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 the whole of it is just immersed in water is :

Options

A. `5/12pir^(4)pg`
B. `1/2prg`
C. `4/3pir^(3)pg`
D. `2/3pir^(4)pg`

Answer

A

Explanation

When the ball is pushed down, the water gains P.E. Gain in P.E. of water
`=Vprg-V/2p[3/8r].` Now `V=4/3pir^(3)`
`therefore` Gain in P.E.
`=4/3pir^(4)pg-1/4pir^(4)pg=13/12pir^(4)pg`
Loss of P.E. of ball =`4/3pir^(4)`p. g or work done
`=13/12pir^(4)pg-4/3pir^(4)p.g`
`therefore` Work done=`pir^(4)pg[13/12-4/3(p.)/p]`
`=pir^(4)pg[13/12-4/3xx0.5]`
`5/12pir^(4)p.g`
Correct choice is (a).