# Dimensional Equation: Uses and Limitation | Physics Grade XI

### Dimensional Equation # Uses of Dimensional Equation and Its Limitation

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) = [M0LT-2]

Dimension of R.H.S [2π √( l/g ) ] = √[L]/ [M0LT-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:

1. Length of String (L)
2. Mass of bob (m)
3. Gravity (g)

i.e. T α lambgc

T = K lambgc………………….(i)

Where K is proportionality constant and has no dimension,

Now writing dimensional equation in eqn (i),

[M0 L0 T] = [M0 L T0]a [ML0T0]b  [M0 L T-2]c

Or, [M0 L0 T] = [M0 La T0] [Mb L0 T0] [MLc  T-2c]

Or, [M0 L0 T] = MbLa+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 . L1/2 M0 g-1/2

Or T = K . L1/2 / g1/2

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] = [M0 L T-1]a [M L0 T0]b [M0 L T0]c

Or, [M L T-2] = [M0 La T-a] [Mb L0 T0] [M0Lc T0]

Or, [M L T-2] = [MLa+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 V2 M R-1

F = K mv2/r

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

F = mv2/r

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

Let n1 be the numerical value, u1 be the units of physical quantity in one system and n2 be the numerical value, u2 be the units of physical quantity in another system.

Then,

n1u1=n2u2

n1[M1x L1y T1z] = n2[M2x L2y T2z

where x, y, z are dimensions of physical quantity

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

72km/hr  ------------------->  x  m/s

n1 = 72                                                                  n2 = x

L1 = 1km                                                               L2 = 1m

T1 = 1 hr                                                               T2 = 1s

Since the dimensional equation is [M0 L T-1], x=0, y=1 and z=-1

We know that

n1[M1x L1y T1z] = n2[M2x L2y T2z]

n2 = n1 [M1/M2]x [L1/L2]y [T1/T2]z

n2 = 72 [M1/M2]0 [L1/L2]1 [T1/T2]-1

n2 = 72 [1km/1m]1 [1hr/1s]-1

n2 = 72 * [1000/1] * [1s/3600s]

n2 = 20 m/s

Try yourself

• Convert 20m/s into km/hr
• Convert 10N to dynes
• Convert 105 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/R2.

From the given equation,

G = FR2/Mm

Dimension of G = Dimension of F * Dimension of R2/ Dimension of (m*M)

Dimension of G = [M L T-2] [M0 L2 T0] / [M L0 T0] [M L0 T0]

Dimension of G = [M L3 T-2] / [M2]

Dimension of G = [M-1 L3 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/v2)(v-b) = RT.

Limitations of Dimensional Analysis

1. It doesn’t give the information about dimensionless constant.
2. If the dimensional physical quantity depends upon more than three factors (mass, length and time), the exact formula can’t be derived.
3. The exact form of physical quantity cannot be derived if there is more than one term. E.g. v2 = u2 + 2as
4. Dimensional method is not applicable to the trigonometric function.

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