
Given, volume of block =(10×10−2)3=10−3 m3
Let density of block =ρkg/m3
mass of block =ρ×10−3 kg
Buoyant Force (FB)=1000×210−3×10=5 N
F.B.D. of blocks
Balancing torque about point O , we get
$\begin{aligned}
& \operatorname{mg}\left(2 \times 10^{-2}\right)-\mathrm{F}_{\mathrm{B}}\left(2 \times 10^{-2}\right)=0.2 \mathrm{g}\left(25 \times 10^{-2}\right) \
& \rho \times 10^{-3} \times 10 \times 2-10=50 \
& \rho=3000 \mathrm{kg} / \mathrm{m}^3
\end{aligned}$
Hence, mass of block =ρ×10−3
=3000×10−3=3 kg