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

The correct option is V/2.

Step 1: Analyze the initial floating state
Initially, the hollow sphere of volume $V$ floats on the water surface with half of its volume immersed. According to the principle of floatation, the buoyant force acting on the sphere equals its weight. The buoyant force is equal to the weight of the water displaced by the immersed part of the sphere.

Let $m_s$ be the mass of the hollow sphere and ${\rho }_w$ be the density of water. The volume of water displaced is $\frac{V}{2}$.

Weight of the sphere = Buoyant force of displaced water

$m_{s}g=\left(\frac{V}{2}\right){\rho }_{w}g$

Simplifying the equation by dividing both sides by $g$, we get the mass of the sphere:

$m_s=\frac{V{\rho }_w}{2}$

Step 2: Determine the sinking condition
For the sphere to sink (so that it is completely submerged in water), its total weight—which is the weight of the sphere itself plus the weight of the poured water—must be at least equal to the buoyant force when the entire volume $V$ of the sphere is submerged under water.

Let $v$ be the minimum volume of water poured inside the sphere. The mass of this water is $v{\rho }_w$.

When completely submerged, the volume of water displaced by the sphere is $V$. Therefore, the maximum buoyant force is:

$F_b=V{\rho }_{w}g$

Step 3: Calculate the minimum volume of water
To just sink the sphere, we set the total weight of the system equal to the maximum buoyant force:

$\left(m_s+v{\rho }_w\right)g=V{\rho }_{w}g$

Dividing both sides by $g$ yields:

$m_s+v{\rho }_w=V{\rho }_w$

Now, substitute the value of $m_s$ from Step 1 into this equation:

$\frac{V{\rho }_w}{2}+v{\rho }_w=V{\rho }_w$

Divide the entire equation by the density of water ${\rho }_w$:

$\frac{V}{2}+v=V$

Solving for $v$:

$v=V-\frac{V}{2}$

$v=\frac{V}{2}$

Thus, the minimum volume of water that must be poured inside the sphere so that it sinks is $\frac{V}{2}$.