1. The problem statement, all variables and given/known data An aluminum cylinder weighs 1.03 N. When this same cylinder is completely submerged in alcohol, the volume of the displaced alcohol is 3.89 multiplied by 10-5 m3. If the cylinder is suspended from a scale while submerged in the alcohol, the scale reading is 0.732 N. What is the specific gravity of the alcohol? 2. Relevant equations Fscale + FBuoyancy = mg FBouyancy = DVg Specific gravity = Dobject / Dmedium 3. The attempt at a solution First I used Fscale + FBuoyancy = mg (.732)(FB) = 1.03 FB = .298 N Then I used FB = DVg .298 = Dalcohol(3.89e-5)(9.8) Dalcohol = .298 / [(3.89e-5)(9.8)] Dalcohol = 781.7009 kg.m3 Then I used Specific gravity = Dobject / Dmedium, where D = mv. Specific gravity = (m/v) / Dalcohol Specific gravity = [(1.02/9.8)/3.85e-5] / (781.7009) Specific gravity = 3.458 I tried this answer and it was wrong. Can anyone help out? Thank you!