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In an experiment to measure the resistivity of nichrome, the resistance, the diameter and appropriate length of a sample of nichrome wire were measured - Leaving Cert Physics - Question 4 - 2009

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In an experiment to measure the resistivity of nichrome, the resistance, the diameter and appropriate length of a sample of nichrome wire were measured. The followi... show full transcript

Worked Solution & Example Answer:In an experiment to measure the resistivity of nichrome, the resistance, the diameter and appropriate length of a sample of nichrome wire were measured - Leaving Cert Physics - Question 4 - 2009

Step 1

Describe the procedure used in measuring the length of the sample of wire.

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Answer

To measure the length of the nichrome wire accurately, follow these steps:

  1. Straighten the Wire: Ensure the wire sample is taut and straight without any bends or knots.

  2. Measure the Length: Using a ruler or a measuring tape, measure the distance between the points where the resistance was taken. Ensure to record the measurement accurately.

Step 2

Describe the steps involved in finding the average diameter of the wire.

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Answer

To determine the average diameter of the nichrome wire, you should follow these steps:

  1. Use Appropriate Equipment: Utilize a micrometer or digital calipers for accurate measurement.

  2. Initial Placement: Position the wire between the jaws of the caliper. (Indicate if using a digital device.)

  3. Tightening: Carefully tighten the jaws around the wire to keep it in place.

  4. Take Readings: Read the initial measurement and note it down.

  5. Repeat Measurements: Measure at different points along the wire to account for any irregularities in thickness.

  6. Calculate Average: Sum all the diameter readings and divide by the number of measurements to find the average diameter.

Step 3

Use the data to calculate the resistivity of nichrome.

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Answer

To calculate the resistivity of the nichrome wire, use the formula:

ρ=RAl\rho = \frac{R A}{l}

where:

  • ρ\rho is the resistivity,
  • RR is the resistance (7.9 Ω),
  • AA is the cross-sectional area,
  • ll is the length of the wire (in meters).
  1. Calculate the Area: The diameter is 0.31 mm, hence the radius is: r=0.312×103 m=0.155×103 mr = \frac{0.31}{2} \times 10^{-3} \text{ m} = 0.155 \times 10^{-3} \text{ m}

    The area is: A=πr2=π(0.155×103)27.55×108 m2A = \pi r^2 = \pi (0.155 \times 10^{-3})^2 \approx 7.55 \times 10^{-8} \text{ m}^2

  2. Substitute Values: Using the length of wire (54.6 cm = 0.546 m):

    ρ=7.9×7.55×1080.546\rho = \frac{7.9 \times 7.55 \times 10^{-8}}{0.546}

  3. Calculate: ρ1.09×106Ωm\rho \approx 1.09 \times 10^{-6} \Omega m

Step 4

The experiment was repeated on a warmer day. What effect did this have on the measurements?

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Answer

On a warmer day, the following effects can be observed:

  • Resistance Increased: The increase in temperature generally leads to an increase in resistance in metallic conductors.
  • Wire Expansion: The effective length of the wire may expand, leading to an increase in overall diameter.
  • Diameter Measurements: As the wire expands, the diameter may increase, affecting calculations based on measurements taken at lower temperatures.

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