# The percentage error in specific resistance $p = \pi r^2R/ l$, where r = radius of wire = $(0.26 \pm 0.02)$ cm, l = length of wire = $(156.0 \pm 0.1)$ cm, R = resistance of wire = $(64 \pm 2)$ ohm.

## 1 Answer

$\pm18.6 \%$
Hence (A) is the correct answer.
answered Jun 4, 2014