a spherical body of r radius and relative density 0.5 is floating in equlibrium in water with half of it immersed in water. the work done in pushing the ball so that whole of it is immersed in water?

work done by gravity(u)+work done by pushing(v)=work done by buoyant force(w)

w=2u, therefore v=u=(4πr³/3)*0.5*1000*gr

please let me know if you want more explanation

my previous solution is WRONG

There are three forces involved: gravity,pushing and buoyancy

gravity(g) + pushing(p)=buoyancy(b)

thus, p=b-g, so work done by p(wp)=work done by buoyancy(wb)-wg

let mass of ball=m, wg=mgr

wb=∫b(x) dx, from x=0 to x=r, where x is the distance by which the centre of the ball is below the surgace of water

The answer for this question is (5pi/12)dgr⁴ where (d is the density of water and r is the radius of the spherical body) . the answer is completely independent of the relative density of the floating body.