Question

The dimensional formula for Reynold’s number is –
A. \[[{L^0}{M^0}{T^0}]\]
B. \[[{L^1}{M^1}{T^1}]\]
C. \[[{L^{ - 1}}{M^1}{T^1}]\]
D. \[[{L^1}{M^1}{T^{ - 1}}]\]

Answer 1

Hint: To solve this question we have to know that, The Reynolds number is utilized to decide if a liquid is in laminar or fierce stream. In light of the API \[13\] D proposals, it is expected that a Reynolds number not exactly or equivalent to \[2100\] demonstrates a laminar stream, and a Reynolds number more prominent than \[2100\] shows a tempestuous stream.

Complete step by step answer:
We also know that the Reynolds number is the proportion of inertial powers to gooey powers. The Reynolds number is a dimensionless number used to classify the liquids frameworks in which the impact of consistency is significant in controlling the speeds or the stream example of a liquid.
Reynold’s number formula is expressed by \[{R_e} = \dfrac{{\rho VL}}{\mu }\].
The SI unit of density is \[\rho = Kg.{m^{ - 3}}\].
The SI unit of velocity is \[V = m.{s^{ - 1}}\].
The Si unit of length is meter. The SI unit of diameter is Metre.
The SI unit of viscosity of fluid \[\mu = N.s.{m^{ - 2}}\]
We know that, \[N = Kg.m.{s^{ - 2}}\]
Putting the units we get,
\[{R_e} = \dfrac{{kg.{m^{ - 3}}.m.m.{s^{ - 1}}}}{{kg/m.s}}\]
So, the dimensions will be \[\dfrac{{[M{L^{ - 3}}{L^2}{T^{ - 1}}]}}{{[M{T^{ - 1}}{L^{ - 2}}]}}\].

Hence, the right answer is option A.

Note: we also have to know that, The Reynolds number is a valuable instrument for relating how a meter will respond to a variety of liquids from gases to fluids. Since an unimaginable measure of exploration would be needed to test each meter on each liquid we wish to gauge, it is attractive that a connection between liquid elements be known. We can say, Reynolds number is a dimensionless amount that is utilized to decide the sort of stream design as laminar or fierce while coursing through a line. Reynolds number is characterized by the proportion of inertial powers to that of thick powers.