Submersion

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MCAT Physical › Submersion

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An object is at equilibrium when of the total volume is submerged in gasoline. Find the density of the object.

$\small \rho_{gasoline} = 0.8\frac{g}{mL}$

$0.264\frac{g}{mL}$

$0.8\frac{g}{mL}$

$0.736\frac{g}{mL}$

$0.264\frac{g}{L}$

$0.736\frac{g}{L}$

Explanation

To solve, we can equate the volume of the gasoline displaced to the volume of the portion of the object that is submerged.

$V_{gas}=V_{obj}$

$V=\frac{m}{\rho}$

We know that of the object is submerged, thus of the object's total volume will equal the volume of water displaced.

$\frac{m_{gas}}{\rho_{gas}}=0.33*\frac{m_{obj}}{\rho_{obj}}$

$\frac{\rho_{obj}}{\rho_{gas}}=0.33\frac{m_{obj}}{m_{gas}}$

The displaced masses are equal.

$\frac{\rho_{obj}}{\rho_{gas}}=0.33$

$\rho_{obj}=(0.33)(0.8\frac{g}{mL})=0.264\frac{g}{mL}$

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?

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

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

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

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

Aluminum

Copper

Gold

Two or more experience the same buoyant force

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

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.

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?

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

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

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

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

Aluminum

Copper

Gold

Two or more experience the same buoyant force

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

Explanation

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.

An object is at equilibrium when of the total volume is submerged in gasoline. Find the density of the object.

$\small \rho_{gasoline} = 0.8\frac{g}{mL}$

$0.264\frac{g}{mL}$

$0.8\frac{g}{mL}$

$0.736\frac{g}{mL}$

$0.264\frac{g}{L}$

$0.736\frac{g}{L}$

Explanation

To solve, we can equate the volume of the gasoline displaced to the volume of the portion of the object that is submerged.

$V_{gas}=V_{obj}$

$V=\frac{m}{\rho}$

We know that of the object is submerged, thus of the object's total volume will equal the volume of water displaced.

$\frac{m_{gas}}{\rho_{gas}}=0.33*\frac{m_{obj}}{\rho_{obj}}$

$\frac{\rho_{obj}}{\rho_{gas}}=0.33\frac{m_{obj}}{m_{gas}}$

The displaced masses are equal.

$\frac{\rho_{obj}}{\rho_{gas}}=0.33$

$\rho_{obj}=(0.33)(0.8\frac{g}{mL})=0.264\frac{g}{mL}$

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.

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.

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

II and IV

I and III

II only

IV only

V only

Explanation

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

The more volume an object occupies, the more water it can displace.
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.

Ignore the contribution of atmospheric pressure.

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.

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

II and IV

I and III

II only

IV only

V only

Explanation

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

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

A researcher performs an elemental analysis on a compound. He finds that the compound is made up of only carbon, hydrogen, and oxygen atoms. He isolates a pure sample of the compound and finds that this sample contains of carbon, of hydrogen, and of oxygen. The researcher wants to perform further analysis on this compound the next day. Before leaving the lab the researcher creates three stock solutions of varying concentrations of this compound: (solution A), (solution B), and (solution C). He stores these solutions overnight at a temperature of .

Molecular weight of this compound = $300.486\frac{g}{mol}$

Which of the following is true regarding the densities of the three stock solutions?

$\rho_A<\rho_B<\rho_C$

$\rho_A=\rho_B=\rho_C$

$\rho_A>\rho_B>\rho_C$

$\rho_A=\rho_B<\rho_C$

Explanation

Density is a measure of concentration, defined as mass per unit volume.

$\rho=\frac{\text{mass}}{\text{volume}}$

Since the stock solutions contain different concentrations, the density of each solution must be different. The solution with the highest concentration will contain the highest density, and the solution with lowest concentration will contain the lowest density. The order of increasing density is solution A, solution B, and solution C. Solution C is the most concentrated and will contain the greatest amount of mass per unit volume, followed by solution B and solution A.

$\rho_A<\rho_B<\rho_C$

Remember that concentration can be defined in various ways: molarity, molality, mass percent, density, etc. Increasing concentration means you are increasing all of these quantities. Also remember that the density of the compound remains the same in each solution; only the solutions, as a whole, have different densities.

Molecular weight of this compound = $300.486\frac{g}{mol}$

Which of the following is true regarding the densities of the three stock solutions?

$\rho_A<\rho_B<\rho_C$

$\rho_A=\rho_B=\rho_C$

$\rho_A>\rho_B>\rho_C$

$\rho_A=\rho_B<\rho_C$

Explanation

Density is a measure of concentration, defined as mass per unit volume.

$\rho=\frac{\text{mass}}{\text{volume}}$

$\rho_A<\rho_B<\rho_C$

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