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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.
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 =1*V 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 Please feel free to post as many doubts on our discussion forum as you can. If you find any question Difficult to understand - post it here and we will get you the answer and detailed solution very quickly. We are all IITians and here to help you in your IIT JEE preparation. All the best. Regards, Askiitians Experts Badiuddin
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
=1*V
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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