In the experiment of Ohm's law, a potential difference of 5.0 V is applied across the end of a conductor of length 10.0 cm and diameter of 5.00 mm. The measured current in the conductor is 2.00 A. The maximum permissible percentage error in the resistivity of the conductor is :A3.9B8.4C7.5D3.0Check Answer
Question
In the experiment of Ohm's law, a potential difference of 5.0 V is applied across the end of a conductor of length 10.0 cm and diameter of 5.00 mm. The measured current in the conductor is 2.00 A. The maximum permissible percentage error in the resistivity of the conductor is :A3.9B8.4C7.5D3.0Check Answer
Solution
To solve this problem, we first need to understand the formula for resistivity (ρ) which is given by:
ρ = RA/L
where R is the resistance, A is the cross-sectional area of the conductor, and L is the length of the conductor.
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First, we calculate the resistance (R) using Ohm's law (V = IR), where V is the voltage and I is the current. So, R = V/I = 5.0V / 2.0A = 2.5 ohms.
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Next, we calculate the cross-sectional area (A) of the conductor. The conductor is cylindrical in shape, so A = πd²/4, where d is the diameter of the conductor. So, A = π*(5.0mm)²/4 = 19.63 mm². But we need to convert this to m² (since resistivity is typically measured in ohm-meters), so A = 19.63 * 10^-6 m².
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Now we can calculate the resistivity (ρ) using the formula ρ = RA/L. Substituting the values we have, ρ = (2.5 ohms * 19.63 * 10^-6 m²) / 0.1 m = 4.9 * 10^-7 ohm-m.
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The percentage error in the resistivity can be calculated using the formula: Δρ/ρ = ΔR/R + ΔA/A + ΔL/L. Since the errors in R, A, and L are not given, we can assume that they are equal (ΔR = ΔA = ΔL). So, the formula simplifies to: Δρ/ρ = 3ΔR/R.
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Solving for ΔR/R, we get ΔR/R = Δρ/ρ / 3 = 1 / 3 = 0.333 or 33.3%.
So, the maximum permissible percentage error in the resistivity of the conductor is 33.3%. However, this is not one of the options given in the question. It seems like there might be a mistake in the question or in the given options.
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