If the length of a cylinder is l = (4.00 pm 0 | vedclass

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(D) Given:$l = (4.00 \pm 0.01) \; cm \implies \frac{\Delta l}{l} = \frac{0.01}{4.00} = 0.0025$$r = (0.250 \pm 0.001) \; cm \implies \frac{\Delta r}{r}

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$1.21$

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(D) Given:$l = (4.00 \pm 0.01) \; cm \implies \frac{\Delta l}{l} = \frac{0.01}{4.00} = 0.0025$$r = (0.250 \pm 0.001) \; cm \implies \frac{\Delta r}{r} = \frac{0.001}{0.250} = 0.004$$m = (6.25 \pm 0.01) \; g \implies \frac{\Delta m}{m} = \frac{0.01}{6.25} = 0.0016$Density $ho = \frac{m}{V} = \frac{m}{\pi r^2 l}$The relative error in density is given by $\frac{\Delta ho}{ho} = \frac{\Delta m}{m} + 2\frac{\Delta r}{r} + \frac{\Delta l}{l}$Percentage error $= \left( \frac{\Delta m}{m} + 2\frac{\Delta r}{r} + \frac{\Delta l}{l} ight) 	imes 100\%$$= (0.0016 + 2 	imes 0.004 + 0.0025) 	imes 100\%$$= (0.0016 + 0.008 + 0.0025) 	imes 100\%$$= 0.0121 	imes 100\% = 1.21\%$

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 the determination of density.

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