All the derived physical quantities can be expressed in terms of some combination of the seven fundamental or base quantities.

**DIMENSIONAL ANALYSIS**

All
the derived physical quantities can be expressed in terms of some combination
of the seven fundamental or base quantities. These base quantities are known as
dimensions of the physical world, and are denoted with square bracket [ ]. The
three dimensions in mechanics are [L] for length, [m] for mass and [T] for
time. In electricity, [A] is the dimension of electric current. In
thermodynamics, [K] is the dimension for the temperature. In optics [cd] or [Φ]
is the dimension for luminous intensity. The dimension of amount of substance
is [mol].

The
dimensions of a physical quantity are the powers to which the units of base
quantities are raised to represent a derived unit of that quantity.

For
example,

Hence
the dimensions of velocity are 0 in mass, 1 in length and -1 in time.

Dimensional
formula is an expression which shows how and which of the fundamental units are
required to represent the unit of a physical quantity.

For
example, [M^{0}LT^{−2}] is the dimensional formula of
acceleration.

When
the dimensional formula of a physical quantity is expressed in the form of an
equation, such an equation is known as the dimensional equation.

Example,
acceleration = [M^{0}LT−2]. The dimensional formula of various physical
quantities are tabulated in Table 1.11.

On
the basis of dimension, we can classify quantities into four categories.

Physical
quantities, which possess dimensions and have variable values are called
dimensional variables. Examples are length, velocity, and acceleration etc.

Physical
quantities which have no dimensions, but have variable values are called
dimensionless variables. Examples are specific gravity, strain, refractive
index etc.

Physical
quantities which possess dimensions and have constant values are called
dimensional constants. Examples are Gravitational constant, Planck’s constant
etc.

Quantities
which have constant values and also have no dimensions are called dimensionless
constants. Examples are π, e, numbers etc.

The
principle of homogeneity of dimensions states that the dimensions of all the
terms in a physical expression should be the same. For example, in the physical
expression v^{2} = u^{2} + 2as, the dimensions of v^{2},
u^{2} and 2 as are the same and equal to [L^{2}T^{−2}].

**This method is used to**

i.
Convert a physical quantity from one system of units to another.

ii.
Check the dimensional correctness of a given physical equation.

iii.
Establish relations among various physical quantities.

This
is based on the fact that the product of the numerical values (n) and its
corresponding unit (u) is a constant. i.e, n [u] = constant (or) n_{1}[u_{1}]
= n_{2}[u_{2}].

Consider
a physical quantity which has dimension ‘*a’*
in mass, ‘*b’* in length and ‘*c’* in time. If the fundamental units in
one system are M_{1}, L_{1} and T_{1} and the other
system are M_{2}, L_{2} and T_{2} respectively, then we
can write, n_{1} [M_{1}^{a} L_{1}^{b} T_{1}^{c}]
= n_{2} [M_{2}^{a} L_{2}^{b} T_{2}^{c}]

We
have thus converted the numerical value of physical quantity from one system of
units into the other system.

Let
us take the equation of motion

v
= u + at

Apply
dimensional formula on both sides

[LT^{−1}]
= [LT^{−1}] + [LT^{−2}] [T]

[LT^{−1}] = [LT^{−1}] + [LT^{−1}]

(Quantities of same dimension only can be
added)

We
see that the dimensions of both sides are same. Hence the equation is
dimensionally correct.

If
the physical quantity Q depends upon the quantities Q_{1}, Q_{2}
and Q_{3} ie. Q is proportional to Q_{1}, Q_{2} and Q_{3}.

Then,

Q α Q_{1}^{a} Q_{2}^{b}
Q_{3}^{c}_{}

Q
= k Q_{1}^{a} Q_{2}^{b} Q_{3}^{c}_{}

where
k is a dimensionless constant. When the dimensional formula of Q, Q_{1,}
Q_{2 }and Q_{3} are substituted, then according to the
principle of homogeneity, the powers of M, L, T are made equal on both sides of
the equation. From this, we get the values of a, b, c

This
method gives no information about the dimensionless constants in the formula
like 1, 2, ……..π,e, etc.

This
method cannot decide whether the given quantity is a vector or a scalar.

This
method is not suitable to derive relations involving trigonometric, exponential
and logarithmic functions.

It
cannot be applied to an equation involving more than three physical quantities.

It
can only check on whether a physical relation is dimensionally correct but not
the correctness of the relation. For
example using dimensional
analysis, s = ut + 1/3 at^{2} is
dimensionally correct whereas the correct relation is s = ut + 1/2 at^{2}.

**(i) To convert a physical quantity from one system of units to another**

**Example 1.12**

Convert 76 cm of mercury pressure into Nm−2 using the method of dimensions.

*Solution*

In cgs system 76 cm of mercury pressure = 76 × 13.6 × 980 dyne cm−2

The dimensional formula of pressure P is [ML−1T−2]

**Example 1.13**

If the value of universal gravitational constant in SI is 6.6x10−11 Nm2 kg−2, then find its value in CGS System?

*Solution*

Let GSI be the gravitational constant in the SI system and Gcgs in the cgs system. Then

The dimensional formula for G is M−1 *L*3*T* −2

**(ii) To check the dimensional correctness of a given physical equation**

**Example 1.14**

Check the correctness of the equation

using dimensional analysis* *method

*Solution*

Both sides are dimensionally the same, hence the equations

is dimensionally correct.

**(iii) To establish the relation among various physical quantities**

**Example 1.15**

Obtain an expression for the time period T of a simple pendulum. The time period T depends on (i) mass ‘*m’* of the bob (ii) length ‘*l’* of the pendulum and (iii) acceleration due to gravity *g* at the place where the pendulum is suspended. (Constant k = 2π) i.e

*Solution*

Here k is the dimensionless constant. Rewriting the above equation with dimensions

Comparing the powers of M, L and T on both sides, a=0, b+c=0, -2c=1

Solving for a,b and c a = 0, b = 1/2, and c = −1/2

From the above equation T = k. m0 *l*1/2 g−1/2

**Example 1.16**

The force F acting on a body moving in a circular path depends on mass of the body (m), velocity (v) and radius (r) of the circular path. Obtain the expression for the force by dimensional analysis method. (Take the value of k=1)

where k is a dimensionless constant of proportionality. Rewriting above equation in terms of dimensions and taking k = 1, we have

Comparing the powers of M, L and T on both sides

From the above equation we get

Tags : Dimension of Physical Quantities, Application and Limitations, Solved Example Problems , 11th Physics : UNIT 1 : Nature of Physical World and Measurement

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11th Physics : UNIT 1 : Nature of Physical World and Measurement : Dimensional Analysis | Dimension of Physical Quantities, Application and Limitations, Solved Example Problems

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