We are open Saturday and Sunday!

Call Now to Set Up Tutoring:

(888) 888-0446

AP Physics 1 : AP Physics 1

Study concepts, example questions & explanations for AP Physics 1

Create An Account

All AP Physics 1 Resources

7 Diagnostic Tests 170 Practice Tests Question of the Day Flashcards Learn by Concept

Example Questions

Example Question #2 : Resistivity

An electrician wishes to cut a copper wire $(\rho = 1.724 * 10^{-8}\Omega m)$ that has no more than $10 \Omega$ of resistance. The wire has a radius of 0.725mm. Approximately what length of wire has a resistance equal to the maximum $10 \Omega$ ?

Possible Answers:

2.6cm

960m

38m

9.6m

10cm

Correct answer:

960m

Explanation:

To relate resistance R, resistivity $\rho$ , area A, and length L we use the equation $R = \frac{\rho L}{A}$ .

Rearranging to isolate the quantity we wish to solve for, L, gives the equation $L = \frac{RA}{\rho }$ . We must first solve for A using the radius, 0.725mm.

$A = \pi r^{2} = \pi (0.000725m)^{2}=1.65 * 10^{-6}m^{2}$

Plugging in our numbers gives the answer, 960m.

$L = \frac{10\Omega * 1.65 * 10^{-6}m^{2}}{1.724*10^{-8}\Omega m} = 957m \approx960m$

Report an Error

Example Question #2 : Resistivity

What is the current in a circuit with a $2\Omega$ resistor followed by a $3\Omega$ resistor that are both in parallel with a $5\Omega$ resistor? The voltage supplied to the circuit is 5V.

Possible Answers:

1.5A

0.25A

Correct answer:

Explanation:

Resistors in serries add according to the formula:

$R_{f}=R_{1}+R_{2}$

Resistors in parallel add according to the formula:

$\frac{1}{R_{f}}=\frac{1}R_{1}{}+\frac{1}{R_{2}}$

We can find the total equivalent resistance:

$\frac{1}{R_{f}}=\frac{1}{2+3}+\frac{1}{5}=\frac{2}{5}$

$R_f=2.5\Omega$

Now we can use Ohm's law to find the current:

V=IR

$I=\frac{V}{R}$

Solve:

$I = \frac{5 V}{2.5 \Omega }=2 A$

Report an Error

Example Question #1 : Resistivity

Basic circuit

$V_{in}=12V$

$Z_1=3\Omega$

In the circuit above, your aim is to limit the current to , so you must design a resistor to serve in the place of Z_2 .

To make this resistor, you have a spool of mystery metal, which has a cross sectional area of 1mm^2 and a resistivity of, $\rho = 3\cdot 10^{-3}\Omega m$ . What length of wire should you cut?

Possible Answers:

3mm

1mm

150mm

15mm

3000mm

Correct answer:

3mm

Explanation:

First, find out how much total resistance should be in the circuit in order to get the desired current:

$R_{tot}=\frac{V}{I}=\frac{12V}{1A}=12\Omega$

Determine what the second resistance should be.

$Z_2=12\Omega-Z_1$

$Z_2=9\Omega$

Find the necessary length of wire.

$Z_2=\frac{\rho L }{A}$

$L=\frac{Z_2A}{\rho}$

$L=\frac{(9\Omega)(1mm^2)}{3\cdot 10^{-3}\Omega m}=\frac{(9\Omega)(10^{-6}m^2)}{3\cdot 10^{-3}\Omega m}$

$L=3\cdot 10^{-3}m=3mm$

Report an Error

Example Question #1 : Resistivity

What is the resistance of a 100m length of round copper wire with a radius of 0.3mm ?

$\rho_{copper}=1.68\cdot 10^{-8}\Omega m$

Possible Answers:

$118.8\Omega$

$37.7\Omega$

$3.77\Omega$

$5.94\Omega$

$59.4\Omega$

Correct answer:

$5.94\Omega$

Explanation:

Resistance and resistivity are related as follows:

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

$A={\pi}(0.0003m)^2$

$A\approx 2.83\cdot 10^{-7}$

$R\approx \frac{1.68\cdot10^{-8}\Omega m(100m)}{2.83\cdot10^{-7}m^2}$

$R\approx 5.94\Omega$

Report an Error

Example Question #3 : Resistivity

Two students are performing a lab using lengths of wire as resistors. The two students have wires made of the exact same material, but Student B has a wire the has twice the radius of Student A's wire. If Student B wants his wire to have the same resistance as Student A's wire, how should Student B's wire length compare to Student A's wire?

Possible Answers:

Student B's wire should be the same length as Student A's wire

Student B's wire should be $\frac{1}{4}$ the length of Student A's wire

Student B's wire should be $\frac{1}{2}$ the length of Student A's wire

Student B's wire should be twice as long as Student A's wire

Student B's wire should be four times as long as Student A's wire

Correct answer:

Student B's wire should be $\frac{1}{4}$ the length of Student A's wire

Explanation:

Resistance is proportional to length, and inversely proportional to cross-sectional area. Area depends on the square of the radius: $A= \pi r^{2}$ , so Student B's wire has $2^{2}=4$ times the cross-sectional area of Student A's wire. In order to compensate for the increased area, Student B must make his wire $\frac{1}{4}$ the length of Student A's wire. This can be shown mathematically using the equation for resistance:

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

Report an Error

Example Question #2 : Resistivity

By how much will resistivity change if resistance and length are constant, and cross sectional area is doubled?

Possible Answers:

The resistivity will not change

The resistivity will double

The resistivity will be halved

The resistivity will be quadrupled

Correct answer:

The resistivity will double

Explanation:

Recall the formula for resistance is given by

$R=\frac{\rho L}{A}$ , where is the cross sectional area, $\rho$ is resistivity, and is length.

Solve for resistivity:

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

From this, we can tell that resistivity is proportional to cross sectional area by:

$A \propto \rho$

Since is doubled, and resistance and length are constant, resistivity $\rho$ will also be doubled.

Report an Error

Example Question #1 : Resistivity

Ratio is given by:

Resistivity of first resistor: Resistivity of second resistor

What is the ratio of resistivity of 2 resistors with identical resistances and area, where the first resistor is twice the length of the second resistor?

Possible Answers:

1:2

2:1

1:1

4:1

Correct answer:

1:2

Explanation:

Resistivity $\rho$ is given by:

$\rho=\frac{RA}{L}$ , where is the resistance, is the area, and is the length.

Since both have identical resistances and area, the first resistor will have half the resistivity since it has twice the length. Therefore the resistivity relation is

1:2

Report an Error

Example Question #1 : Resistivity

What is the resistance of a copper rod with resistivity of $1.78\cdot 10^{-8} \Omega m$ , diameter of 10mm , and length of 1000m ?

Possible Answers:

$\Omega$

6.9 $\Omega$

$\Omega$

Correct answer:

$\Omega$

Explanation:

The equation for resistance is as follows: $R=\frac{\rho l}{A}$ . Where $\rho$ is resistivity, is the length of the wire, and is the cross section of the wire which can be found using $\frac{\pi}{4} d^{2}$ .

Report an Error

Example Question #11 : Resistivity

You have a very long wire connected to an electric station. Even though you are suppling 120V from the source, by the time it reaches the station, there is a loss of voltage. The wire is 100 meters long.

If 100V reaches the power station, what is the resistivity of the wire? Assume a current of .

Possible Answers:

$1.45\frac{\Omega}{m}$

$0.10\frac{\Omega}{m}$

$2.2\frac{\Omega}{m}$

$0.55\frac{\Omega}{m}$

$0.15\frac{\Omega}{m}$

Correct answer:

$0.10\frac{\Omega}{m}$

Explanation:

The voltage drop from the source to the station (the "load") indicates that there is an internal resistance in the wire. According to the voltage law, the total amount of voltage drop is equal to the total amount of voltage supplied. Since 100V was supplied, and 100V drops at the station, that means that 20V drops along the wire.

$V_{source}=V_{wire}+V_{station}$

$V_{wire}=V_{source}-V_{station}$

$V_{wire}=120V-100V=20V$

Now that the voltage drop across the wire is known, Ohm's law will give the resistance of the wire:

$V=IR\rightarrow R=\frac{V}{I}=\frac{20V}{2A}=10\Omega$

$R=10\Omega$

The resistivity of the wire is equal to the resistance per unit length, therefore, in order to find resistivity you divide the total resistance by the length:

$\rho=\frac{R}{L}=\frac{10\Omega}{100m}=0.10\frac{ \Omega}{m}$

Report an Error

Example Question #11 : Resistivity

Which of the following actions would decrease the resistance of a wire by a factor of ?

Possible Answers:

Tripling the cross-sectional area of the wire

Halving the cross-sectional area of the wire

Doubling the cross-sectional area of the wire

Tripling the length of the wire

Halving the length of the wire

Correct answer:

Doubling the cross-sectional area of the wire

Explanation:

The equation for resistance is as follows:

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

Where is the resistance, $\rho$ is the resistivity of the material, is the length of the material and is the cross-sectional area of the material. Looking at this equation, by doubling the area we effectively reduce the resistance by a factor of two.

Report an Error

View AP Physics 1 Tutors

Bhupinder
Certified Tutor

Panjab University, Bachelor of Engineering, Biotechnology. Hamburg University of Technology, Master of Science, Chemical Engi...

View AP Physics 1 Tutors

Steven
Certified Tutor

University of Louisville, Bachelor of Science, Mechanical Engineering. University of Phoenix-Louisville Campus, Master of Eng...

View AP Physics 1 Tutors

Vikash
Certified Tutor

national institute of technology Jamshedpur , Bachelor, Civil engineering .

All AP Physics 1 Resources

7 Diagnostic Tests 170 Practice Tests Question of the Day Flashcards Learn by Concept

Popular Subjects

Statistics Tutors in Phoenix, Computer Science Tutors in Denver, Computer Science Tutors in Washington DC, SAT Tutors in San Diego, English Tutors in Phoenix, Chemistry Tutors in San Diego, SAT Tutors in Dallas Fort Worth, Math Tutors in Seattle, GMAT Tutors in Miami, Biology Tutors in Atlanta

Popular Courses & Classes

GMAT Courses & Classes in Seattle, ISEE Courses & Classes in Houston, ACT Courses & Classes in Houston, SAT Courses & Classes in Miami, GRE Courses & Classes in San Francisco-Bay Area, ACT Courses & Classes in Seattle, GRE Courses & Classes in Atlanta, SAT Courses & Classes in New York City, MCAT Courses & Classes in Los Angeles, GRE Courses & Classes in Seattle

Popular Test Prep

LSAT Test Prep in Houston, MCAT Test Prep in Miami, GMAT Test Prep in Seattle, GRE Test Prep in Seattle, SAT Test Prep in Chicago, SAT Test Prep in New York City, GRE Test Prep in Houston, MCAT Test Prep in Los Angeles, MCAT Test Prep in Philadelphia, MCAT Test Prep in Boston

DMCA Complaint

If you believe that content available by means of the Website (as defined in our Terms of Service) infringes one or more of your copyrights, please notify us by providing a written notice (“Infringement Notice”) containing the information described below to the designated agent listed below. If Varsity Tutors takes action in response to an Infringement Notice, it will make a good faith attempt to contact the party that made such content available by means of the most recent email address, if any, provided by such party to Varsity Tutors.

Your Infringement Notice may be forwarded to the party that made the content available or to third parties such as ChillingEffects.org.

Please be advised that you will be liable for damages (including costs and attorneys’ fees) if you materially misrepresent that a product or activity is infringing your copyrights. Thus, if you are not sure content located on or linked-to by the Website infringes your copyright, you should consider first contacting an attorney.

Please follow these steps to file a notice:

You must include the following:

A physical or electronic signature of the copyright owner or a person authorized to act on their behalf; An identification of the copyright claimed to have been infringed; A description of the nature and exact location of the content that you claim to infringe your copyright, in \ sufficient detail to permit Varsity Tutors to find and positively identify that content; for example we require a link to the specific question (not just the name of the question) that contains the content and a description of which specific portion of the question – an image, a link, the text, etc – your complaint refers to; Your name, address, telephone number and email address; and A statement by you: (a) that you believe in good faith that the use of the content that you claim to infringe your copyright is not authorized by law, or by the copyright owner or such owner’s agent; (b) that all of the information contained in your Infringement Notice is accurate, and (c) under penalty of perjury, that you are either the copyright owner or a person authorized to act on their behalf.

Send your complaint to our designated agent at:

Charles Cohn Varsity Tutors LLC
101 S. Hanley Rd, Suite 300
St. Louis, MO 63105

Or fill out the form below:

Do not fill in this field

Contact Information

* Full legal name

Company name

* Phone number

* Email address

Address

* Address1

Address2

* City

* State

* Zipcode

* Country

Complaint Details

* URL of copyrighted work (where your original material is located)

* Please describe the copyrighted work so that it may be easily identified

I am the owner, or an agent authorized to act on behalf of the owner of an exclusive right that is allegedly infringed.

I have a good faith belief that the use of the material in the manner complained of is not authorized by the copyright owner, its agent, or the law.

This notification is accurate.

I acknowledge that there may be adverse legal consequences for making false or bad faith allegations of copyright infringement by using this process.

* Signature (your digital signature is legally binding)

HIGH SCHOOL

GRADUATE SCHOOL

K-8

Search 50+ Tests

math tutoring

science tutoring

foreign languages

elementary tutoring

other

Search 350+ Subjects

AP Physics 1 : AP Physics 1

Study concepts, example questions & explanations for AP Physics 1

All AP Physics 1 Resources

Example Questions

Example Question #2 : Resistivity

Example Question #2 : Resistivity

Example Question #1 : Resistivity

Example Question #1 : Resistivity

Example Question #3 : Resistivity

Example Question #2 : Resistivity

Example Question #1 : Resistivity

Example Question #1 : Resistivity

Example Question #11 : Resistivity

Example Question #11 : Resistivity

All AP Physics 1 Resources

Report an issue with this question

DMCA Complaint

Contact Information

Address

Complaint Details

Email address:
Your name:
Feedback: