Solved Example Problems for Application of the Method of Dimensional Analysis

Solved Example Problems 

(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 L3T −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  l1/2 g−1/2