Question

A piece of iron of density 7800 kgm-3 and volume 100 cm3 is completely immersed in water. Calculate its apparent weight in water. (g = 10 S.I unit) *​

Answer 1

Answer:

Apparent weight of iron = Weight of iron in air -  weight of iron in water

   As we know , Density = volumemass

So mass of iron piece in air =  (volume) (density) =(10−4)(7800)kg=0.78kg

Weight of iron piece in air=mg=7.8N

weight of iron piece in water = upthrust  = Vρg = 10−4(1000)10N=1N

so apparent weight of the iron piece =7.8N−1N=6.8N