Question

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

Answer 1

Given:
Mass of the metal piece, m =160 gm = 160 × 10⁻³ kg
Density of the metal piece, ρ_m = 8000 kg/m³
Density of the water, ρ_w = 1000 kg/m^3
​Let R be the normal reaction and U be the upward thrust. 

 

From the diagram, we have:
mg = U + R 

⇒ R = mg −Vρ_wg            [U = Vρ_wg]

\[\Rightarrow \text{R = mg }- \frac{\text{m}}{\rho_\text{m}} \times \rho_\text{w} \times \text{g}\]

\[= 160 \times {10}^{- 3} \times \left( 10 - \frac{{10}^3 \times 10}{8000} \right)\]

\[ = 160 \times {10}^{- 3} \times 10\left( 1 - \frac{1}{8} \right)\]

\[ = 1 . 4 \text{N }\]