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(C) The buoyant force acting on an immersed body is equal to the weight of the fluid displaced by it when the fluid is at rest. When the fluid is acce

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float with $40 \%$ above the water surface

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(C) The buoyant force acting on an immersed body is equal to the weight of the fluid displaced by it when the fluid is at rest. When the fluid is accelerating,the effective gravity $g_{	ext{eff}}$ must be considered.In the first case,the block is in equilibrium: $V_{	ext{immersed}} ho g = V_{	ext{total}} ho_b g$,where $ho$ is the density of water and $ho_b$ is the density of the block.Given $40 \%$ is above the surface,$60 \%$ is immersed,so $V_{	ext{immersed}} = 0.6 V_{	ext{total}}$.Thus,$0.6 V ho g = V ho_b g \implies ho_b = 0.6 ho$.When the lift accelerates upward with $a = g/2$,the effective gravity is $g_{	ext{eff}} = g + a = 1.5 g$.The buoyant force becomes $F_B = V_{	ext{immersed}} ho (1.5 g)$.The weight of the block in the accelerating frame is $W_{	ext{eff}} = V_{	ext{total}} ho_b (1.5 g)$.For equilibrium,$F_B = W_{	ext{eff}} \implies V_{	ext{immersed}} ho (1.5 g) = V_{	ext{total}} ho_b (1.5 g)$.Dividing by $1.5 g$,we get $V_{	ext{immersed}} ho = V_{	ext{total}} ho_b$.Substituting $ho_b = 0.6 ho$,we get $V_{	ext{immersed}} = 0.6 V_{	ext{total}}$.Therefore,the fraction of the volume immersed remains $60 \%$,and the fraction above the surface remains $40 \%$.

$A$ wooden block is floating on water kept in a beaker. $40 \%$ of the block is above the water surface. Now the beaker is kept inside a lift that starts going upward with acceleration equal to $g / 2$. The block will then

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