A piece of gold-alminium alloy weighs 49N. When suspended from a spring balance and submerged in water it weighs 39.2N. What is the weight of Gold in the alloy if the specific gravity of gold is 19.3 and that if aluminium is 2.5?

Answer 1

find the volume
49 N - 39.2 N = weight of water displaced (Says Archimedes) = 9.8 N water weight
water is about 1000 kg / m^3
so
9.8 N = mass of water * g
if g = about 9.8 m/s^2
then mass of water displaced = 1 kg
so water volume = metal volume = 1/1000 meter^3
so you have a rock weighing 49 N (mass = 49/9.8 = 5 kg) of volume 0.001 m^3
Au mass + Al mass = 5 kg
Au vol + Al vol = 0.001 m^3
====================
Au mass = 19.8 Au vol * (1000 kg/m^3) = 19,800 * Au vol
Al mass = 2.5 Al vol * (1000 kg/m^3) = 2,500 *Al vol
so
19,800 Au vol + 2,500 *Al vol = 5
1,000 Au vol + 1,000 Al vol = 1
============================
19,800* Au vol + 2,500 *Al vol = 5
2,500 * Au vol + 2,500 *Al vol = 2.5
-----------------------------------------
17,300 Au vol = 2.5
Au vol = 1.45 * 10 ^-4 meters^3
Au mass = 1.45*10^-4 * 19,800 = 2.86 kg
Au weight = 2.86*9.8 = 28 Newtons on earrth