JEE MainPhysicsMathematics in Physics
In an experiment to determine the specific resistance of a wire, the formula used is = r^2 R l . The measured values are radius r = 0.20 0.01 cm , resistance R = 60 3 , and length l = 150 1.5 cm . The maximum percentage error in the specific resistance is:
Options
- A16 %
- B11 %
- C14 %
- D9 %
Correct answer
A. 16 %
Step-by-step solution
The formula for specific resistance is = r^2 R l . First, calculate the percentage errors for each of the measured quantities: Percentage error in radius r : r r 100 = 0.01 0.20 100 = 5 % Percentage error in resistance R : R R 100 = 3 60 100 = 5 % Percentage error in length l : l l 100 = 1.5 150 100 = 1 % The maximum percentage error in is given by the sum of the individual percentage errors multiplied by their respective exponents: 100 = 2 ( r r 100 ) + ( R R 100 ) + ( l l 100 ) Substituting the calculated percent