Stanton D. answered • 12/16/19

Tutor to Pique Your Sciences Interest

Hi Yogita S.,

Yes, there are some problems in your calculations. You should know that by eyeballing your values. Your value for Fb in water looks good : 0.48 kg x 9.81 m s^-2 . The apparent weight in alcohol should be 0.047 kg x 9.81 m s^-2, nowhere near your value. That 0.89 refers to the density of the wood NOT apparent weight (since the alcohol neutralizes all but 0.047 kg of the 0.48 kg initial mass, the density of the wood is correspondingly slightly proportionately greater than that of alcohol). Do you see how that works? You are substituting in the density of alcohol, and proportionating it up to the total mass-equivalent of the wood. Proportions still work, even though the quality used (density) is a stand-in for another quality (mass). So the calculation would be: 0.79 * (0.48)/(0.48-0.047) = (conceptually:) the part * (the whole/the part) .

So for volume you must divide mass by density. Your numbers will match then.

Cheers, -- Mr. d.