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HOW to apply MLT dimension formula applying in questions?

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

 Since a physical quantity is expressed in terms of appropriate units of the same nature in all systems, the dimensions should remain the same even though the number indicating its magnitude may differ. If a physical quantity of dimensions a, b and c in mass, length and time respectively has magnitude n₁ and n₂ in two systems having fundamental units M₁, L₁, T₁ and M₂, L₂, T₂ respectively then,

Example:

Convert one newton into dynes

They are the units of force in SI and CGS respectively.

We have the dimensional formula of force as MLT ^-2

Let n₁ newton be equal to n₂ dynes.

ii) Check the correctness of an equation.

An equation is correct only if the dimensions of each term on either side of the equation are equal. In this case the equation is dimensionally correct, but, may or may not be the correct equation for the physical quantity. If the homogeneity of dimensions does not hold good, the equation is definitely correct.

Example:

Check the accuracy of the equation,

Dimension of the term S = L

Dimension of the term ut = LT ^-1 × T = L and the dimensions of the term (½)at² = LT ^-2 ×  T² = L.

Thus, we find that each term has the same dimensions. So the equation is dimensionally correct.

iii) derive the correct relationship between physical quantities.

When one physical quantity depends on several physical quantities, then the relationship between the quantities can be derived using dimensional method.

Example:

To derive an expression for the period of oscillation of a simple pendulum.

The period of oscillation t of a simple pendulum may depend on

(a) the length of the pendulum l₁

(b) the mass of the bob m and

(c) the acceleration due to gravity g

Let the form of the equation be t = kl^xm^yg^z

Taking dimensions of both sides, T = L^xM^y(LT ^-2)^z = L^(x+z)M^yT ^-2z    [Since k is dimensionless]

equating dimensions of L, M and T