

n1 = 1

M1 = 1 kg

L1 = 1m

T1 = 1s

In CGS

n2 = ?

M2 = 1 gm

L2 = 1cm

T2 = 1s

Given quantity is force

So,

x = 1, y = 1 and z = -2

From equation (i)

$$\text{or, }n_1 = n_2[\frac{M_2}{M_1}] ^x[\frac{L_2}{L_1}]^y[\frac{T_2}{T_1}]^z $$

$$\text{or, }n_1 = 1[\frac{1 gm}{1 kg}] ^1[\frac{1 cm}{1 m}]^2[\frac{1s}{1s}]^-2 $$

$$\text{or, }n_1 = [\frac{1 gm}{1000 gm}] ^1[\frac{1 cm}{100 cm}]^2[\frac{1s}{1s}]^-2 $$

$$= 1000 \times 100$$

$$\text{or, } n_2 = 10^5$$

Therefore, 1 N = 105 dyne.

Example: Convert 1 erg into Joule.

In SI (MKS)

n1 = 1

M1 = 1 kg

L1 = 1m

T1 = 1s

In CGS

n2 = ?

M2 = 1 gm

L2 = 1cm

T2 = 1s

Given quantity is work

So,

x = 1, y = 2 and z = -2

From equation (i)

$$\text{or, }n_2 = n_1[\frac{M_1}{M_2}] ^x[\frac{L_1}{L_2}]^y[\frac{T_1}{T_2}]^z $$

$$\text{or, }n_2 = 1[\frac{1 kg}{1 gm}] ^1[\frac{1 m}{1 cm}]^1[\frac{1s}{1s}]^-2 $$

$$\text{or, }n_2 = [\frac{1000 gm}{1 gm}] ^1[\frac{100 cm}{1 cm}]^1[\frac{1s}{1s}]^-2 $$

$$= 10^{-3} \times (10^{-2})^2$$

$$ \text{or, }n_1 = 10^{-7}$$

Therefore, 1 ERG= 10-7Joule.

4. To find the dimension of a dimensionless constant.

Example: Find the dimensions of universal gravitational constant.

We have,

$$F = \frac{(Gm_1m_2)}{r^2}$$

$$\text{or,}G = \frac{Fr^2}{Gm_1m_2}$$

$$F = [M^1L^1T^{-2}], m_1 = m_2 = [ML^0T^0], r = [M^0L^1T^0] $$

Here, F = [M1L1T-2], m1= m2 = [ML0T0], r = [ M0L1T0]

$$\text{or,} G = \frac {[M^1L^1T^{-2}][l^2]}{M^2}$$

$$\text{or,} G = [M^{-1}L^3T^{-2}]$$

Dimensions are (-1, 3, -2)

Limitations of dimensional method

  1. This method cannot be used to determine the values of dimensionless constant.
  2. This method cannot be used to derive the formula containing addition and subtraction.
  3. This method cannot be used to derive the relation in which a physical quantity depends on more than three other physical quantities.
  4. This method cannot be used to derive the relations which involve trigonometric, exponential and logarithmic functions.
  5. This method does not give the information that the quantity is scalar or vector.
  6. This method cannot derive the formula containing trigonometric function, the exponential function, logarithmic function etc.

Question: Find the dimensions of η (eta) (coefficient of viscosity) in the formula: V =. (Poiseuiless formula) where ‘V’ is the volume per second, ‘p’ is the pressure, ‘r’ is the radius and l’ is the length.

Here,

The equation is

$$V = \frac{\pi}{8}. \frac{pr^4}{nl}$$

$$\text{or, }\frac{[L^3]}{T} =\frac{[M^1L^2T^{-2}][l^4]}{nl}$$

$$\text{or, } \eta =\frac{[M^1L^2T^{-2}[T^{-1}]][l^4]}{[L^4].[L^2]}$$

$$\eta = [m^1 L^{-1}T^{-1}]$$

Question: Check the correctness of the equation

F = η. A \(\frac vx\)(Newton’s formula) where A is the area, η is the coefficient of viscosity and v is the velocity

Given,

F = η. A

$$F = \eta. A\frac vx$$

$$\text{or,} [M^1L^{-1}T^{-2}] = \frac {[M^1L^{-1}T^{-2} [L^2][L^2T^{-1}]}{[L^1]}$$

$$\text{or,} [M^1L^{-1}T^{-2}] = [M^1L^{-1-1+3}T^{-2}]$$

$$\text{or,} [M^1L^{-1}T^{-2}] = [M^1L^1T^{-2}]$$

All the terms in both sides of the equation have the same dimension. So, this equation is dimensionally correct.

Precision and Significant Figures

Measurement of any physical quantity is never correct. Precision is the degree of exactness or correctness which gives the limitation of measuring instrument. The accuracy depends upon the length of count of measuring instrument. We can measure the diameter more precisely by the vernier callipers than by measuring tape. If a length of a copper cylinder is 5.2 cm measured by the measuring tape, it may be 5.17 measuring by vernier calipers and further 5. 176 if measured by screw gauge. So, length given by screw gauge is the most precise.

Question: Viscous force (F) on a sphere falling through a viscous medium depends on the radius (r) and velocity (v) of the sphere and coefficient of the viscosity (η) of the medium. Find the relation between them.

$$\text{i.e.}F\propto \eta^x. v^y. r^z$$

$$\text{or,}F = K. \eta^x. v^y. r^z\dots(i)$$

where ‘k’ is a dimensionless constant

In dimensional form, equation (i) becomes,

$$[M^1L^1T^{-2}] = [M^1L^{-1}T^{-1}]^x. [L]^y. [L^1T^{-1}]^z$$

$$\text{or, }[M^1L^1T^{-2}] =[M^x][L^{-x+y+z}][T^{-x-z}]$$

Equating powers of similar quantities on both sides of the equation,

$$x = 1\dots(ii)$$

$$ - x + y + z = 1\dots(iii)$$

$$ - 2= -x-z$$

$$ - x + z = 2\dots(iv)$$

Putting value of x in equation (iv)

$$1 + z = 2$$

$$\text{or,} z = 1 $$

Putting value of x and z in equation (iii)

$$-1 + y + 1 = 1$$

$$\text{or,} y= 1 $$

Putting these values in equation (i)

$$F = k\eta r v$$

Question: Find the dimensions of mass in the system in which Force (F), Velocity (V) and Time (T) are the fundamental quantities.

Here,

$$ Mass (M) =\frac {Force(F)}{Acceleration(A)}$$

$$ =\frac {Force(F)}{\frac {Velocity(V)}{Time(T)}}$$

$$ =\frac {[F]}{\frac {[V]}{[T]}}$$

$$=[F^1V^{-1}T^1]$$

The dimensionally correct equation is not necessarily correct mathematically.

Dimensional equations is used to check the correctness of the equation.

Dimensional equations is used to find the relation between the physical quantities involved in the process.

Dimensional equations is used to convert the unit of one physical quantity from one system to another.

Dimensional equations is used to find the relation between the physical quantities involved in the process.

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Very Short Questions

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DISCUSSIONS ABOUT THIS NOTE

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Sujan Poudel

The diameter of steel rod is given as 56.47 /- 0.002.What does it mean?


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Sujan Poudel

What is significant figure?


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makshye narayan sainju

can I find also gravity value ?


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