Question

In a water container, an aluminum piece of volume \( 0.5 \,\text{m}^3 \) 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 statement is correct?

• A. \( F_A > 4F_L \)
• B. \( F_L > 4F_A \)
• C. \( F_A > 2F_L \)
• D. \( F_L > 2F_A \)
• E. \( F_L = F_A \)

Hint:

Buoyant force depends only on displaced fluid, not on object density or mass.

Answer 1

Answer: (E)

Concept:
Buoyant force is given by: \[ F = \rho_{\text{fluid}} V g \]

Step 1: Identify important point
Both objects are completely submerged.

Step 2: Compare volumes
Both have same volume \( V = 0.5 \, m^3 \)

Step 3: Apply formula
\[ F_A = \rho_{\text{water}} V g \] \[ F_L = \rho_{\text{water}} V g \]

Step 4: Final conclusion
\[ F_A = F_L \] \[ \boxed{F_L = F_A} \]