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:
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
