Question

A capillary tube is attached horizontally to a constant heat arrangement. If the radius of the capillary tube is increased by 25 \% , then the rate of flow of liquid will change nearly by:

• A. 100 \%
• B. 112 \%
• C. 124 \%
• D. 144 \%

Hint:

In capillary flow, the rate of flow is proportional to the fourth power of the radius. A small increase in the radius results in a significant increase in the flow rate.

Answer 1

The Correct Option is D

The rate of flow of liquid through a capillary tube is governed by the formula: \[ Q = \frac{\pi r^4 \Delta P}{8 \eta L} \] where: - \( Q \) is the rate of flow, - \( r \) is the radius of the capillary, - \( \Delta P \) is the pressure difference, - \( \eta \) is the viscosity of the liquid, - \( L \) is the length of the capillary tube. 
Step 1: If the radius \( r \) is increased by 25%, the new radius \( r' \) will be: \[ r' = 1.25r \] 
Step 2: The rate of flow \( Q \) is proportional to the fourth power of the radius, so the new rate of flow \( Q' \) will be: \[ Q' = \frac{\pi (r')^4 \Delta P}{8 \eta L} = \frac{\pi (1.25r)^4 \Delta P}{8 \eta L} \] 
Step 3: Simplifying: \[ Q' = Q \times (1.25)^4 \] \[ (1.25)^4 = 2.4414 \] Thus, the new rate of flow will be approximately 144% of the original rate of flow. Hence, the rate of flow will change by \( \boxed{144\%} \).