Answer:
W=2.2 N
F=1.4 N
W'=0.8 N
Step-by-step 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
![m=(4)/(3)\pi r^3* \rho_1\ kg](https://img.qammunity.org/2020/formulas/physics/college/6ajmf8r6orqz9ao7vui6cep7a6ei8kguav.png)
![m=(4)/(3)* \pi* 0.0135^3* 2.14* 10^4\ kg](https://img.qammunity.org/2020/formulas/physics/college/gb44grmss70evnoedo033j6dhg20ipheof.png)
m = 0.22 kg
W= 0.22 x 10 = 2.2 N (↓) ( take g =10 m/s²)
The buoyant force
![F= (4)/(3)\pi r^3* \rho_2* g](https://img.qammunity.org/2020/formulas/physics/college/fcc0jahik2hu0cfh8pcuh0ucexyomyzs1k.png)
![F=(4)/(3)* \pi* 0.0135^3* 1.36* 10^4* 10](https://img.qammunity.org/2020/formulas/physics/college/tno715gxdt5x0g9hokuna9xm42agbxxy0b.png)
F= 1.4 N (↑)
The apparent weight
W' = 2.2 - 1.4 N
W'= 0.8 N