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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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Verified Answer
The correct answer is:
$\frac{V}{2}$
When a body (sphere) is half immersed, then Upthrust $=$ weight of sphere
$\frac{V}{2} \times \rho_{\text {liq }} \times g=\nu \times \rho \times g$
$\therefore \quad \rho=\frac{\rho_{\text {liq }}}{2}$
When body (sphere) is fully immersed then, Upthrust $=$ weight of sphere + weight of water pourd in sphere
$V \times \rho_{\text {liq }} \times g=V \times \rho \times g+V^{\prime} \times \rho_{\text {liq }} \times g$
$V \times \rho_{\text {liq }}=\frac{v \times \rho_{\text {liq }}}{2}+V^{\prime} \times \rho_{\text {liq }}$
$V^{\prime}=\frac{V}{2}$
$\frac{V}{2} \times \rho_{\text {liq }} \times g=\nu \times \rho \times g$
$\therefore \quad \rho=\frac{\rho_{\text {liq }}}{2}$
When body (sphere) is fully immersed then, Upthrust $=$ weight of sphere + weight of water pourd in sphere
$V \times \rho_{\text {liq }} \times g=V \times \rho \times g+V^{\prime} \times \rho_{\text {liq }} \times g$
$V \times \rho_{\text {liq }}=\frac{v \times \rho_{\text {liq }}}{2}+V^{\prime} \times \rho_{\text {liq }}$
$V^{\prime}=\frac{V}{2}$
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