Question Number 83307 by ajfour last updated on 29/Feb/20
Commented by mr W last updated on 01/Mar/20
$${at}\:{equilibrium}\:{position}\:{half}\:{of}\:{the} \\ $$$${object}\:{is}\:{submerged},\:{therefore} \\ $$$$\rho=\frac{\mathrm{1}}{\mathrm{2}}\rho_{\mathrm{0}} \\ $$
Commented by ajfour last updated on 29/Feb/20
$$\mathrm{If}\:\mathrm{a}\:\mathrm{cuboidal}\:\mathrm{piece}\:\mathrm{of}\:\mathrm{wood}\:, \\ $$$$\mathrm{released}\:\mathrm{over}\:\mathrm{the}\:\mathrm{surface}\:\mathrm{of}\:\mathrm{a} \\ $$$$\mathrm{large}\:\mathrm{tank}\:\mathrm{just}\:\mathrm{submerges}\: \\ $$$$\mathrm{completely}\:\mathrm{and}\:\mathrm{then}\:\mathrm{starts} \\ $$$$\mathrm{rising}\:\mathrm{up}\:\mathrm{again},\:\mathrm{and}\:\mathrm{oscillates} \\ $$$$\mathrm{in}\:\mathrm{this}\:\mathrm{manner},\:\mathrm{find}\:\mathrm{its}\:\mathrm{density}\:\rho. \\ $$$$\left(\mathrm{given}\:\mathrm{density}\:\mathrm{of}\:\mathrm{water}\:\mathrm{is}\:\rho_{\mathrm{0}} \right) \\ $$
Commented by mr W last updated on 29/Feb/20
Commented by ajfour last updated on 01/Mar/20
$${Yes}\:{Sir},\:{thanks},\:{quite}\:{straight}! \\ $$