Question

A piece of iron of density 7·8 × 10^-3 kg/m3 and volume 100 cm3 is totally immersed in water Calculate (a) the weight of the iron piece in air (b) the upthrust and (c) apparent weight in water.

Answer 1

(a) weight in air = mg

m = density x volume

so, weight (W) will be ... 7.8 N
$7.8 \times {10}^{ 3} \times 100 \times {10 }^{ - 6} \times 10 = 7.8 \: newton$
(b) Upthrust = weight of water displaced = density of water * volume of iron piece * g
= 1000 x 0.0001 x 10 = 1 N

(c) apparent weight = actual weight - Upthrust = 7.8 - 1 = 6.8 N