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(A) Let $V$ be the total volume of the cubical block and $\rho_b$ be its density. Let $\rho_l$ be the density of the liquid.Initially,the block is in

Vedclass · Accepted Answer

$1 / 4$

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(A) Let $V$ be the total volume of the cubical block and $\rho_b$ be its density. Let $\rho_l$ be the density of the liquid.Initially,the block is in equilibrium,so the weight of the block equals the buoyant force:$V \rho_b g = V_{immersed} \rho_l g$Given $V_{immersed} = V / 4$,we have:$V \rho_b g = (V / 4) \rho_l g \implies \rho_b = \rho_l / 4$When the system accelerates upward with acceleration $a = g / 4$,the effective gravity becomes $g' = g + a = g + g / 4 = 5g / 4$.The new buoyant force $F_B'$ must balance the effective weight $W'$ of the block:$W' = V \rho_b g' = V (\rho_l / 4) (5g / 4)$$F_B' = V_{immersed}' \rho_l g' = V_{immersed}' \rho_l (5g / 4)$Equating $W' = F_B'$:$V (\rho_l / 4) (5g / 4) = V_{immersed}' \rho_l (5g / 4)$$V / 4 = V_{immersed}'$Thus,the fraction of volume immersed remains $1 / 4$.

$A$ cubical block is floating in a liquid with one-fourth of its volume immersed in the liquid. If the whole system accelerates upward with an acceleration of $g / 4$,what will be the fraction of the volume immersed in the liquid?

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