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MCAT Physical : Submersion

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Example Question #4 : Buoyancy And Displacement

Two students in a physics class are conducting an experiment to see how different objects displace water in a container. Student A places a lead spherical marble with a radius of $\small 5cm$ in the container. Student B places a considerably lighter solid glass sphere in the container, with twice the volume of the lead marble. Which of the following statements is true?

I. The amount of displaced water depends on the mass of a fully immersed object

II. The amount of water displaced depends on the volume of a fully immersed object

III. Student A's object will make the water level rise more than Student B's

IV. Student B's object will make the water level rise more than Student A's

V. From the information given, one cannot determine which object would displace more water

Possible Answers:

V only

I and III

II and IV

II only

IV only

Correct answer:

II and IV

Explanation:

This question relates directly to Archimedes' principle considering buoyancy. The take home points of this question are given in the answers:

1) The more volume an object occupies, the more water it can displace.
2) The mass of an object has no relevance to the amount of water displaced as long the object is fully immersed.

From the second of these two points, we can eliminate statement I since the amount of water displaced is independent of mass. We can also eliminate statements III and V, since the marble with greater volume will displace a greater volume of water. This leaves the correct statements: II and IV.

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Example Question #5 : Buoyancy And Displacement

A 100g block of copper, a 100g block of aluminum, and a 100g sphere of gold are placed into a large tub of water. Which object expereiences the greatest buoyant force?

$\small \rho_{Cu}=9.0\frac{g}{cm^{3}}$

$\small \rho_{Al}=2.7\frac{g}{cm^{3}}$

$\small \rho_{Au}=19.3\frac{g}{cm^{3}}$

$\small \rho_{H20}= 1\frac{g}{cm^{3}}$

Possible Answers:

Aluminum

There is not enough information, without knowing the depth of each object

Copper

Gold

Two or more experience the same buoyant force

Correct answer:

Aluminum

Explanation:

Buoyant force depends only on the density of the fluid, and the volume of fluid displaced, according to $\small F_{b}= \rho_{fluid}*V_{displaced}*g$ . Since each object has a density greater than that of water, they will all be completely submerged. So, the volume of fluid displaced equals the volume of the object in each case, meaning that whichever object has the largest volume will experience the greatest buoyant force.

For copper: $\small V_{Cu}=\frac{m_{Cu}}{\rho _{Cu}}=\frac{100g}{9.0\frac{g}{cm^{3}}}=11cm^{3}$

For aluminum: $\small V_{Al}=\frac{m_{Al}}{\rho _{Al}}=\frac{100g}{2.7\frac{g}{cm^{3}}}=37cm^{3}$

For gold: $\small V_{Au}=\frac{m_{Au}}{\rho _{Au}}=\frac{100g}{19.3\frac{g}{cm^{3}}}=5.18cm^{3}$

Since aluminum has the largest volume, and the largest volume of water displaced, it experiences the greatest buoyant force.

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Example Question #6 : Buoyancy And Displacement

When a copper sphere is suspended from a spring scale in air, the scale reads 2.94N . Assuming the density of copper is $9000\frac{kg}{m^3}$ , what will the scale read if the sphere is lowered (completely submerged) in a large beaker of water?

Possible Answers:

3.27N

0.96N

0.33N

2.61N

Correct answer:

2.61N

Explanation:

We need to find the buoyant force on the sphere to see how much the scale reading changes.

Buoyant force is given by $F_{b}= \rho_{fluid} Vg$ , where V is the volume displaced by the sphere.

We need to find this volume, which also equals the sphere's volume. Given the sphere's mass (m = F/g = 2.94/9.8 = 0.30kg) and density, we can find volume with either $\rho_{sphere}=\frac{V}{g}$ or $V = \frac{m}{\rho _{sphere}} = \frac{0.300kg}{9000 \frac{kg}{m^{3}}}=3.33*10^{-5} m^{3}$ .

Plugging this volume into the buoyant force equation, together with the density of water (1000 kg/m³) gives the following.

$F_{b}=(1000 \frac{kg}{m^{3}})(3.33*10^{-5}m^{3})(9.8 \frac{m}{s^{2}}) \approx 0.33 N$

So now the upwards force provided by the scale is reduced by 0.33N.

$F_{scale, final} = F_{scale, original} - F_{b} = 2.94 - 0.33 = 2.61 N$

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Example Question #1 : Buoyancy And Displacement

A soccer ball with a mass of 0.5kg and diameter 0.2m is held under water. What is the acceleration of the ball the instant it is released?

$\rho_{\text{water}}=1 \frac{g}{ml}$

$g=10\frac{m}{s^2}$

Possible Answers:

$73.8\frac{m}{s^2}$

$21.9\frac{m}{s^2}$

$101.8\frac{m}{s^2}$

$40.5\frac{m}{s^2}$

$12.4\frac{m}{s^2}$

Correct answer:

$73.8\frac{m}{s^2}$

Explanation:

To calculate the ball's instantaneous acceleration, we need to calculate the net force on the ball. We can neglect the resistance due to water since we are calculating the net force the moment the ball is released, before it has begun to move. Therefore, there are only two forces on the ball: gravitational force and buoyant force.

Force of gravity:

$F_g = mg = (0.5kg)(10\frac{m}{s^2}) = 5N$

The bouyant force is simply the force of water that is displaced by the ball:

$F_b = V_{\text{ball}}\cdot \rho_{\text{water}}\cdot g = (\frac{4}{3}\pi r^3)\rho g$

$F_b = \frac{4}{3}\pi(0.1m)^3(1000\frac{kg}{m^3})(10\frac{m}{s^2})$

F_b = 41.89 N

Now we can calculate the net force on the ball:

$F_{net} = F_b - F_g = 41.89 N - 5 N = 36.89 N$

We can simply use Newton's second law to calculate the acceleration of the ball:

$F_{net}=ma$

$a = \frac{36.89N}{0.5kg} = 73.8 \frac{m}{s^2}$

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Example Question #21 : Submersion

Ethanol has a specific gravity of 0.780. If a long barometer tube is placed in a dish of ethanol, with the top end open to the atmosphere, to what height in the tube will the ethanol rise?

Atmospheric pressure = 1.01 * 10⁵Pa

g = 9.81 m/s²

Density of water = 1000 kg/m³)

Possible Answers:

7.6cm

12.6cm

8.3m

4.2m

13.2m

Correct answer:

13.2m

Explanation:

The pressure exerted by the air on the top of the ethanol must equal the fluid pressure in the column of liquid at equilibrium. Fluid pressure is the product of density, gravity, and height of the fluid.

$P_{atm} = \rho_{ethanol} gh$

The density of ethanol, $\rho_{ethanol}$ , may be calculated by multiplying specific gravity times the reference density of water, as specific gravity is simply a ratio of a compound's density to that of water.

$\rho_{ethanol} = 0.780*1000\frac{kg}{m^{3}}=780\frac{kg}{m^{3}}$

Using this density in the pressure equation allows us to solve for the height of the ethanol.

$1.01 * 10^{5} Pa = \left ( 780\frac{kg}{m^{3}} \right )\left ( 9.81\frac{m}{s^{2}} \right )h$

After calculating, $h = 13.2 m$

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Example Question #22 : Submersion

A diver in a lake rises from 35m deep, where the gauge pressure reads 2atm , to the surface, where the gauge pressure reads 0atm . Upon reaching the surface, by what percent has the pressure experienced by the diver changed?

Possible Answers:

Correct answer:

Explanation:

When the diver is submerged, the pressure experienced is the sum of the gauge pressure and the atmospheric pressure above the surface.

$P_{35m} = P_{atm} + P_{gauge}=1atm+2atm=3atm$

At the surface, the gauge pressure will be zero.

$P_{surface} = P_{atm} + P_{gauge}=1atm+0atm=1atm$

The percent change in pressure will be equal to the ratio of the change in pressure to the pressure while submerged.

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Example Question #23 : Submersion

A tank is filled to a depth of $\small 3.27m$ with water. Oil with a density of $\small 800\frac{kg}{m^3}$ is added to the tank until the final depth is $\small 5.0m$ . What is the reading of the pressure gauge at the bottom of the tank?

Ignore the contribution of atmospheric pressure.

Possible Answers:

53.8kPa

41.9kPa

45.6kPa

34.2kPa

497Pa

Correct answer:

45.6kPa

Explanation:

The pressure contribution of the liquid can be found using the equation:

$P=SG*h=\rho g*h$

The product of the density, gravity, and height of the column of liquid will give the pressure. Start by calculating the pressure of the water.

$P_{water}=(1000\frac{kg}{m^3})(9.8\frac{m}{s^2})(3.27m)=32046Pa$

Then calculate the pressure of the oil.

$P_{oil}=(800\frac{kg}{m^3})(9.8\frac{m}{s^2})(5.0m-3.27m)$

$P_{oil}=(800\frac{kg}{m^3})(9.8\frac{m}{s^2})(1.73m)=13563Pa$

Sum the pressure of each liquid to find the total pressure at the bottom of the tank.

32046Pa+13563Pa=45609Pa=45.6kPa

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Example Question #24 : Submersion

Two coins lie at the bottom of a pool with an uneven bottom. One is under $\small 5m$ of the water, and the second is under $\small 0.5m$ of water. What is the approximate ratio of total pressure experienced by the first coin to total pressure experienced by the second?

Possible Answers:

10:1

2:1

1:10

50:3

25:3

Correct answer:

25:3

Explanation:

$P_{total} = P_{atm} + P_{gauge} = P_{atm} + (\rho_{water})gh$

Atmospheric pressure always equals $\small 1atm$ on earth, assuming normal elevation.

Gauge pressure always equals $\small \rho gh$ , where $\small \rho$ is the density of the medium, $\small g$ is gravity, and $\small h$ is the distance to the surface.

$P_1=1atm+(1\frac{g}{mL})(9.8\frac{m}{s^2})(5m)=50atm$

$P_2=1atm+(1\frac{g}{mL})(9.8\frac{m}{s^2})(0.5m)=5.9atm$

$P_1:P_2\rightarrow 50:5.9\approx 25:3$

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Example Question #25 : Submersion

Boiling point is the temperature a liquid needs to achieve in order to begin its transformation into a gaseous state. Campers and hikers who prepare food during their trips have to account for differences in atmospheric pressure as they ascend in elevation. During the ascent, the decrease in atmospheric pressure changes the temperature at which water boils.

Further complicating the matter is the observation that addition of a solute to a pure liquid also changes the boiling point. Raoult’s Law can be used to understand the changes in boiling point if a non-volatile solute is present, as expressed here.

$\LARGE P = X_{solv} * P_{solv}$

In this law, $\small X_{solv}$ is the mole fraction of the solvent, $\small P_{solv}$ is the vapor pressure of the pure solvent, and is the vapor pressure of the solution. When this vapor pressure is equal to the local atmospheric pressure, the solution boils.

Like boiling point change, osmotic pressure is a property influenced by the addition of solute to solvent. A scientist is studying a solution in a beaker. Osmotic pressure in this solution pulls __________ from __________ solute concentration to __________ solute concentration.

Possible Answers:

Osmotic pressure only exists across cell membranes

solute . . . high . . . low

water . . . low . . . high

water . . . high . . . low

solute . . . low . . . high

Correct answer:

water . . . low . . . high

Explanation:

Osmotic pressure pulls water, as a pure solvent, into regions of high solute density to even out the solute concentration between different regions of solution.

Osmotic pressure is related to solute concentration, but does not affect the flow of solute.

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Example Question #111 : Ap Physics 2

The law of Laplace states what relationship between that the tension in the wall of a bubble and the radius of the bubble?

Possible Answers:

Tension is directly proportional to the radius

Tension is inversely proportional to the square of the radius

Tension is proportional to the cube of the radius

Tension is directly proportional to the square of the radius

Tension is inversely proportional to the radius

Correct answer:

Tension is directly proportional to the radius

Explanation:

The law of Laplace states that wall tension in a bubble, balloon, pulmonary alveolus, or blood vessel is directly proportional to the radius and the internal pressure.

Law of Laplace: $P=T(\frac{1}{R})$

$T \propto P*R$

In other words, as the size of the bubble increases at constant pressure or as the pressure increases at constant size, the wall tension needed to keep the bubble from increasing in size is increased. In part, this explains why aneurysms on blood vessels burst.

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MCAT Physical : Submersion

Study concepts, example questions & explanations for MCAT Physical

All MCAT Physical Resources

Example Questions

Example Question #4 : Buoyancy And Displacement

Example Question #5 : Buoyancy And Displacement

Example Question #6 : Buoyancy And Displacement

Example Question #1 : Buoyancy And Displacement

Example Question #21 : Submersion

Example Question #22 : Submersion

Example Question #23 : Submersion

Example Question #24 : Submersion

Example Question #25 : Submersion

Example Question #111 : Ap Physics 2

All MCAT Physical Resources

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