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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.
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Solution
Given:
Density of wood, ρw = 0.8 gm/cc
Density of lead, ρpb = 11.3 gm/cc
Weight of the cubical wood block, mw = 200 g
The cubical block floats in water.
Now,
(mw+ mpb) × g = (Vw + Vpb)ρ × g
Here,
ρ = Density of water
Vw = Volume of wood
Vpb = Volume of lead
\[\Rightarrow ( \text{m}_\text{w} + \text{m}_{\text{pb}} ) = \left( \frac{\text{m}_\text{w}}{\rho_\text{w}} + \frac{\text{ m}_{\text{pb}}}{\rho_{\text{pb}}} \right)\rho\]
\[ \Rightarrow (200 + \text{m}_{\text{pb}} ) = \left( \frac{200}{0 . 8} + \frac{\text{m}_{\text{pb}}}{11 . 3} \right) \times 1\]
\[ \Rightarrow \text{m}_{\text{pb}} - \frac{\text{m}_{\text{pb}}}{11 . 3} = 250 - 200\]
\[ \Rightarrow \frac{10 . 3 \text{m}_{\text{pb}}}{11 . 3} = 50\]
\[ \Rightarrow \text{m}_{\text{pb}} = \frac{50 \times 11 . 3}{10 . 3} = 54 . 8 \text{ gm }\]
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