Question

the critical velocity Vc of a liquid flowing through a tube depends upon the coefficient of viscosity n of the liquid , density P of the liquid and radius r of the tube use the method of dimension to obtain the formula for Bc
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Answer 1

Answer:

$V_c = k\frac{\eta}{Dr^2}$

Explanation:

The dimensions for various factors are -

Critical velocity, $[V_c]$ = [L¹T⁻¹]

Coefficient of viscosity, [η] = [M¹L⁻¹T⁻¹]

Density, [D] = [M¹L⁻³]

radius, [r] = [L¹]

Hence, the dimension of critical velocity should be same as the formula for it.

$[V_c] = [\eta]^a[D]^b[r]^c\\\\\[[LT^{-1}]=[ML^{-1}T^{-1}]^a[ML^{-3}]^b[L]^c$

So, M⁰ = Mᵃ⁺ᵇ ⇔ a = -b

And,

$L^1 = L^{-a-3b+c}\\1=-a-3b+c\\c=a+3b\\c=a -3a = -2a$

And,

T⁻¹ = T⁻ᵃ

a = 1

Hence,

a = 1

b = -1

c = -2

So, the formula goes like

$[V_c] = [\eta]^a[D]^b[r]^c\\\\\[[V_c] = [\eta]^1[D]^{-1}[r]^{-2}$

∴$V_c = k\frac{\eta}{Dr^2}$