A wire of length l = (6±0.06) am and radius r = (0.5±0.005)cm has mass m = (0.3±0.003)g. maximum percentage error in density is

Ans: (a)

Sol: $\displaystyle \rho = \frac{m}{\pi r^2 l} $

$\displaystyle \frac{\Delta \rho}{\rho}\times 100 = \pm(\frac{\Delta m}{m} + 2\frac{\Delta r}{r} + \frac{\Delta l}{l})\times 100$

$\displaystyle \frac{\Delta \rho}{\rho}\times 100 = \pm(\frac{0.003}{0.3} + 2\frac{0.005}{0.5} + \frac{0.06}{6})\times 100$

$\displaystyle \frac{\Delta \rho}{\rho}\times 100 = \pm(\frac{1}{100} + 2\times \frac{1}{100} + \frac{1}{100})\times 100$

= 4 %