Question

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.

Answer 1

Answer:

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

Answer 2

Answer:

1. weight of sphere, W = 2.4 N

2. Buoyant force on sphere, U = 1.5 N

3. Apparent weight of sphere , W₁ = 0.9 N

Explanation:

Weight of sphere,W

W = mg where m = mass of sphere = density of sphere,ρ × volume of sphere, V.

ρ = 2.14 × 10⁴ kg/m³, V = 4πr³/3 where r = radius of sphere = 0.0139 m

W = ρgV = 2.14 × 10⁴ kg/m³ × 9.8 m/s² × 4π × 0.0139³/3 = 2.4 N

Buoyant force acting on sphere, U

U = weight of mercury displaced = mg where m = mass of mercury = density of mercury, ρ₁ × volume of sphere

ρ₁ = 1.36 × 10⁴ kg/m³, V = 4πr³/3 where r = radius of sphere = 0.0139 m

W = ρgV = 1.36 × 10⁴ kg/m³ × 9.8 m/s² × 4π × 0.0139³/3 = 1.5 N

The sphere's apparent weight , W₁

The sphere's apparent weight, W₁ = W - U = (2.4 - 1.5) N = 0.9 N