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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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fioat with $40 \%$ above the water surface
The buoyant force acting on an immersed body is equal to the weight of the fluid displaced by it, if the fluid is in rest. In this case, the fluid is accelerating upwards so the buoyant force must also provide the displaced fluid force to accelerate.
Therefore, buoyant force will be \(\mathrm{fV} \rho \mathrm{g}_{\text {eff }}\) where \(\mathrm{V}=\) volume of body and \(\mathrm{f}=\) fraction of volume of body immersed in fluid and \(\mathrm{g}_{\text {eff }}=\mathrm{g}+\mathrm{a}=1.5 \mathrm{~g}\).
When fluid is at rest: \(\mathrm{f}=1-0.4=0.6\),
so \(0.6 \mathrm{~V} \rho \mathrm{g}=\mathrm{V} \rho_{\mathrm{b}} \mathrm{g}\)
where \(\rho_{\mathrm{b}}\) is the density of the block. \(\Rightarrow \rho_{\mathrm{b}}=0.6 \rho .\)
In the second case: \(1.5 \mathrm{fV} \rho \mathrm{g}=\mathrm{V} \rho_{\mathrm{b}} \mathrm{g}+\mathrm{V} \rho_{\mathrm{b}} \mathrm{a}=1.5 \mathrm{~V} \rho_{\mathrm{b}} \mathrm{g} \Rightarrow \mathrm{f} \rho=\rho_{\mathrm{b}} \Rightarrow \mathrm{f}=0.6\) Thus, the fraction of immersed volume remains the same.
Body will float with \(40 \%\) of the volume above water surface.
Therefore, buoyant force will be \(\mathrm{fV} \rho \mathrm{g}_{\text {eff }}\) where \(\mathrm{V}=\) volume of body and \(\mathrm{f}=\) fraction of volume of body immersed in fluid and \(\mathrm{g}_{\text {eff }}=\mathrm{g}+\mathrm{a}=1.5 \mathrm{~g}\).
When fluid is at rest: \(\mathrm{f}=1-0.4=0.6\),
so \(0.6 \mathrm{~V} \rho \mathrm{g}=\mathrm{V} \rho_{\mathrm{b}} \mathrm{g}\)
where \(\rho_{\mathrm{b}}\) is the density of the block. \(\Rightarrow \rho_{\mathrm{b}}=0.6 \rho .\)
In the second case: \(1.5 \mathrm{fV} \rho \mathrm{g}=\mathrm{V} \rho_{\mathrm{b}} \mathrm{g}+\mathrm{V} \rho_{\mathrm{b}} \mathrm{a}=1.5 \mathrm{~V} \rho_{\mathrm{b}} \mathrm{g} \Rightarrow \mathrm{f} \rho=\rho_{\mathrm{b}} \Rightarrow \mathrm{f}=0.6\) Thus, the fraction of immersed volume remains the same.
Body will float with \(40 \%\) of the volume above water surface.
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