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 :

• A. 3.9
• B. 7.5
• C. 8.4
• D. 3.0

Hint:

If absolute errors aren't given, assume the least count is 1 unit of the last decimal place (e.g., 5.00 has $\Delta = 0.01$).

Answer 1

The Correct Option is A

Step 1: Resistivity $\rho = \frac{RA}{L} = \frac{V}{I} \frac{\pi d^2}{4L}$.
Step 2: Relative error: $\frac{\Delta \rho}{\rho} = \frac{\Delta V}{V} + \frac{\Delta I}{I} + 2\frac{\Delta d}{d} + \frac{\Delta L}{L}$.
Step 3: Using least counts from significant figures: $\Delta V = 0.1, \Delta I = 0.01, \Delta d = 0.01, \Delta L = 0.1$. $%$ error $= (\frac{0.1}{5.0} + \frac{0.01}{2.00} + 2\frac{0.01}{5.00} + \frac{0.1}{10.0}) \times 100$. $%$ error $= (0.02 + 0.005 + 0.004 + 0.01) \times 100 = 3.9%$.