the radius of the capillary tube increased 0.2%then the percentage increase in the rate of flow of liquid through it
Answers
If the radius of the capillary tube is increased by 0.2%, then the percentage increase in the rate of flow of liquid through it is 0.8%.
Explanation:
Let the rate of flow of the liquid be denoted as “Q”.
According to the Poiseuille’s Law formula, we get
The rate of flow of a liquid through the capillary tube per unit time as,
Q = dV/dt = [πρr⁴] / [8ɳl]
Since in the question it is given that only the radius is increased by 0.2% so, all the other terms in the above formula remain constant, so, we can write
Q ∝ r⁴
⇒ Q₂/Q₁ = [r₂/r₁]⁴…… (i)
Where,
Q₂ = final rate of flow of liquid
Q₁ = initial rate of flow of liquid
r₁ = initial radius of the capillary tube
r₂ = final radius of the capillary tube = [1 + (0.2/100)]r₁ = 1.002 r₁
Substituting the value of r₂ in eq. (i), we get
Q₂/Q₁ = [(1.002 r₁) / r₁]⁴
⇒ Q₂/Q₁ = 1.0080
⇒ Q₂ = 1.0080 Q₁ …… (ii)
Thus,
The percentage increase in the rate of flow of liquid through the capillary tube is,
= [(Q₂ – Q₁)/Q₁] * 100
= [(1.0080 - 1)Q₁ / Q₁] * 100
= 0.0080 * 100
= 0.8%
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