
Understanding Resistivity and Its Measurement

Interactive Video
•
Physics
•
9th - 10th Grade
•
Hard
Jennifer Brown
FREE Resource
10 questions
Show all answers
1.
MULTIPLE CHOICE QUESTION
30 sec • 1 pt
What is the relationship between resistance, length, and area of a wire?
Resistance is inversely proportional to length and directly proportional to area.
Resistance is directly proportional to length and inversely proportional to area.
Resistance is inversely proportional to both length and area.
Resistance is directly proportional to both length and area.
2.
MULTIPLE CHOICE QUESTION
30 sec • 1 pt
What is the purpose of the resistor in the experimental circuit?
To connect the voltmeter in parallel.
To measure the resistance of the wire.
To limit the current for safety and prevent overheating.
To increase the voltage across the wire.
3.
MULTIPLE CHOICE QUESTION
30 sec • 1 pt
Why is a micrometer screw gauge used to measure the wire's diameter?
Because it is easier to use than other tools.
Because it can measure the length of the wire.
Because the wire's diameter is typically smaller than a millimeter.
Because it is more accurate than a ruler.
4.
MULTIPLE CHOICE QUESTION
30 sec • 1 pt
Why are multiple measurements taken along the wire?
To measure the resistance at different points.
To calculate the average length of the wire.
To check for variations in diameter along the wire.
To ensure the wire is straight.
5.
MULTIPLE CHOICE QUESTION
30 sec • 1 pt
How is resistance calculated in this experiment?
By dividing the voltage by the current.
By subtracting the current from the voltage.
By multiplying the voltage by the current.
By adding the voltage and current.
6.
MULTIPLE CHOICE QUESTION
30 sec • 1 pt
What does the gradient of the resistance vs. length graph represent?
The voltage across the wire.
The cross-sectional area of the wire.
The resistivity of the wire.
The current through the wire.
7.
MULTIPLE CHOICE QUESTION
30 sec • 1 pt
What is the formula for the area of a circle used in this experiment?
Area = π × diameter
Area = π × diameter^2
Area = π × radius
Area = π × (diameter/2)^2
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