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Units and Dimensions: Building of Concepts Part 3
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Usee of dimensional analysis... Concept building

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
very nice effort...keep going... ????????
  1. Units and Dimensions USES OF DIMENSIONAL ANALYSIS 2

  2. Uses of Dimensional Analysis The third primary use of Dimensional Analysis is to check the consistency of a Physical Formula. 3.) Checking the correctness of a formula:- * The formula or an expression that represents a particular Physical quantity must have the same dimensions as that of the Physical Quantity For example , if we are writing 'F = ma, then the dimensions of the Physical quantity on LHS must match with the Dimensions of the Physical Formula on RHS.

  3. We already know that the dimensional formula for Force is MLT-2 whereas the dimensional formula for mass is M110To and for acceleration is Mo1T-2 Now R.H.S. over here has Dimensions (M1L0po) x (M011T-2) = M1L1T-2 Now, as the dimensions on R.HS. and L.H.S are same, therefore F = ma may be the correct formula for force. (Take a note that the word 'may' is stressed ) One major limitation while utilizing this technique is that even if a formula is dimensionally correct, even then it may be Physically incorrect if the formula involves dimensionless quantities.

  4. For example, in case of frequency of a simple pendulum, all the formulae, 2as well as or an 4VT or Are dimensionally correct but the only physically accurate and correct formula is This is because dimensionless quantities lack dimensions and their consistency cannot be tested via Dimensional Analysis. However, if a formula is dimensionally incorrect, then for sure it is wrong.

  5. Similarly, we can test the dimensional consistency if the equation v = u + at Dimn e n sional form ula of v = [MoLT-1] Dimensional formula of u = [MoLT-1] Dimensional formula of a = [M1oLT-2] Dimensional formula of t [MOLOTI LHS = Dimensional formula of v = [MAoLT-11 RHS Dimensional formula of u+Dimensional form ula of a x Dimensional f orm ula of t [MoLT-1] + [MoLT-2] x [MoLoT] [MoLT-1] + [MoLT-1] = = 2 [MoLT-1] Since, 2 is a dimensionless constant. Therefore, LHSRHS Hence, the given relationship equation) is dimensionally correct