If dimensions of critical velocity, v c of a liquid flowing through a tube are expressed as η x ρ y r z , where, η , ρ and r are the coefficient of viscosity of liquid, density of liquid and radius of

Question

If dimensions of critical velocity, v c of a liquid flowing through a tube are expressed as η x ρ y r z , where, η , ρ and r are the coefficient of viscosity of liquid, density of liquid and radius of the tube, respectively, then, the values of x , y and z are given byNEET 2015 (Phase 2)

TestHub · Accepted Answer

Dimensions of velocity, coefficient of viscosity and density are,

$[v] = [{LT}^{- 1}]$

$[η] = \frac{[F]}{[A \frac{d v}{d x}]} = \frac{[{MLT}^{- 2}]}{[L^{2}] [T^{- 1}]} = [{ML}^{- 1} T^{- 1}]$

$[ρ] = [{ML}^{- 3}]$
Now, $[v] = {[η]}^{x} {[ρ]}^{y} {[r]}^{z}$
$\Rightarrow [{LT}^{- 1}] = {[{ML}^{- 1} T^{- 1}]}^{x} {[{ML}^{- 3}]}^{y} {[L]}^{z}$
Now, by using dimensional homogeneity, we will get,
$M^{0} = M^{(x + y)}$
$\Rightarrow y = - x$
For $T$ ,
$x = 1$
$\Rightarrow y =- x =- 1$
For $L$ ,
$1 = - x - 3 y + z$
$\Rightarrow z = - 1$

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