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AP Physics 2 : AP Physics 2

Study concepts, example questions & explanations for AP Physics 2

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Example Questions

Example Question #861 : Ap Physics 2

You have a 4cm long copper wire with a radius of 0.5mm. You have experimentally determined the resistance of the wire to be $0.8709m\Omega$ . What is the resistivity of copper?

Possible Answers:

None of the other answers is correct

$0.011\Omega \cdot m$

$1.71\cdot10^{-8}\Omega\cdot m$

$139.344\ \Omega \cdot m$

$18.5\cdot10^{-8}\Omega\cdot m$

Correct answer:

$1.71\cdot10^{-8}\Omega\cdot m$

Explanation:

The resistivity of a material is how much the material resists the flow of charge through it. Metals have low resistivities (which makes them good conductors), while things like glass or plastic have high resistivities.

The equation for resistivity is as follows:

$R=\frac{\rho L}{A}$

is the length of the wire, is the cross-sectional area, is the resistance, and $\rho$ is the resistivity. We have values for and , and we're given the radius of the wire so we can find , so we're trying to solve for $\rho$ . If we rearrange the equation, we end up with this:

$\frac{RA}{L}=\rho$

We're given the radius of the wire, so to find the area, we square the radius and multiply it by $\pi$ .

$A=\pi r^2$

$A=\pi (0.0005)^2$

$A=7.854\cdot 10^{-7}$

Now, we can plug in our values.

$\frac{RA}{L}=\rho$

$\frac{(0.8709\cdot 10^{-3}) \cdot (7.854\cdot 10^{-7})}{0.04}=\rho$

$1.71\cdot 10^{-8}=\rho$

Therefore, the answer is $1.71\cdot10^{-8}\Omega\cdot m$ .

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Example Question #1 : Resistivity

You create a cylinder of an unknown material that has a diameter of 2cm , and a height of 1cm . You attach the electrodes to the faces of the cylinder and find the sample has a resistance of $25\Omega$ .

What is the resistivity of the material?

Possible Answers:

$78.5\Omega \cdot cm$

$69.5\Omega \cdot cm$

$31.0\Omega \cdot cm$

$35\Omega \cdot cm$

$58.5\Omega \cdot cm$

Correct answer:

$78.5\Omega \cdot cm$

Explanation:

Use the equation for resistivity:

$R= \rho \frac{L}A{}$

First, find the cross sectional area.

$A=\pi r^{2}$

A=3.14cm^2

Plug in known values.

$R= \rho \frac{L}A{}$

$\rho = \frac{25\Omega\cdot 3.14cm^2}{1cm}$

$\rho =78.5\Omega \cdot cm$

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Example Question #1 : Resistivity

You have a sample that is a cube of volume 1cm^3 . You attach electrodes to opposite faces and find the resistance to be $3\Omega$ .

What is the resistivity of the material?

Possible Answers:

$1\Omega \cdot cm$

$5\Omega \cdot cm$

$2\Omega \cdot cm$

$3\Omega \cdot cm$

$4\Omega \cdot cm$

Correct answer:

$3\Omega \cdot cm$

Explanation:

Since we have a cube of volume 1cm^3 , the length of the material must be 1cm . Use the equation for resistivity.

$R=\rho \frac{L}{A}$

Rearrange the equation and plug in known values.

$\rho = \frac{RA}{L}$

$\rho=\frac{3\Omega\cdot 1cm^2}{1cm}=3\Omega\cdot cm$

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Example Question #213 : Electricity And Magnetism

You have a material with resistivity $3500\Omega\cdot m$ . You build a component of a fuel cell out of this material with a cross sectional area of 0.01m^2 , and a thickness of 0.01m .

What will be the resistance of this component?

Possible Answers:

$3500\Omega$

$350\Omega$

$3.5\Omega$

None of these

$0.35\Omega$

Correct answer:

$3500\Omega$

Explanation:

Use the resistivity equation:

$R= \rho \frac{L}{A}$

$R= 3500\Omega \cdot m \cdot \frac{0.01m}{0.01m^2}$

$R=3500\Omega$

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Example Question #48 : Circuits

You are researching new materials for usage in spacecraft electronics.

You have a new material, known as "Type F."

You carve out a cylinder of the material. It is 4.5 cm tall, with a radius of 2.65 cm .

You put electrodes on each face of the cylinder.

You determine the resistance to be $935\Omega$ .

What is the resistivity of "Type F?"

Possible Answers:

None of these

$45.8\Omega\cdot meters$

$30.9 \Omega\cdot meters$

$39.1 \Omega\cdot meters$

$12.1 \Omega\cdot meters$

Correct answer:

$45.8\Omega\cdot meters$

Explanation:

We will use the relationship

$\rho=R\frac{A}{l}$

Where $\rho$ is the resistivity, is the resistance, is the surface area of the face the current is coming in or out of, and is the length from one face to the other.

Remember that the area of a circle is

$A=\pi\cdot r^2$

Combining our equations we get

$\rho=R\frac{\pi\cdot r^2}{l}$

We then need to plug in our values

$\rho=935\frac{\pi\cdot{.0265}^2}{.045}$

$45.8\Omega\cdot meters=\rho$

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Example Question #862 : Ap Physics 2

You are researching new materials for usage in spacecraft electronics.

You have a new material, known as "Type G."

You carve out a cylinder of the material. It is 2.5 cm tall, with a radius of 1.5 cm .

You put electrodes on each face of the cylinder.

You determine the resistance to be 9.53 ohms .

What is the resistivity of "Type G?"

Possible Answers:

None of these

$2.70\times10^{-3}\Omega\cdot meters$

$4.45\times10^{-3}\Omega\cdot meters$

$1.3\times10^{3}\Omega\cdot meters$

$1.9\times10^{-3}\Omega\cdot meters$

Correct answer:

$2.70\times10^{-3}\Omega\cdot meters$

Explanation:

We will use the relationship

$\rho=R\frac{A}{l}$

Where $\rho$ is the resistivity, is the resistance, is the surface area of the face the current is coming in or out of, and is the length from one face to the other.

Remember that the area of a circle is

$A=\pi\cdot r^2$

Combining our equations we get

$\rho=R\frac{\pi\cdot r^2}{l}$

We then need to plug in our values

$\rho=9.53\frac{\pi\cdot .015^2}{.025}$

$2.70\times 10^{-3}\Omega\cdot meters=\rho$

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Example Question #1 : Resistivity

You are researching new materials for usage in spacecraft electronics.

You have a new material, known as "Type Z."

You carve out a cylinder of the material. It is 2.2 cm tall, with a radius of .2 cm .

You put electrodes on each face of the cylinder.

You determine the resistance to be .22 ohms .

What is the resistivity of "Type Z?"

Possible Answers:

$1.75\times 10^{-4} \Omega\cdot meters$

$1.26\times 10^{-4}\Omega\cdot meters$

$1.11\times 10^{-4}\Omega\cdot meters$

None of these

$1.99\times 10^{-4} \Omega\cdot meters$

Correct answer:

$1.26\times 10^{-4}\Omega\cdot meters$

Explanation:

We will use the relationship

$\rho=R\frac{A}{l}$

Where $\rho$ is the resistivity, is the resistance, is the surface area of the face the current is coming in or out of, and is the length from one face to the other.

Remember that the area of a circle is

$A=\pi\cdot r^2$

Combining our equations we get

$\rho=R\frac{\pi\cdot r^2}{l}$

We then need to plug in our values

$\rho=.22\frac{\pi\cdot.002^2}{.022}$

$\rho=1.26\times10^{-4}\Omega\cdot meters$

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Example Question #51 : Circuits

A researcher is testing the electrical properties of a circuit that contains a resistor of unknown material. The resistor is a 3cm cylinder with a radius of 0.5cm . The researcher measures the resistance of the resistor to be $0.3\Omega$ . What is the resistivity of this unknown material?

Possible Answers:

$9.1*10^{-4}\Omega\cdot m$

$3.75\Omega\cdot m$

$8.75*10^{-3}\Omega\cdot m$

$1.6*10^{-2}\Omega\cdot m$

$7.85*10^{-4}\Omega\cdot m$

Correct answer:

$7.85*10^{-4}\Omega\cdot m$

Explanation:

Use the equation for resistivity:

$\rho=R\frac{A}{L}$

Here, $\rho$ is the resistivity, is the resistance, is the cross-sectional area, and is the length of the resistor.

Begin by finding the cross-sectional area of the material, which is circular:

$A=\pi r^2$

$A=\pi (0.005m)^2$

$A\approx 0.0000785m^2$

Now plug in all known values and solve for resistivity.

$\rho=0.3\Omega*\frac{0.0000785m^2}{0.03m}$

$\rho = 0.000785\Omega\cdot m$

$\rho = 7.85*10^{-4}\Omega\cdot m$

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Example Question #1 : Resistivity

A scientist carves out a cylinder of a new material she developed. It is 4.5 cm tall, with a radius of .60 cm . She put electrodes on each face of the cylinder. She determined the resistance to be $.75\Omega$ . What is the resistivity of this material?

Possible Answers:

$3.70 *10^{ -3} \Omega \cdot m$

None of these

$1.88 *10^{-3} \Omega\cdot m$

$4.50 *10^{ -3} \Omega \cdot m$

$1.70 *10^{ -3} \Omega \cdot m$

Correct answer:

$1.88 *10^{-3} \Omega\cdot m$

Explanation:

Use the relationship:

$\rho =R\frac{A}{l}$

Here, $\rho$ is the resistivity, is the resistance, is cross sectional area, and is the length.

The area of a circle is:

$A=\pi r^2$

Substitute:

$\rho =R\frac{\pi r^2}{l}$

Plug in given values and solve

$\rho =.75\Omega \frac{\pi (.006m)^2}{.045m}$

$\rho = 1.88 *10^{-3} \Omega\cdot m$

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Example Question #5 : Resistivity

There is a sample of an unknown material under testing. It is 4.75 cm tall, with a radius of .65 cm . Electrodes are placed on each face of the cylinder. It is determined that the resistance to 10.5 ohms .

What is the resistivity?

Possible Answers:

$.0375 \Omega*meters$

$.0333 \Omega*meters$

$.0293 \Omega*meters$

None of these

$.0750 \Omega*meters$

Correct answer:

$.0293 \Omega*meters$

Explanation:

Using the relationship

$\rho=R\frac{A}{l}$

Where $\rho$ is the resistivity, is the resistance, is the cross sectional area, and is the length from one face to the other.

The area of a circle is

$A=\pi*r^2$

Combining equations

$\rho=R\frac{\pi*r^2}{l}$

Plugging in values

$\rho=10.5\frac{\pi*.0065^2}{.0475}$

$.0293 \Omega*meters=\rho$

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AP Physics 2 : AP Physics 2

Study concepts, example questions & explanations for AP Physics 2

All AP Physics 2 Resources

Example Questions

Example Question #861 : Ap Physics 2

Example Question #1 : Resistivity

Example Question #1 : Resistivity

Example Question #213 : Electricity And Magnetism

Example Question #48 : Circuits

Example Question #862 : Ap Physics 2

Example Question #1 : Resistivity

Example Question #51 : Circuits

Example Question #1 : Resistivity

Example Question #5 : Resistivity

All AP Physics 2 Resources

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