Answer to Question #81891 in Mechanics | Relativity for devachowhury

Question

A piece of gold weighs 10g in air and 9g in water.what is the volume of cavity

Answer 1

F arh. = Pair – P water, where Pair and P water is body weight in air and in water respectively

Pair = m air*g

Pwater = m water * g

Pair = 0.01*10 = 0.1 N

Pwater = 0.009*10 = 0.09 N

F arh. = 0.1-0.09 = 0.01 N

F arh. = ρ*g*V, where ρ is gold density

V = F arh./ ρ*g

V = 0.01/ 19320*10 = 5.18*10^-8 m^3