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A hollow sphere of volume $V$ is floating on water surface with half immersed in it. What should be the minimum volume of water poured inside the sphere so that the sphere now sinks into the water
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The correct answer is:
$V / 2$
When body (sphere) is half immersed, then upthrust = weight of sphere
$\Rightarrow \frac{V}{2} \times \rho_{\text {liq }} \times g=V \times \rho \times g \quad \therefore=\frac{\rho_{\text {liq }}}{2}$
When body (sphere) is fully immersed then, Upthrust = wt. of sphere + wt. of water poured in sphere
$\Rightarrow V \times \rho_{\text {liq }} \times g=V \times \rho \times g+V \times \rho_{\text {liq }} \times g$
$\Rightarrow V \times \rho_{\text {liq }}=\frac{V \times \rho_{\text {liq }}}{2}+V \times \rho_{\text {liq }} \Rightarrow V^{\prime}=\frac{V}{2}$
$\Rightarrow \frac{V}{2} \times \rho_{\text {liq }} \times g=V \times \rho \times g \quad \therefore=\frac{\rho_{\text {liq }}}{2}$
When body (sphere) is fully immersed then, Upthrust = wt. of sphere + wt. of water poured in sphere
$\Rightarrow V \times \rho_{\text {liq }} \times g=V \times \rho \times g+V \times \rho_{\text {liq }} \times g$
$\Rightarrow V \times \rho_{\text {liq }}=\frac{V \times \rho_{\text {liq }}}{2}+V \times \rho_{\text {liq }} \Rightarrow V^{\prime}=\frac{V}{2}$
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