Question

the total weight of a piece of wood is 6kg. in the floating state in water 1/3 part remains inside the water. on this floating solid what maximum weight is to be put such that the whole piece of woid is to be drowned into the water.

Answer 1

I think you have a typo where the "8" is and you didn't finish the sentence, so I'm going to assume that the wood starts 1/3 part inside the water and that you want it to be entirely in the water.  The important part is that you pick up the principles necessary to solve this type of problem.

 

The force of buoyancy is equal to the weight of the fluid displaced.  Thus, it is proportional to the volume displaced.  Therefore, if we triple the volume submerged, we triple the force of buoyancy.  In order for it to be stable at this point, we need to also triple the downward force of gravity.  We started with 6 kg, so we need to add 12 kg to get to 18 kg total.  If we put less than 12 kg, part of the wood will still be out.

 

If you have additional questions on this, please let me know.

Answer 2

If a wooden block is floating,

          weight of the wooden block = buoyant force

                                                    $mg =U$------ (1)

But Archimedes says that,

                       buoyant force = weight of the fluid that the body displaced

                                            $U =Vdg$------ (2)    

Where, m = mass of the block

            g = acceleration due to gravity = 10 m/s²

            V = weight of the fluid that the body displaced

            d = density of water = 1000 kg/m³

           V' = volume of the block

           M = mass of the extra block  

Substituting equation (2) to equation (1)

                                          $mg = Vdg$

• if 1/3 volume of the block is submerged in water,

                                        $6 * g = (1/3)* V' *1000 *g$

                                              $6 = 1000V'/3$

                                             $V' = 18 /1000$ ------ (3)

• if whole volume of the block is submerged in water,

                                       $Mg + 6g = V' dg$

                                            $M + 6 = V' * 1000$

                                            $M + 6 = (18/1000) *1000$     (by (3))

                                            $M + 6 = 18$

              Answer :                   $M = 12 kg$