Question Number 221264 by fantastic last updated on 28/May/25
$${A}\:{wooden}\:{block}\:{and}\:{a}\:{solid}\:{lead}\:{ball}\:{are}\: \\ $$$${placed}\:{in}\:{a}\:{container}\:{fully}\:{filled}\:{with}\:{water} \\ $$$${Now}\:{if}\:{the}\:{lead}\:{ball}\:{is}\:{placed}\:{on}\:{top}\:{of}\:{the} \\ $$$${floating}\:{wooden}\:{block},{what}\:{is}\:{the}\:{change} \\ $$$${in}\:{water}\:{level}\:{in}\:{the}\:{container}?? \\ $$
Commented by mr W last updated on 28/May/25
$${the}\:{water}\:{level}\:{will}\:{rise}\:{when}\:{the} \\ $$$${lead}\:{ball}\:{is}\:{placed}\:{on}\:{the}\:{top}\:{of}\:{the} \\ $$$${floating}\:{wooden}\:{block}. \\ $$
Commented by fantastic last updated on 29/May/25
$${I}\:{know}\:{that}\:{answer}\:{but}\:{can}\:{you}\:{please}\:{explain}\:{me}\:{using}\:{mathematics} \\ $$
Commented by mr W last updated on 29/May/25
$${we}\:{don}'{t}\:{need}\:{complex}\:{mathematical} \\ $$$${calculation}\:{but}\:{only}\:{some}\:{basic} \\ $$$${physics}\:{knowledge}\:{to}\:{understand} \\ $$$${the}\:{question}.\:{the}\:{easiest}\:{explanation} \\ $$$${for}\:{me}\:{is}: \\ $$$${in}\:{both}\:{cases}\:{the}\:{bottom}\:{of}\:{the}\: \\ $$$${container}\:{carries}\:{the}\:{total}\:{weight} \\ $$$${from}\:{the}\:{lead}\:{ball},\:{the}\:{wooden}\:{block} \\ $$$${and}\:{the}\:{water}.\:{the}\:{total}\:{weight}\:{from} \\ $$$${them}\:{is}\:{both}\:{cases}\:{the}\:{same}. \\ $$$${in}\:{the}\:{first}\:{case}\:{the}\:{lead}\:{ball}\:{lies}\: \\ $$$${directly}\:{on}\:{the}\:{bottom}\:{of}\:{container},\: \\ $$$${so}\:{the}\:{bottom}\:{of}\:{the}\:{container}\: \\ $$$${carries}\:{a}\:{part}\:{of}\:{its}\:{weight}\:{directly}\: \\ $$$${and}\:{the}\:{rest}\:{weight}\:{through}\:{water}\: \\ $$$${pressure}.\:\: \\ $$$${in}\:{the}\:{second}\:{case}\:{the}\:{bottom}\:{of}\:{the} \\ $$$${container}\:{carries}\:{the}\:{total}\:{weight} \\ $$$${through}\:{water}\:{pressure}.\:{that}\:{means} \\ $$$${the}\:{water}\:{pressure}\:{in}\:{the}\:{second} \\ $$$${case}\:{must}\:{be}\:{larger}\:{than}\:{that}\:{in}\:{the} \\ $$$${first}\:{case},\:{i}.{e}.\:{the}\:{water}\:{level}\:{will} \\ $$$${rise}\:{when}\:{the}\:{lead}\:{ball}\:{is}\:{moved}\:{from} \\ $$$${the}\:{ground}\:{of}\:{the}\:{container}\:{onto}\: \\ $$$${the}\:{floating}\:{wooden}\:{boat}. \\ $$
Commented by mr W last updated on 29/May/25
$${now}\:{the}\:{explanation}\:{with} \\ $$$${mathematical}\:{calculation}: \\ $$
Commented by mr W last updated on 29/May/25
Commented by mr W last updated on 29/May/25
Commented by fantastic last updated on 29/May/25
$${thanks}\:{sir}.{I}\:{understand}\:{it}\:{now} \\ $$
Commented by mr W last updated on 29/May/25
$$\begin{array}{|c|c|c|c|}{}&\hline{{volume}}&\hline{{density}}\\{{water}}&\hline{{V}_{{W}} }&\hline{\rho_{{W}} }\\{{wooden}\:{block}}&\hline{{V}_{{B}} }&\hline{\rho_{{B}} <\rho_{{W}} }\\{{lead}\:{ball}}&\hline{{V}_{{L}} }&\hline{\rho_{{L}} >\rho_{{W}} }\\\hline\end{array} \\ $$$$\underline{{case}\:\mathrm{1}:} \\ $$$${say}\:{the}\:{volume}\:{of}\:{the}\:{submerged} \\ $$$${part}\:{of}\:{wooden}\:{block}\:{is}\:{V}_{{BS}_{\mathrm{1}} } \\ $$$${V}_{{BS}_{\mathrm{1}} } \rho_{{W}} ={V}_{{B}} \rho_{{B}} \\ $$$$\Rightarrow{V}_{{BS}_{\mathrm{1}} } =\frac{\rho_{{B}} {V}_{{B}} }{\rho_{{W}} } \\ $$$${the}\:{lead}\:{ball}\:{is}\:{completely}\:{submerged} \\ $$$${in}\:{water}. \\ $$$${say}\:{the}\:{cross}−{section}\:{area}\:{of}\:{the} \\ $$$${container}\:{is}\:{A},\:{then} \\ $$$${h}_{\mathrm{1}} {A}={V}_{{W}} +{V}_{{L}} +{V}_{{BS}_{\mathrm{1}} } ={V}_{{W}} +{V}_{{L}} +\frac{\rho_{{B}} }{\rho_{{W}} }{V}_{{B}} \\ $$$$\underline{{case}\:\mathrm{2}:} \\ $$$${say}\:{the}\:{volume}\:{of}\:{the}\:{submerged} \\ $$$${part}\:{of}\:{wooden}\:{block}\:{is}\:{V}_{{BS}_{\mathrm{2}} } \\ $$$${V}_{{BS}_{\mathrm{2}} } \rho_{{W}} ={V}_{{B}} \rho_{{B}} +{V}_{{L}} \rho_{{L}} \\ $$$$\Rightarrow{V}_{{BS}_{\mathrm{2}} } =\frac{{V}_{{B}} \rho_{{B}} +{V}_{{L}} \rho_{{L}} }{\rho_{{W}} } \\ $$$${h}_{\mathrm{2}} {A}={V}_{{W}} +{V}_{{BS}_{\mathrm{2}} } ={V}_{{W}} +\frac{{V}_{{B}} \rho_{{B}} +{V}_{{L}} \rho_{{L}} }{\rho_{{W}} } \\ $$$$\:\:\:\:\:\:\:\:\:={V}_{{W}} +\frac{\rho_{{L}} }{\rho_{{W}} }{V}_{{L}} +\frac{\rho_{{B}} }{\rho_{{W}} }{V}_{{B}} \\ $$$$\:\:\:\:\:\:\:\:\:>{V}_{{W}} +{V}_{{L}} +\frac{\rho_{{B}} }{\rho_{{W}} }{V}_{{B}} ={h}_{\mathrm{1}} {A} \\ $$$$\Rightarrow{h}_{\mathrm{2}} >{h}_{\mathrm{1}} \\ $$$${i}.{e}.\:{the}\:{water}\:{level}\:{rises}. \\ $$
Commented by mr W last updated on 29/May/25
$${for}\:{the}\:{same}\:{reason},\:{when}\:{heavy} \\ $$$${things}\:{are}\:{thrown}\:{from}\:{the}\:{flooting} \\ $$$${boat}\:{into}\:{the}\:{water},\:{the}\:{water}\:{level} \\ $$$${in}\:{container}\:{will}\:{sink}. \\ $$