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Question Number 5351 by sanusihammed last updated on 10/May/16
A cylindrical iron rod 8cm and 6cm in diameter stands in a  cylindrical tin of 12cm in diameter. Water is poured into the  tin until it depth is 8cm. How far would the level drop when the   rod is removed ?
$${A}\:{cylindrical}\:{iron}\:{rod}\:\mathrm{8}{cm}\:{and}\:\mathrm{6}{cm}\:{in}\:{diameter}\:{stands}\:{in}\:{a} \\ $$$${cylindrical}\:{tin}\:{of}\:\mathrm{12}{cm}\:{in}\:{diameter}.\:{Water}\:{is}\:{poured}\:{into}\:{the} \\ $$$${tin}\:{until}\:{it}\:{depth}\:{is}\:\mathrm{8}{cm}.\:{How}\:{far}\:{would}\:{the}\:{level}\:{drop}\:{when}\:{the}\: \\ $$$${rod}\:{is}\:{removed}\:? \\ $$
Commented by Yozzii last updated on 11/May/16
I am guessing that the rod and tin share  the same height of 8cm, and that the  height of the upper end of the rod that  is above the top edge of the tin, when  placed to stand in the tin, is negligible  to give an approximate answer.  So, the volume left for the water is  given by V=πh(r_t ^2 −r_r ^2 )=π×8(6^2 −3^2 )=216π cm^3 .  When the water is added, its volume  is approximately 216πcm^3 . So when the  rod is removed the water settles, taking  up a volume of about 216πcm^3  still with  base radius equal to 6cm.  So, for some h⇒πhr_t ^2 =216π⇒h=((216)/6^2 )=6cm.
$${I}\:{am}\:{guessing}\:{that}\:{the}\:{rod}\:{and}\:{tin}\:{share} \\ $$$${the}\:{same}\:{height}\:{of}\:\mathrm{8}{cm},\:{and}\:{that}\:{the} \\ $$$${height}\:{of}\:{the}\:{upper}\:{end}\:{of}\:{the}\:{rod}\:{that} \\ $$$${is}\:{above}\:{the}\:{top}\:{edge}\:{of}\:{the}\:{tin},\:{when} \\ $$$${placed}\:{to}\:{stand}\:{in}\:{the}\:{tin},\:{is}\:{negligible} \\ $$$${to}\:{give}\:{an}\:{approximate}\:{answer}. \\ $$$${So},\:{the}\:{volume}\:{left}\:{for}\:{the}\:{water}\:{is} \\ $$$${given}\:{by}\:{V}=\pi{h}\left({r}_{{t}} ^{\mathrm{2}} −{r}_{{r}} ^{\mathrm{2}} \right)=\pi×\mathrm{8}\left(\mathrm{6}^{\mathrm{2}} −\mathrm{3}^{\mathrm{2}} \right)=\mathrm{216}\pi\:{cm}^{\mathrm{3}} . \\ $$$${When}\:{the}\:{water}\:{is}\:{added},\:{its}\:{volume} \\ $$$${is}\:{approximately}\:\mathrm{216}\pi{cm}^{\mathrm{3}} .\:{So}\:{when}\:{the} \\ $$$${rod}\:{is}\:{removed}\:{the}\:{water}\:{settles},\:{taking} \\ $$$${up}\:{a}\:{volume}\:{of}\:{about}\:\mathrm{216}\pi{cm}^{\mathrm{3}} \:{still}\:{with} \\ $$$${base}\:{radius}\:{equal}\:{to}\:\mathrm{6}{cm}. \\ $$$${So},\:{for}\:{some}\:{h}\Rightarrow\pi{hr}_{{t}} ^{\mathrm{2}} =\mathrm{216}\pi\Rightarrow{h}=\frac{\mathrm{216}}{\mathrm{6}^{\mathrm{2}} }=\mathrm{6}{cm}. \\ $$$$ \\ $$$$ \\ $$

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