Answer:
W=2.2 N
F=1.4 N
W'=0.8 N
Explanation:
Given that
Radius ,r = 0.0135 m
Density of the platinum ,ρ₁ = 2.14 x 10⁴ kg/m³
Density of the mercury ,ρ₂ = 1.36 x 10⁴ kg/m³
The weight of the sphere
W= m g
mass = m = volume x density
[tex]m=\dfrac{4}{3}\pi r^3\times \rho_1\ kg[/tex]
[tex]m=\dfrac{4}{3}\times \pi\times 0.0135^3\times 2.14\times 10^4\ kg[/tex]
m = 0.22 kg
W= 0.22 x 10 = 2.2 N (↓) ( take g =10 m/s²)
The buoyant force
[tex]F= \dfrac{4}{3}\pi r^3\times \rho_2\times g[/tex]
[tex]F=\dfrac{4}{3}\times \pi\times 0.0135^3\times 1.36\times 10^4\times 10[/tex]
F= 1.4 N (↑)
The apparent weight
W' = 2.2 - 1.4 N
W'= 0.8 N
