If the percentage errors in measuring the length and the diameter of a wire are 0.1\% each. The percentage error in measuring its resistance will be:
- 0.144%
- 0.2%
- 0.1%
- 0.3%
Answer
(4) 0.3%
Solution
R= LA= 4 L d2
in a product the percentage errors add, each multiplied by its power
RR 100= LL 100+2 dd 100
=0.1\%+2(0.1\%)=0.3\%
Answer: (4)