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AP Physics 2 : AP Physics 2

Study concepts, example questions & explanations for AP Physics 2

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All AP Physics 2 Resources

6 Diagnostic Tests 149 Practice Tests Question of the Day Flashcards Learn by Concept

Example Questions

Example Question #52 : Circuits

A sample of a new material is being tested. It is 3.8 cm tall, with a radius of .38 cm . Electrodes are placec on each face of the cylinder. The resistance is determined to be 1.35 ohms .

What is the resistivity of this material?

Possible Answers:

$1.42*10^{-3}\Omega*meters$

$1.61*10^{-3}\Omega*meters$

None of these

$3.49*10^{-3}\Omega*meters$

$2.22*10^{-3}\Omega*meters$

Correct answer:

$1.61*10^{-3}\Omega*meters$

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

$\rho=1.35*\frac{\pi*.0038^2}{.038}$

$1.61*10^{-3}\Omega*meters=\rho$

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Example Question #53 : Circuits

A cylinder of an unknown material is being tested. It is 5.75 cm tall, with a radius of .95 cm . Electrodes are placed on each face of the cylinder. It is determined that the resistance is 59 ohms . What is the resistivity?

Possible Answers:

$.375 \Omega*meters$

$.762 \Omega*meters$

$.931\Omega*meters$

$.291\Omega*meters$

None of these

Correct answer:

$.291\Omega*meters$

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

$\rho=59\frac{\pi*.0095^2}{.0575}$

$\rho=.291\Omega*meters$

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Example Question #51 : Circuit Properties

A cylinder of an unknown material is being tested for physical properties. It is 5.5 cm tall, with a radius of .65 cm . Electrodes are placed on each face of the cylinder. The resistance is $.35 \Omega$ . What is the resistivity?

Possible Answers:

None of these

$5.85*10^{-4} \Omega\cdot m$

$6.11*10^{-4} \Omega\cdot m$

$8.45*10^{-4} \Omega\cdot m$

$4.85*10^{-4}\Omega\cdot m$

Correct answer:

$8.45*10^{-4} \Omega\cdot m$

Explanation:

Use the relationship:

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

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:

$A=\pi*r^2$

Combine equations, convert to and plug in values:

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

$\rho=\frac{.35*\pi*.0065^2}{.055}$

$\rho=8.45*10^{-4} \Omega\cdot m$

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Example Question #55 : Circuits

A cylinder of a new material is under testing. It is 2.0 cm tall, with a radius of .2 cm . Electrodes are placed on each face of the cylinder. The resistance is $75 \Omega$ . What is the resistivity?

Possible Answers:

$2.3*10^{-4}\Omega\cdot m$

None of these

$3.4*10^{-4} \Omega\cdot m$

$1.9*10^{-4}\Omega\cdot m$

$4.7*10^{-4}\Omega\cdot m$

Correct answer:

$4.7*10^{-4}\Omega\cdot m$

Explanation:

Use the relationship:

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

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:

$A=\pi*r^2$

Combine equations, convert to and plug in values.

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

$\rho=\frac{75*\pi*.002^2}{.02}$

$\rho=4.7*10^{-4}\Omega\cdot m$

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Example Question #56 : Circuits

A sample of a new material is being tested for electrical resistivity. It is 1.5 cm tall, with a radius of .5 cm . Electrodes are placed on each face of the cylinder. Testing determines the resistance is $22 \Omega$ . What is the resistivity?

Possible Answers:

None of these

$1.15 \Omega\cdot m$

$1.11 \Omega\cdot m$

$.95 \Omega\cdot m$

$.67 \Omega\cdot m$

Correct answer:

None of these

Explanation:

Use the relationship:

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

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:

$A=\pi*r^2$

Combine these equations, convert to and plug in values:

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

$\rho=\frac{22*\pi*.005^2}{.015}$

$\rho=.115 \Omega\cdot m$

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Example Question #61 : Circuits

A cylindrical resistor is 3cm long with a diameter of 4mm . Determine the resistivity of the material if the resistance is $50\Omega$ .

Possible Answers:

$.805\Omega*m$

$.0714\Omega\cdot m$

$.0209\Omega\cdot m$

$.0626\Omega\cdot m$

$.567\Omega \cdot m$

Correct answer:

$.0209\Omega\cdot m$

Explanation:

Use the following relationship:

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

Where $\rho$ is the resistivity

is the resistance

is the length

is the cross sectional area

Convert and to and plug in values:

$\rho=50*\frac{.002^2*\pi}{.03}$

$.0209\Omega\cdot m=\rho$

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Example Question #222 : Electricity And Magnetism

Suppose that the resistance of an unknown conductive material is found to be $3\: \Omega$ . Considering that the cross-sectional area of this material is $7.85\cdot10^{-7}\:m^{2}$ and its length is $4.5\cdot 10^{-2}\:m$ , what is the resistivity of this unknown material?

Possible Answers:

$9.68\cdot 10^{-5}\: \Omega\cdot m$

$5.23\cdot 10^{-5}\: \Omega\cdot m$

$3.14\cdot 10^{-5}\: \Omega\cdot m$

$4.38\cdot 10^{-5}\: \Omega\cdot m$

$6.20\cdot 10^{-5}\: \Omega\cdot m$

Correct answer:

$5.23\cdot 10^{-5}\: \Omega\cdot m$

Explanation:

In this question, we're presented with the length and area of a given material, as well as the resistance of the material. We're being asked to find the resistivity of this material.

First, it's important to distinguish between resistance and resistivity. Resistance represents the impediment to the flow of charge that is caused by such factors as the length and cross-sectional area of the material. Resistivity is a value that is intrinsic to each material; changing the cross-sectional area or the length will not affect the resistivity.

To solve this problem, we'll need to utilize the equation that relates resistance and resistivity:

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

From this equation, we can see that as the length of the conductive material increases, so too does the resistance of that material. However, as the area increases, the resistance decreases.

We can go ahead and rearrange this equation in order to isolate the resistivity term, $\rho$ .

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

Next, we can plug in the values that we have in order to solve for our answer:

$\rho=3\: \Omega(\frac{7.85\cdot10^{-7}\: m^{2}}{4.5\cdot 10^{-2}\: m})$

$\rho=5.23\cdot 10^{-5}\: \Omega\cdot m$

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Example Question #62 : Circuit Properties

A single resistor of radius 1mm and length 1.5cm is placed into a circuit with a single voltage source of . If the current is 2.3A , determine the resistivity of the resistor.

Possible Answers:

$9.09*10^{-4}\Omega\cdot m$

None of these

$6.26*10^{-4}\Omega\cdot m$

$7.14*10^{-4}\Omega\cdot m$

$2.73*10^{-4}\Omega\cdot m$

Correct answer:

$2.73*10^{-4}\Omega\cdot m$

Explanation:

Using

V=IR

and

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

Combining equations:

$V=I*\rho *\frac{l}{A}$

Solving for $\rho$ :

$\rho=\frac{V*A}{I*l}$

Converting and to and plugging in values:

$\rho=\frac{3*.001^2*\pi}{2.3*.015}$

$\rho=2.73\Omega\cdot m$

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Example Question #63 : Circuit Properties

A resistor has cross sectional area $2\textup{ mm}^2$ and length $1\textup{ cm}$ . When placed in series with a $3\textup{ V}$ battery, a current of $340 \textup{ mA}$ is produced. Determine the resistivity.

Possible Answers:

$8.221*10^{-3}\ \Omega\cdot \textup{m}$

None of these

$9.996*10^{-3}\ \Omega\cdot \textup{m}$

$1.764*10^{-3}\ \Omega\cdot \textup{m}$

$3.445*10^{-3}\ \Omega\cdot \textup{m}$

Correct answer:

$1.764*10^{-3}\ \Omega\cdot \textup{m}$

Explanation:

Using Ohm's law:

V=IR

Converting to and plugging in values

3=.340*R

Solving for resistance:

$R=8.82\Omega$

Using the equation for resistivity:

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

Converting mm^2 to and to and plugging in values:

$\rho=8.82\frac{2*10^{-6}}{.01}$

$\rho=1.764*10^{-3}\Omega\cdot m$

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Example Question #64 : Circuit Properties

A resistor has cross sectional area $16\textup{ mm}^2$ and length $2\textup{ cm}$ . When placed in series with a $3\textup{ V}$ battery, a current of $56.0\textup{ mA}$ is produced. Determine the resistivity.

Possible Answers:

None of the above

$\rho=1.213*10^{-3}\ \Omega\cdot \textup{m}$

$\rho=3.751*10^{-3}\ \Omega\cdot \textup{m}$

$\rho=1.764*10^{-3}\ \Omega\cdot \textup{m}$

$\rho=2.028*10^{-3}\ \Omega\cdot \textup{m}$

Correct answer:

$\rho=1.764*10^{-3}\ \Omega\cdot \textup{m}$

Explanation:

Using Ohm's law:

V=IR

Converting to and plugging in values

3=.340*R

Solving for resistance:

$R=8.82\Omega$

Using the equation for resistivity:

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

Converting mm^2 to and to and plugging in values:

$\rho=8.82\frac{2*10^{-6}}{.01}$

$\rho=1.764*10^{-3}\Omega*m$

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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 #52 : Circuits

Example Question #53 : Circuits

Example Question #51 : Circuit Properties

Example Question #55 : Circuits

Example Question #56 : Circuits

Example Question #61 : Circuits

Example Question #222 : Electricity And Magnetism

Example Question #62 : Circuit Properties

Example Question #63 : Circuit Properties

Example Question #64 : Circuit Properties

All AP Physics 2 Resources

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