Question

A cubical block of wood weighing 200 g has a lead piece fastened underneath. Find the mass of the lead piece which will just allow the block to float in water. Specific gravity of wood is 0.8 and that of lead is 11.3. 

Answer 1

Given:
Density of wood, ρ_w = 0.8 gm/cc
Density of lead, ρ_pb = 11.3 gm/cc
Weight of the cubical wood block, m_w  = 200 g  

The cubical block floats in water.
Now,
(m_w+ m_pb) × g = (V_w + V_pb)ρ × g
Here,
ρ = Density of water
V_w = Volume of wood
V_pb = Volume of lead

\[\Rightarrow ( \text{m}_\text{w} + \text{m}_{\text{pb}} ) = \left( \frac{\text{m}_\text{w}}{\rho_\text{w}} + \frac{\text{ m}_{\text{pb}}}{\rho_{\text{pb}}} \right)\rho\]

\[ \Rightarrow (200 + \text{m}_{\text{pb}} ) = \left( \frac{200}{0 . 8} + \frac{\text{m}_{\text{pb}}}{11 . 3} \right) \times 1\]

\[ \Rightarrow \text{m}_{\text{pb}} - \frac{\text{m}_{\text{pb}}}{11 . 3} = 250 - 200\]

\[ \Rightarrow \frac{10 . 3 \text{m}_{\text{pb}}}{11 . 3} = 50\]

\[ \Rightarrow \text{m}_{\text{pb}} = \frac{50 \times 11 . 3}{10 . 3} = 54 . 8 \text{ gm }\]