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If the length of a cylinder is $l=(4.00 \pm 0.01) cm$, radius $r =(0.250 \pm 0.001) \;cm$ and mass $m =6.25 \pm 0.01\; g$. Calculate the percentage error in determination of density.
$12.1$
$0.121$
$12.18 $
$1.21$
Solution
$\,{l}\,\, = \,\,(4.00\,\, \pm \,\,0.01\,\,)\,cm$
$\frac{{\Delta {l}}}{{l}}\, = \,\frac{{0.01}}{{4.0}}\, = \,0.0025$
$r\,\, = \,\,(0.250\,\, \pm \,\,0.001\,)\,\,cm\,\,$
$\frac{{\Delta r}}{r}\, = \,\frac{{0.001}}{{0.250}}\, = \,0.004$
$m\,\, = \,\,(6.25\,\, \pm \,\,0.01)\,\,g\,\,\,\,$
$\frac{{\Delta m}}{m} = \frac{{0.01}}{{6.25}}\, = \,0.0016$
density = mass/volume $ = \,\,\frac{m}{{\pi {r^2}{l}}}\,$
$ \frac{{\Delta \rho }}{\rho }\, = \,\frac{{\Delta m}}{m}\, + \,\frac{{2\Delta r}}{r} = \frac{{\Delta {l}}}{{l}}$
$\frac{{\Delta \rho }}{\rho } \times \,100\,\% $ $ = \left( {\frac{{\Delta m}}{m} \times \,100} \right)\% + $ $2\left( {\frac{{\Delta r}}{r} \times 100} \right)\% + \left( {\frac{{\Delta {l}}}{{l}} \times \,100} \right)\% $
$ = \,(0.0025\, \times \,100)\,\% \,\, + \,\,2(0.004\, \times \,100)\% \,\, + \,\,(0.0016\, \times 100)\,\% $
$\, = \,0.25\% \, + \,0.8\% + \,0.16\,\% = \,1.21\% $
