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 :

Solution

Physics sir · 3 steps

Write ρ in terms of V, I, L and d, read each error from the last digit, then add the percentages.

1. Formula for resistivity. R = V/I and R = (ρ L)/A with A = (π d²)/4. Together these give ρ = (π d² V)/4IL.

   Δρ/ρ× 100 = ((Δ V)/V + (Δ I)/I + (Δ L)/L + 2 (Δ d)/d)× 100

2. Errors from data. No errors are given, so each error is one unit in the last digit written. For example, 5.0 V means Δ V = 0.1 V and 5.00 mm means Δ d = 0.01 mm.

   Quantity	Value	Error	Contribution to error in ρ
   V	5.0 V	0.1 V	0.1/5.0 = 2%
   L	10.0 cm	0.1 cm	0.1/10.0 = 1%
   I	2.00 A	0.01 A	0.01/2.00 = 0.5%
   d (squared)	5.00 mm	0.01 mm	2 × 0.01/5.00 = 0.4%

3. So. Maximum error = 2 + 1 + 0.5 + 0.4 = 3.9%, which is option (a).

The idea to keep. When no errors are given, take the least count from the last significant digit. In a product or quotient the percentage errors always add, each multiplied by its power.

Answer from the NTA final answer key. Solution by PrepWiser.