In a water container, an aluminium piece of volume 0.5 m3 is lowered through an external force, until it is completely submerged. In another identical water container, a lead piece of same volume was similarly submerged using the same amount of external force. The mass density of lead is 4 times larger than the mass density of the aluminum. If FA and FL are the buoyancy forces acting on aluminum and lead respectively, then which of the following statements is correct?

Question

In a water container, an aluminium piece of volume 0.5 m3 is lowered through an external force, until it is completely submerged. In another identical water container, a lead piece of same volume was similarly submerged using the same amount of external force. The mass density of lead is 4 times larger than the mass density of the aluminum. If FA and FL are the buoyancy forces acting on aluminum and lead respectively, then which of the following statements is correct? (A) FA > 4 FL (B) FL > 4 FA (C) FA > 2 FL (D) FL > 2 FA

Tardigrade Admin · Accepted Answer

Buoyancy force on a body is given as Fb=ρ V g where, ρ= fluid density, V= Volume of liquid, which displaced by body. ∵ g=9.806 m / s 2 So, for aluminium piece FA=ρ liq (0.5 m 3) 9.8 ....(i) Similarly, for lead FL=ρ text liq (0.5 m 3) 9.8 ...(ii) Hence, from Eqs. (i) and (ii), we get FA=FL

$F_{A} > 4 F_{L}$

$F_{L} > 4 F_{A}$

$F_{A} > 2 F_{L}$

$F_{L} > 2 F_{A}$

$F_{L} = F_{A}$

FA​>4FL​

FL​>4FA​

FA​>2FL​

FL​>2FA​

FL​=FA​

Buoyancy force on a body is given as Fb​=ρVg where, ρ= fluid density, V= Volume of liquid, which displaced by body. ∵g=9.806m/s2 So, for aluminium piece FA​=ρliq​(0.5m3)9.8....(i) Similarly, for lead FL​=ρliq ​(0.5m3)9.8...(ii) Hence, from Eqs. (i) and (ii), we get FA​=FL​

$F_{A} > 4 F_{L}$

$F_{L} > 4 F_{A}$

$F_{A} > 2 F_{L}$

$F_{L} > 2 F_{A}$

$F_{L} = F_{A}$