A spherical ball of radius r and relative density 0.5 is floating in equilibrium in water with half of it immersed in water. The work done in pushing the ball down so that whole of it is just immersed in water is : (where ρ is the density of water)

The correct option is 5π r⁴ρg/12.

Let us solve the problem step-by-step from scratch.

Consider a spherical ball of radius ^r and density $\sigma$ floating in water of density $\rho$. The relative density of the ball is given as 0.5, which means:
$\sigma =0.5\rho$
Initially, the ball is floating in equilibrium with half of its volume immersed in water. This is the equilibrium state where the buoyant force equals the gravitational force on the ball.

Now, let the ball be pushed down vertically by a distance ^x from its equilibrium position. When the ball is pushed down by ^x, the depth of the center of the sphere below the original waterline increases, causing more volume of the sphere to be submerged. The additional submerged volume increases the upward buoyant force, creating a restoring force.

Let us set up a coordinate system where ^y is the downward displacement of the center of the sphere from the position where the top of the sphere is just touching the water surface.
When the ball is floating in equilibrium, half of it is immersed. This means the water level passes exactly through the center of the sphere. Thus, in the equilibrium state, the depth of the center of the sphere below the water surface is 0.
When the ball is pushed down by a distance ^x, the center of the sphere is at a depth ^x below the water surface.
The volume of a spherical cap of height ^h is given by the formula:
$V\left(h\right)=\frac{\pi }{3}{h}^2(3r-h)$
At any displacement ^x (where $0\le x\le r$), the depth of the center of the sphere is ^x below the surface. The portion of the sphere that is submerged has a height:
$h=r+x$
Therefore, the submerged volume of the ball when it is pushed down by ^x is:
$V_{\text{sub}}\left(x\right)=\frac{\pi }{3}{(r+x)}^2(3r-(r+x\left)\right)=\frac{\pi }{3}{(r+x)}^2(2r-x)$

Expanding this expression:
${(r+x)}^2(2r-x)=(r^2+2rx+x^2)(2r-x)$
$=2r^3-r^{2}x+4r^{2}x-2r{x}^2+2r{x}^2-x^3$
$=2r^3+3r^{2}x-x^3$
So, the submerged volume is:
$V_{\text{sub}}\left(x\right)=\frac{\pi }{3}(2r^3+3r^{2}x-x^3)$

The upward buoyant force acting on the ball is:
$F_B\left(x\right)=V_{\text{sub}}\left(x\right)\rho g=\frac{\pi \rho g}{3}(2r^3+3r^{2}x-x^3)$

The downward gravitational force acting on the ball is:
$W=V_{\text{total}}\sigma g=\left(\frac{4}{3}\pi r^3\right)\left(0.5\rho \right)g=\frac{2}{3}\pi r^3\rho g$

The net downward external force ^F required to hold the ball at a displacement ^x is:
$F\left(x\right)=F_B\left(x\right)-W$
$F\left(x\right)=\frac{\pi \rho g}{3}(2r^3+3r^{2}x-x^3)-\frac{2}{3}\pi r^3\rho g$
$F\left(x\right)=\frac{\pi \rho g}{3}(3r^{2}x-x^3)$

To find the total work done in pushing the ball down until it is just fully submerged, we integrate the force ^F(x) from the equilibrium position ($x=0$) to the fully submerged position ($x=r$):
$\text{Work Done}={\int }_0^{r}F\left(x\right)dx$
$\text{Work Done}={\int }_0^r\frac{\pi \rho g}{3}(3r^{2}x-x^3)dx$
$\text{Work Done}=\frac{\pi \rho g}{3}{[\frac{3}{2}}_0^r$
$\text{Work Done}=\frac{\pi \rho g}{3}\left(\frac{3}{2}\right)$
$\text{Work Done}=\frac{\pi \rho g}{3}\left(\frac{6-1}{4}{r}^4\right)=\frac{\pi \rho g}{3}\left(\frac{5}{4}\right)$
$\text{Work Done}=\frac{5\pi r^4\rho g}{12}$