# 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?

Chetan Mandayam Nayakar
312 Points
11 years ago

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

Chetan Mandayam Nayakar
312 Points
11 years ago

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

Swapnil Saxena
102 Points
11 years ago

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.