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(A) Let $V$ be the total volume of the cubical block,$\rho_s$ be the density of the solid block,and $\rho_L$ be the density of the liquid.Initially,th

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$1/2$

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(A) Let $V$ be the total volume of the cubical block,$\rho_s$ be the density of the solid block,and $\rho_L$ be the density of the liquid.Initially,the block is floating in equilibrium. The weight of the block is balanced by the buoyant force:$V \rho_s g = V_{sub} \rho_L g$Given that half of the volume is immersed,$V_{sub} = V/2$. Thus,$V \rho_s g = (V/2) \rho_L g$,which implies $\rho_s / \rho_L = 1/2$.When the system accelerates upwards with acceleration $a = g/3$,the effective gravity becomes $g_{eff} = g + a = g + g/3 = 4g/3$.The new equilibrium condition is:$V \rho_s g_{eff} = V'_{sub} \rho_L g_{eff}$Dividing both sides by $V \rho_L g_{eff}$,we get:$V'_{sub} / V = \rho_s / \rho_L$Since $\rho_s / \rho_L = 1/2$,the fraction of the volume immersed remains $1/2$. The fraction is independent of the acceleration of the system.

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

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