Question

If the length of a cylinder is $l=(4.00\pm 0.01)cm$, radius $r=(0.250\pm 0.001)cm$ and mass $m=6.25\pm 0.01g$. Calculate the percentage error in determination of density.

Answer 1

Given that,

Mass $m=(6.25\pm 0.01)g$

Radius $r=(0.250\pm 0.001)cm$

Length $l=(4.00\pm 0.01)cm$

We know that,

$\frac{\Delta \rho }{\rho }=\frac{\frac{\Delta m}{m}}{2\frac{\Delta r}{r}+\frac{\Delta l}{l}}$

$\frac{\Delta \rho }{\rho }=\frac{\Delta m}{m}-2\times \frac{\Delta r}{r}-\frac{\Delta l}{l}$

$\frac{\Delta \rho }{\rho }=\frac{0.01}{6.25}-2\times \frac{0.001}{0.250}-\frac{0.01}{4.00}$

$\frac{\Delta \rho }{\rho }=1.6\times {10}^{-3}-8\times {10}^{-3}-2.5\times {10}^{-3}$

$\frac{\Delta \rho }{\rho }\times 100=0.16-(0.8+0.25)$

$\frac{\Delta \rho }{\rho }\times 100=0.89$%

Hence, the percentage error in determination of density is $0.89$%