A wooden block floats in a liquid with 40% of | vedclass

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(D) Let $V$ be the total volume of the wooden block and $\rho_b$ be its density. Let $\rho_l$ be the density of the liquid.
In the initial state (at

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$40$

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(D) Let $V$ be the total volume of the wooden block and $\rho_b$ be its density. Let $\rho_l$ be the density of the liquid.
In the initial state (at rest), the block floats, so the weight equals the buoyant force:
$V \rho_b g = V_{in} \rho_l g$
Given $V_{in} = 0.4 V$, we have $0.4 V \rho_l g = V \rho_b g$, which implies $\rho_b = 0.4 \rho_l$.
When the vessel accelerates upwards with $a = g/2$, the effective acceleration due to gravity becomes $g' = g + a = g + g/2 = 3g/2$.
The new buoyant force is $V'_{in} \rho_l g'$, where $V'_{in}$ is the new submerged volume.
Equating the weight and the new buoyant force:
$V \rho_b g' = V'_{in} \rho_l g'$
Since $g'$ appears on both sides, it cancels out:
$V \rho_b = V'_{in} \rho_l$
Substituting $\rho_b = 0.4 \rho_l$:
$V (0.4 \rho_l) = V'_{in} \rho_l$
$V'_{in} = 0.4 V$
Thus, the percentage of volume inside the liquid remains $40\%$.

$A$ wooden block floats in a liquid with $40\%$ of its volume inside the liquid. When the vessel containing the liquid starts rising upwards with an acceleration $a = g/2$, the percentage of volume inside the liquid is ......... $\%$

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