A platinum sphere with radius 0.0139 m is totally immersed in mercury. Find the weight of the sphere, the buoyant force acting on the sphere, and the sphere's apparent weight. The densities of platinum and mercury are 2.14 × 10 4 kg/m3 and 1.36 × 10 4 kg/m3, respectively.

A) W = 0.59 N

B) Buoyant Force = 0.37N

C) Apparent Weight = 0.22N

Explanation:

Volume of a sphere is givwn by;

V= (4/3)πr³

We are given radius r = 0.0139 m

Thus, V = (4/3)π(0.0139³)

V = 2.81237 x 10^(-6) m³

A) Weight of sphere, W= mg

where mass, m can be expressed as; m = Volume x density of platinum

m = 2.81237 x 10^(-6) x 2.14 × 10⁴ = 6.0185 x 10^(-2) kg

So, W = 6.0185 x 10^(-2) x 9.8

W = 0.59 N

B) Buoyant force = mg

where mass of displaced mercury m can be expressed as; m = Volume x density of mercury

m = 2.81237 x 10^(-6) x 1.36 × 10⁴

m = 3.825 x 10^(-2) kg

Thus, Buoyant force = 3.825 x 10^(-2) x 9.8 = 0.37N

C) Apparent weight = Weight of sphere - Buoyant force

Thus, apparent weight = 0.59 - 0.37 = 0.22N

that would be because of both gravity and friction. so     c