**Uses of Dimensional Equation**

**a. To check the correctness of a physical quantity**

A dimensional equation of a physical quantity can be used to verify whether the given physical relation is correct. Suppose, a physical relation is given as;

T = 2π √( l/g )

Dimension of time period (T) = [T]

Dimension of Length (l) = [L]

Dimension of gravity (g) = [M^{0}LT^{-2}]

Dimension of R.H.S [2π √( l/g ) ] = √[L]/ [M^{0}LT^{-2}] = [T]

According to the principle of homogeneity, The dimensions of each term on the two side of correct physical relation must be same.

Here, in relation T = 2π √( l/g ), the dimension of each term on the two sides is same. So, the given relation is dimensionally correct.

**b. To derive the relation between physical quantities**

We can use dimensional formula to derive physical relation if we know the factors on which quantity depends upon.

** Proof:**

i. Derive the relation for time period of simple pendulum.

From above figure, we find that, the time period of simple pendulum depends upon:

- Length of String (L)
- Mass of bob (m)
- Gravity (g)

i.e. T α l^{a}m^{b}g^{c}

T = K l^{a}m^{b}g^{c}………………….(i)

Where K is proportionality constant and has no dimension,

Now writing dimensional equation in eqn (i),

[M^{0} L^{0} T] = [M^{0} L T^{0}]^{a} [ML^{0}T^{0}]^{b} [M^{0} L T^{-2}]^{c}

Or, [M^{0} L^{0} T] = [M^{0} L^{a} T^{0}] [M^{b} L^{0} T^{0}] [M^{0 }L^{c} T^{-2c}]

Or, [M0 L0 T] = M^{b}L^{a+c}+T^{-2c}

Equating equal bases, we get

b = 0 ……(a)

-2c = 1

Or, c = -1/2 …………(b)

a+c = 0

a = 1/2 ……………(c)

Putting value of a, b, c in eqn (i), we get,

T = K . L^{1/2} M^{0} g^{-1/2}

Or T = K . L^{1/2 }/ g^{1/2}

Or, T = K (L/g)^{1/2}

Or, T = K √( l/g )

The value of K was found to be 2π, then the relation becomes,

T = 2π √( l/g ) …..(ii)

ii. Relation for centripetal force.

Here, for centripetal force we have,

Where K is proportionality constant and doesn’t have dimension.

Now writing dimensional equation of eqn (i),

[M L T^{-2}] = [M^{0} L T^{-1}]^{a} [M L^{0} T^{0}]^{b} [M^{0} L T^{0}]^{c}

Or, [M L T^{-2}] = [M^{0} L^{a} T^{-a}] [M^{b} L^{0} T^{0}] [M^{0}L^{c} T^{0}]

Or, [M L T^{-2}] = [M^{b }L^{a+c} T^{-a}]

Equating equal bases,

b = 1 ……(a)

-a = -2

or a = 2 ………(b)

a+c = 1

2 + c = 1

c = -1 ……..(c)

Putting values of a, b, c in eqn (i), we get,

F = K V^{2} M R^{-1}

F = K mv^{2}/r

The value of k was experimentally found to be 1, So relation be

F = mv^{2}/r

**c. To convert the value of physical quantity from one system to another**

Let n_{1} be the numerical value, u_{1} be the units of physical quantity in one system and n_{2} be the numerical value, u_{2} be the units of physical quantity in another system.

Then,

n_{1}u_{1}=n_{2}u_{2}

n_{1}[M_{1}^{x} L_{1}^{y} T_{1}^{z}] = n_{2}[M_{2}^{x} L_{2}^{y} T_{2}^{z}]

where x, y, z are dimensions of physical quantity

i. Convert 72km/hr to m/s.

72km/hr -------------------> x m/s

n_{1} = 72 n2 = x

L_{1} = 1km L_{2} = 1m

T_{1} = 1 hr T_{2} = 1s

Since the dimensional equation is [M^{0} L T^{-1}], x=0, y=1 and z=-1

We know that

n_{1}[M_{1}^{x} L_{1}^{y} T_{1}^{z}] = n_{2}[M_{2}^{x} L_{2}^{y} T_{2}^{z}]

n_{2} = n_{1} [M_{1}/M_{2}]^{x} [L_{1}/L_{2}]^{y} [T_{1}/T_{2}]^{z}

n_{2} = 72 [M_{1}/M_{2}]^{0} [L_{1}/L_{2}]^{1} [T_{1}/T_{2}]^{-1}

n_{2} = 72 [1km/1m]^{1} [1hr/1s]^{-1}

n_{2} = 72 * [1000/1] * [1s/3600s]

n_{2} = 20 m/s

**Try yourself**

- Convert 20m/s into km/hr
- Convert 10N to dynes
- Convert 10
^{5}dyne to N - Convert 10erg to Joule

**d.To find the dimension of constant in a given physical relation.**

Using dimensional equation, we can also find the dimension of a constant.

i. Find the dimension of G from the relation F = GMm/R^{2}.

From the given equation,

G = FR^{2}/Mm

Dimension of G = Dimension of F * Dimension of R^{2}/ Dimension of (m*M)

Dimension of G = [M L T-^{2}] [M^{0} L^{2} T^{0}] / [M L^{0} T^{0}] [M L^{0} T^{0}]

Dimension of G = [M L^{3} T-^{2}] / [M^{2}]

Dimension of G = [M^{-1} L^{3} T^{-2}]

**Try yourself**

- Find the dimension of coefficient of viscosity (η) from the relation F = 6πrvη
- Find the dimension of plank’s constant (h) from relation, E = hf.
- Find the dimension of ‘a’ and ‘b’ from relation (p + a/v
^{2})(v-b) = RT.

**Limitations of Dimensional Analysis**

- It doesn’t give the information about dimensionless constant.
- If the dimensional physical quantity depends upon more than three factors (mass, length and time), the exact formula can’t be derived.
- The exact form of physical quantity cannot be derived if there is more than one term. E.g. v
^{2}= u^{2}+ 2as - Dimensional method is not applicable to the trigonometric function.

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