Question

A hollow sphere of volume $V$ is floating on the 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?
A. $\dfrac{V}{2}$
B. $\dfrac{V}{3}$
C. $\dfrac{V}{4}$
D. $V$

Answer 1

This question is based on variation in pressure with depth. When the body is half immersed in the water, the upthrust force will be equal to the weight of sphere and when the body is fully immersed then we will equate the upthrust with the sum of weight of sphere and weight of water poured in sphere to get the minimum volume of water poured inside the sphere so that the sphere sinks.

Complete step by step answer:
The difference in pressure between the object's top and bottom causes it to go upward. This is referred to as upthrust.The upthrust on an item in a fluid, according to Archimedes' principle, is equal to the weight of the fluid displaced, and the volume of the object multiplied by the density of the fluid.

When a body (spherical) is half immersed, the following happens:
Upthrust = weight of the sphere
$\dfrac{V}{2} \times {\rho _{liq}} \times g = v \times \rho \times g$
Therefore, from here we will deduce the value of $\rho$
$\rho = \dfrac{{{\rho _{liq}}}}{2}$

When body (sphere) is fully immersed then,
Upthrust = weight of sphere + weight of water poured in sphere
$\Rightarrow V \times {\rho _{liq}} \times g = V \times \rho \times g + V' \times {\rho _{liq}} \times g \\\ \therefore V' = \dfrac{V}{2}$
Therefore, the minimum volume of water poured inside the sphere so that the sphere now sinks into the water $\dfrac{V}{2}$.

So, the correct option is (A).

Note: The surface tension (capillarity) acting on the body is not taken into account by Archimedes' principle, but this additional force simply changes the amount of fluid displaced and the spatial distribution of the displacement, so the principle that buoyancy = weight of displaced fluid remains valid.