A group of students investigated the flow of a viscous liquid through capillary tubes. The… — ACT

A group of students investigated the flow of a viscous liquid through capillary tubes. They measured the volumetric flow rate, $Q$ (in $\text{cm}^3/\text{s}$ ), under different conditions by varying the pressure difference ( $\Delta P$ , in $\text{kPa}$ ) across the tube, the tube length ( $L$ , in $\text{cm}$ ), and the tube radius ( $r$ , in $\text{mm}$ ). The results of their trials are recorded in the table below:

Trial	Pressure Difference ( $\Delta P$ , $\text{kPa}$ )	Tube Length ( $L$ , $\text{cm}$ )	Tube Radius ( $r$ , $\text{mm}$ )	Flow Rate ( $Q$ , $\text{cm}^3/\text{s}$ )
1	100	10	1.0	0.20
2	100	20	1.0	0.10
3	200	10	1.0	0.40
4	100	10	2.0	3.20

Based on the trends shown in the table, if the students were to conduct a fifth trial using a pressure difference of $150\text{ kPa}$ , a tube length of $5\text{ cm}$ , and a tube radius of $3.0\text{ mm}$ , what would be the expected flow rate of the liquid?

$48.60\text{ cm}^3/\text{s}$ Answer
B
$12.15\text{ cm}^3/\text{s}$
C
$5.40\text{ cm}^3/\text{s}$
D
$1.80\text{ cm}^3/\text{s}$

Answer

The expected flow rate of the liquid is

48.60\text{ cm}^3/\text{s}

The correct answer is

48.60\text{ cm}^3/\text{s}

. Comparing the trials shows that the flow rate (

Q

) is directly proportional to the pressure difference (

\Delta P

), inversely proportional to the tube length (

L

), and directly proportional to the fourth power of the radius (

r^4

). When compared to Trial 1, Trial 5 has

1.5

times the pressure difference, half the length, and

3

times the radius. Therefore, the new flow rate is

0.20\text{ cm}^3/\text{s} \times 1.5 \times 2 \times 3^4 = 48.60\text{ cm}^3/\text{s}

Step-by-Step Solution

Determine the relationship between flow rate (

Q

) and pressure difference (

\Delta P

) using Trial 1 and Trial 3.

Q \propto \Delta P

(direct proportionality). When the pressure difference is doubled from

100\text{ kPa}

200\text{ kPa}

while other variables are kept constant, the flow rate doubles from

0.20\text{ cm}^3/\text{s}

0.40\text{ cm}^3/\text{s}

To identify how changes in pressure affect flow rate.

Determine the relationship between flow rate (

Q

) and tube length (

L

) using Trial 1 and Trial 2.

Q \propto \frac{1}{L}

(inverse proportionality). When the tube length is doubled from

10\text{ cm}

20\text{ cm}

while other variables are kept constant, the flow rate is halved from

0.20\text{ cm}^3/\text{s}

0.10\text{ cm}^3/\text{s}

To identify how changes in tube length affect flow rate.

Determine the relationship between flow rate (

Q

) and tube radius (

r

) using Trial 1 and Trial 4.

Q \propto r^4

(fourth-power direct proportionality). When the tube radius is doubled from

1.0\text{ mm}

2.0\text{ mm}

while other variables are kept constant, the flow rate increases by a factor of

16

(

3.20 / 0.20 = 16

), which corresponds to

2^4

To identify how changes in tube radius affect flow rate.

Calculate the expected flow rate for Trial 5 by comparing its parameters to Trial 1.

The expected flow rate is

48.60\text{ cm}^3/\text{s}

. The pressure difference increases by a factor of

1.5

(

150 / 100

), the tube length is halved (

5 / 10 = 0.5

, which doubles

Q

due to inverse proportionality), and the tube radius is tripled (

3.0 / 1.0 = 3.0

, which increases

Q

by a factor of

3^4 = 81

). Thus,

Q_5 = 0.20 \times 1.5 \times 2 \times 81 = 48.60\text{ cm}^3/\text{s}

To compute the final flow rate based on all combined proportional relationships.

Key Concept

Direct and Inverse Proportionality

Question

Answer

Step-by-Step Solution

Key Concept

Common Mistakes

Topics

More questions from this topic