In figure-I is shown a sphere of mass `m` and radius `r` resting at the bottom of a large container filled with water. Depth of the container is `h`. Density of material of the sphere is the same as that of water. Now hte whole sphere is slowly pulled out of water as shown in figure-II Work done by the agent in pulling the sphere is equal to

Options

A. `mgr`
B. `0.5mgr`
C. `mg(5r+h)`
D. `mg(r+h)`

Answer

A

Explanation

`costheta=(r)/(R)impliesr=Rcosthetaimpliesdr=-Rsintheta.dtheta` net force on immersed body's part `=0`
`(becauserho_("body")=rho_("water"))`
`:. dw=` workdone in shifting above part through
dr height `=V_("above")rhog(dr)to(1)`
Determination of `V_(underline("above")):-Omega=2pi(1-costheta)implies` volume related to surface area above water
level is `=(R^(3))/(3)Omega [{:( :' 4pi rarr (4)/(3) piR^(3)),(Omega rarr (4)/(3)piR^(3) ((Omega)/(4pi))):}]`
`:. V_("above")=(R^(3))/(3)2pi(1-costheta)to(2)`
`becausedw=(R^(3))/(3)[2pi-2picostheta]rhog-(-R)sinthetadtheta`
`becausedw=-(R^(4)rhog)/(3)2pisinthetadtheta+(R^(4)rhog)/(3)2picostheta.sintheta.dtheta`
`=(R^(4)rhoghpi)/(3)[sintheta2thetadtheta-2sinthetadtheta]impliesw=(R^(4)rhoghpi)/(3)(4)`
`[because` after integration limits of `theta:o` to `180^(@)`]
`=(4pi)/(3)R^(3)rhoghR=mgR`