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(A) Let the volume of the block be $V$, the density of the block be $\rho$, and the density of the liquid be $\sigma$.Initially, the block is in equil

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$0.5$

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(A) Let the volume of the block be $V$, the density of the block be $\rho$, and the density of the liquid be $\sigma$.Initially, the block is in equilibrium, so the weight of the block equals the buoyant force:$V \rho g = V_{in} \sigma g$Given that half the volume is immersed, $V_{in} = V/2$, so:$V \rho g = (V/2) \sigma g \implies \rho = \sigma/2$.When the system accelerates upwards with acceleration $a = g/3$, the effective gravity becomes $g' = g + a = g + g/3 = 4g/3$.The new equilibrium condition is:$V \rho g' = V'_{in} \sigma g'$$V \rho (4g/3) = V'_{in} \sigma (4g/3)$$V \rho = V'_{in} \sigma$Since $\rho = \sigma/2$, we have $V(\sigma/2) = V'_{in} \sigma$, which gives $V'_{in} = V/2$.The fraction of volume immersed remains $V'_{in}/V = 0.5$. Thus, the fraction of volume immersed does not change with acceleration.

$A$ cubical block is floating in a liquid with half of its volume immersed in the liquid. When the whole system accelerates upwards with an acceleration of $g/3$, the fraction of volume immersed in the liquid will be

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