Density measurement of irregular shapes using water and dry weights?

Question

I was pretty sure there was a method for measuring the density of a solid by comparing the weight in water with the dry weight, but I cannot find exactly what conversion etc. to use or how to get the density using these weights.

Answer 1

Archimedes' principle states that the weight of an object in water = weight dry - weight of displaced water.
Assume the object has volume V and specific gravity R> 1 ( so it sinks ). The density of water = 1 gm/cc or S.G. = 1.

W (in H2O) = RV - V = (R-1)V; R-1 = W(in H2O) / V
V = W(dry)/R; R-1 = W(in H2O) R/W(dry)
R {1 - W(water)/W(dry) } = 1
R = 1 / [{1 - W(water)/W(dry) } ]

Answer 2

Try taking a lead weight and primary discovering its quantity via displacement. Attach the result in the floating item and permit it sink after which discover the whole quantity displaced, subtract out the lead quantity for the quantity of the abnormal item. Alternatively you might push the abnormal item under the water floor with an overly skinny needle (of negligible quantity) to get it fully submerged. This might obviate the desire for a stinky, flammable solvent of much less density than water. If the item was once nonetheless not up to methanol or kerosene, the ones beverages might no longer be priceless.

Answer 3

Let Wa the dry weight and Ww the weight in water. Then:
Wa = d V g ; Ww = (d - dw) Vg where d is the density of a solid, dw= density of water (d > dw), V is the volume of the solid.
Ww/Wa = 1 - dw/d ---> d = dw Wa/(Wa - Ww)