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Units and Dimensions: Building of Concepts Part 5
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Units and Dimensions theory 5 for the building of concepts to further solve numericals based upon these concepts.

Nikhil Mishra
Nikhil Mishra is an engineer and has been a Faculty of Physics in some of the most premier institutes of the country and has been involved

Unacademy user
Brackish water is a mixture of fresh water and salty water. Sir please make video on sanctuaries and parks with mnemonics .thank you
sir in the 2 nd example of this vedio .. how we can calculate value of a b & c ..????
  1. Units and Dimensions USES OF DIMENSIONAL ANALYSIS 4

  2. Uses of Dimensional Analysis The Fifth primary use of Dimensional Analysis is to determine the physical formula of a quantity by comparing the powers ( and dimensions) on LHS and RHS. For this purpose, we utilize the Dimensional Consistency of the equation. Let us for an example consider a case analysis. It is mentioned to us that the frequency of a Simple Pendulum depends upon the length of the pendulum, and acceleration due to gravity and we are required to determine the probable formula of the frequency of the simple pendulum in terms of its length(1) and the acceleration due to gravity (g)

  3. Here, it is mentioned to us that the frequency depends upon the length of the pendulum and acceleration due to gravity but we do not know how does it exactly depend. Therefore, let us say that f o l" and fox gb Combining we can write, f ox 1agb Dimensionally, we can write, MoLoT_1=(MoL1 Toy. (MoLT-2)b Therefore, M0L0T-1-M0L9ToM0LbT-2b=M"La+"T-2b So, M LOT-1MOLa+T-2b, Now comparing L.H.S and R.H.S, we have a+b=0 and -2b=-1 Solving we get, b-1/2 and a =-1/2

  4. Therefore, f 1-1/2g1/2 So, f = k./T where k is a dimensionless constant which can be described experimentally only. (for your knowledge, k here is 2 Therefore, f= 21t We can study some more examples.

  5. Illustration For a particle to move in a circular orbit uniformly, centripetal force is required which infact depends upon mass (m), velocity (v), and radius (r) of the circle. Express centripetal force in terms of these quantities. Sol. According to the provided information, Foc m'vr' b, c are the constant powers of m, v, r; respectively. of like quantities on both the sides, we have Using (ii), (iii), and (iv), we have a = 1, b = 2, and c =-1 where k is the dimensionless constant of proportionality and a, Now using the principle of homogeneity, comparing the power b+c=1 a=1 (ii) (iii) and b=2 (iv) Using these values in (i), F = k miv2,-2 2 mv F=K which is the desired relatiorn