Question

A block of aluminum that has dimensions 2.07 cm by 2.55 cm by 5.00 cm is suspended from a spring scale. The density of aluminum is 2702 kg/m^3.

(a) What is the weight of the block?
0.699 N

(b) What is the scale reading when the block is submerged in oil with a density of 817 kg/m^3?

Please help I don't know how to do part b; Thanks.

Answer 1

mg = ρ•V •g =
=(2.07•2.55•5) •10^-6•2702•9.8 = 0.699 N
buoyancy force Mg
mg - Mg =
=ρ•V •g – ρ1•V•g = (ρ – ρ1) •V•g =
=(2702 – 817) •(2.07•2.55•5) •10^-6•9.8 = 0.48 N.