a lead bullet is placed in a pool of mercury. what fractional part of the volume of the bullet is submerged?

Answer 1

well, the lead displaces mercury such that the weight of the mercury is equal to the weight of the lead.

weight of mercury displaced= densitymercury*volumeleadsubmerged

weight of lead= densitylead*volumelead

so set them equal

Volumesubmerged/totalvolume= densitylead/densitymercury

Answer 2

Perfect!
Since the density of lead is 11.3 x 10^3 kg/m^3 and the density of mercury is 13.6 x 10^3 kg/m^3, the answer is 0.831.
Thanks much!