Density and Specific Gravity - AP Physics 2

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Question

What is the mass of air inside a small freezer with dimensions by by ?

$\rho_{air}=1.29\frac{kg}{m^3}$

Answer

Write the density formula.

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

The density of air is: $1.29:\frac{kg}{m^3}$

Find the volume of the freezer in meters. Converting the dimensions from to gives.

$V_{freezer}=(0.6)(0.85)(1)= 0.51 \textup{m}^3$

Solve for the mass of the air.

$m=\rho V=(1.29\frac{kg}{m^3})(0.51 \textup{m}^3)= 0.6579 \textup{kg}$

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Question

Suppose that a cube composed of a mysterious substance is hanging from a string while also submerged in water. If the cube is not moving and has a side length of 8cm and the tension in the string is equal to 15N, what is the specific gravity of this mysterious substance?

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

Answer

To answer this question, it's best to start by drawing a free body diagram so that we can identify the forces acting on the cube. Since the cube is submerged in water but not accelerating, we know that we are dealing with a situation in which there is no net force acting on the cube. In the upward direction we have the force caused by the tension in the string as well as the buoyant force of the displaced water. In the downward direction we have the weight of the cube itself.

$\Sigma F_{y}=0=T+F_{B}-mg$

$mg=T+F_{B}$

$m=\frac{T+F_{B}}{g}=\frac{T}{g}+\frac{F_{B}}{g}$

In order to calculate the buoyant force, we'll need to calculate the volume of the cube in order to determine the amount of water displaced. Since the cube is fully submerged, the volume of the cube will be exactly equal to the volume of displaced water.

$F_{B}=\rho _{water}V_{cube}g$

$V_{cube}=(8cm)^{3}=512cm^{3}$

$F_{B}=(1* 10^{-3} \frac{kg}{cm^{3}})(512 cm^{3})g=(0.512kg)g$

Next, plug this value into the above equation for the mass of the cube.

$m=\frac{T}{g}+\frac{F_{B}}{g}=\frac{15 N}{9.8 \frac{m}{s^{2}}}+\frac{(0.512kg)g}{g}$

Now that we've found the mass of the cube, we can divide this value by its volume in order to obtain its density.

$\rho _{cube}=\frac{m}{V}=\frac{2.04 kg}{512 cm^{3}}=0.00398\frac{kg}{cm^{3}}$

And lastly, since we have the density of the cube, we can calculate its specific gravity by dividing its density by the density of water.

$SG=\frac{\rho _{cube}}{\rho _{water}}=\frac{0.00398 \frac{kg}{cm^{3}}}{0.001 \frac{kg}{cm^{3}}}=3.989$

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Question

A raft is observed floating in a fresh-water lake with $\frac{9}{10}$ of its volume submerged. If the raft were floated in the Great Salt Lake, how much of its volume would be submerged? The density of fresh water is $\rho=1000 \frac{kg}{m^3}$ . Due to its high salt content, the density of the Great Salt Lake is $\rho=1100 \frac{kg}{m^3}$ .

Answer

Use the information given to find the density of the raft. Since the raft is at rest, the forces on it add to zero. Therefore the magnitude of the gravity force is equal to the magnitude of the buoyant force:

$mg=F_B=\rho V g$

The mass of the raft is equal to its density times its volume. We start using subscripts to keep track of which densities and volumes we mean:

$\rho_R V_Rg=\rho _{H_2O}, V_{submerged},g$

The 's cancel, and the is $\frac{9}{10}$ of the raft's total volume:
$\rho_R,V_R=\rho_{H_2O},0.9V_R$

$\rho_R=0.9\rho_{H_2O}=900\frac{kg}{m^3}$

Setting the magnitudes equal again at the Great Salt Lake:

$\rho_R V_Rg=\rho _{GSL}, V_{submerged},g$ . Cancel the 's and solve for the ratio of submerged volume to total volume of the raft:

$\frac{V_{submerged}}{V_R} =\frac{\rho _{R} }{\rho_{GSL}}=\frac{900}{1100}=\frac{82}{100}$

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Question

What are the units for specific gravity?

Answer

Specific gravity is a ratio of a material's density to that of the density of water. Density units can be in any system or scale, but when dividing one unit by itself, the number becomes dimensionless.

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Question

A baseball of diameter and mass lands in water. Will it float or not?

Answer

Using density equation:

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

For a sphere:

$V=\frac{4\pi r^3}{3}$

If diameter is , then radius is .

Converting to and plugging in values:

$\rho=\frac{145}{\frac{4\pi*3.734^3}{3}}$

$\rho=.665\frac{g}{cm^3}$

This is less than the density of water, so it will float.

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Question

Suppose that an object has a density of $0.65\ \frac{\textup{g}}{\textup{cm}^{3}}$ . When floating in water, what fraction of the object will be above the surface of the water?

$\rho_{\text{water}}=1.0\ \frac{\textup{g}}{\textup{cm}^{3}}$

Answer

In this question, we're presented with a scenario in which an object of a given density is floating in water. Based on the information given, we're asked to solve for the volume of the object that is above the surface of the water while floating.

To begin, it is useful if we consider a free-body diagram of the object floating on water. In such a diagram, we need not concern ourselves with the horizontal forces, since these will cancel each other out. However, the vertical forces will provide us with the information we need.

Acting in the downward direction, we have the weight of the object, . In the upward direction, we have the buoyant force caused by the displacement of water by the object. This upward buoyant force is equivalent to $\rho _{water}V_{water}g$ . Thus, we can write an equation for the vertical forces.

$\Sigma F_{y}=\rho _{water}V_{water}g-m_{object}g$

Furthermore, recall that the object is floating. Since it is floating, we know that it is not accelerating in any direction, and thus the net force on it must be zero. Thus, we can set the above expression equal to zero.

$\Sigma F_{y}=\rho _{water}V_{water}g-m_{object}g=0$

Next, we can rearrange the equation to obtain the following:

$\rho _{water}V_{water}g=m_{object}g$

$\rho _{water}V_{water}=m_{object}$

Next, it's important for us to recognize that we can rewrite the expression for the mass of the object. Since density is equal to mass divided by volume, we can rearrange this in terms of density and volume as follows:

$\rho_{object}=\frac{m_{object}}{V_{object}}$

$m_{object}=\rho _{object}V_{object}$

$\rho _{water}V_{water}=\rho _{object}V_{object}$

We can then make one final rearrangement

$\frac{V_{water}}{V_{object}}=\frac{\rho _{object}}{\rho _{water}}$

Now let's think about this for a moment. The volume of water that has been displaced by the object is also the volume of the object that is submerged in the water. Thus, the $\frac{V_{water}}{V_{object}}$ term shown above gives us the fraction of the object that is under the water. Therefore, the remaining fraction of the object must be the portion of it that is above the water. If we take as representing the entire volume of the object, we can solve for the fraction of the object that is above the water like so:

$1-\frac{\rho _{object}}{\rho _{water}}=Volume: of: object: above: water$

$1-\frac{0.65: \frac{g}{cm^{3}}}{1.0: \frac{g}{cm^{3}}}=1-0.65=0.35$

Thus, of the object's volume is above the water.

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Question

A spherical ball submerged in water is held in place. Upon release, the ball has an instantaneous upward acceleration of $4.5\frac{m}{s^2}$ . What is the density of the ball?

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

$\rho_w = 1.0\frac{kg}{L}$

Answer

We can start with Newton's 2nd law for this problem:

$F_{net} = ma$

Upon release, we only have two forces, gravity and buoyancy:

$F_b = m_{w,displaced}g$

Since the ball is accelerating upward, let's designate an upward force as positive. Then, the first expression becomes:

$m_{w,displaced}g-m_bg=m_ba$

Now we need to determine the mass of water displaced. To do this, we will need to choose an arbitrary volume for the sphere. To keep things simple, let's just make it .

Therefore we can say:

$m_b = \rho_b V_b$

Substituting this into our last expression, we get:

$m_wg-\rho_bV_bg=\rho_bV_ba$

Rearranging for density:

$m_wg=\rho_bV_b(a+g)$

$\rho_b = \frac{m_wg}{V_b(a+g)}$

We have values for each of these except for the mass of water displaced. Since the ball has a volume of , we know that of water is also displaced. Then we can say:

$m_{w,displaced} = \rho_w\cdot V_w$

$m_{w,displaced} = \left ( 1.0\frac{kg}{L}\right )(1L) = 1kg$

Now plugging in values to the density expression, we get:

$\rho_b = \frac{(1kg)\left ( 10\frac{m}{s^2}\right )}{1L(4.5\frac{m}{s^2}+10\frac{m}{s^2})}$

$\rho_b = 0.69\frac{kg}{L}$

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Question

What is the density of a cube that is submerged in water.

$\rho_{water}=1000\frac{kg}{m^3}$

Answer

Since the block is submerged by percent we know that it has a density that is percent that of water. We know that the density of water is $1 \frac{kg}{L}$ so percent of that will give us the answer of:

$.75\frac{kg}{L}$

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Question

An cube weighs when submerged in water, but in air. What is the specific gravity of the cube?

Answer

We know that the weight of the object in the air is . This means that the buoyant force will be subtracted by the cube's weight in water (). This gives us a buoyant force of .

The ratio of the weight of the object to the buoyant force in water will give us the specific gravity of the cube in the following equation:

$SG=\frac{Weight}{F_{b}}=\frac{21N}{14N}=1.5$

Therefore the correct answer is

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Question

A block of material has a density . A second block of equal mass has three times the volume of the first. What is the density of the second block?

Answer

Equation of density is $p=\frac{m}{V}$ so multiplying the volume by three would decrease the density by $\frac{1}{3}$ .

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Question

What is the mass of air inside a small freezer with dimensions by by ?

$\rho_{air}=1.29\frac{kg}{m^3}$

Answer

Write the density formula.

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

The density of air is: $1.29:\frac{kg}{m^3}$

Find the volume of the freezer in meters. Converting the dimensions from to gives.

$V_{freezer}=(0.6)(0.85)(1)= 0.51 \textup{m}^3$

Solve for the mass of the air.

$m=\rho V=(1.29\frac{kg}{m^3})(0.51 \textup{m}^3)= 0.6579 \textup{kg}$

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Question

Suppose that a cube composed of a mysterious substance is hanging from a string while also submerged in water. If the cube is not moving and has a side length of 8cm and the tension in the string is equal to 15N, what is the specific gravity of this mysterious substance?

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

Answer

To answer this question, it's best to start by drawing a free body diagram so that we can identify the forces acting on the cube. Since the cube is submerged in water but not accelerating, we know that we are dealing with a situation in which there is no net force acting on the cube. In the upward direction we have the force caused by the tension in the string as well as the buoyant force of the displaced water. In the downward direction we have the weight of the cube itself.

$\Sigma F_{y}=0=T+F_{B}-mg$

$mg=T+F_{B}$

$m=\frac{T+F_{B}}{g}=\frac{T}{g}+\frac{F_{B}}{g}$

In order to calculate the buoyant force, we'll need to calculate the volume of the cube in order to determine the amount of water displaced. Since the cube is fully submerged, the volume of the cube will be exactly equal to the volume of displaced water.

$F_{B}=\rho _{water}V_{cube}g$

$V_{cube}=(8cm)^{3}=512cm^{3}$

$F_{B}=(1* 10^{-3} \frac{kg}{cm^{3}})(512 cm^{3})g=(0.512kg)g$

Next, plug this value into the above equation for the mass of the cube.

$m=\frac{T}{g}+\frac{F_{B}}{g}=\frac{15 N}{9.8 \frac{m}{s^{2}}}+\frac{(0.512kg)g}{g}$

Now that we've found the mass of the cube, we can divide this value by its volume in order to obtain its density.

$\rho _{cube}=\frac{m}{V}=\frac{2.04 kg}{512 cm^{3}}=0.00398\frac{kg}{cm^{3}}$

And lastly, since we have the density of the cube, we can calculate its specific gravity by dividing its density by the density of water.

$SG=\frac{\rho _{cube}}{\rho _{water}}=\frac{0.00398 \frac{kg}{cm^{3}}}{0.001 \frac{kg}{cm^{3}}}=3.989$

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Question

A raft is observed floating in a fresh-water lake with $\frac{9}{10}$ of its volume submerged. If the raft were floated in the Great Salt Lake, how much of its volume would be submerged? The density of fresh water is $\rho=1000 \frac{kg}{m^3}$ . Due to its high salt content, the density of the Great Salt Lake is $\rho=1100 \frac{kg}{m^3}$ .

Answer

Use the information given to find the density of the raft. Since the raft is at rest, the forces on it add to zero. Therefore the magnitude of the gravity force is equal to the magnitude of the buoyant force:

$mg=F_B=\rho V g$

The mass of the raft is equal to its density times its volume. We start using subscripts to keep track of which densities and volumes we mean:

$\rho_R V_Rg=\rho _{H_2O}, V_{submerged},g$

The 's cancel, and the is $\frac{9}{10}$ of the raft's total volume:
$\rho_R,V_R=\rho_{H_2O},0.9V_R$

$\rho_R=0.9\rho_{H_2O}=900\frac{kg}{m^3}$

Setting the magnitudes equal again at the Great Salt Lake:

$\rho_R V_Rg=\rho _{GSL}, V_{submerged},g$ . Cancel the 's and solve for the ratio of submerged volume to total volume of the raft:

$\frac{V_{submerged}}{V_R} =\frac{\rho _{R} }{\rho_{GSL}}=\frac{900}{1100}=\frac{82}{100}$

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Question

What are the units for specific gravity?

Answer

Specific gravity is a ratio of a material's density to that of the density of water. Density units can be in any system or scale, but when dividing one unit by itself, the number becomes dimensionless.

Compare your answer with the correct one above

Question

A baseball of diameter and mass lands in water. Will it float or not?

Answer

Using density equation:

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

For a sphere:

$V=\frac{4\pi r^3}{3}$

If diameter is , then radius is .

Converting to and plugging in values:

$\rho=\frac{145}{\frac{4\pi*3.734^3}{3}}$

$\rho=.665\frac{g}{cm^3}$

This is less than the density of water, so it will float.

Compare your answer with the correct one above

Question

Suppose that an object has a density of $0.65\ \frac{\textup{g}}{\textup{cm}^{3}}$ . When floating in water, what fraction of the object will be above the surface of the water?

$\rho_{\text{water}}=1.0\ \frac{\textup{g}}{\textup{cm}^{3}}$

Answer

In this question, we're presented with a scenario in which an object of a given density is floating in water. Based on the information given, we're asked to solve for the volume of the object that is above the surface of the water while floating.

To begin, it is useful if we consider a free-body diagram of the object floating on water. In such a diagram, we need not concern ourselves with the horizontal forces, since these will cancel each other out. However, the vertical forces will provide us with the information we need.

Acting in the downward direction, we have the weight of the object, . In the upward direction, we have the buoyant force caused by the displacement of water by the object. This upward buoyant force is equivalent to $\rho _{water}V_{water}g$ . Thus, we can write an equation for the vertical forces.

$\Sigma F_{y}=\rho _{water}V_{water}g-m_{object}g$

Furthermore, recall that the object is floating. Since it is floating, we know that it is not accelerating in any direction, and thus the net force on it must be zero. Thus, we can set the above expression equal to zero.

$\Sigma F_{y}=\rho _{water}V_{water}g-m_{object}g=0$

Next, we can rearrange the equation to obtain the following:

$\rho _{water}V_{water}g=m_{object}g$

$\rho _{water}V_{water}=m_{object}$

Next, it's important for us to recognize that we can rewrite the expression for the mass of the object. Since density is equal to mass divided by volume, we can rearrange this in terms of density and volume as follows:

$\rho_{object}=\frac{m_{object}}{V_{object}}$

$m_{object}=\rho _{object}V_{object}$

$\rho _{water}V_{water}=\rho _{object}V_{object}$

We can then make one final rearrangement

$\frac{V_{water}}{V_{object}}=\frac{\rho _{object}}{\rho _{water}}$

Now let's think about this for a moment. The volume of water that has been displaced by the object is also the volume of the object that is submerged in the water. Thus, the $\frac{V_{water}}{V_{object}}$ term shown above gives us the fraction of the object that is under the water. Therefore, the remaining fraction of the object must be the portion of it that is above the water. If we take as representing the entire volume of the object, we can solve for the fraction of the object that is above the water like so:

$1-\frac{\rho _{object}}{\rho _{water}}=Volume: of: object: above: water$

$1-\frac{0.65: \frac{g}{cm^{3}}}{1.0: \frac{g}{cm^{3}}}=1-0.65=0.35$

Thus, of the object's volume is above the water.

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Question

A spherical ball submerged in water is held in place. Upon release, the ball has an instantaneous upward acceleration of $4.5\frac{m}{s^2}$ . What is the density of the ball?

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

$\rho_w = 1.0\frac{kg}{L}$

Answer

We can start with Newton's 2nd law for this problem:

$F_{net} = ma$

Upon release, we only have two forces, gravity and buoyancy:

$F_b = m_{w,displaced}g$

Since the ball is accelerating upward, let's designate an upward force as positive. Then, the first expression becomes:

$m_{w,displaced}g-m_bg=m_ba$

Now we need to determine the mass of water displaced. To do this, we will need to choose an arbitrary volume for the sphere. To keep things simple, let's just make it .

Therefore we can say:

$m_b = \rho_b V_b$

Substituting this into our last expression, we get:

$m_wg-\rho_bV_bg=\rho_bV_ba$

Rearranging for density:

$m_wg=\rho_bV_b(a+g)$

$\rho_b = \frac{m_wg}{V_b(a+g)}$

We have values for each of these except for the mass of water displaced. Since the ball has a volume of , we know that of water is also displaced. Then we can say:

$m_{w,displaced} = \rho_w\cdot V_w$

$m_{w,displaced} = \left ( 1.0\frac{kg}{L}\right )(1L) = 1kg$

Now plugging in values to the density expression, we get:

$\rho_b = \frac{(1kg)\left ( 10\frac{m}{s^2}\right )}{1L(4.5\frac{m}{s^2}+10\frac{m}{s^2})}$

$\rho_b = 0.69\frac{kg}{L}$

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Question

What is the density of a cube that is submerged in water.

$\rho_{water}=1000\frac{kg}{m^3}$

Answer

Since the block is submerged by percent we know that it has a density that is percent that of water. We know that the density of water is $1 \frac{kg}{L}$ so percent of that will give us the answer of:

$.75\frac{kg}{L}$

Compare your answer with the correct one above

Question

An cube weighs when submerged in water, but in air. What is the specific gravity of the cube?

Answer

We know that the weight of the object in the air is . This means that the buoyant force will be subtracted by the cube's weight in water (). This gives us a buoyant force of .

The ratio of the weight of the object to the buoyant force in water will give us the specific gravity of the cube in the following equation:

$SG=\frac{Weight}{F_{b}}=\frac{21N}{14N}=1.5$

Therefore the correct answer is

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Question

A block of material has a density . A second block of equal mass has three times the volume of the first. What is the density of the second block?

Answer

Equation of density is $p=\frac{m}{V}$ so multiplying the volume by three would decrease the density by $\frac{1}{3}$ .

Compare your answer with the correct one above

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