Question

the radius of the capillary tube increased 0.2%then the percentage increase in the rate of flow of liquid through it​

Answer 1

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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