Question

Check the correctness of the equation h = 2Scosθ / rρg

Answer 1

Answer:

Explanation:

Given that,

h= 2scosθ/rρg  

The units of s,r,g and density

Surface tension S=[M][T]^{-2}S=[M][T] −2

Radius R= [L]R=[L]

Density \rho=[M][L]^{-3}ρ=[M][L] −3

Gravity due to acceleration g=[L][T]^{-2}g=[L][T] −2

h= \dfrac{r\rho g}{S}h= Srρg

h= \dfrac{ [L]\times [M][L]^{-3}\times[L][T]^{-2}}{[M][T]^{-2}}h

= [M][T] −2 [L]×[M][L] −3 ×[L][T] −2  

h=[L^{-1}]h=[L −1 ]

This relation is incorrect.The correct relation is define as:h= \dfrac{2scos\theta}{r\rho g}h= rρg2scosθ  .....(I)Now put the all value in equation (I)h= \dfrac{S}{r\rho g}h= rρgS  h= \dfrac{[M][T]^{-2}}{ [L]\times [M][L]^{-3}\times[L][T]^{-2}}h= [L]×[M][L] −3 ×[L][T] −2 [M][T] −2   h=[L]h=[L]This relation is correct.Hence, Given relation is incorrect

Read more on Brainly.in - https://brainly.in/question/12732161#readmore