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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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$\frac{1}{2}$

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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,$V_{sub} = V/2$. Therefore:$V \rho_s g = (V/2) \rho_L g \implies \rho_s = \rho_L / 2$.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 $g_{eff}$ gives:$V \rho_s = V'_{sub} \rho_L \implies \frac{V'_{sub}}{V} = \frac{\rho_s}{\rho_L}$.Substituting $\rho_s = \rho_L / 2$,we get:$\frac{V'_{sub}}{V} = \frac{\rho_L / 2}{\rho_L} = \frac{1}{2}$.Thus,the fraction of the volume immersed remains unchanged regardless 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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