Question

In a water container, an aluminium piece of volume $0.5 \; m^3$ 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 $F_A$ and $F_L$ are the buoyancy forces acting on aluminum and lead respectively, then which of the following statements is correct?

• A. $F_A > 4 F_L$
• B. $F_L > 4 F_A$
• C. $F_A > 2 F_L$
• D. $F_L > 2 F_A$
• E. $F_L = F_A$

Answer 1

Answer: (E)

Buoyancy force on a body is given as
$F_{b}=\rho V g$
where, $\rho=$ fluid density, $V=$ Volume of liquid, which displaced by body.
$\because g=9.806 m / s ^{2}$
So, for aluminium piece
$F_{A}=\rho_{ liq }\left(0.5 m ^{3}\right) 9.8\,....(i)$
Similarly, for lead
$F_{L}=\rho_{\text {liq }}\left(0.5\, m ^{3}\right) 9.8\,...(ii)$
Hence, from Eqs. (i) and (ii), we get
$F_{A}=F_{L}$