A metal piece of mass 160 g lies in equilibrium inside a glass of water. The piece touches the glass at small number of points. If the density of the metal is 8000 kg / m³, then the normal force exerted by the bottom of the glass on the metal piece is

As the piece touches the glass at small number of points, water would be there at the bottom of the piece also and hence buoyancy acts on it. 

From equilibrium condition, $\mathrm{mg}={\mathrm{F}}_{\mathrm{B}}+\mathrm{N}$

        $\begin{array}{l}{\mathrm{F}}_{\mathrm{B}}={\mathrm{V\rho }}_{\text{water }}\mathrm{g}=\frac{\mathrm{m}}{{\mathrm{\rho }}_{\text{metal }}}\times {\mathrm{\rho }}_{\text{water }}\times \mathrm{g}\\ =\frac{0.16}{8}\times 10=0.2\mathrm{N}\end{array}$

So, $\mathrm{N}=\mathrm{mg}-{\mathrm{F}}_{\mathrm{B}}=\frac{160}{1000}\times 10-0.2=1.4\mathrm{N}$