PRELIMINARY STAGE 2018
Koforidua Sec. Tech: 48 points
Armed Forces SHTS: 38 points
Awudome SHS: 31 points
ROUND 2 - SPEED RACE
QUESTION
The gauge pressure at a point of depth 5.0 m below the surface of an incompressible fluid is 20 kPa.
Find the change in gauge pressure at a point 1.0 m above that point.
ANSWER: $-4.0$ kPa
SOLUTION:
Using ratios based on the proportional relationship of depth change:
$\dfrac{\Delta P}{P_1} = \dfrac{\Delta h}{h_1}$
Given that moving up $1\text{ m}$ means $\Delta h = -1\text{ m}$:
$\Delta P = P_1 \times \dfrac{\Delta h}{h_1}$
$\Delta P = 20 \times \left(-\dfrac{1}{5}\right)$
Cancel down the fraction:
$\Delta P = 4 \times (-1)$
$\Delta P = -4.0 \text{ kPa}$
---
PRACTICE QUESTIONS
1. QUESTION: The gauge pressure at a point 4.5 m below the surface of an incompressible liquid is 36 kPa. Find the change in gauge pressure at a point 1.5 m above that point.
ANSWER: −12 kPa or $-1.2\times10^{1}$ kPa
SOLUTION:
Gauge pressure is proportional to depth, $P = \rho gh$, so:
$\dfrac{\Delta P}{P_1} = \dfrac{\Delta h}{h_1}$
Moving up 1.5 m means $\Delta h = -1.5\text{ m}$:
$\Delta P = P_1 \times \dfrac{\Delta h}{h_1}$
$\Delta P = 36 \times \dfrac{-1.5}{4.5}$
$\Delta P = -12\text{ kPa}$
2. QUESTION: The gauge pressure at a point 1.2 m below the surface of an incompressible liquid is 96 kPa. Find the change in gauge pressure at a point 2.0 m below that point.
ANSWER: 160 kPa or $1.6\times10^{2}$ kPa
SOLUTION:
Gauge pressure is proportional to depth, $P = \rho gh$, so:
$\dfrac{\Delta P}{P_1} = \dfrac{\Delta h}{h_1}$
Moving down 2.0 m means $\Delta h = +2.0\text{ m}$:
$\Delta P = P_1 \times \dfrac{\Delta h}{h_1}$
$\Delta P = 96 \times \dfrac{2.0}{1.2}$
$\Delta P = \dfrac{96 \times 2}{12\times10^{-1}}$
$\Delta P = 8 \times 2 \times10^{1}$
$\Delta P = 1.6\times10^{2}\text{ kPa}$
3. QUESTION: The gauge pressure at a point 2.5 m below the surface of an incompressible liquid is 25 kPa. Find the change in gauge pressure at a point 0.80 m above that point.
ANSWER: −8.0 kPa
SOLUTION:
Gauge pressure is proportional to depth, $P = \rho gh$, so:
$\dfrac{\Delta P}{P_1} = \dfrac{\Delta h}{h_1}$
Moving up 0.80 m means $\Delta h = -0.80\text{ m}$:
$\Delta P = P_1 \times \dfrac{\Delta h}{h_1}$
$\Delta P = 25 \times \dfrac{-0.80}{2.5}$
$\Delta P = -8.0\text{ kPa}$
4. QUESTION: The gauge pressure at a point 8.0 m below the surface of an incompressible liquid is 64 kPa. Find the change in gauge pressure at a point 3.5 m above that point.
ANSWER: −28 kPa or $-2.8\times10^{1}$ kPa
SOLUTION:
Gauge pressure is proportional to depth, $P = \rho gh$, so:
$\dfrac{\Delta P}{P_1} = \dfrac{\Delta h}{h_1}$
Moving up 3.5 m means $\Delta h = -3.5\text{ m}$:
$\Delta P = P_1 \times \dfrac{\Delta h}{h_1}$
$\Delta P = 64 \times \dfrac{-3.5}{8.0}$
$\Delta P = -28\text{ kPa}$
5. QUESTION: The gauge pressure at a point 1.5 m below the surface of an incompressible liquid is 15 kPa. Find the change in gauge pressure at a point 0.60 m below that point.
ANSWER: 6.0 kPa
SOLUTION:
Gauge pressure is proportional to depth, $P = \rho gh$, so:
$\dfrac{\Delta P}{P_1} = \dfrac{\Delta h}{h_1}$
Moving down 0.60 m means $\Delta h = +0.60\text{ m}$:
$\Delta P = P_1 \times \dfrac{\Delta h}{h_1}$
$\Delta P = 15 \times \dfrac{0.60}{1.5}$
$\Delta P = \dfrac{6 \times 15}{15}$
$\Delta P = 6.0\text{ kPa}$
6. QUESTION: The gauge pressure at a point 5.0 m below the surface of an incompressible liquid is 15 kPa. Find the change in gauge pressure at a point 0.40 m above that point.
ANSWER: −1.2 kPa
SOLUTION:
Gauge pressure is proportional to depth, $P = \rho gh$, so:
$\dfrac{\Delta P}{P_1} = \dfrac{\Delta h}{h_1}$
Moving up 0.40 m means $\Delta h = -0.40\text{ m}$:
$\Delta P = P_1 \times \dfrac{\Delta h}{h_1}$
$\Delta P = -15 \times \dfrac{0.40}{5.0}$
$\Delta P = -\dfrac{15 \times 4\times10^{-1}}{5}$
$\Delta P = -3 \times 4 \times10^{-1}$
$\Delta P = -1.2\text{ kPa}$
7. QUESTION: The gauge pressure at a point 3.5 m below the surface of an incompressible liquid is 42 kPa. Find the change in gauge pressure at a point 1.2 m below that point.
ANSWER: 14 kPa or $1.4\times10^{1}$ kPa
SOLUTION:
Gauge pressure is proportional to depth, $P = \rho gh$, so:
$\dfrac{\Delta P}{P_1} = \dfrac{\Delta h}{h_1}$
Moving down 1.2 m means $\Delta h = +1.2\text{ m}$:
$\Delta P = P_1 \times \dfrac{\Delta h}{h_1}$
$\Delta P = 42 \times \dfrac{1.2}{3.5}$
$\Delta P = \dfrac{12 \times 42}{35}$
$\Delta P = \dfrac{12 \times 6}{5}$
$\Delta P \approx 14\text{ kPa}$
8. QUESTION: The gauge pressure at a point 2.5 m below the surface of an incompressible liquid is 24 kPa. Find the change in gauge pressure at a point 2.5 m below that point.
ANSWER: 24 kPa or $2.4\times10^{1}$ kPa
SOLUTION:
Gauge pressure is proportional to depth, $P = \rho gh$, so:
$\dfrac{\Delta P}{P_1} = \dfrac{\Delta h}{h_1}$
Moving down 2.5 m means $\Delta h = +2.5\text{ m}$:
$\Delta P = P_1 \times \dfrac{\Delta h}{h_1}$
$\Delta P = 24 \times \dfrac{2.5}{2.5}$
$\Delta P = \dfrac{24 \times 25}{25}$
$\Delta P = 24\text{ kPa}$
9. QUESTION: The gauge pressure at a point 3.0 m below the surface of an incompressible liquid is 27 kPa. Find the change in gauge pressure at a point 0.70 m below that point.
ANSWER: 6.3 kPa
SOLUTION:
Gauge pressure is proportional to depth, $P = \rho gh$, so:
$\dfrac{\Delta P}{P_1} = \dfrac{\Delta h}{h_1}$
Moving down 0.70 m means $\Delta h = +0.70\text{ m}$:
$\Delta P = P_1 \times \dfrac{\Delta h}{h_1}$
$\Delta P = 27 \times \dfrac{0.70}{3.0}$
$\Delta P = \dfrac{7\times10^{-1} \times 27}{3}$
$\Delta P = 7 \times 9 \times10^{-1}$
$\Delta P = 6.3\text{ kPa}$
10. QUESTION: The gauge pressure at a point 8.0 m below the surface of an incompressible liquid is 32 kPa. Find the change in gauge pressure at a point 0.80 m above that point.
ANSWER: −3.2 kPa
SOLUTION:
Gauge pressure is proportional to depth, $P = \rho gh$, so:
$\dfrac{\Delta P}{P_1} = \dfrac{\Delta h}{h_1}$
Moving up 0.80 m means $\Delta h = -0.80\text{ m}$:
$\Delta P = P_1 \times \dfrac{\Delta h}{h_1}$
$\Delta P = -32 \times \dfrac{0.80}{8.0}$
$\Delta P = -\dfrac{32 \times 8\times10^{-1}}{8}$
$\Delta P = -4 \times 8 \times10^{-1}$
$\Delta P = -3.2\text{ kPa}$
11. QUESTION: The gauge pressure at a point 6.0 m below the surface of an incompressible liquid is 12 kPa. Find the change in gauge pressure at a point 0.60 m below that point.
ANSWER: 1.2 kPa
SOLUTION:
Gauge pressure is proportional to depth, $P = \rho gh$, so:
$\dfrac{\Delta P}{P_1} = \dfrac{\Delta h}{h_1}$
Moving down 0.60 m means $\Delta h = +0.60\text{ m}$:
$\Delta P = P_1 \times \dfrac{\Delta h}{h_1}$
$\Delta P = 12 \times \dfrac{0.60}{6.0}$
$\Delta P = \dfrac{12 \times 6\times10^{-1}}{6}$
$\Delta P = 2 \times 6 \times10^{-1}$
$\Delta P = 1.2\text{ kPa}$
12. QUESTION: The gauge pressure at a point 9.5 m below the surface of an incompressible liquid is 95 kPa. Find the change in gauge pressure at a point 3.2 m above that point.
ANSWER: −32 kPa or $-3.2\times10^{1}$ kPa
SOLUTION:
Gauge pressure is proportional to depth, $P = \rho gh$, so:
$\dfrac{\Delta P}{P_1} = \dfrac{\Delta h}{h_1}$
Moving up 3.2 m means $\Delta h = -3.2\text{ m}$:
$\Delta P = P_1 \times \dfrac{\Delta h}{h_1}$
$\Delta P = 95 \times \dfrac{-3.2}{9.5}$
$\Delta P = -32\text{ kPa}$