the weight of theicebergis supported by thebuoyant forceof the water; the buoyant force is equal to the weight of the displaced water

weight of displaced water =density of waterx volume displaced x g

since the water is displaced by the iceberg, the volume of water displaced is the volume of the iceberg beneath the surface

weight of iceberg = density of iceberg x volume of iceberg x g

since these are equal, we have

rho(water) x Volume (submerged) x g = rho(ice) Volume (ice) x g

canceling out the g's and rearranging, we have

volume(submerged)/volume(ice) = rho(ice)/rho(water)

the left hand side is just the fraction of the iceberg that is submerged

the right hand side is 920/1030 = 0.89

this means that 89% of the iceberg is submerged beneath the water===============================A small metal weight weighs 110 N in water.

When we tie a wood to the metal, since the wood floats in water, it will tend to reduce the weight of the metal since there is now a greater upward force.Hence the combined weight must be less than 110 NBut in the problem it is given that the total weight is 140 NHence the object cannot be wood. And the density of the object must be greater than that of water.================================A small metal weight weighs 110 N in waterWhen an object is tied to it the weight is 140 NIf the object alone weighed in water it will weigh 140 -110 = 30 NIts weight in air is 100 NHence its loss of weight in water = 100-30 = 70 NThe loss is due to the upward force equal to the weight of equal volume of waterHence70 = V*1000*gVg =0.07Its weight in air = V g = 100 N =100 / (Vg) = 100 / 0.07 = 1428.57kg /m^3