Question

Two tubes of radii r1 and r2, and lengths l1 and l2, respectively are connected in series and a liquid flows through each of them in stream line conditions. p1 and p2 are pressure differences across the two tubes. if p2 is 4p1 and l2 is , then the radius r2 will be equal to

Answer 1

use Poiseuille's law , according to him,
Volume rate = πPr⁴/8ηL
Here P is the pressure difference , r is the radius , η is the coefficient of viscosity and L is the length of tube .
Because volume rate is constant so,
P₁r₁⁴/L₁ = P₂r₂⁴/L₂

Here , P₂ = 4P₁ , L₂ = L₁/4
Then, P₁r₁⁴ /L₁ = 4P₁r₂⁴/(L₁/4) = 16P₁r₂⁴/L₁
so, r₂⁴ = r₁⁴/16 ⇒r₂ = r₁/2

Hence, answer is r₂ = r₁/2

Answer 2

According to  Poiseuille's law:
Q=πPr⁴/8ηL 
where P=Pressure gradient across tube
r=radius of tubing
n=coefficient viscosity of fluid
l= length 
According to question, volume rate is given :
P₁r₁⁴/L₁ = P₂r₂⁴/L₂ 
As given :
P2=4P1
 L2=L1/4
Then  P₁r₁⁴ /L₁ = 4P₁r₂⁴/(L₁/4) =16P1r₂⁴/L₁
r₂⁴=r₁⁴/16r₂=r₁/2
∴ r2=r₁/2