$A$ cubical block of density $\rho$ is floating on the surface of water. Out of its height $L$,a fraction $x$ is submerged in water. The vessel is in an elevator accelerating upward with acceleration $a$. What is the fraction immersed?

(A) Let the density of water be $\rho_{w}$. The block of height $L$ floats on it. Let $x$ be the height of the block submerged in water.
Volume of the block $V = L^{3}$.
Mass of the block $m = V \rho = L^{3} \rho$.
Weight of the block $W = mg = L^{3} \rho g$.
Case $1$: When the elevator is at rest (or moving with constant velocity),the weight of the block is balanced by the buoyant force.
Weight of the block = Weight of the water displaced.
$L^{3} \rho g = (x L^{2}) \rho_{w} g$.
Therefore,the fraction submerged is $\frac{x}{L} = \frac{\rho}{\rho_{w}}$.
Case $2$: When the elevator is accelerating upward with acceleration $a$,the effective acceleration becomes $g' = (g + a)$.
In this frame,the weight of the block becomes $W' = m(g + a) = L^{3} \rho (g + a)$.
The buoyant force also changes because the effective gravity acting on the fluid changes: $F_{B}' = V_{submerged} \rho_{w} (g + a) = (x' L^{2}) \rho_{w} (g + a)$,where $x'$ is the new submerged height.
Equating the two: $L^{3} \rho (g + a) = x' L^{2} \rho_{w} (g + a)$.
Canceling $(g + a)$ from both sides,we get $L^{3} \rho = x' L^{2} \rho_{w}$.
Therefore,the new fraction submerged is $\frac{x'}{L} = \frac{\rho}{\rho_{w}}$.
Conclusion: The fraction of the block submerged remains unchanged regardless of the acceleration of the elevator.