Question 1

A rock is suspended from a scale reads 10.0 N. A beaker of water is raised up so the rock is totally submerged in the water. The scale now reads 6.25 N.What is the density of the rock?

Accepted Answer

The buoyant force is 10.0 N - 6.25 N = 3.75 N, which equals the weight of the water displaced. The weight of the rock is 10.0 N. The density of the rock is the weight of the rock divided by the buoyant force, which is 10.0 N / 3.75 N = 2.67 times the density of water.

Question 2

A hot air balloon plus its cargo has a mass of 400 kg, and it holds 700 m³ of hot air. It is floating at a constant height in air with a density of 1.29 kg/m³. What is the density of the hot air in the balloon?

Accepted Answer

The buoyant force acting on the balloon equals the weight of the displaced air. Therefore, we have: Buoyant force = Weight of air displaced ρ_air* V_balloon * g = m_balloon * g + m_cargo * g ρ_air = (m_balloon + m_cargo) / V_balloon ρ_air = 400 kg / 700 m³ = 0.57 kg/m³ Therefore, the density of the hot air in the balloon is 0.57 kg/m³. Answer choice B is correct.

Question 3

In a certain one-lane road there are no entrances or exits. As the road passes through open country and populated areas, the speed limit changes. Assume that cars travel at the posted speed limit. In the open road, where the speed limit is 55 mph, the cars are spaced so there are 51 cars per mile. If the flow of cars is steady, how many cars per mile are there in a stretch of road where the speed limit is 40 mph?

Accepted Answer

The answer of In a certain one-lane road there are...

Question 4

A person who weighs 550 N empties her lungs as much as possible and is then completely immersed in water while suspended from a harness. Her apparent weight is now 21.2 N. What is her density?

Accepted Answer

The apparent weight is the actual weight minus the buoyant force. The buoyant force is equal to the weight of the water displaced, which is 550 N - 21.2 N = 528.8 N. The density of water is 1000 kg/m³, so the volume of water displaced is 528.8 N / (9.8 m/s² * 1000 kg/m³) = 0.0539 m³. The person's mass is 550 N / 9.8 m/s² = 56.12 kg. The density is mass/volume = 56.12 kg / 0.0539 m³ = 1040 kg/m³.

Question 5

A cylinder, semicircular in cross-section, 10.0 m long and 5.00 m in radius, is completely submerged in water. What is the buoyant force on the cylinder?

Accepted Answer

The buoyant force is equal to the weight of the displaced water. The volume of the cylinder is (1/2)πr²L = (1/2)π(5.00 m)²(10.0 m). The density of water is 1000 kg/m³, and g = 9.81 m/s². Calculating gives the buoyant force as approximately 3.85 × 10⁶ N.

Question 6

The bathyscaphe Trieste consists of a cylindrical hull, 15.2 m long and 3.66 m in diameter, filled with gasoline with a density of 880 kg/m³ to make it buoyant. If it is totally submerged in salt water with a density of 1025 kg/m³ and neither rising nor sinking, what is the weight of the craft?

Accepted Answer

The answer of The bathyscaphe Trieste consists of a cylindrical...

Question 7

A 12,000-N car is raised using a hydraulic lift, which consists of a U-tube with arms of unequal areas, filled with oil and capped at both ends with tight-fitting pistons. The wider arm of the U-tube has a radius of 18.0 cm and the narrower arm has a radius of 5.00 cm. The car rests on the piston on the wider arm of the U-tube. The pistons are initially at the same level. What is the initial force that must be applied to the smaller piston in order to start lifting the car? For purposes of this problem, neglect the weight of the pistons.

Accepted Answer

The answer of A 12,000-N car is raised using a...

Question 8

A 12,000-N car is raised using a hydraulic lift, which consists of a U-tube with arms of unequal areas, filled with oil with a density of 800 kg/m³ and capped at both ends with tight-fitting pistons. The wider arm of the U-tube has a radius of 18.0 cm and the narrower arm has a radius of 5.00 cm. The car rests on the piston on the wider arm of the U-tube. The pistons are initially at the same level. What is the force that must be applied to the smaller piston in order to lift the car after it has been raised 1.20 m? For purposes of this problem, neglect the weight of the pistons.

Accepted Answer

The pressure at any point in the U-tube is given by the equation:

$P = \frac{F}{A}$

where P is the pressure, F is the force applied, and A is the area of the piston.

The pressure is constant throughout the U-tube, so we can set the pressure at the level of the pistons to be equal:

$P_1 = P_2$

where the subscripts 1 and 2 refer to the wider and narrower arms of the U-tube, respectively.

We can use the equation for pressure to relate the forces and areas of the two pistons:

$\frac{F_1}{A_1} = \frac{F_2}{A_2}$

We can solve for $F_2$ in terms of $F_1$:

$F_2 = \frac{A_2}{A_1}F_1$

We can plug in the values for the areas:

$F_2 = \frac{\pi r_2^2}{\pi r_1^2}F_1 = \frac{r_2^2}{r_1^2}F_1$

Now we need to find $F_1$. We know the weight of the car is 12,000 N, and we can use the density of the oil to find the weight of the oil displaced by the car:

$w_{oil} = V_{displaced} 	imes ho_{oil} = (\pi r_1^2 h) 	imes (ho_{oil})$

where $V_{displaced}$ is the volume of oil displaced by the car and $h$ is the height of the oil column in the wider arm of the U-tube. The height of the oil column in the narrower arm of the U-tube is $1.2	ext{ m} - h$.

Setting the weight of the car equal to the weight of the displaced oil, we can solve for $h$:

$12,000	ext{ N} = (\pi r_1^2 h) 	imes (ho_{oil})$

$h = \frac{12,000	ext{ N}}{\pi r_1^2 ho_{oil}}$

We can plug in the values and solve for $h$:

$h = \frac{12,000	ext{ N}}{\pi (0.18	ext{ m})^2 (800	ext{ kg/m}^3)} \approx 0.274	ext{ m}$

Now we can plug in the values for $r_1$, $r_2$, $h$, and $F_1$ to solve for $F_2$:

$F_2 = \frac{(0.05	ext{ m})^2}{(0.18	ext{ m})^2}F_1 \approx 0.287F_1$

Setting $P_1 = P_2$, we can use the equation for pressure to relate $F_1$ to the weight of the displaced oil:

$\frac{F_1}{A_1} = P = ho_{oil}gh$

$F_1 = (ho_{oil}gh)(A_1) = (ho_{oil}g)(\frac{12,000	ext{ N}}{\pi r_1^2 ho_{oil}})(\pi r_1^2) = 12,000	ext{ N}$

Now we can plug in $F_1$ to find $F_2$:

$F_2 = 0.287(12,000	ext{ N})\approx 3,444	ext{ N} \approx 1.96	ext{ kN}$

Therefore, the answer is (C) 1.96 kN.

Question 9

FIGURE 15-8 
-A cylinder, semicircular in cross-section, 10.0 m long and 5.00 m in radius, is completely submerged, with its flat 5.00 m × 10.0 m side horizontal and 8,00 m below the water surface, as shown in Figure 15-8. What is the force on the curved surface of the half-cylinder? Neglect atmospheric pressure.

Accepted Answer

The answer of FIGURE 15-8 
-A cylinder, semicircular in cross-section,...

Question 10

FIGURE 15-5 -A reservoir manometer consists of a U-tube in which the two arms have different diameters, as shown in Figure 15-5. The diameter on the narrow side is 10.0 mm and the diameter on the reservoir side is 25.0 mm. The liquid in the manometer is oil with a density of 830 kg/m³ and the fluids in contact with the oil are gases of negligible density. The dotted line indicates the level of the oil when the pressures p₁ and p₂ are equal. What is the value of h when p₁ - p₂= 1000 Pa?

Accepted Answer

The pressure difference is given by the hydrostatic pressure formula: ΔP = ρgh. Solving for h gives h = ΔP / (ρg). Substituting the given values: h = 1000 Pa / (830 kg/m³ * 9.81 m/s²) = 0.0123 m or 12.3 cm.

Question 11

FIGURE 15-3 -The U-tube shown in Figure 15-3 contains mercury with a density of 13,600 kg/m³. It is open on one end and the other end is connected to a boiler, which contains water at high pressure. The water from the boiler pushes directly on the mercury column. What is the gauge pressure at point A in the boiler?

Accepted Answer

The gauge pressure at point A in the boiler can be calculated using the formula P = rho*g*h, where rho is the density, g is the acceleration due to gravity and h is the height of the mercury column. Since the U-tube is open at one end, the height of the mercury column is the same on both sides. Therefore, we only need to consider the difference in height between the mercury column and the water level in the boiler, which is 0.035 m. Using the given values, we get P = (13600 kg/m^3 * 9.81 m/s^2) * 0.035 m = 4.41 kPa. Therefore, the correct answer is B.

Question 12

A barge with vertical sides is 10.0 m wide and 60.0 m long and is floating in fresh water. How much deeper into the water does the barge sink when 300,000 kg of coal are loaded on the barge?

Accepted Answer

The volume of water displaced equals the weight of the coal divided by the density of water. The density of water is 1000 kg/m³. The volume displaced is 300,000 kg / 1000 kg/m³ = 300 m³. The area of the barge is 10 m * 60 m = 600 m². The depth the barge sinks is 300 m³ / 600 m² = 0.5 m or 50.0 cm.

Question 13

FIGURE 15-7 -The two water reservoirs shown in Figure 15-7 are open to the atmosphere. The manometer contains oil with a density of 820 kg/m³. What is the difference in elevation h if the manometer reading m is 25.0 cm?

Accepted Answer

The difference in elevation h is calculated using the formula h = m * (density of oil / density of water). Given m = 0.25 m, density of oil = 820 kg/m³, and density of water = 1000 kg/m³, h = 0.25 * (820/1000) = 0.205 m.

Question 14

A cubical box 25.00 cm on each side is immersed in a fluid. The pressure at the top surface of the box is 109.4 kPa and the pressure on the bottom surface is 112 kPa. What is the density of the fluid?

Accepted Answer

The pressure difference between the top and bottom surfaces is due to the weight of the fluid column, which is 2.6 kPa (112 kPa - 109.4 kPa). The height of the box is 0.25 m. Using the formula for pressure difference, ΔP = ρgh, where ρ is the density, g is the acceleration due to gravity (9.81 m/s²), and h is the height, we can solve for ρ: ρ = ΔP / (gh) = 2600 Pa / (9.81 m/s² * 0.25 m) = 1060 kg/m³.

Question 15

FIGURE 15-4 -The circular pipes shown in Figure 15-4 both contain water, but at different pressures. They are connected by means of an inverted U-tube, which contains oil with a density of 800 kg/m³. The heights are: h_A = 7.0 cm, h_B = 20.0 cm, and h_C = 13.0 cm. What is the pressure difference (P_B - P_A)?

Accepted Answer

The pressure difference is calculated using the hydrostatic pressure difference due to the oil column: ΔP = ρgh, where ρ is the density of the oil, g is the acceleration due to gravity, and h is the height difference. The effective height difference is (h_B - h_C) - h_A = (20 cm - 13 cm) - 7 cm = 0 cm. Therefore, the pressure difference is due to the oil column between B and C: ΔP = 800 kg/m³ * 9.81 m/s² * 0.07 m = 255 Pa.

Question 16

A U-tube is open to the atmosphere at both ends. Water is poured into the tube until the water rises part-way along the straight sides, and then some oil with a density of 920 kg/m³ is poured into one end. This causes the water surface on that side of the tube to go down by 5.50 mm and the surface on the other side to go up by the same amount. How much higher is the top surface of the oil on that side of the tube compared with the surface of the water on the other side of the tube?

Accepted Answer

The pressure at the bottom of the U-tube is the same on both sides. The pressure due to the water is the same on both sides. The pressure due to the oil is greater on the side with the oil. The difference in pressure is due to the difference in height of the oil and water. The difference in height is 5.50 mm. The density of the oil is 920 kg/m³. The pressure due to the oil is 920 kg/m³ * 9.81 m/s² * 5.50 mm = 51.8 Pa. The difference in height of the oil and water is 51.8 Pa / (920 kg/m³ * 9.81 m/s²) = 0.96 mm.

Question 17

The circular pipes shown in Figure 15-4 both contain water, but there is a pressure difference between them, (P_B - P_A) = 1000 Pa. They are connected by means of an upside-down U-tube, which contains oil with a density of 800 kg/m³. The heights are: h_A = 5.0 cm and h_B = 40.0 cm. What is h_C?

Accepted Answer

The answer of The circular pipes shown in Figure 15-4...

Question 18

A spar buoy consists of a circular cylinder, which floats with its axis oriented vertically. One such buoy has a radius of 1.00 m, a height of 2.00 m, and weighs 40.0 kN. What portion of it is submerged when it is floating in fresh water?

Accepted Answer

The answer of A spar buoy consists of a circular...

Question 19

A barge is 10.0 m wide and 60.0 m long and has vertical sides. The bottom of the hull is 1.20 m below the water surface. What is the weight of the barge and its cargo, if it is floating in fresh water?

Accepted Answer

The answer of A barge is 10.0 m wide and...

Question 20

FIGURE 15-6 -The two water reservoirs shown in Figure 15-6 are open to the atmosphere. The manometer contains mercury with a density of 13,600 kg/m³. What is the difference in elevation h if the manometer reading m is25.0 cm?

Accepted Answer

The answer of FIGURE 15-6 
-The two water reservoirs shown...

Quiz 15: Fluids