WBCHSE Class 11 Physics Dimensions Of Physical Quantities Notes

Measurement And Dimension Of Physical Quantity

Dimensions Of Physical Quantities

The dimension of a physical quantity is its relationship with the seven quantities, each of which has been assigned, by convention, a base unit.

These seven quantities, shown earlier are

1. Length (l),
2. Mass (m),
3. Time (t),
4. Electric current (l),
5. Thermodynamic temperature (T or θ),
6. Amount of substance (n), and
7. Luminous intensity (I_v).

The dimension of each of them is expressed as a single symbolic factor, as shown.

Dimensions Of The Base Quantities

A short and compact notation for expressing the dimension of a quantity is as follows : [length] = L or [l] = L, which is read as “the dimension of length is L”. Here L is a factor that may have multiplication or division with other similar factors, as and when demanded by the relationships among different physical quantities.

Dimensions Of The Base Quantities Example:

Volume (V): v = length x breadth x height. Breadth and height are quantities equivalent to length; each has a dimension L.

So, the dimension of volume = L x L x L, i.e., [V] = L³.

Density (ρ): \(\rho=\frac{\text { mass }}{\text { volume }} \text {. Then, }[\rho]=\frac{[\text { mass }]}{[\text { volume }]}\)

Now, [mass] = M and [volume] = L³.

So, the dimension of density, i.e., \([\rho]=\frac{M}{L^3}=M L^{-3} \text {. }\)

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Dimensions From Units:

1. The SI unit of density is kg/m³. This unit itself shows that density is actually \(\frac{\text { mass }}{(\text { length })^3}\), i.e., its dimension is \(\frac{\mathrm{M}}{\mathrm{L}^3} \text { or } \mathrm{ML}^{-3}\). There are many quantities, like density, for which the dimensions may directly be obtained from the units.

2. The SI unit of force is Newton (N). This derived unit, however, does not show its direct relationship with the base units. To get that, some convenient physical relationship is to be used. Here, a useful relation is force = mass x acceleration

So, the unit of force = unit of mass x unit of acceleration

= kg x m/s² (in SI)

Then, we know that force actually is \(\frac{\text { mass } \times \text { length }}{(\text { time })^2}\)

Hence its dimension is \(\frac{M L}{T^2}\) or MLT^-2.

The last example shows that the dimension always clearly relates a derived quantity with the base quantities, whereas its unit may or may not. A useful physical formula is often necessary to get this dimensional relationship.

In essence, the connection of all derived physical quantities with the base ones is explicitly displayed by the dimensions, but not always by the conventional units.

Dimensionless Quantities: Some physical quantities are actually ratios of other quantities that have the same dimensions. As a result, the ratio becomes dimensionless.

Dimensionless Quantities Example

1. Angle (θ): The angle subtended by a circular arc at its center is defined as,

θ = \(\frac{\text { length of the circular arc }}{\text { radius of the circle }}\)

Here, both the length and the radius have the dimension of length, i.e., L.

So, the dimension of angle is, \([\theta]=\frac{L}{L}=L^0=1\)

Dimension 1 actually indicates that angle θ is a dimensionless quantity.

2. Specific gravity (s): By definition, the specific gravity of the material of a body is,

s = \(\frac{\text { mass of the body }}{\text { mass of an equal volume of water }}\)

So, the dimension of specific gravity, [s] = \(\frac{M}{M}\) = M⁰ = 1. This means that specific gravity is a dimensionless quantity.

If a quantity is dimensionless, its dimension is written as 1. However, expressions like L⁰, M⁰, L⁰M⁰T^0, etc. are equally valid.

Even a dimensionless quantity may have units. Such units are to be assigned to denote different methods of scaling the quantity. For example, an angle θ is dimensionless; but radian and degree are two popular units for the measurement of θ (there are also other, mostly obsolete, units of angle). They actually correspond to the following scalings:

Angle in radian \(\left(\theta^c\right)=\frac{\text { arc }}{\text { radius }} ;\)

Angle in degree \(\left(\theta^{\circ}\right)=\frac{180}{\pi} \times \frac{\text { arc }}{\text { radius }}\)

It is to be noted that,

1. All real numbers are dimensionless;
2. A few physical constants (like π, the ratio between the circumference and diameter of any circle) are dimensionless, whereas some others (like the velocity of light or gravitational constant)

Physical Quantities Of The Same Dimension: Every physical quantity has a definite dimension. But the converse is not true a dimensional expression alone cannot identify the corresponding physical quantity. There are many examples where different quantities have the same dimension. A few of them are given in the following table.

Dimensional Analysis: Any theoretical probe involving the dimensions of different fundamental and derived physical quantities is called dimensional analysis. Usually, from this analysis,

1. The dimensional correctness of a physical expression or equation can be checked
2. The value of a measured quantity can be converted from one system of units to another;
3. Relations among different physical quantities can be determined.

Dimensional analysis stands on the following basic principle:

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The Principle Of Dimensional Homogeneity: This principle states that, in any mathematical expression or equation involving physical quantities, each term in the expression or each term on either side of the equation must have the same dimension.

For example, let us take an expression a + be, or an equation d = a+bc. The principle asserts that, each term—a, be, and d—in the expression or the equation has the same dimension.

This is obvious, because, for example, some mass cannot be added with some length, or some work can never be equal to some density, etc.

Further, in a polynomial series of form 1 + ax + bx² + cx³ + …., the variable x must be dimensionless.

Otherwise, the different terms of the expression would have different dimensions and they cannot be added. It is to be noted that functions like e^x, sinx, cosx, etc. are widely used in physics each of them is actually a polynomial series; hence, x should be dimensionless.

Dimensional Correctness Of An Equation: From the principle of dimensional homogeneity, by analyzing the dimensions of both sides of an equation, the dimen¬sional correctness of an equation may be checked.

Dimensional Correctness Of An Equation Example: Let us check the dimensional correctness of the equation of motion, s = ut + \(\frac{1}{2}\) at².

On LHS, \([s]=\mathrm{L}\)

On RHS, \([u]=\mathrm{LT}^{-1},[t]=\mathrm{T} and [a]=\mathrm{LT}^{-2}\)

i.e., RHS = \([u t]+\left[\frac{1}{2} a t^2\right]\)

= \(L T^{-1} \times \mathrm{T}+1 \times \mathrm{LT}^{-2} \times \mathrm{T}^2=\mathrm{L}+\mathrm{L}=\mathrm{L}\)

Thus, the dimension of physical quantities on both sides of the equation is L. So the equation is dimensionally correct

Conversion Between Unit Systems: The following examples are sufficient to illustrate the method:

1. Relation Between Joule And Erg: 1 joule (J) and 1 erg are the SI and CGS units of work, respectively. Let, 1 J = n erg.

The dimension of work is ML²T^-2. Using the SI and CGS fundamental units directly as per the dimension, we have

1 J = \(n \mathrm{erg} \quad \text { or, } 1 \mathrm{~kg} \cdot \mathrm{m}^2 \cdot \mathrm{s}^{-2}\)

= \(n \mathrm{~g} \cdot \mathrm{cm}^2 \cdot \mathrm{s}^{-2}\)

or, n = \(\frac{\mathrm{kg} \cdot \mathrm{m}^2 \cdot \mathrm{s}^{-2}}{\mathrm{~g} \cdot \mathrm{cm}^2 \cdot \mathrm{s}^{-2}}=\left(\frac{\mathrm{kg}}{\mathrm{g}}\right) \cdot\left(\frac{\mathrm{m}}{\mathrm{cm}}\right)^2=1000 \times(100)^2\)

= \(10^7 \quad[\mathrm{As} 1 \mathrm{~kg}=1000 \mathrm{~g} \text { and } 1 \mathrm{~m}=100 \mathrm{~cm}]\)

Thus, 1 J = \(10^7\) erg and 1erg = \(10^{-7} \mathrm{~J}\).

2. Conversion Of The Value Of Young’s Modulus From CGS System To SI: The value of Young’s modulus (Y) of iron is 2 x 10¹² dyn · cm^-2. The corresponding SI unit is N · m^-2. Let, 2 x 10¹² dyn · cm^-2 = n N · m^-2.

The dimension of Y is ML^-1T^-2.

Using the fundamental units directly in the dimension, we get

2 x \(10^{12} \mathrm{dyn} \cdot \mathrm{cm}^{-2}=n \mathrm{~N} \cdot \mathrm{m}^{-2}\)

or, \(2\times 10^{12} \mathrm{~g} \cdot \mathrm{cm}^{-1} \cdot \mathrm{s}^{-2}=n \mathrm{~kg} \cdot \mathrm{m}^{-1} \cdot \mathrm{s}^{-2}\)

or, n= \(2 \times 10^{12} \times \frac{\mathrm{g} \cdot \mathrm{cm}^{-1} \cdot \mathrm{s}^{-2}}{\mathrm{~kg} \cdot \mathrm{m}^{-1} \cdot \mathrm{s}^{-2}}\)

= \(2 \times 10^{12} \times\left(\frac{\mathrm{g}}{\mathrm{kg}}\right) \times\left(\frac{\mathrm{m}}{\mathrm{cm}}\right)\)

= \(2 \times 10^{12} \times \frac{1}{1000} \times 100=2 \times 10^{11}\)

So, \(Y=2 \times 10^{11} \mathrm{~N} \cdot \mathrm{m}^{-2}\)

Relationship Among Different Physical Quantities

1. Dependence of time period (T) of a simple pendulum on its effective length (t), acceleration due to gravity (g), and mass of the pendulum bob (m):

Let us assume,

T ∝ m^x [when T and n are constant]

T ∝ l^y [when l and n are constant]

T ∝ g² [when l and T are constant]

∴ \( T \propto m^x D_{g^2}\)

or, T = \(k m^x D g^2\), where k is a dimensionless constant of proportionality and x,y, z are numeric indices.

Now, [m] = \(\mathrm{M},[l]=\mathrm{L}\) and \([g]=\mathrm{LT}^{-2}\).

Expressing LHS and RHS in terms of dimensions, \(M^0 L^0 T=M^x \times L^y \times\left(L^{-2}\right)^z\)

or, \(M^0 L^0 T=M^x L^{y+z}-2 z\)

Equating powers of the same bases from both sides we get, x=0

y + z = 0 or, y = -z

-2 z=1 or, z = \(-\frac{1}{2}\)

∴ y = \(+\frac{1}{2} \quad therefore T=k \sqrt{\frac{l}{g}} .\)

We cannot determine the value of k from this analysis. It is also important to note that, x = 0 means the time period does not depend on the mass of the pendulum.

2. Dependence of frequency of vibration (n) of a stretched string on its length (l), mass per unit length (μ), and tension in the string (T):

Let us assume,

n \(\propto l^x\) (when T and μare constant)

n  ∝ T^y (when l and μ are constant)

n \(\propto \mu^z\) (when l and T are constant)

∴ \(n \propto \mu^x T_{\mu^2}\)

or, \(n=k l^x T^y \mu^z\), where k is a dimensionless constant of proportionality and x, y, z are numeric indices.

Here, \([n]=\mathrm{T}^{-1}, \quad[l]=\mathrm{L}, \quad[T]=\mathrm{MLT}^{-2} \quad and [\mu]=\mathrm{ML}^{-1}\).

Then, \(\mathrm{M}^0 \mathrm{~L}^0 \mathrm{~T}^{-1}=\mathrm{L}^x \cdot\left(\mathrm{MLT}^{-2}\right)^y \cdot\left(\mathrm{ML}^{-1}\right)^z\)

or, \(\mathrm{M}^0 \mathrm{~L}^0 \mathrm{~T}^{-1}=\mathrm{L}^{x+y-z} \cdot \mathrm{M}^{y+z} \cdot \mathrm{T}^{-2 y}\)

Equating the power of same bases, x + y – z = 0, -2y = -1

So, y = \(x\frac{1}{2}\)

Then, z = –\(frac{1}{2}\) and x = -1

Hence, n = \(\frac{k}{l} \sqrt{\frac{T}{\mu}}\).

3. Dependence of viscous force (F) on the radius of a ball falling through a viscous fluid (a), coefficient of viscosity of the fluid (η), and terminal velocity (v) of the ball:

Let us assume,

F ∝ a^x [when 77 and v are constant]

F ∝ η^y [when a and v are constant]

F ∝ v^z [when a and η are constant]

∴ \(F \propto a^x \eta^y \nu^z\)

or, \(F \propto a^x \eta^y \nu^z\), where k is a dimensionless constant of proportionality and x, y, z are numeric indices.

Here, [F] = \(\mathrm{MLT}^{-2}, \quad[a]=\mathrm{L}, \quad[\eta]=\mathrm{ML}^{-1} \mathrm{~T}^{-1} and [\nu]=\mathrm{LT}^{-1}\).

So, \(M L T^{-2}=L^x \cdot\left(M L^{-1} T^{-1}\right)^y \cdot\left(\mathrm{LT}^{-1}\right)^z\)

or, \(\mathrm{MLT}^{-2}=\mathrm{L}^{x-y+z} \cdot \mathrm{M}^y \cdot \mathrm{T}^{-y-z}\)

Equating powers of same bases, x-y+z =1; y =1

and -y-z = -2 or, y+z = 2

Then, z = 1 and x = 1

∴ F = kaηv

Detailed dimensional analysis shows that F does not depend on the densities of the fluid and of the material of the ball. We omitted those details here.

Limitations Of Dimensional Analysis

1. The value of a constant in an equation cannot be determined.
   • Example: In a simple pendulum the time period of the bob depends on length and acceleration due to gravity.
   • The relation is \(T=k \sqrt{\frac{l}{g}}\). We cannot determine the value of constant k from dimensional analysis.
2. No relation, containing a constant that is not dimensionless, can be established.
   • Example: We cannot determine the nature of the depen¬dence of force on mass and distance in Newton’s law of gravitation, as it contains a constant G which is not dimensionless.
3. If any relation contains a dimensionless quantity, we cannot determine its nature of dependence on others present in the relation.
   • Example: If a body is displaced by a distance s when a force is acting on it, then work done by the force depends on the applied force, displacement of the body, and the angle between the force and displacement.
   • This is expressed by the relation W = Fs cos θ, where θ is the angle between the direction of applied force and that of displacement. As θ is a dimensionless quantity we cannot determine the relation by dimensional analysis.
4. If a physical quantity depends on different quantities having the same dimension, the relation among them cannot be obtained.
   • Example: The volume of a right cylinder depends on its radius and on its length. Here, the radius and the length are entirely different quantities related to the cylinder, but they have the same dimension the dimension of length. So, the relation V ∝ r²l cannot be obtained from dimensional analysis.

Measurement And Dimension Of Physical Quantity Numerical Examples

Example 1. What will be the conversion factor when you change a value expressed in Newton to dyne?
Solution:

N and dyn are the units of force in SI and CGS systems respectively.

The dimension of force, [F] = MLT^-2

Let 1 N = n dyn

Then using the base units in the dimension, \(1 \mathrm{~kg} \cdot \mathrm{m} \cdot \mathrm{s}^{-2}=n \mathrm{~g} \cdot \mathrm{cm} \cdot \mathrm{s}^{-2}\)

or, n = \(\frac{1 \mathrm{~kg}}{1 \mathrm{~g}} \times \frac{1 \mathrm{~m}}{1 \mathrm{~cm}} \times\left(\frac{1 \mathrm{~s}}{1 \mathrm{~s}}\right)^{-2}\)

= \(\frac{10^3 \mathrm{~g}}{1 \mathrm{~g}} \times \frac{10^2 \mathrm{~cm}}{1 \mathrm{~cm}} \times 1=10^5\)

∴ \(1 \mathrm{~N}=10^5 \mathrm{dyn}\)

So, the conversion factor is \(10^5\).

Example 2. Taking electric potential V as a fundamental quantity instead of electric current, find the dimension of electric current in terms of the dimension of the electric potential.
Solution:

Given

Taking electric potential V as a fundamental quantity instead of electric current

Since V (electric potential) = \(\frac{P(\text { power })}{I(\text { electric current })}\)

[I] = \(\frac{[P]}{[V]}=\frac{\mathrm{ML}^2 \mathrm{~T}^{-3}}{\mathrm{~V}}=\mathrm{ML}^2 \mathrm{~T}^{-3} \mathrm{~V}^{-1} .\)

Example 3. Find the dimension of the universal gravitational I constant, taking the units of density D, velocity V; and work W as base units instead of those of mass, distance, and time.
Solution:

Given

Taking the units of density D, velocity V; and work W as base units instead of those of mass, distance, and time.

From Newton’s law of gravitation, F = \(G \frac{M_1 M_2}{r^2}\) we have

[G] = \(\frac{[F]\left[r^2\right]}{\left[m^2\right]}=\frac{M L T^{-2} \times L^2}{M^2}=M^{-1} L^3 T^{-2}\)

Let \(\mathrm{M}^{-1} \mathrm{~L}^3 \mathrm{~T}^{-2}\) = \(\mathrm{D}^x \mathrm{~V}^y \mathrm{~W}^z\)

= \(\left(\mathrm{ML}^{-3}\right)^x \times\left(\mathrm{LT}^{-1}\right)^y \times\left(\mathrm{ML}^2 \mathrm{~T}^{-2}\right)^z\)

= \(\mathrm{M}^{x+z} \times \mathrm{L}^{-3 x+y+2 z} \times \mathrm{T}^{-y-2 z}\)

Equating powers of same bases, x + z = -1

-3x + y + 2z = 3

and -y – 2z = -2

Solving for x, y, z, we get \(x=-\frac{1}{3}, y=\frac{10}{3} and z=-\frac{2}{3}\)

∴ [G] = \(\mathrm{D}^{-\frac{1}{3}} \mathrm{~V}^{\frac{10}{3}} \mathrm{~W}^{-\frac{2}{3}}\)

Example 4. In a system of units, the unit of length is defined as the distance traveled by light in space in 1 s and the unit of time is the time taken by the earth to revolve around the sun once. Find the velocity unit in the CGS system.
Solution:

Given

In a system of units, the unit of length is defined as the distance traveled by light in space in 1 s and the unit of time is the time taken by the earth to revolve around the sun once.

In this system, unit distance = 3×10¹⁰ cm and unit time = 365.25 d = 3.156 x 10⁷ s

∴ Velocity unit in this system

= \(\frac{3 \times 10^{10}}{3.156 \times 10^7}=950.57 \mathrm{~cm} \cdot \mathrm{s}^{-1}\)

Velocity unit in this system = 950.57 cm⋅s^-1