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				   the weight of a hollow body,made of gold,is 36.670g in air,while it is 33.865g in water. If the density of gold is 19.3 g/cc, find the volume of the hollow part of the body.(Density of water is 1 g/cc.

7 years ago


Answers : (1)


Dear mikir

weight if body =36.67 g

weight of body in water =33.865 g

this reduction in weight is dut to boyancy force

Fb = 36.67 g -33.865 g

    = 2.805 g

but Fb=wait of water replace by body



V= 2.805cc

but this total volume

so volume of hollow part + volume of solid part =V=2.805 cc

for volume of solid part

  volume of solid part*density of gold=weight of body in air

       volume of solid part=36.670/19.3 

                                    =1.9 cc

 so volume of hollow part =2.805-1.9

                                    =.905 cc

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7 years ago

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